M a t h e m a t ic a l P ro o fs A Transition to Advanced Mathematics T H I R D
GARY CHARTRAND
E D I T I O N
ALBERT D. POLIMENI
PING ZHANG
T h i r cl E d i t i o n
Mathematical Proofs A Transition to Advanced Mathematics Gary Chartrand W estern M ichigan U niversity
Albert D. Polim eni State U niversity o f N ew York at Fredonia
Ping Zhang W estern M ichigan University
PEARSON Boston Columbus Indianapolis New York San Francisco Upper Saddle River Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montreal Toronto Delhi Mexico City Sao Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo
Editor in Chief: Deirdre Lynch Senior Acquisitions Editor: William Hoffman Assistant Editor: Brandon Rawnsley Executive Marketing Manager: Jeff Weidenaar Marketing Assistant: Caitlin Crain Senior Production Project Manager: Beth Houston Manager, Cover Visual Research and Permissions: Jayne Conte Cover Designer: Suzanne Behnke Cover Art: Shutterstock.com FullService Project Management: Kailash Jadli, Aptara®, Inc. Composition: Aptara®, Inc. Printer/Binder: Courier Westford Cover Printer: Lehigh/Phoenix Credits and acknowledgments borrowed from other sources and reproduced, with permission, in this textbook appear on the appropriate page within text. Copyright © 2013, 2008, 2003 by Pearson Education, Inc. All rights reserved. Manufactured in the United States of America. This publication is protected by Copyright, and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. To obtain permission(s) to use material from this work, please submit a written request to Pearson Education, Inc., Permissions Department, One Lake Street, Upper Saddle River, New Jersey 07458, or you may fax your request to 2012363290. Many of the designations by manufacturers and sellers to distinguish their products are claimed as trademarks. Where those designations appear in this book, and the publisher was aware of a trademark claim, the designations have been printed in initial caps or all caps. Library of Congress CataloginginPublication Data Chartrand, Gary. Mathematical proofs : a transition to advanced mathematics / Gary Chartrand, Albert D. Polimeni, Ping Zhang.  3rd ed. p. cm. Includes bibliographical references and index. ISBN13: 9780321797094 ISBN10: 0321797094 1. Proof theory—Textbooks. I. Polimeni, Albert D.. 1938 II. Zhang, Ping, 1957 III. Title. QA9.54.C48 2013 511.3'6— dc23
. 2012012552
10 9 8 7 6 5 4 3 2 —CW— 16 15 14 13 12
PEARSON
ISBN13: 9780321797094 ISBN10: 0321797094
To the memory o f my mother and father
G.C.
the memory o f my uncle Joe and my brothers John and Rocky my mother and the memory o f my father
P.Z.
A.D.P.
Contents
0
Communicating Mathematics L earning M athem atics
1
2
W hat O thers Have Said A bout W riting M athem atical W riting U sing Sym bols
4
5
6
W riting M athem atical E xpressions
8
Com m on W ords and Phrases in M athem atics
10
Som e Closing C om m ents A bout W riting
1
Sets l.i
14
D escribing a Set
1.2
Subsets
1.3
Set O perations
18
1.4
Indexed Collections o f Sets
1.5
Partitions o f Sets
1.6
Cartesian Products o f Sets
21
E xercises for C hapter 1
2
12
27 29
Logic
37
2.1
Statem ents
37*
2 .2
The N egation of a Statem ent
2 .3 2 .4
The D isjunction and Conjunction o f Statem ents The Im plication 42
2 .5
M ore on Im plications
39
44
2 .6
The Biconditional
2 .7
Tautologies and Contradictions
47 49
41
2 .8 2 .9 2 .1 0
L ogical Equivalence
51
Som e Fundam ental Properties o f L ogical Equivalence Q uantified Statem ents 55
2 .1 1 C haracterizations o f S tatem ents E xercises for C hapter 2 64
3
63
Direct Proof and Proof by Contrapositive 3 .1
Trivial and Vacuous Proofs
3 .2
D irect Proofs
3 .3
P roof by C ontrapositive
3 .4
P roof by Cases
3 .5
P roof Evaluations
E xercises for C hapter 3
4
78
80 84
89 92 93
More on Direct Proof and Proof by Contrapositive 4.1
Proofs Involving D ivisibility o f Integers
4 .2 4 .3 4 .4
Proofs Involving C ongruence o f Integers Proofs Involving R eal N um bers 105 Proofs Involving Sets 108
4 .5
F undam ental Properties o f Set O perations
99 103
111
4 .6 Proofs Involving Cartesian Products o f Sets Exercises for C hapter 4 114
5
113
Existence and Proof by Contradiction 5 .1
C ounterexam ples
5 .2
P roof by C ontradiction
120
5 .3
A Review o f Three P ro o f Techniques
5 .4
E xistence Proofs
124 130
132
5 .5 D isproving E xistence Statem ents Exercises for C hapter 5 137
6
53
136
Mathematical Induction 6.1
The Principle of M athem atical Induction
6 .2
A M ore G eneral Principle o f M athem atical Induction
142
6 .3
P roof by M inim um C ounterexam ple
151
158
6 .4 The Strong Principle o f M athem atical Induction E xercises for C hapter 6 165
161
[7
Prove or Disprove 7.1
Conjectures in M athem atics
7 .2
R evisiting Q uantified Statem ents
7 .3
Testing Statem ents 185
Equivalence Relations 8 .1
R elations
8.2
Properties of R elations
8 .3
196 Equivalence R elations Properties o f Equivalence Classes 202 C ongruence M odulo n
8 .4 8 .5
192 193
207
The Integers M odulo n 8 .6 Exercises for C hapter 8 210
9
Functions 9.1
The D efinition o f F unction
9 .2
The Set o f A ll Functions from A to fi
9 .3
O netoO ne and O nto Functions
9 .4
Bijective Functions
216
9 .5
C om position o f Functions
9 .6
Inverse Functions
9 .7
Perm utations
219 220
222 225
229 232
E xercises for C hapter 9
flo
173
178
E xercises for C hapter 7
8
170
234
Cardinalities of Sets 1 0 .1
N um erically Equivalent Sets
10.2
D enum erable Sets
1 0 .3
U ncountable Sets
10 .4
Com paring Cardinalities of Sets
1 0 .5
243
244 250
T he SchroderB ernstein T heorem
Exercises for C hapter 10
255 258
262
Proofs in Number Theory 11.1
D ivisibility Properties o f Integers
1 1 .2
The D ivision A lgorithm
267
266
Contents 1 1 .3 11.4
G reatest C om m on D ivisors The Euclidean A lgorithm
271 272
1 1 .5
R elatively Prim e Integers
275
1 1 .6
The F undam ental T heorem o f A rithm etic
1 1 .7 C oncepts Involving Sum s o f D ivisors E xercises for C hapter 11 281
12
288
12.1 1 2 .2
L im its o f Sequences Infinite Series 295
288
12 .3
L im its o f Functions
1 2 .4 12.5
Fundam ental Properties o f Lim its o f Functions Continuity 312
12.6
D ifferentiability
300
Binary O perations
317
322
322
13 .2
G roups
1 3 .3
Perm utation G roups
326
1 3 .4 1 3.5
F undam ental Properties o f G roups Subgroups 336
1 3 .6
Isom orphic G roups
E xercises for C hapter 13
330
340 344
Proofs in Ring Theory (Online) 14 .1
Rings
14 .2 1 4 .3
E lem entary Properties o f Rings Subrings
1 4 .4 1 4 .5
Integral D om ains Fields
E xercises for C hapter 14
115
307
314
Proofs in Group Theory 1 3 .1
14
277 280
Proofs in Calculus
Exercises for C hapter 12
13
vii
Proofs in Linear Algebra (Online) 15 .1 1 5 .2
Properties o f Vectors in 3Space Vector Spaces
1 5 .3
M atrices
1 5 .4
Som e Properties of Vector Spaces
333
viii
Contents 1 5 .5 1 5 .6
Subspaces Spans o f Vectors
1 5 .7
L inear D ependence and Independence
1 5 .8
Linear Transform ations
1 5 .9
Properties o f L inear T ransform ations
E xercises for C hapter 15
16
Proofs in Topology (Online) 16 .1 16.2
M etric Spaces O pen Sets in M etric Spaces
1 6 .3
C ontinuity in M etric Spaces
1 6 .4
Topological Spaces
16 .5
C ontinuity in Topological Spaces
E xercises for C hapter 16
Answers and Hints to Selected OddNumbered Exercises in Chapters 1416 (online)
Answers and Hints to OddNumbered Section Exercises References Index of Symbols Index
351 394 395 396
PREFACE TO THE THIRD EDITION
As w e m entioned in the prefaces o f the first two editions, because the teaching o f calculus in m any colleges and universities has becom e m ore problem oriented w ith added em phasis on the use o f calculators and com puters, the theoretical gap betw een the m aterial presented in calculus and the m athem atical background expected (or at least hoped for) in m ore advanced courses such as abstract algebra and advanced calculus has w idened. In an attem pt to narrow this gap and to better prepare students for the m ore abstract m athem atics courses to follow, m any colleges and universities have introduced courses that are now com m only called transition courses. In these courses, students are intro duced to problem s w hose solution involves m athem atical reasoning and a know ledge of pro o f techniques and w riting clear proofs. Topics such as relations, functions and cardi nalities o f sets are encountered throughout theoretical m athem atics courses. In addition, transition courses often include theoretical aspects o f num ber theory, abstract algebra, and calculus. This textbook has been w ritten for such a course. The idea for this textbook originated in the early 1980s, long before transition courses becam e fashionable, during the supervising o f undergraduate m athem atics re search projects. We cam e to realize that even advanced undergraduates lack a sound understanding of proof techniques and have difficulty w riting correct and clear proofs. A t that tim e, a set o f notes was developed for these students. This w as follow ed by the introduction o f a transition course, for w hich a m ore detailed set o f notes was w ritten. The first edition o f this book em anated from these notes, w hich in turn has led to a second edition and now this third edition. W hile understanding proofs and p ro o f techniques and w riting good proofs are m ajor goals here, these are not things that can be accom plished to any great degree in a single course during a single semester. These m ust continue to be em phasized and p racticed in succeeding m athem atics courses.
Our Approach Since this textbook originated from notes that w ere w ritten exclusively for undergradu ates to help them understand pro o f techniques and to w rite good proofs, this is the tone
Preface to the Third Edition
in w hich all editions o f this book have been w ritten: to be studentfriendly. N um erous exam ples o f proofs are presented in the text. Follow ing com m on practice, w e indicate the end o f a proof with the square sym bol ■. O ften we precede a p ro o f by a discussion, referred to as a p ro o f strategy, w here we think through w hat is needed to present a proof o f the result in question. O ther tim es, w e find it useful to reflect on a p ro o f we have just presented to point out certain key details. We refer to a discussion o f this type as a p r o o f analysis. Periodically, problem s are p resented and solved, and we m ay find it convenient to discuss some features o f the solution, w hich w e refer to sim ply as an analysis. For clarity, we indicate the end o f a discussion o f a pro o f strategy, p ro o f analysis, analysis or solution o f an exam ple w ith the diam ond sym bol # . A m ajor goal o f this textbook is to help students learn to construct proofs of their ow n that are not only m athem atically correct but clearly w ritten. M ore advanced m athe m atics students should strive to present proofs that are convincing, readable, notationally consistent, and gram m atically correct. A secondary goal is to have students gain suffi cient know ledge o f and confidence w ith proofs so that they will recognize, understand, and appreciate a proof that is properly w ritten. As w ith the first two editions, the third edition o f this book is intended to assist the student in m aking the transition to courses that rely m ore on m athem atical p ro o f and reasoning. We envision students w ould take a course based on this book after they have had a year o f calculus (and possibly another course, such as elem entary linear algebra). It is likely that, prior to taking this course, a student’s training in m athem atics consisted prim arily o f doing patterned problem s; that is, students have been taught m ethods for solving problem s, likely including some explanation as to w hy these m ethods worked. Students m ay very w ell have had exposure to som e proofs in earlier courses but, m ore than likely, were unaw are o f the logic involved and o f the m ethod o f p ro o f being used. There m ay have even been tim es w hen the students w ere not certain w hat was being proved.
O utline of the Contents Since w riting good proofs requires a certain degree o f com petence in w riting, w e have devoted C hapter 0 to w riting m athem atics. The em phasis o f this chapter is on effective and clear exposition, correct usage o f sym bols, w riting and displaying m athem atical expressions, and using key w ords and phrases. A lthough every instructor will em phasize w riting in his or her ow n way, we feel that it is useful to read C hapter 0 periodically throughout the course. It w ill m ean m ore as the student progresses through the course. A m ong the additions to and changes in the second edition that resulted in this third edition are the follow ing. ® M ore than 250 exercises have been added, m any of w hich require m ore thought to solve. ® N ew exercises have been added dealing w ith conjectures to give students practice w ith this im portant aspect of m ore advanced m athem atics. ® A dditional exam ples have been provided to assist in understanding and solving new exercises. ® In a num ber o f instances, expanded discussions o f a topic have been given to provide added clarity. In particular, the im portant topic of quantified statem ents is introduced in Section 2.10 and then review ed in Section 7.2 to enhance o n e’s understanding o f this.
Preface to the T hird Edition
xi
• A discussion o f cosets and L agrange’s theorem has been added to C hapter 13 (Proofs in G roup Theory). Each chapter is divided into sections and the exercises for each chapter occur at the end of the chapter, divided into sections in the same way. There is also a final section of exercises for the entire chapter. C hapter 1 contains a gentle introduction to sets, so that everyone has the same back ground and is using the same notation as w e prepare for w hat lies ahead. N o proofs in volving sets occur until C hapter 4. M uch o f C hapter 1 m ay very w ell be a review for many. C hapter 2 deals exclusively w ith logic. The goal here is to present w hat is needed to get into proofs as quickly as possible. M uch o f the em phasis in C hapter 2 is on statem ents, im plications, and quantified statem ents, including a discussion of m ixed quantifiers. Sets are introduced before logic so that the student's first encounter w ith m athem atics here is a fam iliar one and because sets are needed to discuss quantified statem ents properly in C hapter 2. The tw o p ro o f techniques o f direct p ro o f and p ro o f by contrapositive are introduced in C hapter 3 in the fam iliar setting o f even and odd integers. P roof by cases is discussed in this chapter as well as proofs o f “if and only if ” statem ents. C hapter 4 continues this discussion in other settings, nam ely divisibility o f integers, congruence, real num bers, and sets. The technique o f proof by contradiction is introduced in C hapter 5. Since existence proofs and counterexam ples have a connection w ith p ro o f by contradiction, these also occur in C hapter 5. The topic o f uniqueness (of an elem ent w ith specified properties) is also addressed in C hapter 5. P roof by m athem atical induction occurs in C hapter 6 . In addition to the Principle o f M athem atical Induction and the Strong Principle o f M athem atical Induction, this chapter includes proof by m inim um counterexam ple. The m ain goal o f C hapter 7 (Prove or D isprove) concerns the testing o f statem ents o f unknow n truth value, w here it is to be determ ined, w ith justification, w hether a given statem ent is true or false. In addition to the challenge o f determ ining w hether a statem ent is true or false, such problem s provide added practice w ith counterexam ples and the various pro o f techniques. Testing statem ents is preceded in this chapter w ith an historical discussion o f conjectures in m athem atics and a review of quantifiers. C hapter 8 deals w ith relations, especially equivalence relations. M any exam ples in volving congruence are presented and the set o f integers m odulo n is described. C hapter 9 involves functions, with em phasis on the properties o f onetoone and onto. This gives rise to a discussion o f bijective functions and inverses o f functions. The w elldefined property o f functions is discussed in m ore detail in this edition. In addition, there is a discussion o f im ages and inverse im ages o f sets w ith regard to functions and a num ber o f added exercises involving these concepts. Chapter 10 deals w ith infinite sets and a discussion o f cardinalities o f sets. This chapter includes an historical discussion o f infinite sets, beginning with C antor and his fascination and difficulties with the SchroderB ernstein Theorem , then to Zerm elo and the A xiom o f C hoice, and ending w ith a p ro o f of the S chroderB ernstein Theorem . A ll o f the p ro o f techniques are used in C hapter 11 w here num erous results in the area o f num ber theory are introduced and proved. C hapter 12 deals with proofs that occur in calculus. B ecause these proofs are quite different than those previously encountered but are often m ore predictable in nature, m any illustrations are given that involve lim its of
x ii
Preface to the T hird Edition sequences and lim its of functions and their connections w ith infinite series, continuity, and differentiability. The final C hapter 13 deals with m odem algebra, beginning with binary operations and m oving into proofs that are encountered in the area of group theory.
Web Site for M athem atical Proofs T hree additional chapters, Chapters 1416 (dealing w ith proofs in ring theory, linear algebra, and topology), can be found on the W eb site: http://w w w .aw .com /info/chartrand.
T eaching a Course from T his Text A lthough a course using this textbook could be designed in m any w ays, here are our view s on such a course. As we noted earlier, w e think it is useful for students to reread (at least portions of) C hapter 0 throughout the course as we feel that w ith each reading, the chapter becom es m ore m eaningful. T he first part o f C hapter 1 (Sets) w ill likely be fam iliar to m ost students, although the last part m ay not. Chapters 2 6 w ill probably be part o f any such course, although certain topics could receive varying degrees of em phasis (with perhaps proof by m inim um counterexam ple in C hapter 6 possibly even om itted). O ne could spend little or m uch tim e on C hapter 7, depending on how m uch tim e is used to discuss the large num ber o f â€œprove or disproveâ€? exercises. We think that m ost o f Chapters 8 and 9 w ould be covered in such a course. It w ould be useful to cover some o f the fundam ental ideas addressed in C hapter 10 (C ardinalities o f Sets). A s tim e perm its, portions o f the later chapters could be covered, especially those o f interest to the instructor, including the possibility o f going to the Web site for even m ore variety in the three online chapters.
E xercises T here are num erous exercises for Chapters 113 (as w ell as for Chapters 1416 on the Web site). The degree o f difficulty o f the exercises ranges from routine to m edium difficulty to m oderately challenging. As m entioned earlier, the third edition contains m ore exercises in the m oderately challenging category. There are exercises that present students w ith statem ents, asking them to decide w hether they are true or false (with justification). There are proposed proofs o f statem ents, asking if the argum ent is valid. T here are proofs w ithout a statem ent given, asking students to supply a statem ent o f w hat has been proved. A lso, there are exercises that call upon students to m ake conjectures of their own and possibly to provide proofs o f these conjectures. C hapter 3 is the first chapter in w hich students w ill be called upon to w rite proofs. A t such an early stage, we feel that students need to (1) concentrate on constructing a valid p ro o f and not be distracted by unfam iliarity w ith the m athem atics, (2) develop som e selfconfidence w ith this process, and (3) learn how to w rite a proof properly. W ith this in m ind, m any o f the exercises in C hapter 3 have been intentionally structured so as to be sim ilar to the exam ples in that chapter. In general, there are exercises for each section at the end o f a chapter (section exercises) and additional exercises for the entire chapter (chapter exercises). A nsw ers or
Preface lo the Third Edition
xiii
hints to the oddnum bered section exercises appear at the end of text. O ne should also keep in m ind, however, that proofs o f results are not unique in general.
A cknow ledgm ent s It is a pleasure to thank the review ers o f the third edition: D aniel A costa, Southeastern L ouisiana University Scott A nnin, C alifornia State University, Fullerton J. M arshall Ash, D ePaul U niversity A ra B asm ajian, H unter College o f CU N Y M atthias B eck, San Francisco State U niversity R ichard B elshoff, M issouri State U niversity Jam es Braw ner, A rm strong A tlantic State U niversity M anav D as, U niversity o f Louisville D avid D em psey, Jacksonville State U niversity Cristina D om okos, C alifornia State University, Sacram ento Jose D. Flores, U niversity of South D akota Eric G ottlieb, Rhodes College R ichard H am m ack, V irginia Com m onw ealth U niversity A lan Koch. A gnes Scott College M. H arper L angston, Courant Institute o f M athem atical Sciences, N ew York U niversity M aria N ogin, C alifornia State University, Fresno D aniel N ucinkis, U niversity o f Southam pton Thom as Polaski, W inthrop U niversity John Randall, Rutgers U niversity Eileen T. Shugart, V irginia Tech Brian A. Snyder, Lake Superior State U niversity M elissa Sutherland, SU N Y G eneseo M .B. Ulmer, U niversity of South C arolina U pstate M ike W inders, W orcester State U niversity We also thank R enato M irollo, B oston College and Tom W eglaitner for giving a final reading o f portions of the third edition. We have been m ost fortunate to receive the enthusiastic support from m any at Pearson. First, w e w ish to thank the editorial team , as w ell as others at Pearson who have been so helpful and supportive: G reg Tobin, Publisher, M athem atics and StatisÂ tics; W illiam H offm an, Senior A cquisitions Editor; Jeff W eidenaar, Executive M arketÂ ing M anager, M athem atics; and B randon Rawnsley, A ssociate Editor, A rts & Sciences, H igher Education. O ur thanks to all o f you. Finally, thank you as w ell to Beth H ouston, Senior Production Project M anager; K ailash Jadli, Project M anager, Aptara, Inc.; and M ercedes H eston, Copy Editor, for guiding us through the final stages o f the third edition. G ary C hartrand A lbert D. Polim eni Ping Zhang
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0 C om m unicating M athem atics
uite likely, the m athem atics you have already encountered consists o f doing probÂ lems using a specific approach or procedure. These m ay include solving equations sim plifying algebraic expressions, verifying trigonom etric identities, using certain rules to find and sim plify the derivatives o f functions and setting up and evaluatÂ ing a definite integral that give the area o f a region or the volum e o f a solid. A ccom plishing all o f these is often a m atter o f practice. M any of the m ethods that one uses to solve problem s in m athem atics are based on results in m athem atics that w ere discovered by people and show n to be true. This kind o f m athem atics m ay very w ell be new to you and, as w ith anything th a tâ€™s new, there are things to be learned. B ut learning som ething new can be (in fact should be) fun. There are several steps involved here. The first step is discovering som ething in m athem atics that w e believe to be true. H ow does one discover new m athem atics? This usually com es about by considering exam ples and observing that a pattern seem s to be occurring with the exam ples. This m ay lead to a guess on our part as to w hat appears to be happening. We then have to convince ourselves that our guess is correct. In m athem atics this involves constructing a p roof o f w hat we believe to be true is, in fact, true. B ut this is not enough. We need to convince others that w e are right. So w e need to w rite a p ro o f that is w ritten so clearly and so logically that people w ho know the m ethods o f m athem atics w ill be convinced. W here m athem atics differs from all other scholarly fields is that once a proof has been given o f a certain m athem atical statem ent, there is no longer any doubt. This statem ent is true. Period. There is no other alternative. O ur m ain em phasis here w ill be in learning how to construct m athem atical proofs and learning to w rite the proof in such a m anner that it w ill be clear to and understood by others. E ven though learning to guess new m athem atics is im portant and can be fun, w e will spend only a little tim e on this as it often requires an understanding of m ore m athem atics than can be discussed at this tim e. But w hy w ould we w ant to discover new m athem atics? W hile one possible answ er is that it com es from the curiosity that m ost m athem aticians possess, a m ore com m on explanation is that we have a problem to solve that requires know ing that som e m athem atical statem ent is true.
1
2
Chapter 0
Communicating Mathematics
L earning M athem atics O ne of the m ajor goals of this book is to assist you as you progress from an individual w ho uses m athem atics to an individual who understands m athem atics. Perhaps this will m ark the beginning o f you becom ing som eone w ho actually develops m athem atics of your own. This is an attainable goal if you have the desire. T he fact that you've gone this far in your study o f m athem atics suggests that you have ability in m athem atics. This is a real opportunity for you. M uch o f the m athem atics that you w ill encounter in the future is based on w hat you are about to learn here. The better you learn the m aterial and the m athem atical thought process now, the m ore you w ill understand later. To be sure, any area o f study is considerably m ore enjoyable when you understand it. B ut getting to that point will require effort on your part. T here are probably as m any excuses for doing poorly in m athem atics as there are strategies for doing w ell in m athem atics. We have all heard students say (som etim es, rem arkably, even w ith pride) that they are not good at m athem atics. T h a t’s only an alibi. M athem atics can be learned like any other subject. E ven som e students who have done w ell in m athem atics say that they are not good w ith proofs. This, too, is unacceptable. W hat is required is determ ination and effort. To have done w ell on an exam w ith little or no studying is nothing to be proud of. Confidence based on being w ellprepared is good, however. H ere is some advice that has w orked for several students. First, it is im portant to understand w hat goes on in class each day. This m eans being present and being prepared for every class. A fter each class, recopy any lecture notes. W hen recopying the notes, express sentences in your ow n w ords and add details so that everything is as clear as possible. If you run into snags (and you w ill), talk them over w ith a classm ate or your instructor. In fact, it’s a good idea (at least in our opinion) to have som eone w ith w hom to discuss the m aterial on a regular basis. N ot only does it often clarify ideas, it gets you into the habit o f using correct term inology and notation. In addition to learning m athem atics from your instructor, solidifying your under standing by careful notetaking and by talking w ith classm ates, your text is (or at least should be) an excellent source as well. R ead your text carefully w ith pen (or pencil) and paper in hand. M ake a serious effort to do every hom ew ork problem assigned and, eventually, be certain that you know how to solve them . If there are exercises in the text that have not been assigned, you m ight even try to solve these as well. A nother good idea is to try to create your ow n problem s. In fact, w hen studying for an exam , try creating your own exam . If you start doing this for all o f your classes, you m ight be surprised at how good you becom e. Creativity is a m ajor part of m athem atics. D iscovering m athe m atics not only contributes to your understanding o f the subject but has the potential to contribute to m athem atics itself. Creativity can com e in all form s. The follow ing quote is due to the w ellknow n w riter J. K. R ow ling (author o f the H arry Potter novels). Som etim es ideas ju s t com e to me. O ther tim es I have to sw eat a n d alm ost bleed to m ake ideas come. I t ’s a m ysterious p ro c ess, but I hope I never fin d out exactly how it works.
Learning M athematics
3
In the book D efying G ravity: The Creative C areer o f Stephen Schw artz fro m G odspell to Wicked, the author C arol de G iere w rites a biography o f S tephen Schw artz, one o f the m ost successful com poserlyricists, in w hich she discusses not only creativ ity in m usic but how an idea can lead to som ething special and interesting and how creative people m ay have to deal w ith disappointm ent. Indeed, de G iere dedicates her book to the creative spirit within each o f us. W hile he w rote the m usic for such fam ous show s as G odspell and W icked, Schw artz discusses creativity headon in the song “The Spark o f C reation” he w rote for the m usical Children o f E den. In her book, de G iere writes: In m any ways, this song expresses the them e o f Stephen S chw artz’s life— the naturalness and im portance o f the creative urge w ithin us. A t the sam e time he created an anthem fo r artists. In m athem atics our goal is to seek the truth. Finding answ ers to m athem atical questions is im portant, but we cannot be satisfied w ith this alone. We m ust be certain that w e are right and that our explanation for w hy we believe w e are correct is convincing to others. The reasoning w e use as w e proceed from w hat we know to w hat we w ish to show m ust be logical. It m ust m ake sense to others, not ju st to ourselves. There is jo in t responsibility here. As w riters, it is our responsibility to give an accurate clear argum ent w ith enough details provided to allow the reader to understand w hat w e have w ritten and to be convinced. It is the rea d er’s responsibility to know the basics o f logic and to study the concepts involved so that a w ellpresented argum ent w ill be understood. Consequently, in m athem atics w riting is im portant, very im portant. Is it really im portant to w rite m athem atics w ell? A fter all, isn ’t m athem atics m ainly equations and sym bols? N ot at all. It is not only im portant to w rite m athem atics well, it is im portant to w rite well. You will be w riting the rest o f your life, at least reports, letters and em ail. M any people w ho never m eet you w ill know you only by w hat you w rite and how you write. M athem atics is a sufficiently com plicated subject that we d o n 't need vague, hazy and boring w riting to add to it. A teacher has a very positive im pression o f a student w ho hands in w ellw ritten and w ellorganized assignm ents and exam inations. You w ant people to enjoy reading w hat y o u ’ve w ritten. It is im portant to have a good reputation as a writer. I t’s part of being an educated person. Especially with the large num ber of em ail letters that so m any o f us write, it has becom e com m onplace for w riting to be m ore casual. A lthough all people w ould probably subscribe to this (since it is m ore efficient), w e should know how to write w ell form ally and professionally w hen the situation requires it. You m ight think that considering how long y o u ’ve been w riting and that yo u 're set in your ways, it w ill be very difficult to im prove your w riting. N ot really. If you w ant to im prove, you can and will. E ven if you are a good w riter, your w riting can always be im proved. Ordinarily, people d o n ’t think m uch about their w riting. O ften ju st thinking about your w riting is the first step to w riting better.
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Chapter 0
Communicating Mathematics
W hat Others Have Said about WritingM any people w ho are w ell know n in their areas o f expertise have expressed their thoughts about w riting. H ere are quotes by som e o f these individuals. A nything that helps com m unication is good. A nything that hurts it is bad. I like words m ore than num bers, and I alw ays did— conceptual more than com putational. Paul H alm os, m athem atician W riting is easy. A ll you have to do is cross out all the wrong words. M ark Twain, author (The A dventures o f H uckleberry Finn) You d o n ’t w rite because you w ant to say som ething; yo u w rite because yo u ’ve got som ething to say. F. Scott Fitzgerald, author (The G reat G atsby) W riting com es m ore easily i f you have som ething to say. Scholem A sch, author E ith er w rite som ething worth reading or do som ething worth writing. Benjam in Franklin, statesm an, writer, inventor W hat is written w ithout effort is in general read w ithout pleasure. Sam uel Johnson, w riter E asy reading is dam ned hard writing. N athaniel H aw thorne, novelist (The Scarlet Letter) E verything that is w ritten m erely to p lea se the author is worthless. The last thing one know s when writing a b ook is w hat to p u t first. I have m ade this letter longer because I lack the time to m ake it short. B laise Pascal, m athem atician and physicist The best w ay to becom e acquainted w ith a subject is to w rite a book about it. B enjam in D israeli, prim e m inister o f E ngland In a very real sense, the w riter w rites in order to teach him self, to understand h im self, to satisfy him self; the publishing o f his ideas, though it brings gratification, is a curious anticlim ax. A lfred K azin, literary critic The skill o f w riting is to create a context in which other peo p le can think. E dw in Schlossberg, exhibit designer A w riter needs three things, experience, observation, and im agination, any two o f which, at tim es any one o f w hich, can supply the lack o f the other. W illiam Faulkner, w riter (The Sound and the Fury) I f confusion runs ram pant in the passage ju s t read, It m ay very w ell be that too m uch has been said.
Mathematical Writing
5
So that's w hat he m eant! Then w hy d id n ’t he say so? Frank Harary, m athem atician A m athem atical theory is not to be considered com plete until yo u have m ade it so clear that you can explain it to the first man whom you m eet on the street. D avid H ilbert, m athem atician E verything should be m ade as sim ple as possible, but not simpler. A lbert Einstein, physicist N ever let anything you w rite be p ub lish ed w ithout having had others critique it. D onald E. K nuth, com puter scientist and w riter Som e books are to be tasted, others to be swcdlowed, and som e fe w to be chewed and digested. R eading m aketh a fu ll m an, conference a ready m an, and w riting an exact mcm. Francis B acon, w riter and philosopher fu d g e an article not by the quality o f w hat is fra m e d a n d hanging on the wall, but by the quality o f w hat 's in the w astebasket. A nonym ous (Quote by Leslie Lam port) We are all apprentices in a craft where noone ever becom es a master. Ernest Hem ingway, author (For W hom the B ell Tolls) There are three rules fo r writing a novel. Unfortunately, no one know s w hat they are. W. Som erset M augham , author (O f H um an Bondage)
M athem atical W riting M ost o f the quotes given above p ertain to w riting in general, not to m athem atical w riting in particular. H ow ever these suggestions for w riting apply as w ell to w riting m athem at ics. For us, m athem atical w riting m eans w riting assignm ents for a m athem atics course (particularly a course w ith proofs). Such an assignm ent m ight consist o f w riting a single proof, w riting solutions to a num ber o f problem s or perhaps w riting a term paper which, m ore than likely, includes definitions, exam ples, background and proofs. W e’ll refer to any o f these as an assignm ent. Your goal should be to w rite correctly, clearly and in an interesting manner. Before you even begin to w rite, you should have already thought about a num ber o f things. First, you should know w hat exam ples and proofs you plan to include if this is appropriate for your assignm ent. You should not be overly concerned about w riting good proofs on your first attem pt— but be certain that you do have proofs. As y o u ’re w riting your assignm ent, you m ust be aw are o f your audience. W hat is the target group for your assignm ent? O f course, it should be w ritten for your instructor. But it should be w ritten so that a classm ate w ould understand it. As you grow m athem atically, your audience w ill grow w ith you and you w ill adapt your w riting to this new audience. G ive yourself enough tim e to w rite your assignm ent. D o n ’t try to put it together ju st a few m inutes before it’s due. The disappointing result w ill be obvious to your
6
Chapter 0
Communicating Mathematics instructor. A nd to you! F ind a place to w rite that is com fortable for you: your room , an office, a study room , the library and sitting at a desk, at a table, in a chair. Do w hat w orks best for you. Perhaps you w rite best w hen it’s quiet or w hen there is background music. N ow that y o u ’re com fortably settled and have allow ed enough tim e to do a good job, le t’s p ut a plan together. If the assignm ent is fairly lengthy, you m ay need an outline, w hich, m ost likely, w ill include one or m ore o f the following: 1.
B ackground and motivation
2.
The definitions to be presented and possibly the notation to be used
3.
The exam ples to include
4.
The results to be presented (w hose proofs have already been w ritten, probably in rough form)
5.
R eferences to other results you intend to use
6.
The order o f everything m entioned above.
If the assignm ent is a term paper, it m ay include extensive background m aterial and m ay need to be carefully m otivated. The subject o f the paper should be placed in perspective. W here does it fit in w ith w hat w e already know ? M any w riters w rite in spirals. Even though you have a plan for your assignm ent w hich includes an ordered list of things you w ant to say, it is likely that you w ill reach som e point (perhaps sooner than you think) w hen you realize that you should have in cluded som ething earlier— perhaps a definition, a theorem , an exam ple, som e notation. (This happened to us m any tim es w hile w riting this textbook.) Insert the m issing m aterial, start over again and write until once again you realize that som ething is m issing. It is im portant, as you reread, that you start at the beginning each tim e. T hen repeat the steps listed above. We are about to give you som e advice, som e pointers, about w riting m athem atics. Such advice is necessarily subjective. N ot everyone subscribes to these suggestions on w riting. Indeed, w riting experts d o n ’t agree on all issues. For the present, your instructor will be your best guide. B ut w riting does not follow a list o f rules. As you m ature m athem atically, perhaps the best advice about your w riting is the sam e advice given by Jim iny Cricket to D isney’s Pinocchio: A lw ays let your conscience be yo u r guide. You m ust be yourself. A nd one additional piece o f advice: Be careful about accepting advice on w riting. O riginality and creativity d o n ’t follow rules. U ntil you reach the stage of being com fortable and confident w ith your own w riting, however, w e believe that it is useful to consider a few w riting tips. Since a num ber o f these w riting tips m ay not m ake sense (since, after all, w e d o n ’t even have anything to w rite as yet), it w ill probably be m ost useful to return to this chapter periodically as you proceed through the chapters that follow.
U sing Sym bols Since m athem atics is a sym boloriented subject, m athem atical w riting involves a m ixture o f w ords and sym bols. H ere are several guidelines to w hich a num ber o f m athem aticians subscribe.
Using Symbols 1.
7
N ever start a sentence with a symbol. W riting m athem atics follow s the same practice as w riting all sentences, nam ely that the first w ord should be capitalized. This is confusing if the sentence w ere to begin w ith a sym bol since the sentence appears to be incom plete. A lso, in general, a sentence sounds better if it starts w ith a word. Instead o f w riting: x 2 — 6x + 8 = 0 has tw o distinct roots. write: The equation x 2 — 6x + 8 = 0 has tw o distinct roots.
2.
Separate sym bols not in a list by words if possible. Separating sym bols by w ords m akes the sentence easier to read and therefore easier to understand. The sentence: E xcept for a, b is the only root o f (x — a) (x — b) = 0. w ould be clearer if it w ere w ritten as: E xcept for a, the num ber b is the only root o f (x — a ) (x — b) — 0.
3.
E xcept when discussing logic, avoid writing the fo llo w in g sym bols in your assignm ent: —>. V, 3, 9 , etc. The first four sym bols stand for "im plies”, “for every”, “there exists” and “ such that”, respectively. You m ay have already seen these sym bols and know w hat they m ean. If so, this is good. It is useful w hen taking notes or w riting early drafts o f an assignm ent to use shorthand sym bols but m any m athem aticians avoid such sym bols in their professional w riting. (We will visit these sym bols later.)
4.
Be careful about using i.e. and e.g. These stand for that is a n d /o r exam ple, respectively. There are situations w hen w riting the w ords is preferable to w riting the abbreviations as there m ay be confusion w ith nearby sym bols. For exam ple, y —1 and ( 1Y lim 1 H— n—>oo \ nJ num bers.
5.
are not rational num bers, that is, i and e are not rational
W rite out integers as words when they are used as adjectives and when the num bers are relatively sm all or are easy to describe in words. W rite out num bers num erically when they specify the value o f som ething. There are exactly two groups o f order 4. Fifty m illion Frenchm en c a n ’t be wrong. There are one m illion positive integers less than 1,000,001.
6.
D o n 't m ix words and sym bols improperly. Instead o f w riting: Every integer > 2 is a prim e or is com posite.
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Chapter 0
Communicating Mathematics it is preferable to write: Every integer exceeding 1 is a prim e or is com posite. or If n > 2 is an integer, then n is prim e or com posite. A lthough Since (.v — 2)(x — 3) = 0, it follow s that x = 2 or 3. sounds correct, it is not w ritten correctly. It should be: Since (x — 2)(.v — 3) = 0, it follow s that x = 2 or x = 3. 7.
A void using a sym bol in the statem ent o f a theorem when it's not needed. D o n 't write: T heorem E very bijective fu n ctio n f has an inverse. D elete “ / ” . It serves no useful purpose. The theorem does not depend on w hat the function is called. A sym bol should not be used in the statem ent o f a theorem (or in its proof) exactly once. If it is useful to have a nam e for an arbitrary bijective function in the p ro o f (as it probably w ill be), then “ / ” can be introduced there.
8.
Explain the m eaning o f every sym bol that you introduce. A lthough w hat you intended m ay seem clear, d o n ’t assum e this. For exam ple, if you w rite n = 2 k + \ and k has never appeared before, then say that k is an integer (if indeed k is an integer).
9.
Use “fro zen sym bols" properly. If m and n are typically used for integers (as they probably are), then d o n ’t use them for real num bers. If A and B are used for sets, then d o n ’t use these as typical elem ents o f a set. If / is used for a function, then d o n ’t use this as an integer. W rite sym bols that the reader w ould expect. To do otherw ise could very w ell confuse the reader.
10.
Use consistent symbols. U nless there is som e special reason to the contrary, use sym bols that “fit” together. O therw ise, it is distracting to the reader. Instead o f w riting If x and y are even integers, then x = 2a and y = 2 r for some integers a and r. replace 2r by 2 b (where then, o f course, we write “for som e integers a and b ”). On the other hand, you m ight prefer to w rite x — 2r and y = 2s.
W riting M athem atical E xpressions T here w ill be num erous occasions w hen you w ill w ant to write m athem atical expres sions in your assignm ent, such as algebraic equations, inequalities, and form ulas. If
Writing M athem atical Expressions
9
these expressions are relatively short, then they should probably be w ritten w ithin the text of the p ro o f or discussion. (W e’ll explain this in a m om ent.) If the expressions are rather lengthy, then it is probably preferred for these expressions to be w ritten as displays. For exam ple, suppose that we are discussing the B inom ial Theorem . (It’s not im portant if you d o n ’t recall w hat this theorem is.) I t’s possible that w hat w e are w riting includes the follow ing passage: For exam ple, if we expand (,a + b)4, then we obtain (a + b)4 = a 4 + 4 a 3b + 6 a 2 b 2 + 4 a b 3 + b4. It w ould probably be better to w rite the expansion o f (a + b ) 4 as a d isp lay , w here the m athem atical expression is placed on a line or lines by itself and is centered. This is illustrated below. For exam ple, if we expand (a + b )4 , then w e obtain (a + b )4 = a 4 + 4 a 3b + 6 a 2 b 2 + 4 a b 3 + b4. If there are several m athem atical expressions that are linked by equal signs and inequality sym bols, then we w ould alm ost certainly write this as a display. For exam ple, suppose that we w anted to write n 3 + 3 n 2 — n + 4 in term s o f k, w here n = 2 k + 1. A possible display is given next: Since n = 2 k + 1, it follow s that n 3 + 3 n~  n + 4 = (2k + l )3 + 3(2 k + i f  (2k + 1) + 4 = (8£3 + 12k 2 +
6
k + 1) + 3(4k 2 + 4 k + 1)  2 k  1 + 4
= 8/r3 + 24k2 + 16k + 7 = 8 k 3 + 2 4 k2 + 16 k + 6 + 1 = 2(4 k 3 + 12k 1 + 8Ar + 3) + 1. N otice how the equal signs are lined up. (We w rote two equal signs on one line since that line w ould have contained very little m aterial otherw ise, as well as to balance the lengths o f the lines better.) L et’s return to the expression (a + b ) 4 — a 4 + 4a 3b + 6a 2 b 2 + 4a b 3 + b 4 for the m om ent. If we w ere to w rite this expression in the text o f a paragraph (as we are doing) and if we find it necessary to write portions o f this expression on tw o separate lines, then this expression should be broken so that the first line ends w ith an operation or com parative sym bol s u c h a s + , —, < , > o r = . I n other w ords, the second line should not begin with one o f these sym bols. The reason for doing this is that ending the line w ith one o f these sym bols alerts the reader that m ore w ill follow; otherw ise, the reader m ight conclude (incorrectly) that the portion o f the expression appearing on the first line is the entire expression. Consequently, write For exam ple, if we expand (a + b)4, then we obtain (a + b)4 = a 4 + 4a 3b + 6 a 2 b 2 + 4a b 3 + b4. and not For exam ple, if we expand (a + b)4, then we obtain (a + b )4 = a 4 + 4a 3b + 6 a 2 b 2 + 4 a b 3 + b4.
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Chapter 0
Commimicating Ma lhematics If there is an occasion to refer to an expression that has already appeared, then this expression should have been w ritten as a display and labeled as below: (ia + b ) 4 = a 4 + 4 a 3b + 6 a 2 b 2 + 6 a b 2 + b4.
(1)
Then w e can sim ply refer to expression (1) rather than w riting it out each tim e.
C om m on W ords and P h rases in M athem atics There are some w ords and phrases that appear so often in m athem atical w riting that it is useful to discuss them. 1.
I
We
One
L e t’s I w ill now show that n is even. We will now show that n is even. One now show s that n is even. L et's now show that n is even.
T hese are four ways that w e m ight write a sentence in a proof. W hich of these sounds the best to you? It is not considered good practice to use “I” unless you are w riting a personal account o f som ething. O therw ise, “I ” sounds egotistical and can be annoying. U sing “one” is often awkward. U sing “w e” is standard practice in m athem atics. T his word also brings the reader into the discussion w ith the author and gives the im pression o f a team effort. T he w ord “le t’s” accom plishes this as well but is m uch less form al. There is a danger of being too casual, however. In general, your w riting should be balanced, m aintaining a professional style. O f course, there is the possibility o f avoiding all o f these words: The integer n is now show n to be even. 2.
Clearly O bviously O f course C ertainly T hese and sim ilar w ords can turn a reader off if w h at’s w ritten is n ot clear to the reader. It can give the im pression that the author is putting the reader down. These w ords should be used sparingly and w ith caution. If they are used, then at least be certain that w hat you say is true. There is also the possibility that the w riter (a student?) has a lack of understanding o f the m athem atics or is not being careful and is using these w ords as a coverup. T his gives us even m ore reasons to avoid these words.
3.
A ny
Each
E very This statem ent is true for any integer n.
Does this m ean that the statem ent is true for som e integer n or all integers «? Since the w ord any can be vague, perhaps it is best to avoid it. If by any, we m ean each or every, then use one o f these two w ords instead. W hen the w ord any is encountered, m ost o f the tim e the author m eans each or every.
Common Words and Phrases in Mathematics
4.
11
Since • • •, then ■■■ A num ber o f people connect these tw o w ords. You should use either “If ■■•, then • ■•” (should this be the intended m eaning) or “Since ■• •, it follow s that • • •” or, possibly, “ Since • • w e have • • For exam ple, it is correct to write If n 2 is even, then n is even. or Since n 2 is even, it follow s that n is even. or perhaps Since n 2 is even, n is even. but avoid Since n 2 is even, then n is even. In this context, the w ord since can be replaced by because.
5.
Therefore Thus H ence C onsequently So It fo llo w s that This im plies that This is tricky. M athem aticians cannot survive w ithout these words. Often w ithin a proof, we proceed from som ething w e ’ve ju st learned to som ething else that can be concluded from it. There are m any (m any!) openings to sentences w hich attem pt to say this. A lthough each o f the w ords or phrases Therefore Thus H ence C onsequently So It fo llo w s that This im plies that is suitable, it is good to introduce some variety into your w riting and not use the sam e w ords or phrases any m ore often than necessary.
6.
That Which These w ords are often confused w ith each other. Som etim es they are interchangeable; m ore often they are not. The solution to the equation is the num ber less than 5 that is positive.
(2)
The solution to the equation is the num ber less than 5 w hich is positive.
(3)
W hich of these two sentences is correct? The sim ple answ er is: Both are correct— or, at least, both m ight be correct. For exam ple, sentence (2) could be the response to the question: W hich o f the num bers 2, 3, and 5 is the solution o f the equation? Sentence (3) could be the response to the question: W hich of the num bers 4.9 and 5.0 is the solution o f the equation? The w ord that introduces a restrictive clause and, as such, the clause is essential to the m eaning o f the sentence. T hat is, if sentence (2) w ere w ritten only as “The solution to the equation is the num ber less than 5” then the entire m eaning is changed. Indeed, we no longer know w hat the solution o f the equation is. On the other hand, the w ord w hich does n ot introduce a restrictive clause. It intro duces a nonrestrictive (or parenthetical) clause. A nonrestrictive clause only provides additional inform ation that is not essential to the m eaning o f the sentence. In sentence (3)
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Communicating Mathematics
the phrase “w hich is positive" sim ply provides m ore inform ation about the solution. This clause m ay have been added because the solution to an earlier equation is negative. In fact, it w ould be m ore appropriate to add a com ma: The solution to the equation is the num ber less than 5, w hich is positive. For another illustration, consider the follow ing two statem ents: I always keep the m ath text that I like w ith me.
(4)
I always keep the m ath text w hich I like w ith me.
(5 )
W hat is the difference betw een these tw o sentences? In (4), the w riter o f the sentence clearly has m ore than one m ath text and is referring to the one that he/she likes. In (5), the w riter has only one m ath text and is providing the added inform ation that he/she likes it. The nonrestrictive clause in (5) should be set off by com m as: I always keep the m ath text, w hich I like, w ith me. A possible guideline to follow as you seek to determ ine w hether that or which is the proper w ord to use is to ask yourself: D oes it sound right if it reads “w hich, by the w ay”? In general, that is norm ally used considerably m ore often than which. H ence the advice here is: Bew are o f w icked w hich’s! W hile we are discussing the w ord that, w e m ention that the w ords assum e and suppose often precede restrictive clauses and, as such, the w ord that should im m edi ately follow one o f these words. O m itting that leaves us w ith an im plied that. M any m athem aticians prefer to include it rather than om it it. In other w ords, instead o f writing: A ssum e N is a norm al subgroup. m any w ould write A ssum e that N is a norm al subgroup.
Som e C losing Com m ents about W riting 1.
U se good English. W rite in com plete sentences, ending each sentence w ith a period (or a question m ark when appropriate) and capitalize the first w ord of each sentence. (Rem em ber: No sentence begins w ith a sym bol!)
2.
C apitalize theorem and lem m a as in T heorem 1 and L em m a 4.
3.
M any m athem aticians do not hyphenate w ords containing the prefix non, such as
4.
M any w ords that occur often in m athem atical w riting are com m only m isspelled. A m ong these are:
nonem pty,
nonnegative,
nondecreasing,
nonzero.
com m utative (independent o f order) com plem ent (supplem ent, balance, rem ainder) consistent (conform ing, agreeing)
Some Closing Comments about Writing
feasible (suitable, attainable) its (possessive, not â€œit isâ€?) occurrence (incident) parallel (nonintersecting) preceding (foregoing, form er) principle (postulate, regulation, rule) proceed (continue, m ove on) and, o f course, corollary, lem m a, theorem . There are m any pairs o f w ords that fit together in m athem atics (while interchanging w ords am ong the pairs do not). For exam ple, We We We We We
ask questions. pose problem s. present solutions. prove theorem s. solve problem s, and We conclude this chapter.
u
Sets
n this initial chapter, you w ill be introduced to, or m ore than likely be rem inded of, a fundam ental idea that occurs throughout m athem atics: sets. Indeed, a set is an object from w hich every m athem atical structure is constructed (as w e w ill often see in the succeeding chapters). A lthough there is a form al subject called set theory in w hich the properties o f sets follow from a num ber o f axiom s, this is neither our interest nor our need. It is our desire to keep the discussion o f sets inform al w ithout sacrificing clarity. It is alm ost a certainty that portions o f this chapter w ill be fam iliar to you. N evertheless, it is im portant that w e understand w hat is m eant by a set, how m athem aticians describe sets, the notation used w ith sets, and several concepts that involve sets. Y ou’ve been experiencing sets all your life. In fact, all o f the follow ing are exam ples o f sets: the students in a particular class w ho have an iPod, the item s on a shopping list, the integers. As a sm all child, you learned to say the alphabet. W hen you did this, you were actually listing the letters that m ake up the set w e call the alphabet. A set is a collection o f objects. The objects that m ake up a set are called its elem ents (or m em bers). The elem ents o f a softball team are the players: w hile the elem ents o f the alphabet are letters. It is custom ary to use capital (upper case) letters (such as A, B, C , S, X , Y ) to designate sets and low er case letters (for exam ple, a, b, c, s, x , y ) to represent elem ents o f sets. If a is an elem ent o f the set A, then w e w rite a e A; if a does not belong to A, then w e write a £ A.
I
1.1 D escrib in g a Set T here w ill be m any occasions w hen w e (or you) w ill need to describe a set. The m ost im portant requirem ent w hen describing a set is that the description m akes it clear precisely w hich elem ents belong to the set. If a set consists o f a sm all num ber o f elem ents, then this set can be described by explicitly listing its elem ents betw een braces (curly brackets) w here the elem ents are separated by com m as. Thus .S' — {I. 2, 3} is a set, consisting o f the num bers 1, 2 and 3. The order in w hich the elem ents are listed do esn ’t matter. Thus the set S ju st m entioned could be w ritten as S = {3, 2, 1} or S — {2, 1, 3}, for exam ple. They describe the same
14
1.1
Describing a Set
15
set. If a set T consists o f the first five letters o f the alphabet, then it is not essential that w e w rite T = {a, b, c, d , e}\ that is, the elem ents o f T need not be listed in alphabet ical order. O n the other hand, listing the elem ents o f T in any other order m ay create unnecessary confusion. The set A o f all people who signed the D eclaration o f Independence and later becam e president o f the U nited States is A = {John A dam s, T hom as Jefferson} and the set B o f all positive even integers less than 20 is B = {2, 4, 6 , 8 , 10, 12, 14, 16, 18}. Some sets contain too m any elem ents to be listed this way. Perhaps even the set B ju st given contains too m any elem ents to describe in this manner. In such cases, the ellipsis or “three dot notation” is often helpful. F or exam ple, X = { 1 ,3 ,5 .........49} is the set o f all positive odd integers less than 50, w hile Y = {2, 4 , 6 , . . . } is the set o f all positive even integers. The three dots m ean “and so on” for Y and “and so on up to ” for X . A set need not contain any elem ents. A lthough it m ay seem peculiar to consider sets w ithout elem ents, these kinds o f sets occur surprisingly often and in a variety o f settings. For exam ple, if S is the set o f real num ber solutions o f the equation A'2 + 1 = 0 , then S contains no elem ents. There is only one set that contains no elem ents, and it is called the em pty set (or som etim es the null set or void set). The em pty set is denoted by 0. We also write 0 = { }. In addition to the exam ple given above, the set o f all real num bers x such that .v" < 0 is also empty. The elem ents o f a set m ay in fact be sets them selves. The sym bol # below indicates the conclusion of an example.
Example 1.1
The set S = {1, 2, {1, 2}, 0} consists o f fo u r elem ents, two o f which are sets, namely, {1, 2} and 0. I f we w rite C = {1, 2}, then we can also w rite S = {1, 2, C, 0}. The set T = {0, {1, 2, 3}, 4, 5} also has fo u r elem ents, namely, the three integers 0, 4 and 5 and the set { 1, 2, 3}. Even though 2 € {1 ,2 , 3}, the num ber 2 is n o ta n elem ent i t f f i t h a t is, 2 f T . ♦
O ften sets consist o f those elem ents satisfying som e condition or possessing som e speci fied property. In this case, we can define such a set as S — {a : p (x )} , w here, by this, we m ean that S consists o f all those elem ents a satisfying som e condition p ( x ) concerning a . Some m athem aticians write S — [x  p(x)}; that is, som e prefer to w rite a vertical line rather than a colon (which, by itself here, is understood to m ean “such that”). For exam ple, if we are studying real num ber solutions o f equations, then S = {a : (a  I )i'.v + 2 )(a + 3) = 0} is the set o f all real num bers x such that (a — 1)(a + 2)(.v + 3) = 0; that is, S is the solution set o f the equation (x — 1 )(.v + 2)(x + 3) = 0. We could have w ritten S = {1, —2, —3}; however, even though this w ay o f expressing S is apparently sim pler, it does not tell us that w e are interested in the solutions o f a particular equation. The absolute value a  o f a re a l num ber a is a if a > 0; w hile \x\ = —a if a < 0. Therefore, T = {x : a  = 2} is the set o f all real num bers having absolute value 2, that is, T = {2, —2}. In the sets S and T that w e have ju st described, w e understand that “x ” refers to a real num ber x . If
16
Chapter 1
Sets
there is a possibility that this w ouldn’t be clear to the reader, then we should specifically say that x is a real number. W e’ll say m ore about this soon. The set P — [x : x has been a president o f the U nited States} describes, rather obviously, all those individuals w ho have been president o f the U nited States. So A braham L incoln belongs to P but B enjam in Fran k lin does not.
Example 1.2
L et A = {3, 4, 5, . . . , 20}. If B denotes the set consisting o f those elem ents o f A that are less than 8 , then we can write B = { x e A : x < 8} = {3, 4, 5, 6 , 7}.
#
Som e sets are encountered so often that they are given special notation. We use N to denote the set o f all positive integers (or natural num bers); that is, N = {1, 2, 3, . . .}. The set o fa ll integers (positive, negative, and zero) is denoted by Z. So Z = {___  2 ,  1 , 0, 1,2, . . .}. W ith the aid o f the notation w e’ve ju st introduced, w e can now describe the set E = { . . . , —4, —2, 0, 2 , 4 , . . . } of even integers by E = {y : y is an even integer} or E — {2x : x is an integer}, or as E — {y '■ y = 2 x for som e x e Z ) or E = {2x : x e Z }. Also, S = {;v“ : x is an integer} = {x 2 : x c Z} = {0, 1, 4, 9, ...} describes the set o f squares o f integers. The set o f real num bers is denoted by R, and the set o f positive real num bers is denoted by R + . A real num ber that can be expressed in the form f , w here m . n e Z and n / 0, is called a rational num ber. For exam ple, §, ^ , 17 = ^ and £ are rational num bers. O f course, 4 /6 = 2 /3 . The set o f all rational num bers is denoted by Q. A real num ber that is not rational is called irrational. The real num bers +/2, V 3 , $ 2 , n and e are know n to be irrational; that is, none o f these num bers can be expressed as the ratio of tw o integers. It is also know n that the real num bers w ith (infinite) nonrepeating decim al expansions are precisely the irrational num bers. There is no com m on sym bol to denote the set o f irrational num bers. We w ill often use I for the set o f all irrational num bers, however. Thus, s/2 e R and \ /2 Q ; so \/2 e I. For a set S, we w rite S  to denote the num ber o f elem ents in S. The num ber 5  is also referred to as the cardinal num ber or cardinality o f S. I f A = { 1 ,2} and B = {1, 2, {1, 2}, 0}, then Aj = 2 and B  = 4 . A lso, 0 = 0. A lthough the notation is identical for the cardinality o f a set and the absolute value o f a real number, w e should have no trouble distinguishing betw een the two. A set S is finite if ,S' = n for some nonnegative integer n. A set S is infinite if it is not finite. For the present, w e w ill use the notation S  only for finite sets S. In C hapter 10, w e will discuss the cardinality o f infinite sets. L et’s now consider a few exam ples o f sets that are defined in term s of the special sets w e have ju st described.
1.1
Example 1.3
Solution
17
L et D = [n € N : n < 9}, E = {x e Q : x < 9}, H = {x e R : x 2 — 2 = 0} a n d J = {x e Q : x 2 — 2 — 0 }. (a)
D escribe the set D by listing its elem ents.
(b)
Give an exam ple o f three elem ents that belong to E but do not belong to D.
(c)
D escribe the set H by listing its elem ents.
(d)
D escribe the set J in another manner.
(e)
D eterm ine the cardinality o f each set D , H and J .
(a)
D = {1,2, 3 ,4 , 5 ,6 , 7, 8 , 9J.
(b)
2 ,0 ,  3 .
(c)
H = {y/ 2 ,  7 2 ) .
(d)
J = 0.
(e)
Example 1.4
Describing a Set
/)  9 . / /!  2 and ./ j = 0 .
In which o f the fo llo w in g sets is the integer —2 an elem ent? 51 = {  1 ,  2 , {  1}, { 2 } , {  1, —2 }}, S2 = {x e N :  x e N}, S3 = {x e Z : x 2 — 2X}, S 4 = [x G Z : 'x  = —x},
Ss = { {  1.  2}, {  2 ,  3}, {  1,  3}}. Solution
The integer —2 is an elem ent o f the sets Si and S4 . For S4 ,  — 2 = 2 = —( —2). T he set 5 2 = 0. Since (—2 ) 2 = 4 and 2 ~ 2= 1/4 , it follow s that —2 ^ S 3 . B ecause each elem ent o f S5 is a set, it contains no integers. # A com plex num ber is a num ber o f the form a + b i, w here a, b e R and i = < /T A com plex num ber a + bi w here b = 0, can be expressed as a + 0/ or, m ore simply, as a. H ence a + Of — a is a real number. Thus every real num ber is a com plex number. Let C denote the set o f com plex num bers. If K — [x e C : x 2 + 1 — 0}, then K = {/, —/}, O f course, i f / . — {x C R : x 2 + 1 = 0 } , then L = 0. You m ight have seen that the sum o f two com plex num bers a + hi and c + d i is (a + c) + (b + d )i. w hile their product is
(<a + b i) ■(c + d i ) — ac + a d i + bci + b d i2 = (ac — bd) + (ad + bc)i. The special sets that w e've ju st described are now sum m arized below: sym bol N Z Q I R C
fo r th e set of natural num bers (positive integers) integers rational num bers irrational num bers real num bers com plex num bers
18
Chapter 1
Sets
1.2 Subsets A set A is called a su b se t o f a set B if every elem ent o f A also belongs to B . If A is a subset o f B , then w e w rite A c B . If A, B and C are sets such that A c B and B C C, then A c C . To see w hy this is so, suppose that som e elem ent x belongs to A. Because A c B , it follow s that x e B . B ut B C ( '. w hich im plies that x 6 C . Therefore, every elem ent that belongs to A also belongs to C and so A c C . This property o f subsets m ight rem ind you of the property o f real num bers w here if a, b, c e R such that if a < b and b < c, then a < c. For the sets A = {1, 3, 6 } and Y = { 1 ,2 ,3 . 5, 6 ), we have X C  Y . Also, N c Z and Q c R. In addition, R c C. Since Q c R and R c C , it therefore follow s that Q C C . M oreover, every set is a subset o f itself. Example 1.5 S o lu tio n
F ind two sets A and B such that A is both an elem ent o f and a subset o f B . Suppose that w e seek two sets A and B such that A e B and A c B . L et’s start w ith a sim ple exam ple for A, say A = {1}. Since w e w ant A e B , the set B m ust contain the set {1} as one o f its elem ents. On the other hand, we also require that A C S , so every elem ent o f A m ust belong to B . Since 1 is the only elem ent o f A, it follow s that B m ust also contain the num ber 1. A possible choice for B is then B = {1, {1}}, although B = {1 ,2 . {1}} w ould also satisfy the conditions. f In the follow ing exam ple, w e w ill see how we arrives at the answ er to a question asked there. This is a prelude to logic, w hich w ill be discussed in C hapter 2.
Example 1.6
Two sets A and B have the property that each is a subset o f {I, 2. 3, 4, 5} a n d A  = i?  = 3. Furtherm ore, (a) (b) (c) (d) (e)
1 belongs 2 belongs 3 belongs 4 belongs 5 belongs
to to to to to
A but not to B . B but not to A. exactly one o f A and B . exactly one o f A and B . at least one o f A and B .
W hat are the p ossibilities f o r the set A ? S o lu tio n
B y (a) and (b), 1 e A and 1 £ B, w hile 2 e B and 2 € A. By (c), 3 belongs to A or B but not both. By (d), 4 belongs to A or B but not both. If 3 and 4 belong to the same set, then either 3 and 4 both belong to A or 3 and 4 both belong to B. Should it occur that 3 £ A and 4 e A, then 1 ^ B, 3 ^ B and 4 0 B . This m eans that B  7^ 3. On the other hand, if 3 e B and 4 e B , then 3 ^ A and 4 ^ A. T herefore, A contains none of 2, 3 and 4 and so A  ^ 3 . We can therefore conclude that 3 and 4 belong to different sets. The only way that A  = B  = 3 is for 5 to belong to both A and B and so either A = {1,3. 5} or A = { 1 .4 ,5 } . # If a set C is not a subset o f a set D , then we w rite C 2 D . In this case, there m ust be som e elem ent o f C that is not an elem ent o f D . O ne consequence of this is that the em pty set 0 is a subset o f every set. If this w ere not the case, then there m ust be some
1.2
Subsets
19
set A such that 0 £ A . B ut this w ould m ean there is som e elem ent, say x , in 0 that is not in A. However, 0 contains no elem ents. So 0 C A for every set A. Example 1.7
L e t S = {1, {2}, {1,2}}.
Solution
(a)
D eterm ine w hich o f the follow ing are elem ents o f S : 1, { 1} ,2 ,{ 2 } ,{ 1, 2 },{{1, 2 }}.
(b)
D eterm ine w hich o f the follow ing are subsets o f S : {1}, {2 } ,{ 1, 2 },{{1} ,2 } ,{ 1,{ 2 }},{{1},{2 }},{{1, 2}}.
(a)
The follow ing are elem ents o f S: 1, {2}, {1, 2}.
(b)
The follow ing are subsets o f S: {1}, {1, {2}}, {{1, 2}}.
$
In a typical discussion o f sets, we are ordinarily concerned w ith subsets o f some specified set U , called the universal set. For exam ple, w e m ay be dealing only with integers, in w hich case the universal set is Z, or we m ay be dealing only w ith real num bers, in w hich case the universal set is R. O n the other hand, the universal set being considered m ay be neither Z nor R. Indeed, U m ay not even be a set o f num bers. Som e frequently encountered subsets o f R are the socalled “intervals,” w hich you have no doubt encountered often. For a , b e R and a < b, the open interval (a , b) is the set (a , b) — {x e R : a < x < b}. Therefore, all o f the real num bers f , \/5 , e, 3, n , 4.99 belong to (2, 5), but none o f the real num bers V 2, 1.99, 2, 5 belong to (2, 5). For a , b e R and a < b, the closed interval [a, b] is the set [a, b] = {x e R : a < x < b}. W hile 2, 5 ^ (2, 5), we do have 2 ,5 e [2, 5], The “interval” [a, a ] is therefore {a}. Thus, for a < b, w e have (a , b) c [a, b ]. F o ra , b e R a n d a < b, the halfopen or halfclosed intervals [a, b ) and (a, b] are defined as expected: [</. b) — {.v € R : a < x < b } and (a, b] = {x € R : a < x < b}. For ( i e R , the infinite intervals (  o o , a ), (—oo, a }, (a, oo) and [a , oo) are defined as (—oo, a) = {x e R : x < a}, ( —oo, </] — {.v R : x < (a, oo) = {x e R : x > a}, [a, oo) = {x e R : x > a}.
a},
The interval (—oo, oo) is the set R. N ote that the infinity sym bols oo and —oo are not real num bers; they are only used to help describe certain intervals. T herefore, [1. oo], for exam ple, has no meaning. Two sets A and B are equal, indicated by w riting A — I), if they have exactly the same elem ents. A nother way o f saying A = B is that every elem ent o f A is in B and every elem ent o f B is in A, that is, A C B and B c A. In particular, w henever some elem ent x belongs to A, then x e B because A c B. A lso, if )> is an elem ent o f B , then because B C A, it follow s that y e A. T hat is, w henever an elem ent belongs to one of these sets, it m ust belong to the other and so A = B . This fact will be very useful to us
Chapter 1
Sets
Figure 1.1
Venn diagrams for two sets A and B
in C hapter 4. If A / B , then there m ust be som e elem ent belonging to one o f A and B but not to the other. It is often convenient to represent sets by diagram s called Venn diagram s. For exam ple, Figure 1.1 shows Venn diagram s for tw o sets A and B . T he diagram on the left represents tw o sets A and B that have no elem ents in com m on, w hile the diagram on the right is m ore general. T he elem ent x belongs to A but not to B, the elem ent y belongs to B but not to A, the elem ent z belongs to both A and B, w hile w belongs to neither A nor B . In general, the elem ents o f a set are understood to be those displayed w ithin the region that describes the set. A rectangle in a Venn diagram represents the universal set in this case. Since every elem ent under consideration belongs to the universal set, each elem ent in a Venn diagram lies w ithin the rectangle. A set A is a proper subset o f a set B if A c B but A ^ B . If A is a proper subset o f B, then we w rite A C B. For exam ple, if S = {4, 5, 7} and T = {3, 4, 5, 6 , 7}, then S C T . (A lthough w e w rite A C B to indicate that A is a proper subset o f B, it should be m entioned that som e prefer to write A c fi to indicate that A is a proper subset of B. Indeed, there are som e who w rite A C B, rather than A C S , to indicate that A is a subset o f B . We w ill follow the notation introduced above, however.) T he set consisting o f all subsets o f a given set A is called the pow er set o f A and is denoted by V( A) . Example 1.8
For each set A below, determ ine V{ A) . In each case, determ ine  A a nd \P (A )\. (a) A = 0,
Solution
(b) A = [a, b},
(c) A = {1,2, 3).
(a)
V ( A ) : = {0 }. In this case., A = 0 and \V (A )\ = 1.
(b)
V ( A ) : = {0 , {a}, {b}, {a, b }}. In this case, A = 2 and \P {A )\ = 4.
(c)
V I A ) : = {0 , {l},{2},{3}, {1,2}, {1,3}, {2, 3}, {1,2,3}}. In this case, A = 3 and W( A ) \ = 8.
N otice that for each set A in Exam ple 1.8, w e have \ V( A)\ = 2 /4'. In fact, if A is any finite set, w ith n elem ents say, then V { A ) has 2" elem ents; that is, V ( A ) \ â€” 2 1A for every finite set A. (Later we w ill explain w hy this is true.)
1.3
Example 1.9
Set Operations
21
I f C = {0, {0}}, then P (C ) = {0, {0 } ,{ { 0 } } ,{ 0 , {0}}}. It is im portant to note that no two o f the sets 0, {0} and {{0}} are equal. (An em pty box and a box containing an em pty box are not the sam e.) For the set C above, it is therefore correct to write 0 c C, 0 C C , 0 e C, {0} c C , {0} c C , {0} 6 C . as w ell as
{{0}} Q C . {{0}} £ C , {{0}} e V(C).
♦
1.3 Set O perations Just as there are several ways o f com bining two integers to produce another integer (addition, subtraction, m ultiplication and som etim es division), there are several w ays to com bine two sets to produce another set. The union o f tw o sets A and B . denoted by A U B, is the set o f all elem ents belonging to A or B, that is, A U B = {„v : l e A o r i e B } . The use o f the w ord “or” here, and in m athem atics in general, allow s an elem ent of A U B t o belong to both A and B . T hat is, x is in A U B if x is in A or x is in B or x is in both A and B . A Venn diagram for A U B is show n in F igure 1.2. Example 1.10
For the sets A \ = {2, 5, 7, 8}, A 2 = {1, 3, 5} a n d A 3 = {2, 4, 6 , 8 }, we have Aj U A2 = {1,2, 3, 5, 7, 8}, A t U A 3 = {2,4, 5 ,6 , 7 ,8 }, A2 U A 3 = { 1 ,2 , 3 ,4 , 5 ,6 , 8 } . A lso. N U Z = Z and Q U I = R.
♦
The intersection o f two sets A and B is the set o f all elem ents belonging to both A and B . The intersection o f A and B is denoted by A fl B . In sym bols, A H B = {x : x G A and x e B}. A Venn diagram for A fl B is show n in Figure 1.3.
Figure 1.2
A Venn diagram for A U B
Chapter 1
Example 1.11
Sets
For the sets A •. A i an d A 3 described in E xam ple 1.10, A i n A2 — {5}, A \ fl A3 = {2, 8 } and A2 Pi A3 — 0. A lso, N f l Z = N and Q fl R = Q.
4
For every tw o sets A and B , it follow s that A n B c A U B. To see w hy this is true, suppose that x is an elem ent belonging t o A H B . Then x belongs to both A and B . Since x e A, for exam ple, x e A U B and so A fl £> C A U 5 . If tw o sets A and B have no elem ents in com m on, then A fl B = 0 and A and B are said to be d isjo in t. Consequently, the sets A 2 and A 3 described in E xam ple 1.10 are disjoint; however, A 1 and A 3 are not disjoint since 2 and 8 belong to both sets. A lso, Q and I are disjoint. The d iffe ren ce A — B o f tw o sets A and B ( also w ritten as A \ B by som e m athe m aticians) is defined as A — B = {x : .v c A a n d x £ B }. A Venn diagram for A — B is show n in Figure 1.4. Example 1.12
For the sets A \ = {2, 5, 7, 8} and A 2 = {1, 3, 5} in E xam ples 1.10 and 1.11, A; — A2 — {2, 7, 8} and A 2 — A \ — {1, 3}. F urtherm ore, R — Q = I. 4
Figure 1.4
A Venn diagram for A — B
1.3
Set Operations
23
U
Figure 1.5 Example 1.13
A Venn diagram for A
For A = {x e R : x  < 3}, R = [x e R : \x\ > 2} a nd C = j.v e R : \x — 1
< 4}:
(a) Express A , B and C using interval notation. (b) D eterm ine A C \ B , A — B , B n C , B U C , B — C a n d C — B.
Solution (a) A = [  3 , 3], B = (  o o ,  2 ) U (2, oo) and C = [  3 , 5], (b) A n B = [  3 ,  2 ) U (2, 3], A  B = [  2 , 2 ] , B T C = [  3 ,  2 ) U (2, 5], B U C = (  o c , oo), B  C = (  o o ,  3 ) U (5, oo) and C  B = [  2 , 2], 4 Suppose that w e are considering a certain universal set U , that is, all sets being discussed are subsets o f U . For a set A, its com plem ent is A — U — A = {x : x e U and x <? A}. H U = Z, then N = {0, —1, —2 , .. .} ; w hile if U = R, then Q = I. A Venn diagram for A is show n in Figure 1.5. The set difference A — B is som etim es called the relative com plem ent o f B in A. Indeed, from the definition, A — B = [x : x e A and x £ B). The set A  B can also be expressed in term s o f com plem ents, namely, A — B = A O B. This fact w ill be established later. Example 1.14
L et U = { 1 ,2 .........10} be the universal set, A — {2, 3, 5, 7} and B = {2, 4, 6 , 8 , 10}. D eterm ine each o f the follow ing:
(a) B, Solution
Example 1.15
(b) A  B,
(c)
An
(a)
B
(b)
A  B ■= {3,5,7}.
(c)
An
(d)
B = B  {2,4, 6, 8, 10}.
=
b
(d) B.
{1.,3, 5,7,9}. b
= {3,5,7} = A  B.
L et A = {0, {0}, {0, {0}}}. (a)
D eterm ine w hich o f the follow ing are elem ents o f A: 0, {0}, {{0}}.
(b)
D eterm ine A.
24
Chapter 1
Sets (c)
D eterm ine w hich of the follow ing are subsets o f A: 0, {0}, {{0}}. For (d)(i), determ ine the indicated sets.
(d)
{0}n A
(e)
{{0}} n A
(f)
{{{0 }}} n a
(g)
{0} U A
(h) {{0}} U A (i) {{{0}}} U A.
S o lu tio n
W hile 0 and {0} are elem ents o f A, {{0}} is not an elem ent of A. (b) The set A has three elem ents: 0, {0}, {0, {0}}. Therefore,  A j = 3. (c) The integer 0 is not a set and so cannot be a subset o f A (or a subset o f any other set). Since 0 e A and {0} e A, it follow s that {0} c A and {{0}} C A. (d) Since 0 is the only elem ent that belongs to both {0}and A, it follow s that {0 } n a = {0 }. (e)
Since {0} is the only elem ent that belongs to both {{0}} and A, it follow s that {{0}} n A = {{0}}.
(f)
Since {{0}} is not an elem ent o f A, it follow s that {{{0})} and A are disjoint sets and so {{{0}}} n A = 0.
(g)
Since 0 6 A, it follow s that (0) U A = A.
(h)
Since {0} e A, it follow s that {{0}} U A = A.
(i)
Since {{0}} ÂŁ A, it follow s that {{{0}}} U A = {0, {0}, {{0}}, {0, {0}}}.
â™Ś
1=4 Indexed C ollections of Sets We w ill often encounter situations w here m ore than tw o sets are com bined using the set operations we described earlier. In the case o f three sets A, B and C , the standard Venn diagram is show n in Figure 1.6.
1.4
Indexed Collections of Sets
25
The union A U B U C is defined as U />’
C — .\ : x e A or x e 6 or x e C}.
Thus, in order for an elem ent to belong to A U B U C , the elem ent m ust belong to at least one o f the sets A, B and C . Because it is often useful to consider the union of several sets, additional notation is needed. The union o f the n > 2 sets A t, A z, . . *, A„ is denoted by A \ U A 2 U • • • U A„ or U "=1 A, and is defined as n
A/ = {a : x e A/ for som e i . 1 < / < n}. i= 1
Thus, for an elem ent a to belong to U "=1 A ,, it is necessary that a belongs to at least one o f the sets A t, A 2, . . . , A„. Example 1.16
L et B \ = {1, 2}, B 2 = (2, 3 } ,.. . , Bio  {10, 11}; that is, Bj = {i, i + 1} fo r i = 1, 2, . . . . 10. D eterm ine each o f the fo llo w in g : 5
10
7
(a) ( J B/.
(b) [ J B ,.
(c) ( J Bj.
i= 1
Solution
i= 1
(a)  J / > ';  { 1. 2 .........6 }. i= 1 7 (c) ( J B/ = { 3 , 4 , . . . , 8}.
k
(d)
i —3
B,, where 1 < j < k < 10. i= j
(b) I J b , = {1, 2 , . . . , 11} ;= 1 k
(d)  J Bf = { j, j + 1, . . . , k + 1}.
We are often interested in the intersection o f several sets as well. The intersection o f the n > 2 sets A i , A2, A„ is expressed as A t H A2 H ■■• fl A n Or A, and is defined by n A, = (a : x e A, for every i , 1 < i < n), i=1
The next exam ple concerns the sets m entioned in Exam ple 1.16. Example 1.17
L et Bj = {i, i + 1} fo r i = 1, 2, . . . , 10. D eterm ine the fo llo w in g : 10 (a) P ] Bj. /=i
(b) Bj n />’/ ..).
j +1 (c) p  Bj. w here 1 < j < 10. i =j
k
(d)
 Bj w here 1 < j < k < 10. '=j
So lu tio n
10 (a) f ] Bj = 0. Z= 1
(b) Bj n B ,+ i = {; + 1}.
j+ l (c) f ] Bj = { j + 1}.
k
(d) p  Bj = [ j + 1} i f * = j + 1; While P  Bj = 0 i f k > j + l .
26
Chapter 1
Sets T here are instances w hen the union or intersection o f a collection of sets cannot be described conveniently (or perhaps at all) in the m anner m entioned above. For this reason, we introduce a (nonem pty) set I , called an in d e x set, w hich is used as a m echanism for selecting those sets w e w ant to consider. F or exam ple, for an index set I , suppose that there is a set Sa for each a e l . We w rite {S„}o,e/ to describe the collection of all sets Sa , w here a e l . Such a collection is called an in d ex ed collection o f sets. We define the union o f the sets in {Sa }aei by [ J Sa = {.v : x e Sa for some a e l } , ael
and the intersection o f these sets by
H ence an elem ent a belongs to U s e / Sa ® a belongs to at least one o f the sets in the collection {.S',, w hile a belongs to H a e / ^ a belongs to every set in the collection {5ff}„e/. We refer to U a e / Sa as the union o f the collection {Sia }'a s / and f \ e/ Sa as the intersection o f the collection {Sff}„6/. Just as there is nothing special about our choice o f i in U "=i (that is, w e could ju st as w ell describe this set by U "= i A j , say), there is nothing special about a in U ffe/ Sacould also describe this set by U r e / ■&“ The variables i and a above are dum m y variables and any appropriate sym bol could be used. Indeed, we could w rite J or som e other sym bol for an index set. Example 1.18
For n e N, define Sn = {«, 2n}. For exam ple, Si = {1, 2}, S% — {2, 4} and S 4 = {4. 8}. Then Si U S% U S4 = (1, 2, 4, 8). We can also describe this set by m eans o f an index set. I f we let I = { 1 , 2 , 4}, then
Example 1.19
For each n e N, define A„ to be the closed interval [—4, i ] o f real num bers; that is,
So A \ = [—1, 1], A 2 — [ — t , 2], ^3 = [ — 3 , 5] and so on. We have now defined the sets A 1, A 2 , A 3 , . . . . The union o f these sets can be w ritten as A \ U A 2 U A 3 U • • • or U “ i A,. Using N as an index set, w e can also write this union as U « sn Since A n c A] = [—1, I] f o r every n e N, it fo llo w s that U neN ~ A n fo r every n e N ; in fa c t, n , !eN 4 * = (0}♦
Example 1.20
L et A denote the set o f the letters o f the alphabet, that is, A = { a , b , . . . , zj. For a e A, let A a consist o f a an d the two letters that fo llo w a . So A a = {a, b. c} a nd A* = {/>. c, cl}. B y A y , w e w ill m ean the se t { y , z , a] a nd A :  {z, a, b}. H ence A „ = 3fo r every a e A. Therefore U aeA A a — A. Indeed, if B — [a, d, g, j, m, p, s, v, y},
1.5
Example 1.21
Partitions of Sets
27
then l j as£ A a — A as well. On the other hand, i f I = { p , q , r } , then U « e /
=
[p. q. r, s, t] w hile □„<=/
♦
= I''}
L et 5 = { 1 ,2 .........10}. Each o f the sets
Si = {1, 2 , 3, 4}, S2 = {4, 5, 6 , 7, 8 } a n d S3 • {7, 8 , 9, 10} is a subset o f S. A lso, S\ U S 2 U S3 = S .T h is union can be described in a num ber o f ways. D efine / — {1 .2 .3 } and J — {S i, S2, S3}. Then the union o f the three sets belonging to J is precisely S\ U S 2 U S3, which can also be written as
S = Si u
s2u
3
S3 = U Si = U s a = y 1= 1
ae/
X.
♦
XeJ
1.5 P artitions o f Sets Recall that tw o sets are disjoint if their intersection is the em pty set. A collection S of subsets o f a set A is called p a irw ise d isjo in t if every tw o distinct subsets that belong to S are disjoint. For exam ple, let A = {1, 2, . . ., 7}, B — {1, 6 }, C = {2, 5}, D = {4, 7} and S = {B, C, D}. Then S is a pairw ise disjoint collection o f subsets o f A since B H C — B n D = C n D = 0. On the other hand, let A ' = {1, 2, 3}, B' = {1,2}, (”  {1. 3}, D' = {2, 3} and S' = {S', C ', D'}. A lthough S' is a collection o f subsets o f A' and B ' f l C ' n D ' — 0, the set S' is not a pairw ise disjoint collection of sets since B ' f l C ' / 0 , for exam ple. Indeed, B ' Pi D ’ and C ' P D ' are also nonem pty. We w ill often have the occasion (especially in C hapter 8) to encounter, for a nonem pty set A, a collection S o f pairw ise disjoint nonem pty subsets o f A w ith the added property that every elem ent o f A belongs to som e subset in S . Such a collection is called a p a r titio n o f A. A p a r titio n o f A can also be defined as a collection S of nonem pty subsets of A such that every elem ent o f A belongs to exactly one subset in S . Furtherm ore, a partition o f A can be defined as a collection S o f subsets o f A satisfying the three properties:
Example 1.22
( 1)
X f  0 for every set X e 5 ;
(2)
for every tw o sets X , Y e S , either X — Y or X C\Y =
(3)
U x& s X — A .
C onsider the fo llo w in g collections o f subsets o f the se t A — { 1 ,2 , 3, 4, 5, 6 }: 51 = {{1,3, 6 }, {2, 4}, {5}}; 52 = { { 1 ,2 ,3 } ,{ 4 } ,0 ,{ 5 ,6 } } ; 53 = {{1, 2 }, {3, 4. 5}, {5, 6}}; & = {{1,4}, {3, 5}, {2}}. D eterm ine which o f these sets are partitions o f A .
28
Chapter 1
S o lu tio n
Sets
The set Si is a partition o f A. The set S i is not a partition o f A since 0 is one o f the elem ents o f S 2 ■T he set S3 is not a partition o f A either since the elem ent 5 belongs to tw o distinct subsets in S3 , namely, {3, 4, 5} and {5, 6 }. Finally, S4 is also not a partition o f A because the elem ent 6 belongs to no subset in S4 . ♦ As the w ord pa rtitio n probably suggests, a partition o f a nonem pty set A is a division o f A into nonem pty subsets. The partition Si o f the set A in Exam ple 1.22 is illustrated in the diagram show n in Figure 1.7. For exam ple, the set Z o f integers can be partitioned into the set of even integers and the set o f odd integers. The set R o f real num bers can be partitioned into the set R o f positive real num bers, the set o f negative real num bers and the set {0 } consisting of the num ber 0. In addition, R can be partitioned into the set Q o f rational num bers and the set I o f irrational num bers.
Example 1.23
So lu tio n
L e t A = { 1 , 2 , , 12}.
(a)
Give an exam ple o f a partition S o f A such that S  = 5 .
(b)
Give an exam ple o f a subset T o f the partition S in (a) such that 7"  = 3 .
(c)
List all those elem ents B in the partition S in (a) such that \B  = 2 .
(a)
We are seeking a partition S o f A consisting o f five subsets. O ne such exam ple is S = {{1, 2}, {3, 4}, {5, 6 }, {7, 8, 9}, {10, 11, 12}}.
(b)
We are seeking a subset T o f S (given in (a)) consisting o f three elem ents. One such exam ple is T = {{1,2}, {3, 4}, {7, 8, 9}}.
(c)
We have been asked to list all those elem ents o f S (given in (a)) consisting of tw o elem ents o f A. These elem ents are: {1, 2}, {3, 4}, {5, 6 }. ♦
1.6 C artesian P roducts o f Sets W e’ve already m entioned that w hen a set A is described by listing its elem ents, the order in w hich the elem ents o f A are listed d o esn 't matter. T hat is, if the set A consists o f two elem ents x and y , then A = {x, y} = [y, x}. W hen we speak o f the ordered pair (x, y ),
Exercises for Chapter 1
29
however, this is another story. The ordered pair (x, y) is a single elem ent consisting of a pair of elem ents in w hich x is the first elem ent (or first coordinate) o f the ordered pair (x, y) and y is the second elem ent (or second coordinate). M oreover, for tw o ordered pairs i.v, v) and fw , z) to be equal, that is, (x, y ) = (w , z), w e m ust have x = w and y = z. So, if x / y, then (x, y) / (v, x). The C artesian product (or sim ply the product) A x B o f two sets A and B is the set consisting o f all ordered pairs w hose first coordinate belongs to A and w hose second coordinate belongs to B . In other words, A x B = {(a, b) : a € A and b e B }. Example 1.24
I f A = {x, y} and B = {1, 2, 3}, then A x B = {(x, 1), (x, 2), (x, 3), (y, 1), (y, 2), (y, 3)}, while B x A = {(1, x ), (1, y), (2, x ), (2, y), (3, x ), (3, y ) . Since, fo r exam ple, (x, 1) € A x B and (x, 1) ^ B x A , these two sets do n ot contain the sam e elem ents; so A x B ^ B x A . Also, A x A — {(x, x ), (x, y), (y, x ), (y, y)} and B x 5 = {( 1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), ( 3 , 1), (3, 2), (3, 3)}.
♦
We also note that if . 4 = 0 o r B = 0 , then A x B = 0. The C artesian product R x R i s the set o f all points in the E uclidean plane. For exam ple, the graph o f the straight line y = 2x + 3 is the set {(x, y) e R x R : y = 2 x + 3}. For the sets A = {x , y} a n d S = { l , 2, 3} given in E xam ple 1.24, A  = 2 and \B\ = 3, w hile  A x B \ = 6 . Indeed, for all finite sets A and B , A x B, = ,A • \B\. Cartesian products w ill be explored in m ore detail in C hapter 7.
E X E R C IS E S FOR C H A PT E R 1 Section 1.1: Describing a Set 1.1. Which of the following are sets? (a) 1 , 2 , 3 (b) {1,2}, 3 (c) {{1},2},3 (d) {1. {2}. 3} (e) {1, 2, a, b }.
30
Chapter 1
Sets
1.2. Let S — {—2, —1, 0, 1, 2, 3}. Describe each of the following sets as {x e S : p(x)}, where p( x) is some condition on x. (a) A = {1,2,3} (b) B = (0, 1,2,3} (c) C = {  2 ,  1 } (d) D = { 2 ,2 ,3 } 1.3. Determine the cardinality of each of the following sets: (a) A= {1,2, 3, 4, 5} (b) B = {0, 2, 4, . . . , 20} (c) C = {25, 26, 27........... 75} (d) D = {{1,2}, {1,2, 3, 4}} (e) E = {0} (f) F = {2 , {2,3,4}} 1.4. Write each of the following sets by listing its elements within braces. (a) A = {n e Z :  4 < n < 4} (b) B
= {/? e Z : n 2
< 5}
(c) C
= {n € N : n 3
< 100}
(d) I) = (e) E
{.vr R:
x 2 — x = 0}
= [x e R : x 2 + 1 = 0}
1.5. Write each of the following sets in the form {jc e Z : p(x)}, where p ( x ) is a property concerning x . (a) A = {  1 ,  2 ,  3 , . . . } (b) B = { 3 ,  2 , . . . , 3} (c) C = { 2 ,  1 , 1,2} 1.6. The set E = [2x : x e Z} can be described by listing its elements, namely E = {___—4,  2 . 0, 2, 4, ...}. List the elements of the following sets in a similar manner. (a) A = [2x (b) 5 = [An (c) C = {iq
1 : x E Z} :e +1
Z} :
q e
Z}
1.7. The set £ = { ...,  4 .  2 , 0, 2. 4. ...} of even integers can be described by means of a defining condition by E = [y = 2,v : x e Z} = {2x : x e Z}. Describe the following sets in a similar manner. (a) A = { ....  4 ,  1 , 2 , 5 ,8 , ...} (b) B =  1 0 ,  5 . 0 , 5, 10, . . .} (c) C = { 1 ,8 ,2 7 ,6 4 , 125, ...} 1.8. Let A = {/; e Z : 2 < \n\ < 4}, B = {x e Q : 2 < x < 4}, C = {x e R : X2  (2 + V 2)x + 2 ^ 2 = 0} and D = {x e Q : x 2  (2 + V 2 )x + 2 s/2 = 0}. (a) Describe the set A by listing its elements. (b) Give an example of three elements that belong to B but do not belong to A. (c) Describe the set C by listing its elements. (d) Describe the set D in another manner. (e) Determine the cardinality of each of the sets A, C and D.
Exercises for Chapter 1
.31
1.9. For A = {2, 3, 5, 7, 8. 10, 13), let B == {x G A : x = y + z, where y, z e A} and C = [r G B : r + ,v G B for some s G 5 ). Determine C .
Section 1.2: Subsets 1.10. Give examples of three sets A , B and C such that (a)
Ac
(b)
A e B, B
B C C
(c)
A G B and A C C.
g
C and A £ C
1.11. Let (a, b)be an open interval of real numbers and let c G (a, b). Describe an open interval / centered at c such that / c (a, b). 1.12. W hich of the A = [n g Z 5 = (« e Z C = {» G Z
following sets are equal? :\n\ < 2} D = jn e Z : i r < 1) :ii3 = n) E = {—1, 0, 1}. :n 1 < n }
1.13. For a universal set U = {1, 2, . . . , 8} and two sets A = {1, 3, 4, 7} and B = {4, 5, 8}, draw a Venn diagram that represents these sets. 1.14. Find V ( A ) and \V(A)\ for (a) A = {1,2}. (b) A = {0, 1, {a}}. 1.15. Find V ( A ) for A = {0, {0}}. 1.16. Find VCP({ 1})) and its cardinality. 1.17. Find V ( A ) and \V (A )\ for A = {0, 0, {0}}. 1.18. For A = {x : x = 0 or x G P({0})}, determine V( A) . 1.19. Give an example of a set S such that (a) S c V(S) (b) S e V( N) (c) S c P (N ) and \S\ = 5 (d) S G P ( N) and S = 5 1.20. Determine whether the following statements are true or false. (a) If {1} G V( A) , then 1 e A but {1} $ A. (b) If A, B and C are sets such that A C V ( B ) C C and A = 2, then C  can
be 5 but C  cannot be 4.
(c) If a set B has one more element than a set A, then V ( B ) has at leasttwo more elements than V{A). (d) If four sets A, B . C and D are subsets of {1, 2, 3} such that A = \B\ = C  = D  = 2, then at least two of these sets are equal. 1.21. Three subsets A, B and C of {1, 2, 3, 4, 5} have the same cardinality. Furthermore, (a) 1 belongs to A and B but not to C . (b) 2 belongs to A and C but not to B . (c) 3 belongs to A and exactly one of B and C. (d) 4 belongs to an even number o f A, B and C.
32
Chapter 1
Sets
(e) 5 belongs to an odd number of A, B and C. (f) The sums of the elements in two of the sets A, B and C differ by 1. W hat is B I
Section 1.3: Set Operations 1.22. Let U = {1, 3 , . . . s, 15} be the universal set, A = {1, 5, 9, 13}, and B = {3, 9, 15}. Determine the following: (a) A U B (b) A n B (c) A  B (d) B  A (e) A ( f ) A f l B. 1.23. Give examples of two sets A and B such that A  B \ = A D B\ = \B  A = 3. Draw the accompanying Venn diagram. 1.24. Give examples of three sets A. B and C such that B
but B — A = C — A.
1.25. Give examples of three sets A, B and C such that (a) A € B, A c C and B £ C (b) B 6 A, B c C and A n C ^ 0 (c) A G B, B c C and A g C. 1.26. Let (7 be a universal set and let A and B be two subsets of U. Draw a Venn diagram for each of the following sets. (a) A U S
(b) I n B
(c) A H B
(d) A U B .
W hat can you say about parts (a) and (b)? parts (c) and (d)? 1.27. Give an example of a universal set U, two sets A and B and accompanying Venn diagram such that  A n B  = \ A  B \ = \B  A = T U S  = 2 . 1.28. Let A, B and C be nonempty subsets of a universal set U . Draw a Venn diagram for each of the following set operations. (a) (C  B ) U A (b) C n ( A  B). 1.29. Let A = {0, {0}, {{0}}}. (a) Determine which of the following are elements of A: 0, {0}, {0, {0}}. (b) Determine  A j. (c) Determine which of the following are subsets of A: 0, {0}, {0, {0}}. For (d)(i), determine the indicated sets. (d) 0 n A (e) {0} n A (f) {0, {0}} n A (g) 0 U A (h) {0} U A (i) {0 ,{ 0 }}UA. 1.30. Let A = (x G R : \x  1 < 2}, B = {x e R : j.v > 1} and C =  x g R :
\x
+ 2\ < 3}.
(a) Express A, B and C using interval notation. (b) Determine each of the following sets using interval notation: A U B, A H B, B n c , B  C . 1.31. Give an example of four different sets A. B , C and D such that ( l ) A U f i = {1,2} and C Pi D = {2, 3} and (2) if B and C are interchanged and U and D are interchanged, then we get the same result.
Exercises for Chapter 1
33
1.32. Give an example of four different subsets A, B , C and D of {1, 2, 3, 4} such that all intersections of two subsets are different. 1.33. Give an example of two nonempty sets A and B such that {A U B. A D B, A — B, B — A) is the power set o f some set. 1.34. Give an example of two subsets A and B of { 1 ^ , 3} such that all of the following sets are different: A U B, A U B, A U B, A U B, A n B, A n B, A n B, A n B. 1.35. Give examples of a universal set U and sets A, B and C such that each of the following sets contains exactly one element: A n B n C, (A n B) — C, (A n C ) — B, (B Pi C) — A, A — (S U C), f i  ( A U C), C — (A U S ) , A U B U C . Draw the accompanying Venn diagram.
Section 1.4: Indexed Collections of Sets 1.36. For a re a l n um bers, define S, to be the interval [r — 1, r + 2]. Let A = {1, 3, 4). Determine IJae/t $<* ancl OareA ' 1.37. Let A = {1, 2, 5}, B = {0, 2, 4}, C = {2, 3, 4} and S = {A, B, C}. Determine U x sS % an^ PlxeS %■ 1.38. For a real number r, define A, = j/'2}, B, as the closed interval [r — 1, r + 1] and C, as the interval (r, oo). For S = {1, 2, 4}, determine (a) UaeS A a and DaeS A a
(b) UaeS Ba and PlaeS Boi (c ) UaeS Ca and PlaeS Ca1.39. Let A = {a, b , . . . , zj be the set consisting of the letters of the alphabet. For a e A, let Aa consist of a and the two letters that follow it, where A y = {v, z, a } and A, = {z, a, b }. Find a set S c A of smallest cardinality such that IJaes A a — A. Explain why your set S has the required properties. 1.40. For i e Z, let A, = {/ — 1, i + 1). Determine the following: 5
5
5
(a) I j A a (b) ( J ( A , n A/+1) (c) ( J ( A 2i_i n A2i+1). i=i i=i i=i 1.41. For each of the following, find an indexed collection {A„}„eN of distinct sets (that is, no two sets are equal) satisfying the given conditions. (a) n ^ i An = {0} and ( J “ , A n = [0 , 1] (b) n r = i An = {  L O , 1} and U “ i A„ = Z 1.42. For each of the following collections of sets, define a set A n for each n e N such that the indexed collection M„},„=n is precisely the given collection of sets. Then find both the union and intersection of the indexed collection of sets. (a) { [ 1 , 2 + 1 ) , [ 1 , 2 + 1 / 2 ) , [ 1 , 2 + 1 / 3 ) , . . . } (b) { (1 , 2), (  3 / 2 , 4), (  5 / 3 , 6), (  7 / 4 , 8) , . . .}. 1.43. F o rr e R +, let A, = [x e R : x < /•}. Determine (JreR+ A, and Q sr+ A, 1.44. Each of the following sets is a subset of A = { 1 , 2 , . . . , 10}: A, = {1, 5, 7, 9, 10}, A2 = {1, 2, 3, 8, 9}, A3 = {2, 4, 6, 8, 9}, A4 = {2, 4, 8}, A5 = {3, 6, 7}, A6 = {3, 8, 10}, A7 = {4, 5, 7, 9}, A8 = {4, 5, 10}, A g = {4, 6 , 8}, A i0 = {5, 6 , 10}, A „ = {5, 8, 9}, A12 = {6, 7, 10}, A 13 = {6 , 8,9}. Find a set / c {1, 2, ___13} such that for every two distinct elements j , k 6 I , A ; fl A* = 0 and ( J /e/ Ai \ is maximum. 1.45. For n € N, let A„ = (  £ , 2  ±, ). Determine ( J n€N A„ and n „ sN A „■
34
Chapter 1
Sets
Section 1.5: Partitions of Sets 1.46. W hich of the following are partitions of A = {a, b, c, d , e, / , g}? For each collection of subsets that is not a partition of A, explain your answer. (a) Si = {{a , c, 4 g}, {b, /} , {d}} (b) S2 = {{a , h. c, d}. [e, /}} (e) S3 = {A} (d) S4 = {{a}, 0, {b , c, d j, {e, / , g}} (e) Ss = {{a, c, d}, {,b, ,!>!. {e}, {ft, /}}. 1.47. W hich of the following sets are partitions of A = {1, 2, 3, 4, 5}? ( a  S i = {{1,3}, {2 , 5}} (b) = {{1, 2}. {3,4.5}} (c) S 3 = {{1, 2}, {2, 3}, {3, 4}, {4, 5}} (d) S4 = A. 1.48. Let A = {1, 2, 3, 4. 5, 6}. Give an example of a partition S of A such that
= 3.
1.49. Give an example of a set A with jAj = 4 and two disjoint partitions Si and Sn of A with .Si = S2] = 3 . 1.50. Give an example of a partition of N into three subsets. 1.51. Give an example of a partition of Q into three subsets. 1.52. Give an example of three sets A, Si and S2 such that Si is a partition of A, S2 is a partition of Si and \Si\ < Si < A. 1.53. Give an example of a partition of Z into four subsets. 1.54. Let A = {1, 2 ........ 12}. Give an example of a partition S of A satisfying the following requirements: (i) S] = 5, (ii) there is a subset T of S such that \T \ = 4 and  UxeT V;  10 and (iii) there is no element B â‚Ź S such that B  = 3. 1.55. A set S is partitioned into two subsets Si and S2. This produces a partition <p\ of S where V \ = {Si, S2} and so \Vi  = 2. One of the sets in V \ is then partitioned into two subsets, producing a partition P 2 of S with IP2I = 3. A total of \Vi  sets in V 2 are partitioned into two subsets each, producing a partition V 3 of S. Next, a total of \V2\ sets in V 3 are partitioned into two subsets each, producing a partition V 4 of S. This is continued until a partition V 6 of S is produced. What is "P6? 1.56. We mentioned that there are three ways that a collection S of subsets of a nonempty set A is defined to be a partition of A. Definition 1 The collection S consists of pairwise disjoint nonempty subsets of A and every element of A belongs to a subset in S . Definition 2 The collection S consists of nonempty subsets of A and every element of A belongs to exactly one subset in S . Definition 3 The collection S consists of subsets of A satisfying the three properties (1) every subset in S is nonempty, (2) every two subsets of A are equal or disjoint and (3) the union of all subsets in S is A. (a)
Show that any collection S of subsets of
(b)
Show that any
collection S of subsets of
A
A
satisfying Definition 1
satisfying Definition 2
satisfies Definiti
(c)
Show that any
collection S of subsets of
A
satisfying Definition 3
satisfies Definiti
Section 1.6: Cartesian Products of Sets 1.57. Let A = {x, y, z } and B = {x, y}. Determine A x B. 1.58. Let
A=
{1, {1}, {{1}}}. Determine A x A.
1.59. For
A=
{ÂŤ, b}, determine A x V(A) .
1.60. For
A=
{0, {0}}, determine A x V{A).
1.61. For
A=
{1,2} and B = {0}, determine A x B an d V( A ) x V( B) .
1.62. Describe the graph of the circle whose equation is x* + y 2 = 4 as a subset o f R x R .
satisfies Definition 2.
Additional Exercises for Chapter 1
1.63. List the elements of the set S = Euclidean x y  plane.
{(a ,
y) e Z x Z :
a 
35
+ [y;j = 3}. Plot the corresponding points in the
1.64. For A — {1,2} and B = {1}, determine V (A x
B).
1.65. For A = {.v £ R : x  11 < 2} and B = {y e in the xyplane belonging to A x B.
R : \y  4]<
1.66. For A = {a e R : \a\ « !} and B = {b e R : \b\ = 1},give jryplane belonging to (A x B ) U (B x A).
2), givea geometric description of the po
a geometric description of the points in the
A D D IT IO N A L E X E R C IS E S FOR C H A PT E R 1 1.67. The set T = {2k + 1 : k e Z} can be described as T = {___ 3 ,  1 , 1. 3 , . ..}. Describe the following sets in a similar manner. (a) A = {4* + 3 ; k £ Z} (b) B = {5k  1 : k e Z}. 1.68. Let S = {—10, —9, . . . , 9. 10}. Describe each of the following sets as {x e S : p(x)}, where p( x) is some condition on x . (a) A = { 1 0 ,  9 , . . .,  1 , 1, . . . , 9 , 10} (b) B = {  1 0 ,  9 . . . . ,  1 , 0 } (c) C = { 5 ,  4 ........ 0 , 1 ..........7} (d) D = {  1 0 ,  9 ........ 4 . 6 , 7 , . . . , 10}. 1.69. Describe each of the following sets by listing its elements within braces. (a) {a e Z : a 3 — 4a = 0} (b) {a e R : a  = 1 } (c) {m e N : 2 < m < 5} (d) {» e N : 0 < n < 3} (e) {k& Q : k 2  4 = 0} (f) {k e Z : 9k 2  3 = 0} (g) {k € Z : 1 < k 2 < 10}. 1.70. Determine the cardinality of each of the following sets. (a) A = {1,2, 3, {1,2, 3},4, {4}} ib) B = {a e R : ;.r =  I} (c) C = {m e N : 2 < m< 5 } (d) l) = {/( C N : n < 0} (e) E = {k e N : 1 < k 2 < 100} I'fi /■• = \k ft Z : 1 < k 2 < 100}. 1.71. For A = {—1,0, 1} and B = {a, )>}, determine Ax B. 1.72. Let U = {1, 2, 3} be the universal set and let A= {1, 2}, B = {2, 3} and C = {1,3}. Determine the following. (a) (A U S )  (S n C ) (b) A__ (c) B U C (d) A x S. 1.73. Let A = {1, 2, . . . , 10}. Give an example of two sets S and S such that S C P (A ), S = 4, S e S and B  = 2. 1.74. For A =
{1} and C = {1,2}. give an example of a set B such that V{ A) c B c V{C).
1.75. Give examples of two sets A and B such that A fl 'P(A) e B and V ( A ) c
A U S.
36
Chapter 1
Sets
1.76. W hich of the following sets are equal? A = {n e Z :  4 < n < 4} D = j i e Z : x 3 = 4x} 5 = {x e N : 2x + 2 = 0} E = { 2 , 0, 2}. C = {x e Z : 3x  2 = 0} 1.77. Let A and B be subsets of some unknown universal set U . Suppose that A = {3, 8, 9}, A — B * {1, 2}, B — A = {8} and A fl B = {5, 7}. Determine U , A and 5 . 1.78. Let / denote the interval [0, oo). For each r S / , define Ar = {(x, y) f R x R : x 2 + y 2 = r 2) By = {(x, j ) e R l R : x 2 + y 2 < r 2} C, = {(x, y) e R x R :x 2+ y 2 > r 2). (a) Determine ( J re/ A,, and
A,..
(b) Determine J (..; Br and
Br.
(c) Determine Q
, C, and f  rg/ C,.
1.79. Give an example of four sets A u A2, A3, A4 such that A, n A j \ = \i  j \ for every two integers i and j with 1 < / < j < 4. 1.80. (a) Give an example of two problems suggested by Exercise 1.79 (above). (b) Solve one of the problems in (a). 1.81. Let A = {1, 2, 3}, B = {1, 2, 3, 4} and C = {1, 2, 3, 4, 5}. For the sets S and T described below, explain whether l^l < 7 , 151 > T  or S = 7+ (a) Let B be the universal set and let S be the set of all subsets X of B for which \X\ / X . Let T be the set of 2element subsets of C. (b)
Let S be the set of all partitions of the set A and let T be the set of 4element subsets of C .
(c)
Let S = {(b, a) . b & B , a £ A, a
b is odd} and let T be the set of all nonempty proper subsets of A.
1.82. Give an example of a set A = {1, 2, . . . , k } for a smallest k G N containing subsets A i, A2, A3 such that  Ai  A j =  A j  Aj\ = \i  j \ for every two integers i and j with 1 < i < j < 3. 1.83. (a) For A = {—3, —2 , . . . , 4} and B = {1, 2 , . . . , 6},determine S = {(a, b) e A x B : a 2 + b 1 = 25}. (b) For C = {a G B \ (a, b) G S} and D = [b determine C x D.
g
A : (a, b) G S}, where A, B , S are the sets in
(a),
1.84. For A = {1, 2, 3}, let B be the set of 2element sets belonging to V ( A ) and let C be the set consisting of the sets that are the intersections of two distinct elements of B. Determine D = V(C). 1.85. For a real number r , let Ar = {/, /• + 1}. L et S = {x £ R : x 2 + 2x  1 = 0}. (a) Determine B = A s x A t for the distinct elements s , t G S , where s < t. (b) Let C = [ab : {a, b) G B\. Determine the sum of the elements of C .
2J Logic
n m athem atics our goal is to seek the truth. A re there connections betw een tw o given m athem atical concepts? If so, w hat are they? U nder w hat conditions does an object possess a particular property? F inding answ ers to questions such as these is im portant, but w e cannot be satisfied only w ith this. We m ust be certain that w e are right and that our explanation for w hy w e believe w e are correct is convincing to others. The reasoning we use as w e proceed from w hat w e know to w hat we w ish to show m ust be logical. It m ust m ake sense to others, not ju st to ourselves. T here is joint responsibility here, however. It is the w riter’s responsibility to use the rules o f logic to give a valid and clear argum ent w ith enough details provided to allow the reader to understand w hat w e have w ritten and to be convinced. It is the read er’s responsibility to know the basics o f logic and to study the concepts involved sufficiently w ell so that he or she will not only be able to understand a w ellpresented argum ent but can decide as w ell w hether it is valid. Consequently, both w riter and reader m ust be fam iliar w ith logic. A lthough it is possible to spend a great deal of tim e studying logic, w e w ill present only w hat we actually need and w ill instead use the m ajority o f our tim e putting w hat w e learn into practice.
I
2.1 Statem ents In m athem atics w e are constantly dealing w ith statem ents. By a statem ent we m ean a declarative sentence or assertion that is true or false (but not both). Statem ents therefore declare or assert the truth o f som ething. O f course, the statem ents in w hich w e w ill be prim arily interested deal w ith m athem atics. For exam ple, the sentences The integer 3 is odd. The integer 57 is prim e. are statem ents (only the first o f w hich is true). Every statem ent has a truth value, nam ely true (denoted by T ) or false (denoted by F). We often use P , Q and R to denote statem ents, or perhaps P \ , Pi, ■■• , Pn if there 37
38
Chapter 2
Logic are several statem ents involved. We have seen that P] : The integer 3 is odd. and P2 : The integer 57 is prime. are statem ents, w here P\ has truth value T and Pi has truth value F. Sentences that are im perative (com m ands) such as Substitute the num ber 2 for .y . Find the derivative o f f ( x ) = e~x cos 2x. or are interrogative (questions) such as Are these sets disjoint? W hat is the derivative o f f ( x ) — e~x cos 2x1 or are exclam atory such as W hat an interesting question! How difficult this problem is! are not statem ents since these sentences are not declarative. It m ay not be im m ediately clear w hether a statem ent is true or false. For exam ple, the sentence “The 100th digit in the decim al expansion o f tt is 7.” is a statem ent, but it may be necessary to find this inform ation in a Web site on the Internet to determ ine w hether this statem ent is true. Indeed, for a sentence to be a statem ent, it is not a requirem ent that we be able to determ ine its truth value. T he sentence “The real num ber r is rational.” is a statem ent provided w e know w hat real num ber r is being referred to. W ithout this additional inform ation, however, it is im possible to assign a truth value to it. This is an exam ple o f w hat is often referred to as an open sentence. In general, an op en se n ten c e is a declarative sentence that contains one or m ore variables, each variable representing a value in som e prescribed set, called the d o m a in of the variable, and w hich becom es a statem ent w hen values from their respective dom ains are substituted for these variables. F or exam ple, the open sentence “3x = 12” w here the dom ain o f x is the set o f integers is a true statem ent only w hen x = 4. A n open sentence that contains a variable x is typically represented by P( x ) , Q ( x ) or R( x) . If P( x ) is an open sentence, w here the dom ain o f x is S, then we say P( x ) is an op en sen ten ce ov er th e d o m a in S. A lso, P ( x ) is a statem ent for each x e S. For exam ple, the open sentence P{ x ) : (x  3)2 < 1 over the dom ain Z is a true statem ent w hen x e {2, 3, 4} and is a false statem ent otherw ise.
Example 2.1
For the open sentence P( x , y ) : \x + 1 +  j  = 1
2.2
P
Q
The Negation of a Statement
P Q
P Q R
T
T
T
T F F T F F
T T T
39
T T T F F T F F
F T T F T F F F T F F F Figure 2.1
Truth tables for one, two and three statements
in tw o variables, suppose that the dom ain o f the variable x is S = {—2, —1, 0, 1} and the dom ain o f the variable y is T — {—1, 0, 1}. Then P {  1 , 1 ) :  (  1 ) + 1  + 1 = 1 is a true statem ent, while P(l,1):
 l + lj + j  l  = l
is a fa lse statem ent. In fa c t. P( x , y) is a true statem ent when (x , y) e { (  2 , 0 ), (  1,  1), (  1, 1), (0 , 0 )}, while P( X, y ) is a fa ls e statem ent fo r all other elem ents (,v. y ) e S x T .
♦
The possible truth values o f a statem ent are often listed in a table, called a tr u th
table. The truth tables for tw o statem ents P and Q are given in Figure 2 .1 . Since there are two possible truth values for each o f P and Q , there are four possible com binations of truth values for P and O . The truth table showing all these com binations is also given in Figure 2.1. If a third statem ent R is involved, then there are eight possible com binations o f truth values for P , Q and R . This is displayed in F igure 2.1 as well. In general, a truth table involving n statem ents P x, P2, ■■■, P„ contains 2" possible com binations of truth values for these statem ents and a truth table show ing these com binations w ould have n colum ns and 2" rows. M uch o f the tim e, w e w ill be dealing w ith tw o statem ents, usually denoted by P and Q ; so the associated truth table w ill have four row s w ith the first two colum ns headed by P and Q . In this case, it is custom ary to consider the four com binations o f the truth values in the order TT, TF, FT, FF, from top to bottom .
2.2 The N egation of a Statem ent M uch of the interest in integers and other fam iliar sets o f num bers com es not only from the num bers them selves but from properties o f the num bers that result by perform ing operations on them (such as taking their negatives, adding or m ultiplying them or
40
Chapter 2
Logic com binations o f these). Similarly, m uch o f our interest in statem ents com es from in Â vestigating the truth or falseness o f new statem ents that can be produced from one or m ore given statem ents by perform ing certain operations on them. O ur first exam ple concerns producing a new statem ent from a single given statem ent. The n eg a tio n o f a statem ent P is the statem ent: n o t P. and is denoted by ~ P . A lthough ~ P could always be expressed as I t is n o t th e case th a t P . there are usually better w ays to express the statem ent
Example 2.2
.
For the statem ent P\ : The integer 3 is odd. described above, we have ~ / >i : It is not the case that the integer 3 is odd. but it w ould be m uch preferred to write ~ / >i : The integer 3 is not odd. or better ye t to write ~ P i : The integer 3 is even. Sim ilarly, the negation o f the statem ent P i : The integer 57 is prim e. considered above is ~ P i : The integer 57 is not prim e. N ote that ~ P i is fa lse, while ~ P i is true.
#
Indeed, the negation o f a true statem ent is always false and the negation o f a false statem ent is alw ays true; that is, the truth value o f ~ P is opposite to that o f P . This is sum m arized in F igure 2.2, w hich gives the truth table for ~ P (in term s o f the possible truth values of P ).
Figure 2.2
P
~P
T F
F T
The truth table for negation
2.3
The Disjunction and Conjunction of Statements
41
2.3 The D isju n ction and C onjunction o f Statem ents For tw o given statem ents P and Q , a com m on w ay to produce a new statem ent from them is by inserting the w ord “or" or “and" betw een P and Q. T he disjunction o f the statem ents P and Q is the statem ent P or Q and is denoted by P V Q . The disjunction P V O is true if at least one o f P and Q is true; otherw ise, P V Q is false. T herefore, P V Q is true if exactly one o f P and Q is true or if both P and Q are true.
Example 2.3
For the statem ents P\ : The integer 3 is odd. and Pi I The integer 57 is prim e. described earlier, the disjunction is the new statem ent Pi V P i: E ither 3 is odd or 57 is prim e. which is true since a t least one o f P\ and Pi is true (nam ely, P\ is true). O f course, in this case exactly one o f P\ and Pi is true. ♦ For tw o statem ents P and Q , the truth table for P V Q is show n in F igure 2.3. This truth table then describes precisely w hen P V Q is true (or false). A lthough the truth o f “P or Q ” allow s for both P and Q to be true, there are instances w hen the use o f “or” does not allow that possibility. For exam ple, for an integer n, if we w ere to say “n is even or n is odd,” then surely it is not possible for both “« is even” and “n is odd” to be true. W hen “or” is used in this m anner, it is called the exclusive or. Suppose, for exam ple, that V = {Si, S i, ■■■, &}, w here k > 2 , is a partition o f a set S and x is som e elem ent o f S. If X € .S'i or x € Si is true, then it is im possible for both x e Si and x e S 2 to be true.
p
Q
PVQ
T
T
T F F T F F
T F
T
Figure 2.3
T
The truth table for disjunction
42
Chapter 2
Logic
P Q P AQ T T T F T F F T F F Figure 2.4
F F
The truth table for conjunction
The conjunction o f the statem ents P and O is the statem ent: P and 0 and is denoted by P A Q. The conjunction P A Q is true only w hen both P and Q are true; otherw ise, P A Q is false. Example 2.4
For P\ : The integer 3 is odd. and P2 : The integer 57 is prim e., the statem ent P\ A Pi : 3 is odd and 57 is prim e,
is fa ls e since Pi is fa ls e and so not both P\ a n d P i are true.
#
The truth table for the conjunction o f tw o statem ents is show n in Figure 2.4.
2.4 The Im plication A statem ent form ed from two given statem ents in w hich we w ill be m ost interested is the im plication (also called the conditional). For statem ents P and Q . the im plication (or conditional) is the statem ent I f P . then Q. and is denoted by P => Q . In addition to the w ording “If P . then Q ," w e also express P Q in w ords as P im plies Q. The truth table for P =>• O is given in Figure 2.5. N otice that P =>• Q is false only w hen P is true and Q is false (P otherw ise). Example 2.5
Q is true
For P\ : The integer 3 is odd. and Pi : The integer 57 is prim e., the im plication P\ =$■ P i I f 3 is an odd integer, then 57 is prim e.
P
Q P=*Q
T T T F F T F F Figure 2.5
T F T T
The truth table for implication
2.4
The Implication
43
is a fa ls e statem ent. The im plication Po
P\ 'If 57 is prim e, then 3 is odd.
is true, however.
♦
W hile the truth tables for the negation ~ P , the disjunction P V Q and the conjunc tion P A Q are probably w hat one w ould expect, this m ay not be so for the im plication P => Q . There is am ple justification, however, for the truth values in the truth table of P => Q. We illustrate this w ith an example. Example 2.6
A student is taking a m ath class (let’s say this one) and is currently receiving a B + . H e visits his instructor a fe w days before the fin a l exam ination and asks her, “Is there any chance that I can get an A in this course?’’ H is instructor looks through her grade book an d says, " I f you earn an A on the fin a l exam, then you w ill receive an A f o r yo u r fin a l grade!' We now check the truth o r fa lsen e ss o f this im plication based on the various com binations o f truth values o f the statem en ts P : You earn an A on the fin a l exam. and Q : You receive an A f o r yo u r fin a l grade, which m ake up the implication.
A n a ly sis
Suppose first that P and Q are both true. T hat is, the student receives an A on his final exam and later learns that he got an A for his final grade in the course. D id his instructor tell the truth? I think w e w ould all agree that she did. So if P and Q are both true, then so too is P => Q , w hich agrees w ith the first row o f the truth table of Figure 2.5. Second, suppose that P is true and Q is false. So the student got an A on his final exam but did not receive an A as a final grade, say he received a B . Certainly, his instructor did not do as she prom ised (as she w ill soon be rem inded by her student). W hat she said was false, w hich agrees w ith the second row o f the table in Figure 2.5. Third, suppose that P is false and Q is true. In this case, the student did not get an A on his final exam (say he earned a B ) but w hen he received his final grades, he learned (and was pleasantly surprised) that his final grade was an A. How could this happen? Perhaps his instructor was lenient. Perhaps the final exam was unusually difficult and a grade o f B on it indicated an exceptionally good perform ance. Perhaps the instructor m ade a m istake. In any case, the instructor did not lie; so she told the truth. Indeed, she never prom ised anything if the student did not get an A on his final exam. T his agrees w ith the third row o f the table in Figure2.5. Finally, suppose that P and Q are both false. T hat is, suppose the student did not get an A on his final exam and he also did not get an A for a final grade. The instructor did not lie here either. She only prom ised the student an A i f he got an A on the final exam. O nce again, she did not prom ise anything if the student did not get an A on the final exam. So the instructor told the truth and this agrees w ith the fourth and final row o f the table. t
44
Chapter 2
Logic In sum m ary then, the only situation for w hich P =# Q is false is w hen P is true and Q is false (so ~ O is true). T hat is, the truth tables for P  > Q ) and P A (~ Q ) are the same. W e’ll revisit this observation again soon. We have already m entioned that the im plication P =>■ 0 can be expressed as both “If P , then Q " and “ P im plies 2  ” In fact, there are several w ays o f expressing P => Q in w ords, nam ely: If P , then Q. g i f P. P im plies Q . P only if Q. P is sufficient for g . Q is necessary for P . It is probably not surpri sing that the first three o f these say the same thing, but perhaps not at all obvious that the last three say the sam e thing as the first three. C onsider the statem ent “ P only i f Q." This says that P is true only under the condition that Q is true; in other words, it cannot be the case that P is true and Q is false. Thus it says that if P is true, then necessarily Q m ust be true. We can also see from this that the statem ent “Q is necessary for P ’ has the same m eaning as “ P only if Q." The statem ent “ P is sufficient for Q ” states that the truth o f P is sufficient for the truth o f Q. In other words, the truth o f P im plies the truth o f Q \ that is, “ P im plies Q V
2.5 M ore on Im plications We have ju st discussed four ways to create new statem ents from one or tw o given statem ents. In m athem atics, however, w e are often interested in declarative sentences containing variables and w hose truth or falseness is only know n once we have assigned values to the variables. The values assigned to the variables com e from their respective dom ains. These sentences are, o f course, precisely the sentences w e have referred to as open sentences. Just as new statem ents can be form ed from statem ents P and Q by negation, disjunction, conjunction or im plication, new open sentences can be constructed from open sentences in the same manner. Example 2.7
C onsider the open sentences P\ ( x) : x = —3. and P.?(.v) : \x\ = 3, where x e R, that is, where the dom ain o f x is R in each case. We can then form the fo llo w in g open sentences: ~ Pi (x) : x ^  3 . Pl'.v) v P 2 (x ) :
X
=  3 or xj = 3.
P\ ( x) A P 2( x ) : x = —3 and x 
= 3.
P i(x ) —> P 2(x) : I f x = —3, then \x\ = 3.
2.5
More on Implications
45
For a specific real num ber x , the truth value o f each resulting statem ent can be deter mined. For exam ple, ~ P ](—3) is a fa ls e statem ent, w hile each o f the rem aining sentences above results in a true statem ent when x = —3. B oth P \(2) v P j(2) a n d P\{2) A P i(2) are fa ls e statem ents. On the other hand, both ~ P\ (2) and P\ (2) => P 2 O.) a re ?rwe state m ents. In fa c t, fo r each real num ber x ^ —3, f/ze im plication P \(x ) =>• A O O w a true statem ent since P \(x ) : x = —3 is a fa ls e statem ent. Thus P \(x ) =>■ /*2(•*•') w true fo r all i e R .W e w ill see that open sentences which result in true statem ents fo r a ll values o f the dom ain w ill be especially interesting to us. L isted below are various w ays o f wording the im plication P \(x ) =>• P i{x ): I f x = —3, then x  = 3. \x\ = 3 i f x = —3. x = —3 im plies that \x\ = 3. x = —3 only i f\ x \ — 3. x — —3 is sufficient fo r \x\ = 3. x  = 3 is necessary fo r x = —3.
♦
We now consider another exam ple, this tim e from geometry. You m ay recall that a triangle is called e q u ila te ra l if the lengths o f its three sides are the sam e, w hile a triangle is isosceles if the lengths o f any tw o o f its three sides are the same. F igure 2.6 shows an isosceles triangle T\ and an equilateral triangle Ti Actually, since the lengths o f any tw o o f the three sides o f 7? are the same, T2 is isosceles as well. Indeed, this is precisely the fact w e w ish to discuss. Example 2.8
For a triangle T , let P (T ) : T is equilateral, and Q (T ) : T is isosceles. Thus, P ( T ) and Q (T ) are open sentences over the dom ain S o f all triangles. C onsider the im plication P (T ) = $ Q (T ), where the dom ain then o f the variable T is the set S. For an equilateral triangle T \, both P ( J \) and Q (T \) are true statem ents and so P{T{) =$■ Q iT i) is a true statem ent as well. I f T 2 is n o t an equilateral triangle, then P {T i) is a fa ls e statem ent and so F i j i ) => Q fT j) is true. Therefore, P ( T ) =4>Q (T ) is a true statem ent fo r a ll T € S. We now state P ( T ) =>• Q (T ) in a variety o f ways: I f T is an equilateral triangle, then T is isosceles. A triangle T is isosceles i f T is equilateral. A triangle T being equilateral im plies that T is isosceles. A triangle T is equilateral only i f T is isosceles. For a triangle T to be isosceles, it is sufficient that T be equilateral. For a triangle T to be equilateral, it is necessary that T be isosceles.
♦
Chapter 2
Logic
N otice that at tim es w e change the w ording to m ake the sentence sound better. In general, the sentence P in the im plication P => O is com m only referred to as the h y p o th e sis or p re m ise o f P =>• Q, w hile Q is called the co n clu sio n of P =» Q. It is often easier to deal w ith an im plication w hen expressed in an “if, then” form. This allow s us to identify the hypothesis and conclusion m ore easily. Indeed, since im plications can be stated in a wide variety o f w ays (even in addition to those m entioned above), being able to rew ord an im plication as “if, then” is especially useful. For exam ple, the im plication P ( T ) =$■ Q {T ) described in Exam ple 2.8 can be encountered in many w ays, including the following: • • ® •
L et T be an equilateral triangle. Then T is isosceles. Suppose that T is an equilateral triangle. Then T is isosceles. Every equilateral triangle is isosceles. W henever a triangle is equilateral, it is isosceles.
We now investigate the truth or falseness o f im plications involving open sentences for values o f their variables. Example 2.9
L e t S — {2, 3, 5} a nd let P (n ) : n 2 — n + 1 is prim e, and Q {n) : n 3  n + 1 is prime. be open sentences over the dom ain S. D eterm ine the truth or fa lsen e ss o f the im plication P (n ) —> <2 (ft) fo r each n e S.
Solution
In this case, w e have the following: P ( 2 ) : 3 is prim e. Q t2) : 7 is prim e,
P (3 ) : 7 is prim e. Q (3) : 25 is prim e.
P ( 5 ) : 21 is prime. Q (5) : 121 is prime.
Consequently, P {2) =>• Q (2) and P ( 5) => Q (5 ) are true, w hile P ( 3) Example 2.10
0 ( 3 ) is false. ♦
L e t S = {1,2} and let T — {• 1. 4}. A lso, let P (x , y ) : [\x + y\  ]x — 3>ll = 2. and Q (x , y) : x ^ 1 = y*. be open sentences, where the dom ain o f the variable x is S and the dom ain o fy is T . D eter m ine the truth or fa lsen e ss o f the im plication P (x , y ) =>■ Q (x , y ) fo r a ll (x, y ) e S x T .
Solution
For (x, y) = (1, —1), w e have P ( 1,  1 ) =#■ Q ( l,  1) : If 2 = 2, then 1 =  1 . w hich is false. For (x, y) = (1, 4), w e have P ( l , 4 ) =► 0 ( 1 ,4 ) : If 2 = 2. then w hich is also false.
1= 4 .
For (x, y) = (2, —1), w e have P (.2,  • ! ) =» 0 ( 2 .  1 ) : If 2 = 2,then 1= 1.
w hich is true; w hile for (x, y) = (2, 4), we have P(.2, 4) =» 2 ( 2 , 4) : If 2 = 4, then 32 = 16. w hich is true.
$
2.6
The Biconditional
47
2.6 The B iconditional For statem ents (or open sentences) P and Q, the im plication Q =>• P is called the converse o f P =>• Q . The converse o f an im plication w ill often be o f interest to us, either by itself or in conjunction w ith the original im plication. Example 2.11
For the statem ents P i : 3 is an odd integer.
P j : 57 is prim e,
the converse o f the im plication P —> P: : I f 3 is an odd integer, then 57 is prim e. is the im plication Pi => lJi '■ I f 57 is prim e, then 3 is an odd integer.
♦
For statem ents (or open sentences) P and Q , the conjunction ( P => G ) a ( G =» P ) o f the im plication P =>■ Q and its converse is called the b ic o n d itio n a l o f P and Q and is denoted by P Q. For statem ents P and Q, the truth table for P 0 can therefore be determ ined. This is given in Figure 2.7. From this table, we see that P $> Q is true w henever the statem ents P and Q are both true or are both false, w hile P Q is false otherw ise. T hat is, P O Q is true precisely w hen P and Q have the same truth values. The biconditional P O Q is often stated as P is eq u iv ale n t to Q.
P Q P=>Q T T T
Q=>P
(P =» Q) A (Q => P)
T
T
F
F
T
F
F T F F
T T
F T
F T
T
P * Q P <£> Q T T T T F F F F Figure 2.7
T F
F T
The truth table for a biconditional
Chapter 2
Logic or P if a n d only if Q . or as P is a n ec essary a n d sufficient co n d itio n fo r Q. For statem ents P and Q, it then follow s that the biconditional “P if and only if Q ” is true only w hen P and Q have the sam e truth values.
Example 2.12
The biconditional 3 is an odd integer i f and only i f 51 is prim e, is fa lse ; while the biconditional 100 is even i f a nd only if 101 is prim e. is true. Furtherm ore, the biconditional 5 is even i f a nd only i f A is odd. is also true.
4
The phrase “if and only if” occurs often in m athem atics and w e shall discuss this at greater length later. For the present, w e consider tw o exam ples involving statem ents containing the phrase “if and only if.” Example 2.13
We noted in E xam ple 2 .7 that fo r the open sentences P\ (.v) : x —  3 . a n d P2(.x) : x  = 3. over the dom ain R , the implication P \{x ) =$■ P i(x ) : I f x = —3, then x  = 3. is a true statem ent fo r each x e R. However, the converse P i(x ) =>• P \(x ) : I f \ x \ = 3, th e n x = —3. is a fa ls e statem ent when x ■= 3 since Pz{?>) is true and P i(3) is fa lse. For all other real num bers x , the im plication P ^ix) =$■ P \(x ) is true. Therefore, the biconditional P \{x)
P i{x) : X = —3 if and only i f \x\ = 3.
is fa ls e when x = 3 a nd is true fo r all other real num bers x . Example 2.14
For the open sentences P ( T ) : T is equilateral, a n d Q ( T ) : T is isosceles.
#
2.7
Tautologies and Contradictions
49
over the dom ain S o f all triangles, the converse o f the im plication P ( T ) =£■ Q ( T ) : I f T is equilateral, then T is isosceles. is the im plication Q ( T ) =>■ P C D : I f T is isosceles, then T is equilateral. We noted' that P ( T ) —> Q ( T ) is a true statem ent'for all triangles T , while Q ( T ) =>■ P ( T ) is a fa ls e statem ent w hen T is an isosceles triangle that is not equilateral. O n the other hand, the second im plication becom es a true statem ent fo r all other triangles T . Therefore, the biconditional P { T ) O Q ( T ) : T is equilateral i f and only i f T is isosceles. is fa ls e fo r all triangles that are isosceles and not equilateral, while it is true fo r all other triangles T . ♦ We now investigate the truth or falseness o f biconditionals obtained by assigning to a variable each value in its domain. Example 2.15
L et S = {0, 1, 4}. C onsider the fo llo w in g open sentences over the dom ain S:
n(n + 1)(2n + 1) .
6
P{n) :  is odd. 3=  n' „3 + 1. Q(n) : (n + 1)J D eterm ine three distinct elem ents a. b, c in S such that P ( a ) P( b) is fa lse, an d P{c) Q(c) is true.
Solution
Q (a ) is fa lse, Q(b)
O bserve that P ( 0) : 0 is odd. 2 (0 ) : 1  1 .
P (l) :
1 is odd.
P ( 4 ) : 30 is odd.
2(1):
8 = 2.
2 ( 4 ) : 125
Thus P ( 0) and P( 4) are false, w hile P{ 1)is true. A lso, Q{ \ ) and 2 ( 4 ) are false, w hile 2 ( 0 ) is true. T h u s /3(1) =>■ 2 ( l ) a n d 2 ( Q ) => P( 0) are false, w hile P (4) 2 ( 4 ) is true. H ence w e m ay take a — 1, b = 0 and c = 4. $
Analysis
N otice in Exam ple 2.15 that both P( 0) 2 ( 0 ) a n d / ’ ( l) o als. H ence the value 4 in S is the only choice for c.
2 ( 1 ) are false bicondition ♦
2 .7 T autologies and C ontradictions T he sym bols ~ , V , A , =>• and are som etim es referred to as logical connectives. From given statem ents, we can use these logical connectives to form m ore intricate statem ents. For exam ple, the statem ent ( P V Q ) A ( P v R) is a statem ent form ed from the given statem ents P , Q and R and the logical connectives V and A . We call (P V Q) A ( P V R) a
50
Chapter 2
Logic
com pound statem ent. M ore generally, a co m p o u n d s ta te m e n t is a statem ent com posed o f one or m ore given statem ents (called co m p o n en t sta te m e n ts in this context) and at least one logical connective. For exam ple, for a given com ponent statem ent P, its negation is a com pound statement. T he com pound statem ent P V ( ~ P), w hose truth table is given in Figure 2.8, has the feature that it is true regardless o f the truth value o f P . A com pound statem ent S is called a ta u to lo g y if it is true for all possible com bina tions o f truth values o f the com ponent statem ents that com prise S. H ence P v ( ~ P) is a tautology, as is ( ~ 2 ) V ( P =>• Q ). This latter fact is verified in the truth table shown in Figure 2.9. L etting P i : 3 is odd. and Pi : 57 is prim e, w e see that not only is 57 is not prim e, or 57 is prim e if 3 is odd. a true statem ent, but (  /V) P2 are being considered.
V
(Pi > P ) is true regardless o f w hich statem ents Pi and
On the other hand, a com pound statem ent S is called a c o n tra d ic tio n if it is false for all possible com binations o f truth values o f the com ponent statem ents that are used to form S. The statem ent P A i ^ P ) is a contradiction, as is show n in Figure 2.10. H ence the statem ent 3 is odd and 3 is not odd. is false. A nother exam ple o f a contradiction is (P in the truth table show n in Figure 2.11.
Q)
a
a
(Q =>• ( ~ P)), w hich is verified
Indeed, if a com pound statem ent S is a tautology, then its negation diction.
P
T
F
T
T
P Q ~ Q T
T F F
PV(vP)
F
Figure 2.8
T
~ P
T
An example of a tautology P=>Q
(~g)v(P^Q )
F
T
T
F
T
F
T
T
F
T
T
F
T
T
T
Figure 2.9
Another tautology
is a contra
2.8 P
Logical Equivalence
51
PA(~P)
T
F
F
F
T
F
Figure 2.10 P
An example of a contradiction
P
Q
~ P
T
T
F
T
Q Q => (~ P ) ( P A Q ) A (Q F
F F
a
T
F
F
F
T
F
T
T
F
T
F
F
F
T
F
T
F
Figure 2.11
P))
Another contradiction
2.8 L ogical E quivalen ce Figure 2.12 shows a truth table for the tw o statem ents P =>• Q and ( ~ P ) V Q. The corresponding colum ns o f these com pound statem ents are identical; in other w ords, these tw o com pound statem ents have exactly the same truth value for every com bination o f truth values o f the statem ents P and Q . L et R and S be tw o com pound statem ents involving the sam e com ponent statem ents. Then R and 5 are called logically eq u iv ale n t if R and S have the sam e truth values for all com binations o f truth values o f their com ponent statem ents. If R and S are logically equivalent, then this is denoted by R = S. H ence P = > Q and ( ~ P ) V Q are logically equivalent and so P = > • Q = ( ~ P ) V Q. A nother, even sim pler, exam ple o f logical equivalence concerns P A Q and Q A P. That P A Q = Q A P is verified in the truth table show n in Figure 2.13. W hat is the practical significance o f logical equivalence? Suppose that R and S are logically equivalent com pound statem ents. T hen we know that R and 5 have the same truth values for all possible com binations o f truth values o f their com ponent statem ents. But this m eans that the biconditional R O S is true for all possible com binations o f truth values o f their com ponent statem ents and hence R O S is a tautology. Conversely, if R <£> S is a tautology, then R and S are logically equivalent. L et R be a m athem atical statem ent that w e w ould like to show is true and suppose that R and som e statem ent S are logically equivalent. If w e can show that S is true, then R is true as well. F or exam ple, suppose that w e w ant to verify the truth o f an
p
Q
~ P
P => Q
(~P)VQ
T
T
F
T
T
T
F
F
F
F
F
T
T
T
T
F
F
T
T
T
Figure 2.12
Verification of P =» Q = ( ~ P ) V Q
Chapter 2
Logic P Q
P AQ
Q AP
T
T
T
T
T
F
F
F
F T F F
F
F
F
F
Figure 2.13
Verification of /’ a Q
Q
a P
im plication P =$■ Q. If we can establish the truth of the statem ent ( ~ / >) V Q , then the logical equivalence o f P =$■ Q and ( ~ P ) v Q guarantees that P Q is true as well. Example 2.16
R eturning to the m athem atics instructor in E xam ple 2 .6 and w hether she kept her prom ise that I f you earn an A on the fin a l exam, then you w ill receive an A fo r the fin a l grade. w e need only know that the student d id not receive an A on the fin a l exam or the student received an A as a fin a l grade to see that she kept her prom ise. ♦ Since the logical equivalence o f P = > ■ Q and (~ .P ) V Q, verified in Figure 2.12, is especially im portant and w e will have occasion to use this fact often, w e state it as a theorem .
Theorem 2.17
L et P and Q be two statem ents. Then
P
= *
Q and
(
~
P
)
V
Q
are logically equivalent. L et’s return to the truth table in Figure 2.13, w here w e show ed that P A Q and Q A P are logically equivalent for any two statem ents P and Q. In particular, this says that (P => Q )
A
(Q => P ) and (Q =» P )
A
(P =► Q)
are logically equivalent. O f course, (P Q ) A (Q =4 P ) is precisely w hat is called the biconditional o f P and Q. Since (P =>• Q ) A (Q => P ) and [Q => P ) A (P =$■ Q ) are logically equivalent, (Q =$■ P ) A (P =} Q ) represents the biconditional o f P and Q as well. Since Q => P can be w ritten as “P if Q ” and P => Q can be expressed as “P only if <2,” their conjunction can be w ritten as “P if Q and P only if Q " or, m ore simply, as P if and only if Q. Consequently, expressing P <S> Q as “P if and only if Q ” is justified. Furtherm ore, since Q => P can be phrased as “P is necessary for Q " and P => Q can be expressed as “P is sufficient for Q ’’ w riting P Q as “P is necessary and sufficient for Q " is likew ise justified.
2.9
Some Fundam ental Properties of Logical Equivalence
53
2.9 Som e F undam en tal P roperties o f L ogical E quivalen ce It probably com es as no surprise that the statem ents P and ~ ( ~ P ) are logically equiv alent. This fact is verified in Figure 2.14. We m entioned in Figure 2.13 that, for two statem ents P and Q , the statem ents P a Q and Q A P are logically equivalent. T here are other fundam ental logical equivalences that w e often encounter as well. Theorem 2.18
For statem ents P , Q a nd R , (1) C om m utative Law s (a) P V Q = Q w P U» p A Q = Q A P (2) A ssociative Law s (a) P v ( ( J v « ) = ( P v Q ) v R (.b) P A (Q A R ) = ( P a Q) a R (3) D istributive Law s {a) P V (Q AR ) = (P V Q) A ( P 0b ) P A (Q v R ) = (P A Q ) v (P (4) D e M organ's Law s (a) ~ ( / > v 2 ) e K ) a H ) (,
b)
~
(
P
A
g
)
s
(
~
P
)
V
(
~
G
)
VR ) a
R)
.
Each p ait o f T heorem 2.18 is verified by m eans o f a truth table. We have already established the com m utative law for conjunction (nam ely P A Q = Q A P ) in F ig ure 2.13. In Figure 2.15 P V (Q A R ) = (P V Q ) A (P V R ) is verified by observing that the colum ns corresponding to the statem ents P V (Q a R ) and (P V Q ) (P V R ) are identical. The laws given in T heorem 2.18, together w ith other know n logical equivalences, can be used to good advantage at tim es to prove other logical equivalences (without introducing a truth table). Example 2.19
Suppose we are asked to verify that ~ { P => Q ) = P a ( ~ 0 ) fo r every two statem ents P and Q. Using the logical equivalence o f P ( ~ P ) v Q fro m Theorem 2.17 and Theorem 2.18(4a), we see that
~ (/> => G) = ~ ( ( ~P ) V Q) = ( {  /> ) ) A (~<2) = P A (~G)t P
T P1 Figure 2.14
r^ j
P r
P
r* J ( r ^ J
P
)
T F
Verification of P = ~ (~ P )
Q and
(2.1)
Chapter 2
Logic P
Q
R
Q
T
T
T
T
T
T
T
T
T
T
F
F
T
T
T
T
T
F
T
F
T
T
T
T
T
F
F
F
T
T
T
T
F
T
T
T
T
T
T
T
F
T
F
F
T
F
F
F
F
F
T
F
F
T
F
F
F
F
F
F
F
F
F
F
Figure 2.15
a
R
P V Q
P v R
Pv(QaR)
(P
v
0
a
(P
v
P)
Verification of the distributive law P . (Q \ R ! r u / ’ V Q) A (P v R i
im plying that the statem ents ~ ( P => Q ) and P A ( ~ Q ) are logically equivalent, which we alluded to earlier. $ It is im portant to keep in m ind w hat w e have said about logical equivalence. For exam ple, the logical equivalence o f P A Q and Q A P allow s us to replace a statem ent o f the type P A Q by Q A P w ithout changing its truth value. As an additional exam ple, according to D e M organ’s Law s in T heorem 2.18, if it is not the case that an integer a is even or an integer b is even, then it follow s that a and b are both odd. Example 2.20
U sing the second o f D e M organ s L aw s and (2.1), we can establish a useful logically equivalent fo r m o f the negation o fP <£> Q by the fo llo w in g string o f logical equivalences: ~ (P
Q ) = ~ ( ( P =► Q ) A (Q => P ))
s(~(P=>G))vM 0=>P» 3 (P A ( ~ G ) ) V (Q A ( ~ P ) ) .
♦
W hat w e have observed about the negation o f an im plication and a biconditional is repeated in the follow ing theorem . Theorem 2.21
For statem ents P a nd Q, (a ) ~ (P =>• Q ) = P A (  0
(61 ~ ( p ^ e ) = ( P A (  e ) ) v ( g A (  p ) ) , Example 2.22
O nce again, let’s return to w hat the m athem atics instructor in E xam ple 2 .6 said: I f you earn an A on the fin a l exam , then you w ill receive an A fo r yo u r fin a l grade. I f this instructor w as not truthful, then it fo llo w s by Theorem 2.21(a) that You earned an A on the fined exam and d id not receive A as y o u r fin a l grade.
2.10
Quantified Statements
55
Suppose, on the other hand, that the m athem atics instructor had said: I f you earn an A on the fin a l exam, then yo u w ill receive an A fo r the fin a l grade— and th a t’s the only w ay that you w ill get an A fo r a fin a l grade. I f this instructor was not truthful, then it fo llo w s by Theorem 2.21(b) that E ither you earned an A on the fin a l exam and d id n 't receive A as yo u r fin a l grade or you received an A fo r yo u r final grade a n d yo u d id n 't get an A on the fin a l exam. $
2.10 Q uantified Statem ents We have m entioned that if P ( x ) is an open sentence over a dom ain S, then P ( x ) is a statem ent for each x e S. We illustrate this again. Example 2.23
I f S = {1, 2, • • •, 7}, then
is a statem ent fo r each n e S. Therefore, P (1) : P ( 2) : P{3 ): P (4 ) :
3 is prim e. 7 is prim e. H i s prim e. 19 is prim e.
are true statem ents; while P ( 5) : 27 is prim e. P ( 6) : 39 is prim e. P ( 7 ) : 51 is prim e. are fa ls e statem ents.
♦
T here are other ways that an open sentence can be converted into a statem ent, nam ely by a m ethod called quantification. L et P ( x ) be an open sentence over a dom ain S. A dding the phrase “F or every x e S ” to P( x ) produces a statem ent called a quantified statem ent. The phrase “for every” is referred to as the universal quantifier and is denoted by the sym bol V. O ther w ays to express the universal quantifier are “for each” and “for all.” This quantified statem ent is expressed in sym bols by Vx e S’, P ( x )
( 2 .2 )
For every x e S, P [x).
(2.3)
and is expressed in w ords by
The quantified statem ent (2.2) (or (2.3)) is true if P ( x ) is true for every x e S, w hile the quantified statem ent (2 .2) is false if P( x ) is false for at least one elem ent x e S.
56
Chapter 2
Logic A nother w ay to convert an open sentence P ( x ) over a dom ain S into a statem ent through quantification is by the introduction o f a quantifier called an existenti al quantifier. Each o f the phrases there exists, there is, fo r som e and fo r a t least one is referred to as an existential quantifier and is denoted by the sym bol 3. The quantified statem ent 3x e S, P( x )
(2.4)
There exists x C S such that P( x ) .
(2.5)
can be expressed in w ords by
The quantified statem ent (2.4) (or (2.5)) is true if P ( x ) is true for at least one elem ent x e S, w hile the quantified statem ent (2.4) is false if P ( x ) is false for all x e S. We now consider tw o quantified statem ents constructed from the open sentence we saw in E xam ple 2.23. Example 2.24
For the open sentence
2«2 + 5 + ( —!)" .
.
P ( n ) :  is prune. over the dom ain S = {1,2, • • •, 7}, the quantified statem ent 2n2 + 5 + (  l ) " V/7 € S., P ( n ) : For every n e S ,  is prim e. is fa ls e since P ( 5) is fa lse, fo r example; while the quantified statem ent 2 n 2 + 5 + (  1 )" 3« e S, P( n) : There exists n e S such t h a t — is prim e. is true since P ( \ ) is true, fo r example. The quantified statem ent Vx e S. P( x ) can also be expressed as If x e S, then P(x) . Consider the open sentence P ( x ) : x 2 > 0. over the set R o f real numbers. Then Vx € R , P( x ) or, equivalently, Vx € R , x 2 > 0 can be expressed as For every real num ber x, x 2 > 0. or If x is a real number, then x 2 > 0 . as w ell as T he square of every real num ber is nonnegative.
§
2.10
Quantified Statements
57
In general, the universal quantifier is used to claim that the statem ent resulting from a given open sentence is true w hen each value o f the dom ain o f the variable is assigned to the variable. Consequently, the statem ent Vx e R. x 2 > 0 is true since x 2 > 0 is true for every real num ber x. Suppose now that we were to consider the open sentence Q( x ) : x 2 < 0. The state m ent Vx e R , Q ( x ) (that is, for every real num ber x , we have x 2 < 0) is false since, for exam ple, 0 ( 1 ) is false. O f course, this m eans that its negation is true. If it w ere not the case that for every real num ber x, w e have x 2 < 0 , then there m ust exist som e real num ber x such that x 2 > 0 . This negation There exists a real num ber x such that x 2 > 0. can be w ritten in sym bols as 3x e R . x 2 > 0 or 3x e R , ~<2(x). M ore generally, if w e are considering an open sentence F i x ) over a dom ain S, then ~ (Vx e S. P( x ) ) = 3x c .S'. ~ P ( x ) . Example 2.25
Suppose that w e are considering the set A = { 1 , 2, 3} a n d its p o w er set V ( A ) , the set o f all subsets o f A . Then the quantified statem ent For every set B e V ( A ) . A  B ^ 0.
(2.6)
is fa ls e since fo r the subset B = A = {1, 2, 3}, w e have A  B = 0. The negation o f the statem ent (2.6) is There exists B e ' P(A) such that A  B = 0.
(2.7)
The statem ent (2.7) is therefore true since fo r B = A e V ( A ) , we have A  B = 0. The statem ent (2.6) can also be written as I f B C A . then A  B ^ 0.
(2.8)
C onsequently, the negation o f (2.8) can be expressed as There exists som e subset B o f A such that A  B = 0.
f
The existential quantifier is used to claim that at least one statem ent resulting from a given open sentence is true w hen the values o f a variable are assigned from its domain. We know that for an open sentence P ( x ) over a dom ain .S', the quantified statem ent 3x e S, P ( x ) is true provided P ( x ) is a true statem ent for at least one elem ent x 6 S. Thus the statem ent 3x € R, x 2 > 0 is true since, for exam ple, x 2 > 0 is true for x = 1. The quantified statem ent 3.v € R, 3x = 12 is therefore true since there is som e real num ber x for w hich 3x = 12, nam ely x = 4 has this property. (Indeed, x = 4 is the only real num ber for w hich 3x = 12.) O n the other hand, the quantified statem ent 3n e Z, 4n — 1 = 0
58
Chapter 2
Logic is false as there is no integer n for w hich An — 1 = 0. (O f course, An — 1 = 0 w hen n — 1/4 , but 1 /4 is not an integer.) Suppose that Q( x ) is an open sentence over a dom ain S. If the statem ent 3x e S, Q { x ) is not true, then it m ust be the case that for every x € .S, Q( x ) is false. T hat is, ~ (3x c S. Q{x) ) = Vx e S, '  Q( x) is true. We illustrate this w ith a specific exam ple.
Example 2.26
The fo llo w in g statem ent contains the existential quantifier: There exists a real num ber x such that x 2 — 3.
(2.9)
I f we let P ( x ) :x 2 = 3, then (2.9) can be rewritten as 3x e R , P( x) . The statem ent (2.9) is true since P ( x ) is true when X = J3 (or when x = —v3J» H ence the negation o f (2.9) is: For every real num ber x , x 2 ^ 3.
(2.10)
The statem ent (2.10) is therefore fa lse.
♦
L et P( x , y) be an open sentence, w here the dom ain o f the variable x is S and the dom ain o f the variable y is T . Then the quantified statem ent For all x e S and y e T, P ( x , _y). can be expressed sym bolically as Vx 6 S , Wy € J , P ( x , y ) .
(2.11)
The negation o f the statem ent (2.11) is ~ (Vx e S, Vy e T. P( x . y)) = 3x e S, ~ (Vv c T, P ( x , v » ee 3x € S , 3y e T , ~ P ( x , y ).
(2.12)
We now consider exam ples o f quantified statem ents involving two variables. Example 2.27
C onsider the statem ent For every W o real num bers x and y, x 2 + v2 > 0.
(2.13)
I f we let P( x . v) : x 2 + _y2 > 0 where the dom ain o f both x and y is R , then statem ent (2.13) can be expressed as Vx G R . Vy e R, P ( x , y)
(2.14)
or as Vx, y c R , P (x , y). Since x 2 > 0 and y 2 > 0 fo r all real num bers x and y, it fo llo w s that x 2 + y2 > 0 and so P( x , y) is true fo r all real num bers x and y. Thus the quantified statem ent (2.14) is true.
2.10
Quantified Statements
59
The negation o f statem ent (2.14) is therefore ~ (V.v e R, Vy c R. P ( x , y)) ss 3* c R. 3y e R, ~ P ( x , y) = 3 x, y e R, ~ /'f.v. y),
(2.15) w h ich , /ÂŤ words, is There exist real num bers x and y such that x 2 + y 2 < 0 . The statem ent (2.16) is therefore fa lse.
(2.16) +
For an open sentence containing tw o variables, the dom ains o f the variables need not be the same. Example 2.28
Consider the statem ent For every s e S and t e T , s t + 2 is a prim e.
(2.17)
where the dom ain o f the variable 5 is S = {1, 3, 5} a n d the dom ain o f the variable t is T = {3, 9}. I f we let Q(s, t ) : s t + 2 is a prim e. then the statem ent (2.17) can be expressed as
(2.18)
Vs e S , V t e I , Q( s , t ) . Since all o f the statem ents 0 ( 1 , 3): 0 ( 5 , 3):
l  3 + 2 /'5 a prim e. 0 ( 3 , 3): 3  3 + 2 / 5 0 prim e. 5 3 + 2 is a prim e.
0 ( 1 ,9 ) : 0 ( 5 , 9):
1 9 + 2 5 9 + 2
is a prim e. is a prim e.
0 ( 3 ,9 ) : 3  9 +
2 is a prim e.
are true, the quantified statem ent (2.18) is true. j45 we saw in (2.12), the negation o f the quantified statem ent (2.18) is  (V.v e S', W e r , 0 (5, 0 ) = 35 e S, 3t e T . ~ Q ( s , t) and so the negation o f (2.17) is There exist s e S a nd t e T such that s t + 2 is not a prim e. The statem ent (2.19) is therefore fa lse.
(2.19) $
A gain, let P( x , y) be an open sentence, w here the dom ain o f the variable x is S and the dom ain o f the variable y is T . The quantified statem ent There exist x e S and y e T such that P ( x , y) can be expressed in sym bols as 3x e S, 3y e T , P (x , y).
(2.20)
Chapter 2
Logic The negation o f the statem ent (2.20) is then M 3 .v e S. 3y c
/>(*, v)> ee Vx e .V. ~ ( 3 y e T, P (x , v)) = Vx € .V. Vv e T , ~ P ( x , y).
(2.21)
We now illustrate this situation. Example 2.29
C onsider the open sentence R i s . t ) : \s  1 + \ t  2\ < 2. where the dom ain o f the variable s is the set S o f even integers a n d the dom ain o f the variable t is the set T o f o dd integers. Then the quantified statem ent 3s c 5 . 3/ £ 7 \ R(s. t).
(2.22)
can be expressed in words as There exist an even integer s a nd an odd integer t such that s — 1 + Jf — 2j < 2 . (2.23) Since R ( 2, 3) : 1 + 1 < 2 .is true, the quantified statem ent (2.23) is true. The negation o f (2.22) is therefore ~(3.v e S, Bt e T
R ( s , /)) = Vs 6 S, Wt e I . ~ R ( s , t).
(2.24)
and so the negation o f (2.22), in words, is For every even integer s and every odd integer i, s — 1 + \t — 2 > 2. The quantified statem ent (2.25) is therefore fa ls e .
(2.25) #
In the next tw o exam ples o f negations o f quantified statem ents, De M o rg an ’s laws are also used. Example 2.30
The negation o f For all integers a and b, if a b is even, then a is even a n d b is even. is There exist integers a and b such that ab is even a n d a or b is odd.
Example 2.31
$
The negation o f There exists a rational num ber r such that r e A = {\ f 2 , j r } or r e B = {—v '2 . v'3,c>}. is For every rational num ber r , both r <£ A and r ^ B .
4
Q uantified statem ents m ay contain both universal and existential quantifiers. W hile w e present exam ples o f these now, we will discuss these in m ore detail in Section 7.2.
2.10 Example 2.32
Quantified Statements
61
Consider the open sentence P( a, b) : ab = 1. where the dom ain o f both a and b is the set Q + o f p o sitive rationed num bers. Then the quantified statem ent Va e Q + , 3b e Q + , P( a, b)
(2.26)
can be expressed in words as For every p ositive rational num ber a, there exists a positive rational num ber b such that a b = 1. It turns out that the quantified statem ent (2.26) is true. I f w e replace Q + b y R, then we have
Va e R. 3b
e R,
P( a, b ).
(2.27)
The negation o f this statem ent is
~ (Va e R, 3b e R, P (a , b)) = 3a e R, ~ (3b e R, P( a, b )) = 3a e R . V k R , ~ P ( a , b ) , which, in words, says that There exists a real num ber a such that fo r every real num ber b , a b ^ 1.
This negation is true since fo r a = 0 and every real num ber b, a b = 0 ^ 1. Thus the quantified statem ent (2.27) is fa lse. ♦ Example 2.33
Consider the open sentence Q(a, b) : a b i s o d d . where the dom ain o f both a and b is the set N o f p o sitive integers. Then the quantified statem ent 3a e N,VZ? e N, Q( a , b ) ,
(2.28)
expressed in words, is There exists a p ositive integer a such that fo r every p o sitive integer b, a b is odd. The statem ent (2.28) turns out to be fa lse. The negation o f (2.28), in sym bols, is ~ (3 a 6 N ,V b € N, Q(a, b)) = Va e N, ~ (Wb € N. Q( a. b)) = Va e N. 3b e N , ~ Q( a. b). In words, this says For every positive integer a, there exists a p ositive integer b such that ab is even. This statem ent, therefore, is true.
♦
Chapter 2
Logic Suppose that P ( x , y ) is an open sentence, w here the dom ain o f X is S and the dom ain o f y is T . Then the quantified statem ent Vx c S , By e /'. P( x , y) is true if 3y e T, P ( x , y) is true for each x e S. This m eans that for every x e S, there is som e y e T for w hich P ( x , y) is true.
Example 2.34
C onsider the open sentence P( x , y ); x + y is prim e. where the dom ain o f x is S = {2, 3} and the dom ain o f y is T = {3, 4}. The quantified statem ent V.v c S , By c T , P ( x , y), expressed in words, is For every x e S, there exists y e T such that x + y is prim e. This statem ent is true. For x = 2, P{ 2, 3) is true a n d fo r x = 3, P (3 , 4) is true.
â™Ś
Suppose that Q ( x , y ) is an open sentence, w here S is the dom ain o f x and T is the dom ain o f y. The quantified statem ent 3x e S, V.v c T , Q( x . y) is true if Vy e T. Q( x , y) is true for som e x e S. This m eans that for some elem ent x in S, the open sentence Q( x , y ) is true for all y ÂŁ T . Example 2.35
Consider the open sentence Q( x , y)\ x + j is prime. where the dom ain o f x is S â€” {3, 5, 7} a n d the dom ain o f y is T = {2, 6 , 8 , 12}. The quantified statem ent 3x e 5, Vy e T , Q( x , y ),
(2.29)
expressed in words, is There exists som e x e S such that fo r every y e T , x + y is prim e. For x = 5, all o f the num bers 5 + 2, 5 + 6 , 5 + 8, and 5 + 12 are prim e. Consequently, the quantified statem ent (2.29) is true. 4
2.11
Characterizations of Statements
63
L et’s review sym bols that we have introduced in this chapter:
V A
=> V 3
negation (not) disjunction (or) conjunction (and) im plication biconditional universal quantifier (for every) existential quantifier (there exists)
2.11 C haracterizations of Statem ents L et's return to the biconditional P Q. Recall that P Q represents the com pound statem ent (P => Q ) A (Q =^> P ). Earlier, we described how this com pound statem ent can be expressed as P if and only if Q. M any m athem aticians abbreviate the phrase “if and only if ” by w riting “iff.” A lthough "iff” is inform al and, o f course, is not a word, its use is com m on and you should be fam iliar w ith it. R ecall that w henever you see P if and only if Q. or P is necessary and sufficient for Q. this m eans If P then Q and if Q then P . Example 2.36
Suppose that P ( x ) : x ~ —3. and Q( x ) : jc] = where x
e
R. Then the biconditional
3.
P ( x ) <£> Q( x ) can be expressed as
X — —3 i f a n d only i f \x\ = 3. or x = —3 is necessary and sufficient f o r .r = 3. or, p erh aps better, as x — —3 is a necessary and sufficient condition fo r \x\ = 3. L e t's now consider the quantified statem ent V x e R , P ( x ) fa ls e because P( 3) O Q ( 3) is fa lse.
Q(x). This statem ent is $
64
Chapter 2
Logic Suppose that som e concept (or object) is expressed in an open sentence P ( x ) over a dom ain S and Q(.x) is another open sentence over the dom ain S concerning this concept. We say that this concept is c h a ra c te riz e d by Q( x ) if V.v e S. P ( x ) Q( x ) is a true statem ent. The statem ent V.v e S, P ( x ) o Q{ x) is then called a c h a ra c te riz a tio n o f this concept. For exam ple, irrational num bers are defined as real num bers that are not rational and are characterized as real num bers w hose decim al expansions are nonrepeating. This provides a characterization o f irrational numbers: A real num ber r is irrational i f and only i f r has a nonrepeating decim al expansion . We saw that equilateral triangles are defined as triangles w hose sides are equal. They are characterized how ever as triangles w hose angles are equal. Therefore, w e have the characterization: A triangle T is equilateral i f and only i f T has three equal angles. You m ight think that equilateral triangles are also characterized as those triangles having three equal sides but the associated biconditional: A triangle T is equilateral if and only if T has three equal sides. is not a characterization o f equilateral triangles. Indeed, this is the definition we gave o f equilateral triangles. A characterization o f a concept then gives an alternative, but equivalent, w ay o f looking at this concept. C haracterizations are often valuable in studying concepts or in proving other results. We w ill see exam ples o f this in future chapters. We m entioned that the follow ing biconditional, though true, is not a characterizaÂ tion: A triangle T is equilateral if and only if T has three equal sides. A lthough this is the definition o f equilateral triangles, m athem aticians rarely use the phrase â€œif and only if â€? in a definition since this is w hat is m eant in a definition. T hat is, a triangle is defined to be equilateral if it has three equal sides. Consequently, a triangle w ith three equal sides is equilateral but a triangle that does not have three equal sides is not equilateral.
E X E R C IS E S FOR C H A PT E R 2 Section 2.1: Statements 2.1. Which of the following sentences are statements? For those that are, indicate the truth value. (a) The integer 123 is prime. (b) The integer 0 is even. (c) Is 5 x 2 = 10? (d) #  4 = 0 . (e) Multiply 5.v + 2 by 3. (f) 5x + 3 is an odd integer. (g) W hat an impossible question!
Exercises for Chapter 2
65
2.2. Consider the sets A, B ,C and D below. Which of the following statements are true? Give an explanation for each false statement. A = {1, 4, 7, 10, 13, 16, . . .} C = {x e Z : x is prime and x # 2} B = {x e Z : x is odd) D = {1, 2, 3,5. 8,13, 21, 34, 5 5 , ...} (a) 2 5 e A
(b) 33 € D
(c) 22 £ A U D
(d) C c
B
(e) 0 e B n D (f) 53
£ C.
2.3. W hich of the following statements are true? Give an explanation for each false statement. (a) 0 e 0 (b) 0 e {0} (c) {1, 3} = {3,1} (d) 0 = {0} (e) 0 C {0} it) 1 C {1}. 2.4. Consider the open sentence P (x j : x(.v — 1) = 6 over the domain R. (a) (b)
For what For what
values of x is values of x is
P( x) a true statement? P ( x ) a false statement?
2.5. For the open sentence P( x) : 3x — 2 > 4 over the domain Z, determine: (a) the values of x for which Fi x ) is true. (b) the values of x for which P( x) is false. 2.6. For the open sentence Pi A) : A C {1, 2, 3} over the domain S = "P({1, 2, 4}), determine: (a) all A G S for which P( A) is true. (b) all A e S for which P ( A ) is false. (c) all A e S for which A fl {1, 2, 3} =
0.
2.7. Let P(n): n and n + 2 are primes, be an open sentence over the domain N. Find six positive integers n for which P(n) is true. If n e N such that P ( n ) is true, then the two integers n, n + 2 are called tw in prim es. It has been conjectured that there are infinitely many twin primes. 2.8. Let P ( n ) : /r ^ y + is even. (a) Find a set Si of three integers such that P(n) is an open sentence over the domain Si and P(n) is true for each n € S i . (b) Find a set S2 of three integers such that P(n) is an open sentence over the domain S2 and Pi n) is false for each n e S2. 2.9. Find an open sentence P(n) over the domain S = {3, 5, 7, 9} such that P(n) is true for half of the integers in S and false for the other half. 2.10. Find two open sentences P(n) and Q{n), both over the domain S = {2, 4, 6 , 8}, such that P ( 2) and <2(2) are both true, P ( 4) and Q( 4) are both false, P( 6) is true and <2(6) is false, while P ( 8) is false and <2(8) is true.
Section 2.2: The Negation of a Statement 2.11. State the negation of each of the following statements. (a) V 2 is a rational number. (b) 0 is not a negative integer. (c) 111 is a prime number. 2.12. Complete the truth table in Figure 2.16. 2.13. State the negation of each of the following statements. (a) The real number r is at most ~J2. (b) The absolute value of the real number a is less than 3. (c) Two angles of the triangle are 45°.
Chapter 2
66
Logic P
Q
T
T
T
F
F
T
F
F
Figure 2.16
~ P
~ Q
The truth table for Exercise 2.12.
(d) The area of the circle is at least 9 tc . (e) Two sides of the triangle have the same length. (f) The point P in the plane lies outside of the circle C . 2.14. State the negation of each of the following statements. (a) (b) (c) (d) (e)
At least two of my library books are overdue, One of my two friends misplaced his homework assignment. No one expected that to happen. It’s not often that my instructor teaches that course. It’s surprising that two students received the same exam score.
Section 2.3: The Disjunction and Conjunction of Statements 2.15. Complete the truth table in Figure 2.17.
P Q ~Q
P a(~Q )
T T T F F T F F Figure 2.17
The truth table for Exercise 2.15
2.16. For the sets A = {1, 2, • ■•, 10} and B = {2, 4, 6, 9, 12, 25}, consider the statements P: A c B.
Q: \ A  B \ = 6.
Determine which of the following statements are true. (a) P V Q (b) P V ( ~ Q) (c)PaQ (d) (~ P ) a Q (e) ( ~ /*) v ( ~ Q). 2.17. Let P: 15 is odd. and Q: 21 is prime. State each of the following in words, and determine whether they are true or false. (a ) P v Q (b) P A Q ( c ) ( ~ P ) v £ (d) P A ( ~ Q). 2.18. Let S = {1,2, . . . , 6} and let P( A) : A n {2, 4. 6} = 0. and Q( A) : A / 0. be open sentences over the domain V(S). (a) Determine all A e V ( S ) for which Pi A)
A
Q( A) is true.
Exercises for Chapter 2
67
(b) Determine all A e V ( S ) for which P ( A ) v (~ Q (A )) is true. (c) Determine all A e V( S ) for which (~ P (A )) A f ~ g (A ) ) is true. S e c t i o n 2 .4 : T h e I m p l i c a t i o n 2.19. Consider the statements P: 17 is even, and Q: 19 is prime. Write each of the following statements in words and indicate whether it is true or false. (a) ~ P (b)PvQ (c) P a Q (d) P =* Q. 2.20. For statements P and Q, construct a truth table for (P => Q) =±> ( ~ P). 2.21. Consider the statements P : s/2 is rational, and Q : 22/7 is rational. Write each of the following statements in words and indicate whether it is true or false. (a ) P = * Q (b ) Q ^ P (c) (~ P) =► (~ Q) ( d ) (  Q ) > (  />). 2.22. Consider the statements: P: V 2 is rational.
Q:  is rational.
R: s/3 is rational.
Write each of the following statements in words and indicate whether the statement is true or false. (a) (P A Q) =>■ R (b) ( P A © 4 ( ~ R) (c) ( (~ P) A Q) => R (d) (P s Q ) R). 2.23. Suppose that {Si, S2} is a partition of a set S and x € S. W hich of the following are true? (a) (b) (c) (d) (e)
If we know that x <£ Si, then x must belong to Sj. It’s possible that .v ^ Si and.r ^ S'i, Either x S\ or x £ S 2 . Either x e Si o r x e S2. It’s possible that v e Si and x e S2.
2.24. Two sets A and B are nonempty disjoint subsets of a set S. I f x e S, then which of the following are true? (a) (b) (c) (d) (e) (f)
It’s possible that x e A fl B . I f x is an element of A, then x can’t be an element of B . I f x is not an element of A, then x must be an element of B . It’s possible that x £ A and x £ B. For each nonempty set C , either x e A D C or x 6 B fl C . For some nonempty set C , both x e A U C and x e B U C .
2.25. A college student makes the following statement: If I receive an A in both Calculus I and Discrete M athematics this semester, then I ’ll take either Calculus II or Computer Programming this summer. For each of the following, determine whether this statement is true or false. (a) The student doesn't get an A in Calculus I but decides to take Calculus II this summer anyway. (b) The student gets an A in both Calculus I and Discrete Mathematics but decides not to take any class this summer. (c) The student does not get an A in Calculus I and decides not to take Calculus II but takes Computer Programming this summer. (d) The student gets an A in both Calculus I and Discrete Mathematics and decides to take both Calculus II and Computer Programming this summer. (e) The student gets an A in neither Calculus I nor Discrete Mathematics and takes neither Calculus II nor Computer Programming this summer.
68
Chapter 2
Logic
2.26. A college student makes the following statement: If I don’t see my advisor today, then I ’ll see her tomorrow. For each of the following, determine whether this statement is true or false. (a) (b) (c) (d)
The The The The
student student student student
doesn't see his advisor either day. sees his advisor both days. sees his advisor on one of the two days. doesn’t see his advisor today and waits until next week to see her.
2.27. The instructor of a computer science class announces to her class that there will be a wellknown speaker on campus later that day. Four students in the class are Alice. Ben, Cindy and Don. Ben says that h e’ll attend the lecture if Alice does. Cindy says that she’ll attend the talk if Ben does. Don says that he will go to the lecture if Cindy does. That afternoon exactly two of the four students attend the talk. Which two students went to the lecture? 2.28. Consider the statement (implication): If Bill takes Sam to the concert, then Sam will take Bill to dinner. Which of the following implies that this statement is true? (a) (b) (c) (d) (e) (f) (g)
Sam takes Bill to dinner only if Bill takes Sam to the concert. Either Bill doesn’t take Sam to the concert or Sam takes Bill to dinner. Bill takes Sam to the concert. Bill takes Sam to the concert and Sam takes Bill to dinner. Bill takes Sam to the concert and Sam doesn’t take Bill to dinner. The concert is canceled. Sam doesn’t attend the concert.
2.29. Let P and Q be statements. Which of the following implies that P V Q is false? (a) ( ~ P) v ( ~ Q) is false, (b) ( ~ P ) v Q is true. (c) ( ~ P } A ( ~ Q) is true, (d) Q => F is true, (e) P A Q is false.
Section 2.5: More on Implications 2.30. Consider the open sentences P(n) : 5;; + 3 is prime, and Q(n) : I n + 1 is prime., both over the domain N. (a) State P (n)
Q(n) in words.
(b) State P ( 2) => Q( 2) in words. Is this statement true or false? (c) State P ( 6) => Q( 6) in words. Is this statement true or false? 2.31. In each of the following, two open sentences P( x) and Q( x) over a domain S are given. Determine the truth value of P{x) =$■ Q(x) for each x e S. (a)
P( x) :X
=
4;Q( x) : x = 4; S = { 4 ,  3 , 1,4,
(b )
P( a )
=
16;
:a 2
(c) P (x ) : a > 3;
0 (a )
Q(x) :  x  =
4;
S
=
{  6 , 4 , 0,
: 4 a  1 > 12; S = {0, 2, 3,
4,
5}. 3 , 4 , 8 ).
6).
2.32. In each of the following, two open sentences P ( x ) and Q( a ) over a domain S are given. Determine all x e S for which P( a ) => Q(x) is a true statement. (a) P( a ) : a  3 = 4 ; Q(x) : x > 8; S = R. (b) P ( a ) :a 2 > 1;2 ( a ) : a > 1; S = R. (c)
P ( a ) :a
2
>
1;Q(.x) : a > 1; S
(d) P ( a ) : x e [  1 , 2]; Q( a ) : a
2
= N.
< 2; S = [  1 , 1],
Exercises for Chapter 2
69
2.33. In each of the following, two open sentences P(x, y) and Q { x , y) are given, where the domain of both x and y is Z. Determine the truth value of P(x, y) =>■ Q (x, y) for the given values of x and v. (a) P(.v, v): x 2 — y 2 = 0. and Q {x, v): x = y. (x, y ) G {(L  1 ) , (3, 4), (5, 5)}. (b) P (x , y): x = y. and Q( x, y): x = y. (x, v) e {(1, 2), (2,  2), (6 , 6)}. (c) P( x, y): x 2 + v2 = 1. and Q(x. y): x 1 y — 1. (x, y) G {(1, —1), (—3. 4), (0, —1), (1, 0)}. 2.34. Each of the following describes an implication. Write the implication in the form “if, then.” (a) Any point on the straight line with equation 2y + x — 3 = 0 whose xcoordinate is an integer also has an integer for its ycoordinate. (b) The square of every odd integer is odd. (c) Let n G Z. Whenever 3n + 7 is even, n is odd. (d) The derivative of the function / ( x ) = cosx is f ' ( x ) ~ —sin x . (e) Let C be a circle of circumference A n . Then the area of C is also A n . (f) The integer /!3 is even only if n is even.
Section 2.6: The Biconditional 2.35. Let P : 18 is odd. and Q : 25 is even. State P <£> Q in words. Is P <s> Q true or false? 2.36. Let P( x) : x is odd. and Q{x) : x 2 is odd. be open sentences over the domain Z. State P( x ) ways: (1) using "if and only if” and (2) using “necessary and sufficient."
Q(x) in two
2.37. For the open sentences P ( X ) : x — 3 < 1. and Q(x) : x e (2, 4). over the domain R, state the biconditional P (x) Q(x) in two different ways. 2.38. Consider the open sentences: P( x) : x = —2. and Q(x) : x 2 = 4. over the domain S = {—2, 0, 2). State each of the following in words and determine all values of x g S for which the resulting statements are true. (a) ~ P {x) (e) Q( x)
(b) P( x) \  Q( x) (c) P( x) P( x) (f) P( x) « Q(x).
A
Q(x)
(d) P{x) => Q(x)
2.39. For the following open sentences P (x ) and Q(x) over a domain S, determine all values o f x £ S for which the biconditional P( x) Q(x) is true, (a) P (x ) : x = 4; Q( x ) : x = 4; S = { 4 ,  3 , 1.4, 5). (b)
P( x ) : x > 3; Q ( x ) : 4x  1 > 12; S = {0, 2, 3, 4, 6}.
(c)
P( x) : x 2 = 16; Q(x) : x 2  Ax = 0; S = {  6,  4 . 0, 3,
4, 8}.
2.40. In each of the following, two open sentences P (x , y ) and Q(x, y) are given, where the domain of both x and y is Z. Determine the truth value of P(x, y) <£> Q(x, y) for the given values of x and y. (a)
P( x, v) : x 2 — y2 = 0 and; Q(x, y) : x = y. (x, y) g {(1, —1). (3,4), (5, 5)}.
(b) P (x , y) : x = y and; Q(x, y) : x = y. ( x, y) g {(1, 2), (2 ,  2), <6 , 6)}. (c) P (x , y) : x 2 + v2 = 1 and; Q (x, y ) : x + y = 1. (x, y) g {(1. —1), (—3, 4), (0, —1), (1, 0)}. 2.41. Determine all values of n in the domain S = {1, 2, 3} for which the following is A necessary and sufficient condition for !Ly R to be even is that !L^S is odd.
a true statement:
70
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2.42. Determine all values of n in the domain S = {2, 3, 4} for which the following is a true statement: The integer is odd if and only if "(n+l) is even. 2.43. Let S = {1, 2, 3}. Consider the following open sentences over the domain S: is o d d .
P ( n ),
Qin): 2"~2 + 3"~2 + 6”~2 > (2.5)” '. Determine three distinct elements a, b, c in S such that P(a) =* Q(a) is false, Q(b) => P(b) is false, and Pic) Q(c) is true. 2.44. Let S = {1, 2, 3, 4}. Consider the following open sentences over the domain S: Put): is even. Qin): 2"~2 — (—2)'!_2 is even. R(n): 5'7_1 + 2" is prime. Determine four distinct elements a, b, c, d in S such that (i) P(a) —> Q ia) is false; (ii) Qib) Pi b) is true; (iii) Pi c) R i f ) is true; (iv) Qi d) R(d) is false. 2.45. Let Pin): 2"  1 is a prime, and Qin): n is a prime, be open sentences over the domain S = {2, 3, 4, 5, 6 . 11}. Determine all values of n e S for which P(n) Q(n) is a true statement.
Section 2.7: Tautologies and Contradictions 2.46. For statements P and Q, show that P =4> (P
v Q ) is a tautology.
2.47. For statements P and Q, show that (P A (~
Q))
a
(F
a
Q) is a contradiction.
2.48. For statements P and Q, show that (P A ( P => Q)) ^ Q is a tautology. Then state ( P A (P =$■ Q)) =$ Q in words. (This isan important logical argument form, called modus ponens.) 2.49. For statements P , Q and R, show that HP =» Q) a (Q => R)) => (P R) is a tautology. Then state this compound statement in words. (This is another important logical argument form, called syllogism.) 2.50. Let R and S be compound statements involving the same component statements. If K is a tautology and S is a contradiction, then what can be said of the following? (a)SvS (b)iiAS (c) R = > S (d) S => R. "
Section 2.8: Logical Equivalence 2.51. For statements P and Q, the implication ( ~ F ) =» (~ Q ) is called the inverse of the implication P =>• Q. (a) Use a truth table to show that these statements are not logically equivalent. (b) Find another implication that is logically equivalent to ( ^ P ) =v ( ^ Q) and verify your answer. 2.52. Let P and Q be statements. (a) Is ~ (P
V
Q) logically equivalent to ( ~ P ) V (~ Q )? Explain.
(b) W hat can you say about the biconditional  ( P
v
Q) o ( ( ~ F ) v (~ () )i?
2.53. For statements P , Q and R , use a truth table to show that each of the following pairs of statements is logically equivalent. (a)
(P A Q) <£> P and P => Q.
(b)
P
( Q v R) and (~<2) =» ( ( ~ P ) V R).
2.54. For statements P and Q, show that ( ~ Q) ^ ( F a ( ~ P ) ) and Q are logically equivalent. 2.55. For statements P , Q and R, show that (P v Q) => R and (P => R ) a (Q => R) are logically equivalent.
Exercises for Chapter 2
71
2.56. Two compound statements S and T are composed of the same component statements P, Q and R. If S and T are not logically equivalent, then what can we conclude from this? 2.57. Five compound statements Si, Sa, S3, S4 and S$ are all composed of the same component statements P and Q and whose truth tables have identical first and fourth rows. Show that at least two of these five statements are togically equivalent.
Section 2.9: Som e Fundam ental Properties of Logical Equivalence 2.58. Verify the following laws stated in Theorem 2 .18: (a) Let P, Q and R be statements. Then P
V
(Q
A
R) and (P
V
Q)
A
(P
V
R) are logically equivalent.
(b) Let P and Q be statements. Then ~G P V
O)
a n d ( — / ’ ,) A ( ~ < 2 ) a r e l o g i c a l l y e q u i v a l e n t .
2.59. Write negations of the following open sentences: (a) Either x = 0 or y = 0. (b) The integers a and b are
both even.
2.60. Consider the implication: I f * and y are even, then x y is even. (a)
State
the
implication using “only if.”
(b)
State
the
converse of the implication.
(c) (d)
State State
the the
implication as a disjunction (see Theorem 2.17). negation of the implication asa conjunction (see Theorem 2.21(a)).
2.61. For a real number x, let P( x) : x 2 = 2. and g (x ) : x = V 5. State the negation of the biconditional P O Q in words (see Theorem 2.21(b)). 2.62. Let P and Q be statements. Show that [(P V Q) A ~ ( P A Q )] = ~ ( P <4 Q). 2.63. Let n e Z. For which implication is its negation the following? The integer 3n + 4 is odd and 5n — 6 is even. 2.64. For which biconditional is its negation the following? n 3 and I n + 2 are odd or n ' and I n + 2 are even.
Section 2.10: Quantified Statements 2.65. Let S denote the set of odd integers and let P( x ) : x 2 + 1 is even,
and Q ( x )
be open sentences over the domain S. State Vx € S,P( x) and
: x 2 is even.
3x e S, Q ( x ) in words.
2.66. Define an open sentence R( x) over some domain S and then state Vx € S, R (x ) and 3x c S, R(x) in words. 2.67. State the negations of the following quantified statements, where all sets are subsets of some universal set U : (a) For every set A, A fl A = 0. (b) There exists a set A such that A c A. 2.68. State the negations of the following quantified statements: (a) For every rational number r, the number \ / r is rational. (b) There exists a rational number r such that r 2 = 2 .
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2.69. Let P(n): (5n  6)/3 is an integer, be an open sentence over the domain Z. Determine, with explanations, whether the following statements are true: (a) Sn e Z, P{n). (b) 3n e Z, P(n). 2.70. Determine the truth value of each of the following statements. (a) 3.v e R, x 2 â€”x = 0. (b) Sn e N, n + 1 > 2 . (c) Vx e R, V x 1 = x. (d) 3.v e Q, 3x2  27 = 0. (e) 3.\ e R. 3y e R, x + y + 3 = 8. (f) Vx, y e R, x + y + 3 = 8. (g) 3x, e R, x 2 + y 2 = 9. (h) Vx e R. Vy e R, x 2 + y 2 = 9. 2.71. The statement For every integer m. either m < 1 or n r > 4. can be expressed using a quantifier as: Sm e Z, m < 1 or m 2 > 4. Do this for the following two statements. (a) There exist integers a and b such that both ab < 0 and a + b > 0. (b) For all real numbers x and y, x / y implies that x 2 + y 2 > 0. (c) Express in words the negations of the statements in (a) and (b). (d) Using quantifiers, express in symbols the negations of the statements in both (a) and (b). 2.72. Let P (x ) and Q(x) be open sentences where the domain of the variable x is S. W hich of the following implies that ( ~ P(x) ) =$ Q(x) is false for some x e S I (a) (b) (c) (d) (e)
P( x) a Q(x) is false for all x e S. P( x) is true for all x e S. Q (x) is true for all x â‚Ź S. P( x) V Q (x) is false for some x e S. P( x ) A ( ~ Q (x)) is false for all x e S.
2.73. Let P( x) and Q ( x ) be open sentences where the domain of the variable x is T . W hich of the following implies that P( x ) Q(x) is true for all x e T1 (a) (b) (c) (d) (e) (f)
P( x) A Q( x) is false for all x e T . Q(x) is true for all x G T. P( x) is false for all x e T . P( x ) A ( ~ (<2(x)) is true for some x e T . P( x ) is true for all x e T. ( ~ P{x)) A ( ~ Q(x)) is false for all x e T .
2.74. Consider the open sentence P (x, v, z) : (x ~ l )2 + {y  2)2 + (z  2)2 > 0. where the domain of each of the variables x,
and z is R.
(a) Express the quantified statement Vx e R. Vy ÂŁ R, V: e R, P(x, y , z) in words. (b) Is the quantified statement in (a) true or false? Explain. (c) Express the negation of the quantified statement in (a)
in symbols.
(d) Express the negation of the quantified statement in (a)
in words.
(e) Is the negation of the quantified statement in (a) true or false? Explain. 2.75. Consider quantified statement For every s e S and t e S, st  2 is prime.
Exercises for Chapter 2
73
where the domain of the variables s and t is S = {3,5, 11}. (a) Express this quantified statement in symbols. (b) Is the quantified statement in (a) true or false? Explain. (c) Express the negation (d) Express the negation
of the quantified statement in (a) in symbols. of the quantified statement in (a) in words.
(e) Is the negation of the quantified statement in (a) true or false? Explain. 2.76. Let A be the set of circles in the plane with center (0, 0) and let B be the set of circles in the plane with center (1, 1). Furthermore, let / ’((.'i • Cz) C i and C 2 have exactly two points in common. be an open sentence where the domain of C\ is A and the domain of C 2 is B. (a) Express the following quantified statement in words: VCj e A, 3C2
g
B , P ( C , , C2).
(2.30)
(b) Express the negation of the quantified statement in (2.30) in symbols. (c) Express the negation of the quantified statement in (2.30) in words. 2.77. For a triangle T, let r ( T ) denote the ratio of the length of the longest side of T to the length of the smallest side of T . Let A denote the set of all triangles and let
'•(•/•,) >r<7,). be an open sentence where the domain of both T\ and T% is A. (a) Express the following quantified statement in words;
3Ti
g
A, Vr2 e A, P(TU 7V).
(2.31)
(b) Express the negation of the quantified statement in (2.31) in symbols. (c) Express the negation of the quantified statement in (2.31) in words. 2.78. Consider the open sentence P{a, b): a / b < 1. where the domain of a is A = {2, 3, 5} and the domain of b is B = {2,4,6}. (a) State the quantified statement Va g A, 3b € B . P(a, b) in words. (b) Show the quantified statement in (a) is true. 2.79. Consider the open sentence Q(a, by. a — b < 0. where the domain of a is A = {3, 5, 8} and the domain of b i s B = {3,6, 10}. (a) State the quantified statement 3b G B . V a e A, Q( a , b) in words. (b) Show the quantified statement in (a) is true.
Section 2.11: Characterizations of Statements 2.80. Give a definition of each of the following and then state a characterization of each. (a) Two lines in the plane are perpendicular. (b) A rational number. 2.81. Define an integer n to be odd if n is not even. State a characterization of odd integers. 2.82. Define a triangle to be isosceles if it has two equal sides. Which of the following statements are characterizations of isosceles triangles? If a statement is not a characterization of isosceles triangles, then explain why.
74
Chapter 2
Logic
(a) If a triangle is equilateral, then it is isosceles. (b) A triangle T is isosceles if and only if T has two equal sides. (c) If a triangle has two equal sides, then it is isosceles. (d) A triangle T is isosceles if and only if T is equilateral. (e) If a triangle has two equal angles, then it is isosceles. (f) A triangle T is isosceles if and only if T has two equal angles. 2.83. By definition, a right triangle is a triangle one of whose angles is a right angle. Also, two angles in a triangle are complementary if the sum of their degrees is 90°. W hich of the following statements are characterizations of a right triangle? If a statement is not a characterization of a right triangle, then explain why. (a) A triangle is a right triangle if and only if two of its sides are perpendicular. (b) A triangle is a right triangle if and only if it has two complementary angles. (c) A triangle is a right triangle if and only if its area is half of the product of the lengths of some pair of its sides. (d) A triangle is a right triangle if and only if the square of the length of its longest side equals to the sum of the squares of the lengths of the two smallest sides. (e) A triangle is a right triangle if and only if twice of the area of the triangle equals the area of some rectangle. 2.84. Two distinct lines in the plane are defined to be parallel if they don’t intersect. W hich of the following is a characterization of parallel lines? (a) Two distinct lines l\ and i 2 are parallel if and only if any line £ 3 that is perpendicular to £1 is also perpendicular to l 2. (b) Two distinct lines #1 and t 2 are parallel if and only if any line distinct from £ x and l 2 that doesn’t intersect also doesn’t intersect l 2. (c) Two distinct lines l\ and % are parallel if and only if whenever a line I intersects in an acute angle a , t h e n  also intersects l 2 in an acute angle a. (d) Two distinct lines l \ and t 2 are parallel if and only if whenever a point P is not on l j , the point P is not on i 2.
A D D IT IO N A L E X E R C IS E S FOR C H A PT E R 2 2.85. Construct a truth table for P
A
2.86. Given that the implication (Q
(Q => ( ~ P)). V
R) => ( ~ P) is false and Q is false, determine the truth values of R and P.
2.87. Find a compound statement involving the component statements P and Q that has the truth table given in Figure 2.18.
P Q
Figure 2.18
~Q
T T
T
F
T
F
T
T
F F
T
F
F
F
T
T
Truth table for Exercise 2.87.
75
Additional Exercises for Chapter 2 2.88. Determine the truth value of each of the following quantified statements: (a) Iv e R. x 3 + 2 = 0. (b) V« e N, 2 > 3 — n. (c) V.t € R. x[ = x. (d) 31 e Q, x 4  4 = 0. (e) 3x, y b R, x + y = n. (f) Vx, y e R, x + y = ^ /x 2 + y 2. 2.89. Rewrite each of the implications below using (1) only if and (2) sufficient. (a) If a function / is differentiable, then / is continuous. ■(b) If x = —5, then x 2 = 25. 2.90. Let Piny. n 2 — n + 5 is a prime, be an open sentence over a domain S. (a) Determine the truth values of the quantified statements V/z £ S, P(n) and 3n g S, ~ P(n) for S = {1,2, 3, 4}. (b) Determine the truth values of the quantified statements V« e S. Pi n) and Hh c: S. ~ Pi n I for S = {1, 2, 3, 4. 5}. (c) How are the statements in ( a) and (b) related? 2.91.
(a) For statements P, Q and R , show that ((P A Q) =* R ) s l ( P A ( /O ) =» ( ~ 0 ) ) . (b) For statements P, Q and R , show that ((P A Q) =» R) = UQ A ( ~ R))
(~P)l
2.92. For a fixed integer /?, use Exercise 2.91 to restate the following implication in two different ways: If n is a prime and n > 2. then n is odd. 2.93. For fixed integers m and n, use Exercise 2.91 to restate the following implication in two different ways: If m is even and n is odd, then m + n is 2.94.
odd.
For a realvalued function / and a real number x, use Exercise 2.91to restate the following implication in two different ways: If / '( x ) = 3x2 — 2x and /( 0 ) = 4, then / ( x ) = x 3 —x 2 + 4.
2.95. For the set S = {1, 2, 3}, give an example of three open sentences Pin), Q(n) and R(n), each over the domain S, such that (1) each of P(n), Q( n ) and R(n) is a true statement for exactly two elements of S, (2) all of the implications F( l) 0 (1 ), <2(2): ■> /\’(2) and R ( 3) => P ( 3) are true, and (3) the converse of each implication in (2) is false. 2.96. Do there exist a set S of cardinality 2 and a set {P(n), Q{n), R(n)} of three open sentences over the domain S such that ( 1) the implications P(a) =► Q(a). Q( b) =► R(b) and R(c) => Pic) are true, where a, b, c e S and (2) the converses of the implications in (1) are false? Necessarily, at least two of these elements a , b and c of S are equal. 2.97. Let A = {1, 2 , ___6} and B = { 1, 2 ..........7}. For x e A, let P( x) : l x + 4 is odd. For y & B, let Q ( y ) : 5y + 9 is odd. Let S = {(P(x), Q n y ) ) : x e A, y e B, P( x) => Q(y) is false}. W hat is ISI? 2.98. Let P(x, y, z) be an open sentence, where the domains of x, y and z are A, B and C, respectively. (a) State the quantified statement Vx e A, Vy e B . 3r e C, P{x, y, z) in words. (b) State the quantified statement Vx e A, Vy 6 B, 3z e C, P( x , y, z) in words for P(x, y, z) : x = yz.
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Chapter 2
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(c) Determine whether the quantified statement in (b) is true when A = {4. 8}, B = {2, 4} and C = {1,2,4}. 2.99. Let P(x, y, z) be an open sentence, where the domains of X'%y and 2 are A, B and C, respectively. (a) Express the negation of Vx e A, Vy e B, 3 z e C, P( x, y, z) in symbols. (b) Express ~ (Vx e A, Vy e B, 3z e C. P( x, y , 2 )) in words. (c) Determine whether ~ (Vx e A, Vy e B, 32 G C, P (x , y»aO') is true when P(x, y, z) : x + 2 = y. for A = {1, 3}, B = {3, 5, 7} and C = {0, 2, 4, 6}. 2.100. Write each of the following using “if, then.” (a) (b) (c ) (d) (e) (f) (g) (h)
A sufficient condition for a triangle to be isosceles is that it has two equal angles. Let C be a circle of diameter s / 2 / n . Then the area of C is 1/2 . The 4th power of every odd integer is odd. Suppose that the slope of a line I is 2. Then the equation of £ is y = 2x + b for some real number b. Whenever a and b are nonzero rational numbers, a / b is a nonzero rational number. For every three integers, there exist two of them whose sum is even. A triangle is a right triangle if the sum of two of its angles is 90°. The number y/3 is irrational.
3 Direct Proof and Proof by Contrapositive
e are now prepared to begin discussing our m ain topic: m athem atical proofs. Initially, w e w ill be prim arily concerned w ith one question: For a given true m athem atical statem ent, how can we show that it is true? In this chapter, you w ill be introduced to tw o im portant proof techniques. A true m athem atical statem ent w hose truth is accepted w ithout p ro o f is referred to as an axiom . For exam ple, an axiom of E uclid in geom etry states that for every line t and point P not on I. there is a unique line containing P that is parallel to t. A true m athem atical statem ent w hose truth can be verified is often referred to as a th e o re m , although m any m athem aticians reserve the w ord theorem for such statem ents that are especially significant or interesting. For exam ple, the m athem atical statem ent “2 + 3 = 5” is true but few, if any, w ould consider this to be a theorem under this latter interpretation. In addition to the w ord theorem , other com m on term s for such statem ents include proposition, result, observation and fact, the choice often depending on the significance o f the statem ent or the degree o f difficulty o f its proof. We w ill use the w ord theorem sparingly, however, prim arily reserving it for true m athem atical statem ents that w ill be used to verify other m athem atical statem ents that w e will encounter later. O therw ise, we will sim ply use the w ord result. For the m ost part then, our results are exam ples used to illustrate proof techniques and our goal is to prove these results. A c o ro lla ry is a m athem atical result that can be deduced from , and is thereby a consequence of, som e earlier result. A le m m a is a m athem atical result that is useful in establishing the truth o f some other result. Some people like to think o f a lem m a as a “helping result.” Indeed, the G erm an w ord for lem m a is hilfsatz, w hose E nglish translation is “helping theorem .” O rdinarily then, a lem m a is n ot o f prim ary im portance itself. Indeed, its very existence is due only to its usefulness in proving another (m ore interesting) result. M ost theorem s (or results) are stated as im plications. We now begin our study of proofs o f such m athem atical statem ents.
W
77
78
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Direct Proof and Proof by Contrapositive
3.1 Trivial and Vacuous Proofs In nearly all o f the im plications P =>• Q that we w ill encounter, P and Q are open sentences; that is, w e w ill actually be considering P ( x ) => Q( x ) or P ( n ) => Q ( n ) or som e related im plication, depending on w hich variable is being used. T he variables x or n (or som e other sym bols) are used to represent elem ents o f som e set S being discussed; that is, S is the dom ain o f the variable. A s w e have seen, for each value o f a variable from its dom ain, a statem ent results. (It is possible, o f course, that P and Q are expressed in term s o f tw o or m ore variables.) W hether P(x') (or Q( x) ) is true ordinarily depends on w hich elem ent x e S w e are considering; that is, it is rarely the case that P ( x ) is true for all x e S (or that P ( x ) is false for all x e S). For exam ple, for P( n ) : 3 n 2 — An + 1 is even w here n e Z, P { \ ) is a true statem ent w hile P ( 2) is a false statem ent. Likew ise, it is seldom the case that Q( x ) is true for all x e S or that Q( x ) is false for all x e S. W hen the quantified statem ent Vx e S, P{ x ) => Q ( x ) is expressed as a result or theorem , w e often write such a statem ent as For
e S, if P ( x ) then Q( x) .
or as Let x e S. If P( x) , then Q{x).
(3.1)
Thus (3.1) is true if P ( x ) => Q( x ) is a true statem ent for each x e S , w hile (3.1) is false if P ( x ) =3> Q( x ) is false for at least one elem ent x e S. For a given elem ent x e S, le t’s recall (see the truth table in Figure 3 .1) the conditions under w hich P{ x ) =>• Q( x ) has a particular truth value. A ccordingly, if Q( x ) is true for all x 6 S or P ( x ) is false for all x e S, then d e term ining the truth or falseness o f (3.1) becom es considerably easier. Indeed, if it can be show n that Q( x ) is true for all x e S (regardless o f the truth value o f P (x )), then, according to the truth table for the im plication (show n in Figure 3.1), (3.1) is true. This constitutes a proof o f (3.1) and is called a trivial proof. A ccordingly, the statem ent L et n e Z. If n 3 > 0, then 3 is odd. is true and a (trivial) proof consists only o f observing that 3 is an odd integer. The follow ing provides a m ore interesting exam ple o f a trivial proof.
P(x)
Q(x)
P ( x ) => Q ( x )
T
T
T
T
F
F
F
T
T
F
F
T
Figure 3.1 The truth table for the implication P( x ) => Q{x) for an element x in its domain
3.1
Result 3.1 Proof
Trivial and Vacuous Proofs
79
L e t x e R. I f x < 0, then x 2 + 1 > 0. Since x 2 > 0 for each real num ber x, it follow s that x 2 + 1 > x2 > 0. H ence x 2 + 1 > 0. C onsider
■
P{ x ) : x < 0 and Q ( x ) : x 2 + 1 > 0 w here x e R. Then Result 3.1 asserts the truth of: For all x e R , P ( x ) =>■ Q (x ). Since we verified that Q( x ) is true for every x e R , it follow s that P ( x ) Q( x ) is true for all x 6 R and so Result 3.1 is true. In this case, w hen considered over the dom ain R, Q (x ) is actually a true statem ent. It is this fact that allow ed us to give a trivial p ro o f of Result 3.1. The proof o f R esult 3.1 does not depend on x < 0 . Indeed, provided that x e R , we could have replaced “x < O’ by any hypothesis (including the m ore satisfying “x e R ”) and the result w ould still be true. In fact, this new result has the same proof. To be sure, it is rare indeed w hen a trivial proof is used to verify an im plication; nevertheless, this is an im portant rem inder o f the truth table in Figure 3.1. The sym bol a that occurs at the end of the pro o f o f Result 3.1 indicates that the proof is com plete. There are definite advantages to using ■ (or som e other sym bol) to indicate the conclusion of a proof. First, as you start reading a proof, you can look ahead for this sym bol to determ ine the length o f the proof. A lso, w ithout this sym bol, you m ay continue to read past the end o f the proof, still thinking that y o u ’re reading a p ro o f of the result. W hen you reach this sym bol, you are supposed to be convinced that the result is true. If you are, this is good! Everything happened as planned. O n the other hand, if yo u 're not convinced, then, to you, the w riter h asn ’t presented a proof. This m ay not be the w riter's fault, however. In the past, the m ost com m on w ay to indicate that a p ro o f has concluded w as to write Q .E.D ., w hich stands for the Latin phrase “quod erat dem onstrandum ,” w hose E nglish translation is “w hich was to be dem onstrated.” Som e still use it. L et P( x ) and Q( x ) be open sentences over a dom ain S. Then Vx e S, P ( x ) —> Q( x ) is a true statem ent if it can be show n that P { x ) is false for all x e S (regardless o f the truth value of <2(x)), according to the truth table for im plication. Such a pro o f is called a v acu o u s p ro o f o f Vx e S, P ( x ) => Q( x) . Therefore, L et n e Z. If 3, is even, then n 3 > 0. is a true statem ent. L e t’s take a look, however, at a m ore interesting exam ple o f a vacuous proof. Result 3.2 Proof
Let x e R. I f x 2 — 2x + 2 < 0, then x 3 > 8 . First observe that x 2 — 2x + 1 = (x — l )2 > 0 . T herefore, x 2 — 2x + 2 = (x — l )2 + 1 > 1 > 0. Thus X2 x G R and the im plication is true.
2x t 2 < 0 is false for all ■
80
Chapter 3
Direct Proof and Proof by Contrapositive For P ( x ) : x 2  2x + 2 < 0 and Q( x ) : x 3 > 8 over the dom ain R , R esult 3.2 asserts the truth o f Vx £ R, P ( x ) =» Q( x) . Since we verified that P ( x ) is false for every ,v c R , it follow s that P (x ) => Q( x ) is true for each x € R. H ence R esult 3.2 is true. In this case, P ( x ) is a false statem ent for each x £ R. T his is w hat perm itted us to give a vacuous p ro o f o f R esult 3.2. In the proof o f R esult 3.2, the truth or falseness o f x 3 > 8 played no role whatsoever. Indeed, had we replaced x 3 > 8 by x 3 < 8, for exam ple, then neither the truth nor the pro o f o f Result 3.2 w ould be affected. W henever there is a vacuous p ro o f o f a result, we often say that the result follow s vacuously. As we m entioned, a trivial p ro o f is alm ost never encountered in m athem atics; however, the sam e cannot be said o f vacuous proofs, as w e will see later. We consider one additional exam ple.
Result 3.3 Proof
Let S = [n e Z : n > 2} and let n e S. I f 2 n + = < 5, then 4n 2 + 4 < 25 n
n~
First, w e observe that if n = 2, then 2n + \ = 5. O f course, 5 < 5 is false. If n > 3, then 2n + y > 2n > 6 . So, w hen n > 3. 2n — 2 < 5 is false as well. Thus 2n +  < 5 ft fl is false for all n £ S. H ence the im plication is true. k In two o f the exam ples that w e presented to illustrate trivial and vacuous proofs, we used the fact (and assum ed it was know n) that 3 is odd. A lso, in the proofs o f Results 3.1 and 3.2, we used the fact that if r is any real num ber, then r 2 > 0. A lthough you are certainly fam iliar w ith this property o f real num bers, it is essential that any facts used w ithin a pro o f are know n to and likely to be recalled by the reader. Facts used w ithin a pro o f should not com e as a surprise to the reader. This subject w ill be discussed in m ore detail shortly. Even though the trivial and vacuous proofs are rarely encountered in m athem atics, they are im portant rem inders o f the truth table for im plication. We are now prepared to be introduced to the first m ajor p ro o f technique in m athem atics.
3 .2 D irect P roofs Typically, w hen we are discussing an im plication P ( x ) =>• Q( x ) over a dom ain S, there is some connection betw een P ( x ) and Q( x) . T hat is, the truth value o f Q( x ) for a particular x £ S often depends on the truth value o f P (x ) for that same elem ent x , or the truth value o f P ( x ) depends on the truth value o f Q( x) . These are the kinds o f im plications in w hich we are prim arily interested and it is the proofs o f these types o f results that will occupy m uch o f our attention. We begin w ith the first m ajor pro o f technique, w hich occurs m ore often in m athem atics than any other technique. L et P ( x ) and Q( x ) be open sentences over a dom ain S. Suppose that our goal is to show that P( X) => Q (x ) is true for every x £ S, that is, our goal is to show that the quantified statem ent Vx £ S, P ( x ) =*■ Q( x ) is true. If P ( x ) is false for som e x £ S , then P CO =>■ Q( x ) is true for this elem ent x. H ence we need only be concerned w ith show ing that P { x ) =► Q ( x ) is true for all x € S for w hich P ( x ) is true. In a d ire c t p ro o f of P (x) => Q{x) for all x £ S, we consider an arbitrary elem ent x e S for w hich P ( x ) is
3.2
Direct Proofs
81
true and show that Q( x ) is true for this elem ent x. To sum m arize then, to give a direct proof o f P ( x ) =£■ Q( x ) for all x 6 S, w e assum e that P ( x ) is true for an arbitrary elem ent x e S and show that Q( x ) m ust be true as w ell for this elem ent JT. In order to illustrate this type o f pro o f (and others as w ell), w e need to deal with m athem atical topics w ith w hich w e ’re all familiar. L e t’s first consider the integers and som e of their elem entary properties. We assum e that you are fam iliar w ith the integers and the follow ing properties o f integers: 1.
The negative o f every integer is an integer.
2.
The sum (and difference) o f every two integers is an integer.
3.
The product o f every tw o integers is an integer.
We w ill agree that we can use any o f these properties. N o justification is required or expected. Initially, we w ill use even and odd integers to illustrate our p ro o f techniques. In this case, however, any properties o f even and odd integers m ust be verified before they can be used. For exam ple, you probably know that the sum o f every tw o even integers is even but this m ust first be proved to be used. We need to lay som e groundw ork before any exam ples o f direct proofs are given. Since w e w ill be w orking w ith even and odd integers, it is essential that we have precise definitions o f these kinds o f num bers. An integer n is defined to be even if n — 2k for som e integer k. For exam ple, 10 is even since 10 = 2  5 (where, o f course, 5 is an integer). A lso,  1 4 = 2 (—7) is even, as is 0 = 2 • 0. The integer 17 is not even since there is no integer k for w hich 17 = 2k. Thus we see that the set o f all even integers is the set S = {2k : k e Z \ = {■■■,  4 ,  2 , 0, 2, 4, • • •}. We could define an integer n to be odd if it’s not even but it w ould be difficult to w ork w ith this definition. Instead, w e define an integer n to be o d d if n = 2k + 1 for some integer k. Now 17 is odd since 17 = 2  8 + 1. A lso,  5 is odd because  5 = 2 (—3) + L On the other hand, 26 is not odd since there is no integer k such that 26 = 2k + 1. In fact, 26 is even. H ence, according to the definition o f odd integers that w e have ju st given, we see that the set o f all odd integers is precisely the set T = { 2 k + \ : k e Z} = {■■■,  5 , —3,  1 , 1, 3, 5, ■• •}. O bserve that S and T are disjoint sets and S LI T = Z; that is, Z is partitioned into S and T . T herefore, every integer is either even or odd. From tim e to tim e, we w ill find ourselves in a position w here we have a result to prove, and it m ay not be entirely clear how to proceed. In such a case, w e need to consider our options and develop a plan, w hich w e refer to as a p ro o f stra te g y . The idea is to discuss a p roof strategy for the result and, from it, construct a proof. At other tim es, we m ay w ish to reflect on a proof that w e have ju st given in order to understand it better. Such a discussion w ill be referred to as a p ro o f an aly sis. A s w ith exam ples, w e conclude both a proof strategy and a proof analysis w ith the sym bol ♦. We are now prepared to illustrate the direct p ro o f technique. We follow the proof by a proof analysis. Result 3.4
I f n is an odd integer, then 3n + 7 is an even integer.
82
Chapter 3 Proof
Direct Proof and Proof by Contrapositive A ssum e that n is an odd integer. Since n is odd, w e can w rite n = 2 k + 1 for som e integer k Now 3n + 7 = 3(2 k + 1) + 7 = 6k + 3 + 7 = 6 k + 1 0 = 2(3 k + 5). Since 3k + 5 is an integer, 3n + 7 is even.
PROOF ANALYSIS
■
First, notice that R esult 3.4 could have been stated as either For every odd integer n, the integer 3n + 7 is even. or Let n be an odd integer. T hen 3n + 7 is even. T hus the dom ain o f the variable n in Result 3.4 is the set o f odd integers. In the proof o f R esult 3.4, the expression 2k + 1 was substituted for n in 3n + 7 and sim plified as 6k + 10. Since our goal w as to show that 3n + 7 is even, w e needed to show that 3 n + 7 can be expressed as tw ice an integer. Consequently, w e factored 2 from 6k + 10 and w rote it as 2(3k + 5). Since 3 and k are integers, so is 3k (the product o f two integers is an integer). Since 3k and 5 are integers, so is 3* + 5 (the sum of tw o integers is an integer). T herefore, 3n + 7 satisfies the definition o f an even integer. O ne other rem ark deserves m ention here. In the second sentence, w e wrote: Since n is odd, we can w rite n = 2k + 1 fo r som e integer k. It w ould be incorrect to write: “I f n is odd” rather than “Since n is odd” because we have already assum ed that n is odd and therefore n is now know n to be odd. ♦ We defined an integer n to be odd if w e can w rite n as 2k + 1 for som e integer k. T his m eans that w henever w e w ant to show that an integer, say m, is odd, w e m ust follow this definition; that is, we m ust show that m — 2 k + \ for some integer k. (O f course, the use o f the sym bol k is not im portant. F or exam ple, an odd integer n can be w ritten as n = 2£ + 1 for som e integer I.) We could have defined an integer n to be odd if it is possible to w rite n = 2k — 1 for som e integer k, but w e d id n ’t. However, if we could prove that an integer n is odd if and only if n can be expressed as 2 k — 1 for some integer k, then w e could use this characterization o f odd integers to show that an integer is odd. This, however, w ould require additional w ork on our part with no obvious benefit. Sim ilarly, we could have defined an integer n to be even if we can w rite n = 2k + 2, or n = 2k — 2 or perhaps n = 2k + 100 for som e integer k. The definitions o f even and odd integers that w e chose are probably the m ost com m only used. Any other definitions that could have been given provide no special advantage to us. The proof given o f R esult 3.4 is an exam ple o f a direct proof. Let Q(n): 3n + 7 is an even integer. over the dom ain o f odd integers. We verified R esult 3.4 by assum ing that n is an arbitrary odd integer and then show ing that Q( n) is true for this elem ent /?. Showing that Q( n) is true essentially required one step on our part. As we venture further into proofs, we w ill see that w e c a n ’t always establish the truth o f the desired conclusion so quickly. It may
3.2
Direct Proofs
83
be necessary to establish the truth o f som e other m athem atical statem ents along the way that can then be used to establish the truth o f Q(n). We w ill see exam ples o f this later. L e t’s consider another exam ple. For variety, w e use an alternative opening sentence and different sym bols in the proof o f the follow ing result. Result 3.5 P roof
I f n is an even integer, then —5m  3 is an odd integer. L et n be an even integer. T hen n = 2x, w here x is an integer. Therefore, —5n  3 =  5 ( 2 x )  3 =  l O x  3 =  l O x  4 + 1 = 2 (  5 x  2) + 1. Since
a
5x  2 is an integer,  5 / ;  3 is an odd integer.
We now consider another exam ple, w hich m ay have a surprise ending. Result 3.6 Proof
I f n is an odd integer, then 4 n 3 + 2n — 1 is odd. A ssum e that n is odd. T hen n — 2_v + 1 for som e integer y . Therefore, 4 n 3+ 2 n  l =
4(2 y + l )3 + 2(2y + 1)  1 = 4(8y3 + 12y2 + 6 y + 1) + 4 y + 2  1 = 3 2y3 + 4 8 y 2 + 28y + 5 = 2(16v3 + 2 4 y 2 + 14y + 2) +
Since 16y3 + 2 4y2 + 14y + 2 is an integer, 4 n 3 + 2n PROOF ANALYSIS
1.
1 is odd.
■
A lthough the direct p ro o f o f Result 3.6 that w e gave is correct, this is not the desired proof. Indeed, had w e observed that 4n 3 + 2n  1 = 4 n3 + 2n  2 + 1 = 2(2n 3 + n  1) + 1 and that 2n 3 + n — 1 6 Z, w e could have concluded im m ediately that 4 n3 + 2n — 1 is odd for every integer n. H ence a trivial p ro o f o f R esult 3.6 could be given and, in fact, is preferred. The fact that 4//3 + 2n  1 is odd does not depend on n being odd. Indeed, it w ould be far better to replace the statem ent o f R esult 3.6 by If n is an integer, then 4/?3 + 2n  1 is odd.
+
We give an additional exam ple o f a som ew hat different type. Result 3.7
t t c ii i ii a i , c U w( « + 3 ) . (n+2)(n5) . L e t S = {1, 2, 3} and let n € S. I f   is even, t h e n  is even. ** 2
Proof
Let n e S such that n(n + 3 )/2 is even. Since n ( n + 3 ) /2 = 2 w hen n = 1, n(n + 3 )/2 = 5 w hen n = 2 and n(n + 3 )/2 = 9 w h e n /i = 3, it follow s that n = 1. W hen n = 1 ,(« + 2) (n  5 )/2 =  6 , w hich is even. Therefore, the im plication is true. m
PROOF ANALYSIS
In the pro o f o f Result 3.7, w e w ere only concerned w ith those elem ents n e S for w hich n(n + 3 )/2 is even. F urtherm ore, it is not initially clear for w hich elem ents n o f S the integer n(n + 3 )/2 is even. Since S consists only o f three elem ents, this can be determ ined
84
Chapter 3
Direct Proof and Proof by Contrapositive rather quickly, w hich is w hat w e did. We saw that only n — 1 has the desired property and this is the only elem ent w e needed to consider. ♦ If our goal is to establish the truth o f P ( x ) =>■ Q{x) for all x in a dom ain S by m eans o f a direct proof, then the proo f begins by assum ing that P ( x ) is true for an arbitrary elem ent x e S. It is often com m on in this situation, however, to om it the initial assum ption that P ( x ) is true for an arbitrary elem ent x e S. It is then understood that w e are giving a direct proof. We illustrate this w ith a short exam ple.
Result 3.8 P roof
I f n is an even integer, then 3n 5 is an even integer. Since n is an even integer, n = 2x for som e integer x . T herefore, 3 n 5 = 3(2x)5 = 3(3 2 x 5) = 9 6 x 5 = 2(4 8 x 5). Since 4 8x5 e Z, the integer 3n 5 is even.
■
For the present, w hen giving a direct p ro o f o f P ( x ) =>• Q( x ) for all x in a dom ain S, we will often include the initial assum ption that P (x ) is true for an arbitrary elem ent x 6 S in order to solidify this technique in your mind.
3 .3
P r o o f b y C o n tra p o sitiv e For statem ents P and Q, the c o n tra p o sitiv e o f the im plication P => Q is the im plication (~<2) ( ' "P ) For exam ple, for P i : 3 is odd and Pi_ : 57 is prim e, the contrapositive of the im plication
P\ => P2 : If 3 is odd, then 57 is prim e. is the im plication ( ~ P 2) => ( ~ P i ) : If 57 is n ot prim e, then 3 is even. The m ost im portant feature o f the contrapositive ( ~ g ) =>• ( ~ / >) is that it is logically equivalent to P => Q. This fact is stated form ally as a theorem and is verified in the truth table show n in Figure 3.2. Theorem 3.9
For every two statem ents P and Q, the im plication P => Q and its contrapositive are logically equivalent; that is,
p
=> Q = (~G) => (~P).
Let
P( x) : x = 2. and Q(x) : x 2 = 4. w here x e R. T he contrapositive o f the im plication
P( x) =^* Q(x) : I f x = 2, then x 2 = 4. is the im plication
3.3
P Q P =» Q ~ P T
85
~ Q
T F T
T F T
F
F
T
F T
T F
F T
F F
T
T
T
T
T F
Figure 3 .2
Proof by Contrapositive
T h e lo g ic a l e q u iv a le n c e o f a n im p lic a tio n a n d its c o n tr a p o s i tiv e
=> C' /^C.v>> : I f x 2 ^ 4, t h e n x ^ 2. Suppose that we w ish to prove a result (or theorem ) w hich is expressed as L et x 6 S. If /J(x). then <2(x )
(3.2)
For all x e S. if P( x) , then Q( x) .
(3.3)
or as
We have seen that a proof o f such a result consists o f establishing the truth o f the im plication P ( x ) =>■ Q( x ) for all x e S. If it can be show n that (~ (X v ) ) >■ ( ~ / J (x)) is true for all x e S, then P ( x ) =$> Q( x ) is true for all x € S. A p ro o f by c o n tra p o sitiv e o f the result (3.2) (or o f (3.3)) is a direct p ro o f o f its contrapositive: L et x € S. If ~<2(x), then ~ P ( x ) . or For all x € S, if ~£>(x), then ~ P ( x ) . Thus to give a proof by contrapositive o f (3.2) (or o f (3.3)), we assum e that ~ g ( x ) is true for an arbitrary elem ent x e S and show that ~ /D(x) is true for this elem ent x. There are certain types o f results w here a p ro o f by contrapositive is preferable or perhaps even essential. We now give som e exam ples to illustrate this m ethod o f proof. Result 3.10 P roof
L et x e Z. I f 5x — 7 is even, then x is odd. A ssum e that x is even. Then x = 2a for som e integer a. So 5x  7 = 5 (2a)  7 = 10a  7 = 10a  8 + 1 = 2(5a  4) + 1. Since 5a — 4 e Z, the integer 5x — 7 is odd.
PROOF ANALYSIS
g
Som e com m ents are now in order. The goal o f R esult 3.10 w as to prove P ( x ) =>• Q( x ) for all x e Z, w here P ( x ) : 5x — 7 is even, and Q( x ) : x is odd. Since w e chose to give a proof by contrapositive, we gave a direct p ro o f o f (~£>(x)) ( ~ P ( j J ) for all x e Z. H ence the proof began by assum ing that x is not odd; that is, x is even. The object then was to show that 5x — 7 is odd. If w e had attem pted to prove R esult 3.10 w ith a direct proof, then w e w ould have begun by assum ing that 5x  7 is even for an arbitrary integer x . T hen 5x  7 = 2a for som e integer a. So x = (2a + 7 )/5 . We then w ould w ant to show that x is odd. W ith the expression w e have for x , it is not even clear that x is an integer, m uch less that x is an odd
Chapter 3
Dircci Proof' and Proof by Contrapositive integer, although, o f course, we were told in the statem ent o f Result 3.10 that the dom ain o f x is the set o f integers. Therefore, it is not only that a p ro o f by contrapositive provides us w ith a rather sim ple m ethod o f proving Result 3.10, it m ay not be im m ediately clear how or w hether a direct proof can be used. H ow did w e know beforehand that it is a p ro o f by contrapositive that w e should use here? T his is not as difficult as it m ay appear. If w e use a direct proof, then we begin by assum ing that 5x — 7 is even for an arbitrary integer x ; w hile if we use a proof by contrapositive, then we begin by assum ing that x is even. T herefore, using a proof by contrapositive allow s us to w ork w ith x initially rather than the m ore com plicated expression 5* — 7. + In all o f the exam ples that w e have seen so far, w e have considered only im plications. N ow we look at a biconditional.
Result 3.11 P roof
L e t x e Z. Then 1 I x — 7 is even i f a nd only i f x is odd. There are tw o im plications to prove here, namely, (1) if x is odd, then 11* — 7 is even and (2) if 1 l x — 7 is even, then x is odd. We begin w ith (1). In this case, a direct p ro o f is appropriate. A ssum e that x is odd. Then x = 2r + 1, w here r e Z. So l l x  7 = 11(2r + 1)  7 = 22r + 1 1  7 = 22r + 4 = 2(11 r + 2). Since H r + 2 is an integer, 1 lx — 7 is even. We now prove (2), w hich is the converse o f (1). We use a p ro o f by contrapositive here. A ssum e that x is even. Then x = 2s, w here s € Z . T herefore, l l x  7 = 11 (2s) — 7 = 22s  7 = 22s  8 + 1 = 2 ( 1 I s — 4 ) + 1. Since 11s — 4 is an integer, l l x — 7 is odd.
■
A com m ent concerning the statem ents of Results 3.10 and 3.11 bears repeating here. These results begin w ith the sentence: L e tx e Z. This, o f course, is inform ing us that the dom ain in this case is Z. That is, w e are being told that x represents an integer. We need not state this assum ption in the proof. The sentence “L et x € Z.” is com m only called an “overriding” assum ption or hypothesis and so x is assum ed to be an integer throughout the proofs o f R esults 3.10 and 3.11. In the pro o f o f R esult 3.11, w e discussed our plan o f attack. N am ely, w e stated that there w ere two im plications to prove, and we specifically stated each. O rdinarily we d o n ’t include such inform ation w ithin the pro o f— unless the proof is quite long, in w hich case a roadm ap indicating the steps w e plan to take m ay be helpful. We give an additional exam ple o f this type, w here this tim e a m ore conventional condensed p ro o f is presented. The follow ing exam ple will be useful to us in the future, thus we refer to it as a theorem . Theorem 3.12
L e t x e Z. Then x 2 is even i f an d only i f x is even.
3.3 Proof
Proof by Coiitrapositive
87
A ssum e th a tM is even. T hen x = 2a for som e integer a. Therefore, x 2 = (2a)2 = 4 a 2 = 2(2 a 2). B ecause 2a 2 e Z, the integer x 2 is even. For the converse, assum e that x is odd. So x — 2b + I, w here b € Z. Then .r2 = (2b + l )2 = 4b2 + 4b + 1 = 2(2 Zr + 2b) + 1. Since 2b 2 + 2b is an integer, x 2 is odd.
■
Suppose now that you w ere asked to prove the follow ing result: L et x £ Z. Then x 2 is odd if and only if x is odd.
(3.4)
How w ould you do this? You m ight think o f proving the im plication “If x is odd, then x 2 is odd.” by a direct proof and its converse “If x 2 is odd, then ,v is odd.” by a p ro o f by contrapositive, w here, o f course, the dom ain o f ,v is Z. If we look at w hat is happening here, w e see that we are duplicating the p ro o f o f T heorem 3.12. This is no surprise whatsoever. T heorem 3.12 states that if x is even, then x 2 is even; and if x 2 is even, then x is even. The contrapositive o f the first im plication is “If x 2 is odd, then x is odd,” w hile the contrapositive o f the second im plication is “If x is odd, then x 2 is odd.” In other w ords, (3.4) sim ply restates T heorem 3.12 in term s o f contrapositives. Thus (3.4) requires no pro o f at all. It is essentially a restatem ent o f T heorem 3.12. A nd speaking of restatem ents o f T heorem 3.12, w e need to recognize that this theorem can be restated in other ways. For exam ple, w e could restate If x is an even integer, then x 2 is even. as The square o f every even integer is even. H ence T heorem 3.12 could be stated as: A n integer is even i f and only i f its square is even. It is not only useful to som etim es restate results in different m anners for variety, it is im portant to recognize w hat a result is saying regardless of the m anner in w hich it m ay be stated. A t this point, it is convenient to pause and discuss how theorem s (or results) can be used and w hy it is that w e may be interested in proving a particular theorem . First, it is only by providing a proof o f a theorem that w e know for certain that the theorem is true and therefore have the right to call it a theorem . A fundam ental reason w hy m athem aticians m ay w ant to give a proof o f som e m athem atical statem ent is that they consider this a challenge— this is w hat m athem aticians do. This, in fact, brings up a question that m any m athem aticians consider o f greater im portance. W here do such statem ents com e from ? O f course, the answ er is that they com e from m athem aticians or students. How these people arrive at such questions does not follow any set rule. B ut this deals w ith the creative aspect o f m athem atics. Some
88
Chapter 3
Direct Proof am i Proof by Contrapositive people are curious and im aginative. Perhaps w hile proving som e theorem , it is realized that the m ethod o f pro o f used could be applied to prove som ething even m ore interesting. (W hat is interesting, o f course, is quite subjective.) M ore than likely however, a person has observed some relationship that exists in an exam ple being considered that appears to occur in a m ore general setting. T his individual then attem pts to show that this is the case by giving a proof. This entire process involves the idea o f conjectures (guesses) and trying to show the accuracy o f a conjecture. W e'll discuss this at greater length later. Suppose that w e have been successful in proving P{ x) =>■ Q(.x) for all x in some dom ain S (by w hatever m ethod). We therefore know that for every x e S for w hich the statem ent P ( x ) is true, the statem ent Q { x ) is true. A lso, for any a e S for w hich the statem ent Q{x) is false, the statem ent P ( x ) is false. For exam ple, since we know that R esult 3.10 is true, if w e should ever encounter an integer n for w hich 5n — 7 is even, then w e know that n is odd. Furtherm ore, if w e should encounter an integer n for w hich n 2 is odd, then we can conclude by statem ent (3.4) or, b etter yet, by T heorem 3.12, that n itself m ust be odd. It is not only know ing that a particular theorem m ight be useful to us in the future, it is perhaps that a theorem seems surprising, interesting or even beautiful. (Yes— to m athem aticians, and hopefully to you as w ell, a theorem can be beautiful.) We next describe a type o f result that we have not yet encountered. C onsider the follow ing result, w hich w e w ould like to prove.
R esult to Prove PROOFSTRATEGY
L et x e Z. If 5 x — 7 is odd, then 9x + 2 is even. This result do esn ’t seem to fit into the kinds o f results w e ’ve been proving. (This is not unusual. A fter learning how to prove certain statem ents, w e encounter new statem ents that require us to . . . think.) If w e attem pt to give either a direct p ro o f or a pro o f by contrapositive o f this result, we m ay be headed for difficulties. There is, however, another approach. E ven though we m ust be very careful about w hat we are assum ing, from w hat we know about even and odd integers, it appears that if 5x — 7 is odd, then x m ust be even. In fact, if w e knew that w henever 5 x — 7 is odd then x is even, this fact w ould be extrem ely helpful. We illustrate this next. D o n ’t forget that our goal is to prove the follow ing result, w hich we w ill refer to as Result 3.14: L et x e Z. If 5 x — 7 is odd, then 9a + 2 is even. The (unusual) num bering o f this result is because we w ill first state and prove a lem m a (Lem m a 3.13) that will aid us in the pro o f o f Result 3.14. ♦
In order to verify the truth o f R esult 3.14, w e first prove the follow ing lem m a. Lemma 3.13 Proof
L et x e Z. I f 5.v — 7 is odd, then x is even. A ssum e that x is odd. T hen J = 2 y + 1, w here y e Z. Therefore, 5 a  7 = 5(2y + 1)  7 = 10y  2 = 2(5y  1). Since 5y — 1 is an integer,
5a
 7 is even.
We are now prepared to give a proo f o f R esult 3.14.
a
3.4
Result 3.14 Proof
Proof by Cases
89
L et x a Z. I f 5 x  7 is o d d , then 9x f 2 is even. Let 5 x — 1 be an odd integer. By L em m a 3.13. the integer x is even. Since x is even, x 2z for some integer z . Thus 9x + 2 = 9(2z) + 2 = 18z + 2 = 2(9z + 1). B ecause 9z + 1 is an integer, 9 x + 2 is even.
*
So, w ith the aid o f L em m a 3.13, w e have produced a very uncom plicated (and, hopefully, easytofollow ) p ro o f of R esult 3.14. The m ain reason for presenting Result 3.14 w as to show how helpful a lem m a can be in producing a proof o f another result. However, having ju st said this, we now show how we can prove Result 3.14 w ithout the aid o f a lem m a, by perform ing a bit of algebraic m anipulation. Alternative Proof o f Result 3.14
A ssum e that 5x — 7 is odd. T hen 5x — 7 = 2n + 1 for som e integer n. O bserve that 9x + 2 = (5x  7) + (4x + 9) = 2n + 1 + 4x + 9 — 2/7 ( 4x f 10 = 2(n H 2x j 5). B ecause n + 2x + 5 is an integer, 9x + 2 is even.
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You m ay prefer one p ro o f o f R esult 3.14 over the other. W hether you do or not, it is im portant to know that two different m ethods can be used. T hese m ethods m ight prove to be useful for future results you encounter. A lso, you m ight think w e used a trick to give the second p roof o f R esult 3.14; but, as we w ill see, if the sam e “trick ” can be used often, then it becom es a technique.
3.4 P roof by Cases W hile attem pting to give a pro o f o f a m athem atical statem ent concerning an elem ent x in som e set 5, it is som etim es useful to observe that x possesses one o f two or m ore properties. A com m on property w hich x m ay possess is that o f belonging to a particular subset of S. If we can verify the truth o f the statem ent for each property that x m ay have, then w e have a proof o f the statem ent. Such a pro o f is then divided into parts called cases, one case for each property that x m ay possess or for each subset to w hich x may belong. This m ethod is called proof by cases. Indeed, it m ay be useful in a p ro o f by cases to further divide a case into other cases, called subcases. For exam ple, in a proof o f V/? e Z. R(n) , it m ight be convenient to use a p ro o f by cases w hose pro o f is divided into the two cases Case 1. n is even, and Case 2. n is odd. O ther possible proofs by cases m ight involve proving Vx e R, P ( x ) using the cases Case 1 .x = 0. Case 2. x < 0 and Case 3. x > 0.
Chapter 3
Direct Proof and Proof by Contrapositive A lso, we m ight attem pt to prove i n e N, P( n) using the cases Case 1. n = 1 and Case 2. n > 2. Furtherm ore, for S = Z — {0}, w e m ight try to prove Vx, y e S, P( x . y) by using the cases: Case 1. x y > 0 and Case 2. x y < 0. C ase 1 could, in fact, be divided into two subcases: Subcase 1.1. x > 0 a n d y > 0. and Subcase 1.2. x < 0 a n d y < 0, w hile Case 2 could be divided into the two subcases: Subcase 2.1. x > 0 and y < 0 . and Subcase 2.2. x < 0 a n d y > 0. L e t’s look at an exam ple o f a proof by cases.
Result 3.15 Proof
I f n e Z, then n 2 + 3n + 5 is an odd integer. We proceed by cases, according to w hether n is even or odd. Case 1. n is even. Then n = 2x for som e x e Z. So n 2 + 3n + 5 = (2 x )2 + 3(2 v) + 5 = 4 x 2 + 6x + 5 = 2 (2 x 2 + 3x + 2) + 1. Since 2 x 2 + 3x + 2 6 Z, the integer n 2 + 3n + 5 is odd. Case 2. n is odd. Then n = 2 y + 1, w here y e Z. Thus n~ + 3n + 5 = (2_y + I )2 + 3(2y + 1) + 5 = 4 y 2 + lOy + 9  2 ( 2 y + 5y + 4) + 1. Because 2 y 2 + 5y + 4 e Z, the integer n 2 + 3n + 5 is odd.
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Two integers x and y are said to be o f th e sa m e p a r ity if x and y are both even or are both odd. The integers x and y are o f o p p o site p a r ity if one o f x and y is even and the other is odd. For exam ple, 5 and 13 are o f the sam e parity, w hile 8 and 11 are o f opposite parity. Because the definition of two integers having the sam e (or opposite) parity requires the tw o integers to satisfy one o f two properties, any result containing these term s is likely to be proved by cases. T he follow ing theorem presents a characterization o f two integers that are o f the same parity. Theorem 3.16 Proof
L et x , y e Z. Then x and y are o f the sam e pa rity i f a n d only if x + y is even. First, assum e that x and y are o f the same parity. We consider tw o cases. C ase 1. x an d y are even. T hen x = 2a and v = 2b for som e integers a and b. So x + y = 2a + 2b = 2(a + b). Since a + b e Z, the integer x + y is even. Case 2. x and y are odd. T hen x = 2a + I and y = 2b + 1, w here a , b e Z. Therefore, x + y = (2 a + 1) + (2b + I) — 2a + 2b + 2 = 2 (a + b + 1). Since a + b + 1 is an integer, x + y is even.
3.4
Proof by Cases
91
For the converse, assum e that x and y are o f opposite parity. A gain, w e consider tw o cases. Case 1. x is even and y is odd. Then x = 2a and y = 2b + 1, w here a, b € Z. Then x + y — 2a + (2b + 1) = 2 (a + b) + 1. Since a + b e Z, the integer x + y is odd. Case 2. x is odd and y is even. The p ro o f is sim ilar to the p ro o f o f the preceding case and is therefore om itted. ■ PROOF ANALYSIS
A com m ent concerning the proof o f T heorem 3.16 is useful here. A lthough there is always som e concern w hen om itting steps or proofs, it should be clear that it is truly a waste o f effort by w riter and reader alike to give a pro o f o f the case w hen x is odd and v is even in T heorem 3.16. Indeed, there is an alternative w hen the converse is considered. For the converse, assum e that x and y are o f opposite parity. W ithout loss o f gener ality, assum e that x is even and y is odd. Then x = 2a and y = 2b + 1, w here a, b e Z. T hen x + y = 2a + (2b + 1) = 2(a + b) + 1. Since a + b e Z, the integer x + y is odd. We used the phrase W LO G ) to indicate that one o f these is needed. similar. We present one
Theorem to Prove PROOFSTRATEGY
♦
w ith o u t loss o f g e n e ra lity (some abbreviate this as W O LO G or the proofs o f the tw o situations are sim ilar, so the p ro o f o f only Som etim es it is rather subjective to say that two situations are additional exam ple to illustrate this.
L et a and b be integers. T hen ab is even if and only if a is even or b is even. B efore w e begin a proof o f this result (Theorem 3.17 below ), le t’s see w hat w e w ill be required to show. We need to prove two im plications, nam ely, (1) If a is even or b is even, then a b is even and (2) if a b is even, then a is even or b is even. We consider (1) first. A direct proof seem s appropriate. H ere, we w ill assum e that a is even or b is even. We could give a proof by cases: (i) a is even, (ii) b is even. O n the other hand, since the proofs o f these cases will certainly be similar, we could say, w ithout loss of generality, that a is even. We w ill see that it is unnecessary to m ake any assum ption about b. If w e w ere to give a direct proof o f (2), then we w ould begin by assum ing that a b is even, say ab = 2k for som e integer k. B ut how could we deduce any inform ation about a and b individually? L e t’s try another approach. If w e use a p ro o f by contrapositive, then we w ould begin by assum ing that it is not the case that a is even or b is even. This is exactly the situation covered by one o f D e M organ’s laws: ~ c p V Q ) is logically equivalent to (~ .P ) A (~<2). It is im portant not to forget this. In this case, w e have P : a is even, and Q : b is even. So the negation o f “a is even or b is even” is “a is odd a n d b is odd.” ♦ L e t’s now prove this result.
92
Chapter 3 Theorem 3.17 P roof
Direct Proof and Proof by Contrapositive L et a and b be integers. Then ab is even i f and only i f a is even or b is even. First, assum e that a is even or b is even. W ithout loss o f generality, let a be even. Then a = 2x for som e integer x . Thus ab = (.2 x )b = 2(xb). Since x b is an integer, a b is even. For the converse, assum e that a is odd and b is odd. T hen a — 2 x + 1 and b = 2y + 1, w here x , y e Z. H ence ab = (2x + l ) ( 2y + 1) = 4 x y + 2x + 2 y + 1 = 2{2 xy + x + y) + 1. Since 2 x y + x + y is an integer, ab is odd.
m
3.5 P roof E valuations We have now stated several results and have given a pro o f o f each result (som etim es preceding a p ro o f by a proof strategy or follow ing the pro o f w ith a pro o f analysis). L e t’s reverse this process by giving an exam ple o f a pro o f o f a result but not stating the result being proved. We w ill follow the p ro o f w ith several options for the statem ents o f the result being proved. Example 3.18 Proof
G iven below is a p ro o f o f a result. A ssum e that n is an odd integer. T hen n = 2 k + \ for som e integer k. Then 3n — 5 = 3(2k + 1)  5 = 6* + 3  5 = 6k  2 = 2(3k  1). Since 3k — 1 is an integer, 3n — 5 is even.
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W hich of the follow ing is proved above? (1) (2) (3) (4)
3n — 5 is an even integer. If n is an odd integer, then 3n — 5 is an even integer. Let n be an integer. If 3n — 5 is an even integer, then n is an odd integer. Let n be an integer. If 3n — 5 is an odd integer, then n is an even integer.
The correct answ ers are (2) and (4). The pro o f given is a direct p ro o f o f (2) and a proof by contrapositive o f (4). T he sentence (I) is an open sentence, not a statem ent, and is only the conclusion o f (2). Statem ent (3) is the converse o f (2). $ W hen learning any m athem atical subject, it is not the least bit unusual to m ake m istakes along the way. In fact, part o f learning m athem atics is to learn from your m istakes and those o f others. For this reason, you will see a few exercises at the end of m ost chapters (beginning w ith this chapter) w here you are asked to evaluate the pro o f of a result. T hat is, a result and a proposed p ro o f o f this result w ill be given. You are then asked to read this proposed p ro o f and determ ine w hether, in your opinion, it is, in fact, a proof. If you d o n ’t believe that the given argum ent provides a p ro o f o f the result, then you should point out the (or a) m istake. We give tw o exam ples o f this. Problem 3.19
E valuate the proposed p r o o f o f the fo llo w in g result.
Exercises for Chapter 3 R e su lt P roof
93
I f x and y are integers o f the sam e parity, then x — y is even. L et x and y be tw o integers o f the sam e parity. We consider tw o cases, according to w hether x and y are both even or are both odd. Case 1. x a n d y are both even. L et x = 6 and y — 2, w hich are both even. T h en * — y = 4, w hich is even. Case 2. x a nd y are both odd. L et x — 7 and y = 1, w hich are both odd. T hen x — y = 6, w hich is even. ■
r o o f E va lu ation
Problem 3.20 R e su lt Proof
A lthough the proof started correctly, assum ing that x and y are two integers of the same parity and dividing the pro o f into these tw o cases, the p ro o f o f each case is incorrect. W hen we assum e that x and y are both even, for exam ple, x and y m ust represent arbitrary even integers, not specific even integers. $ Evaluate the proposed p r o o f o f the fo llo w in g result. I f m is an even integer and n is an odd integer, then 3m + 5n is odd. Let m be an even integer and n an odd integer. Then m = 2k and n = 2k + 1, w here k e Z. Therefore, 3m + 5 n = 3(2 k) + 5(2 k + 1) = 6 k + 10* + 5 = 1 6 * + 5 = 2 ( 8 * + 2 ) + 1. Since 8* + 2 is an integer, 3m + 5n is odd.
P ro o f E va lu a tio n
■
There is a m istake in the second sentence of the proposed proof, w here it is w ritten that m = 2k and n — 2* + 1, w here * e Z. Since the same sym bol * is used for both m and n , w e have inadvertently added the assum ption that n — m + 1. T his is incorrect, however, as it was never stated that m and n m ust be consecutive integers. In other w ords, we should w rite m = 2* and n = 21 + 1, say, w here *, £ e Z. ♦
\E R C I S E S FOR C H A PTER 3 "ection 3.1: Trivial and Vacuous Proofs . Let x e R. Prove that if 0 < x < 1. then x 2 — 2x + 2 / 0 . 2. Let n e N. Prove that if \n V Let r e Q + . Prove that if ~
—
11+ \n + 11 < 1, then \n2 — 11 < 4. < 1, then
< 2.
Let x e R. Prove that if x 3 — 5x — 1 > 0, then (x — 1){x — 3) > —2. 5. Let n e N. Prove that if re + n < 2 ,’ then re2 + \nz < 4.
~.6. Prove that if a, b and c are odd integers such that a + b + c = 0, then abc < 0. (You are permitted to use wellknown properties of integers here.) Prove that if x, y and z are three real numbers such that x 2 + y 2 + z2 < x y + xz + yz, then x + y + z > 0.
94
C h a p te r 3
Direct Proof and Proof by Contrapositive
S e c t i o n 3 .2 : D i r e c t P r o o f s 3.8. Prove that i i i is an odd integer, then 9x + 5 is even. 3.9. Prove that if x is an even integer, then 5x — 3 is an odd integer. 3.10. Prove
that if a and c. are odd integers, then ab + be is even for every integer b.
3.11. Let n
e Z. Prove that if 1 — n2 > 0 , then 3 n — 2 is an even integer.
3.12. Let x € Z. Prove that if 2Zx is an odd integer, then 2~2x is an odd integer. 3.13. Let S
= {0, 1, 2} and let n e S. Prove that if (n + 1)2(n + 2)2/4 is even, then (n + 2f ( n + 3)2/4 is eve
3.14. Let S
= {1. 5, 9}. Prove that if n e S and ,r+ ”~6 is odd, then 2,,3+ln2+n js even. 0 3.15. Let A = {/; e Z : n > 2 and n is odd} and B = [n e Z : n < 11}. Prove that if n e A n B , then n2 —2 is prime.
S e c t i o n 3 .3 : P r o o f b y C o n t r a p o s i t i v e 3.16. Let X e Z. Prove that if l x + 5 is odd, then x is even. 3.17. Let n e Z. Prove that if 15/i is even, then 9n is even. 3.18. Let x € Z. Prove that
5x — 11 is even if and only if x is odd.
3.19. Let x e Z. Use a lemma to prove that if l x + 4 is even, then 3x  11 is odd. 3.20. Let x € Z. Prove that
3x + 1 is even if and only if 5.v — 2 is odd.
3.21. Let n e Z. Prove that
(/? + l )2 — 1 is even if and only if n is even.
3.22. Let S = {2, 3, 4} and let n e S. Use a proof by contrapositive to prove that if n 2(n  l) 2/4 is even, then n 2(n + l) 2/4 is even. 3.23. Let A = {0. 1, 2} and B = {4, 5. 6} be subsets of S = {0, 1_____6}. Let n e S. Prove that if ,i(n~ 1^ n~2) js even, then n e A U B . 3.24. Let n e Z. Prove that 2n2 + n is odd if and only if cos ™ is even. 3.25. Let {A, B } be a partition of the set of 5 = {1, 2............ 7}, where A = {1, 4,5}and B = {2, 3, 6 ,7}. Let n G S. Prove that if ,r +^1~4 is even, then n e A. S e c t i o n 3 .4 : P r o o f b y C a s e s 3.26. Prove that if n e Z, then n 2 — 3n + 9 is odd. 3.27. Prove that if n e Z, then h 3 — n is even. 3.28. Let x , y e
Z. Prove that if xy is odd, then x and v are odd.
3.29. Let a, b e
Z. Prove that if ab is odd, then a 2 + b2 is even.
3.30. Let x , y e
Z. Prove that x —y is even if and only if x and y areof the same parity.
3.31. Let a , b e
Z. Prove that if a + b and ab are of the same parity,then a and b are even.
3.32. (a) Let x and y be integers. Prove that (x + y )2 is even if and only if x and f are of the same parity. (b) Restate the result in (a) in terms of odd integers. 3.33. Let A = {1, 2, 3} and B = {2, 3, 4} be subsets of S = {1,2, 3, 4}.Let n e S. Prove that 2rr — 5n is either (a) positive and even or (b) negative and odd if and only if n A n B. 3.34. Let A = {3, 4} be a subset of 5 = {1,2, — 6}. Let
n e S. Prove that if "2("4H)2iseven, then n
3.35. Prove for every nonnegative integer n that 2" + 6" is an
even integer.
e
A.
Exercises for Chapter 3
95
5.36. A collection o f nonempty subsets of a nonempty set S is called a cover of S if every element of S belongs to at least one of the subsets. (A cover is a partition of S if every element of S belongs to exactly one of the subsets.) Consider the following. R esult
Let a, b e Z. If a is even or b is even, then ab is even.
Proof
Assume that a is even or b is even. We consider the following cases.
Case 1.a is even. Then a = 2k, where k e Z. Thus ab = (2 k)b = 2 (kb). Since kb e Z, it follows that ab is even. Case 2. b is even. Then b = 21, where I e Z. Thus ab = a(2t ) = 2(a t). Since a l e Z, it follows that ab is even. ■ Since the domain is Z for both a and b, we might think of Z x Z being the domain of (a , b). Consider the following subsets of Z x Z: Si = {(a, b) e
Z x Z : a and b are odd}
Si = {(a, b) e
Z x
Z : a is even}
S3 = {(a, b) € Z x
Z : b is even}.
(a) Why is {Si, S2, S3} a cover of Z x Z and not a partition of Z x Z? (b) Why does the set Si not appear in the proof above? (c) Give a proof by cases of the result above where the cases are determined by a partition and not a cover.
Section 3.5: Proof Evaluations 5 37. Below is a proof of a result.
Proof We consider two cases. Case 1. a and b areeven. Then a = 2r and b = 2s for integers r and s. Thus a 2  b 2 = (2r)2  (2s)2 = Ar2 4s 2= 2(2r 2 
2s2).
Since 2r 2 — 2s 2 is an integer, a 2 — b2 is even. Case 2. a and b are odd. Then a = 2r + 1 and b = 2s + 1 for integers r and s.
Thus
a2 — b2 — (2r + l )2  (2s + l )2 = (Ar2 + Ar + 1)  (As2+ As + 1) = 4/2 + Ar  As2  45 = 2(2r2+ 2r  2s 2  2s). Since 2r 2 + 2r
— 2s 2 — 2s
is an integer, a 2 — b2 is even.
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Which of the following is proved? (1) Let a , b e Z. Then a and b are of the same parity if and only if a2— b2 is (2) Let a, b e Z. Then a 2 — b2 is even. (3) Let a, b e Z. If a and b are of the same parity, then a 2 — b2 is even. (4) Let a, b e Z. If a 2 — b2 is even, then a and b are of the same parity. 5 58. Below is given a proof of a result. What result is being proved?
even.
96
C h a p te r 3
Direct Proof and Proof by Contrapositive
P ro o f Assume that x is even. Then x = 2a for some integer a. So 3 xJ  4x  5 = 3(2a)2  4(2a)  5 = 12a2  8a  5 = 2(6a2  4a 
3) + 1.
Since 6a 2 — 4a — 3 is an integer, 3.r2 —4x — 5 is odd. For the converse, assume that x is odd. So x = 2b + 1, where b e Z. Therefore, 3x2  4.v  5 = 3(2/7 + l )2  4(2b + 1)  5 = 3(4i>2 + 4b + 1)  &b 4 
5
= 12ft2 + 4b  6 = 2(6b2 + 2b  3). Since 6b2 + 2b — 3 is an integer, 3x2 — 4x — 5 is even. 3.39. Evaluate the proof of the following result. R esult
Let n e Z. If 3n — 8 is odd, then n is odd.
P ro o f Assume that n is odd. Then n = 2 k + \ for some integer k. Then 3n  8 = 3(2* + 1)  8 = 6k + 3 “ 8 = 6k — 5 = 2(3* — 3) + 1. Since 3k — 3 is an integer, 3n — 8 is odd. 3.40. Evaluate the proof of the following result. R esult
Let a , b e Z. Then a — b i s even if and only if a and b are of the same parity.
P ro o f We consider two cases. Case 1. a and b are o f the same parity. We now consider two subcases. Subcase 1.1. a and b are both even. Then a — 2x and b = 2y, where x , y e Z. Then a — b = 2x — 2y = 2(x — y). Since x — y is an integer, a — b is even. Subcase 1.2. a and b are both odd. Then a = 2x + 1 and b = 2y + 1, where x , y e Z. Then a — b = (2x + 1) — (2y + 1) = 2{.v — y). Since x — y is an integer, a — b is even. Case 2. a and b are o f opposite parity. We again have two subcases. Subcase 2.1. a is odd and b is even. Then a = 2x + 1 and b = 2 y, where x , y e Z. Then a — b = (2x + 1) — 2y = 2(x — y) + 1. Since x — y is an integer, a — b is odd. Subcase 2.2. a is even and b is odd. Then a = 2x and b = 2 y + 1, where x, y e Z. Then a — b — 2x — (2 y + 1) = 2x — 2_v — 1 = 2(x — y — 1) + 1. Since x — y — 1 is an integer, a — b is odd. 3.41. The following is an attempted proof of a result. What is the result and is the attempted proof correct? P ro o f Assume, without loss of generality, that x is even. Then x = 2a for some integer a. Thus x y 2 = {2a)y2 = 2(ay 2). Since a y 2 is an integer, x y 2 is even. 3.42. Given below is a proof of a result. W hat is the result? P ro o f Assume, without loss of generality, that x and y are even. Then x = 2a and y = 2b for integers a and b. Therefore, x y + x z + y z = (2a)(2b) + (2 a)z + (2 b)z = 2(2 ab + az + bz). Since 2ab + az + bz is an integer, xy + x z + y z is even. 3.43. What result is being proved below, and what procedure is being used to verify the result? First, we present the following proof.
Additional Exercises for Chapter 3
Proof
97
Assume that x is even. Then x = 2a for some integer a. Thus 7.v  3 = 7(2a)  3 = 14a  3 = 14a  4 + 1 = 2(7a  2) + 1.
Since l a — 2 is an integer, I x — 3 is odd.
•
We are now prepared to prove our main result.
Proof Assume that I x  3 is even. From the result above, X is odd. So x = 2b + 1 for some integer b. Thus 3,v + 8 = 3(2 b + 1) + 8 = 6b + 11 = 2(3 b + 5) + 1. Since 3b + 5 is an integer, 3x + 8 is odd.
■
5 .44. Consider the following statement. Let n G Z. Then (n  5)(n + 7)(« + 13) is odd if and only if n is even. Which of the following would be an appropriate way to begin a proof of this statement? (a) Assume that (n  5)(n + l)(n + 13) is odd. (b) Assume that (n —5 )(n + l)(n + 13) is even. (c) Assume that n is even. (d) Assume that n is odd. (e) We consider two cases, according to whether n is even or n is odd.
AD D ITIO N AL e x e r c i s e s FO R C H A PT E R 3 5 45. Let x
G
Z. Prove that if I x  8 is even, then x is even.
: 46. Let x
e
Z. Prove that x 3 is even if and only if x is even.
: 47. Let x
e
Z. Use one or two lemmas to prove that 3x3 is even if and only if 5x2is even.
5 48. Give a direct proof of the following: Let x  49. Let x , y
GZ.
Prove that
 50. Let x, y
e Z. Prove that
if x +
GZ.
Z. If 1 lx  5 isodd, then x is even.
is odd, then x and y are of opposite parity.
if 3x + 5y is even, then x and y are of the same parity.
5 51. Let i j g Z . Prove that (x + 52. Let x, y
y
G
l) y 2 is even if and only if x is odd or y is even.
Prove that
if xy and x + y are even, then both x and y are even.
5 53. Prove, for every integer x, that the integers 3x + 1 and 5x + 2 are of opposite parity. : 54. Prove the following two results: (a) Result A: Let k g Z. If rc3 is even, then n is even. (b) Result B: If n is an odd integer, then 5n9 + 13 is even.
5 55. Prove for every two distinct real numbers a and b, either 56. Let X , y
G
> a or
> b.
Z. Prove that if a and b are even integers, then a x + by is even.
5 57. Evaluate the proof of the following result. Result
Let x , y
G
Z and let a and b be odd integers. If a x + by is even, then x and y are of the same parity.
P ro o f Assume that x and y are of opposite parity. Then X = 2p and y = 2q + 1 for some integers p and q. Since a and b are odd integers, a = 2r + 1 and b = 2s + 1for integers r and s. Hence ax + by = (2r + 1)(2 p) + {2s + 1)(2 q + 1) = Apr + 2 p + Aqs + 2s + 2q + 1 = 2(2 p r + p + 2qs + s + q) + 1. Since 2pr + p + 2qs + s + q is an integer, ax + by is odd.
a
Chapter 3
98
Direct Proof and Proof by Contrapositive
3.58. Let S = {a, b, c, d } be a set of four distinct integers. Prove that if either (I) for each .v e S. the integer x and the sum of any two of the remaining three integers of S are of the same parity or (2) for each x G S, the integer x and the sum of any two of the remaining three integers of S are of opposite parity, then every pair of integers of S is of the same parity. 3.59. Prove that if a and b are two positive integers, then a 2(b + 1) + b2(a + 1) > 4ab, 3.60. Let a, b e Z. Prove that if ab = 4, then (a — b )3 — 9(a — b) = 0. 3.61. Let a . b and c be the lengths of the sides of a triangle T where a < b < c. Prove that if T is a right triangle, then
3.62. Consider the following statement. Let n G Z. Then 3/z3 + 4n2 + 5 is even if and only if n is even. Which of the following would be an appropriate way to begin a proof of this statement? (a) (b) (c) (d) (e)
Assume that 3« 3 Assume that 3//3 Assume that n is Assume that n is We consider two cases,
+ An2+ 5 is odd. + 4n 2+ 5 is even. even. odd. according to whether n is even or n is odd.
3.63. Let V = {A, B , C } be a partition of a set S of integers, where A  { n € S: n is odd and n > 0}, B = {n e S : n is odd and n < 0} and C = {n € S : n is even and n > 0}. Prove that if .v and v are elements of S belonging to distinct subsets in V. then xy is either odd, even and greater than 1, or even and less than —1. 3.64. Let /?
G
N. Prove that if n3 — 5n — 10 > 0, then n > 3.
3.65. Prove for every odd integer a that (a 2 + 3)(a2 + 7) = 32b for some integer b. 3.66. Prove for every two positive integers a and b that (a + b) ( '•  ' ) > 4. \a bJ 3.67. Which result is being proved below, and what procedure is being used to verify the result? We begin with the following proof. P ro o f First, assume that x is even. Then x = 2a, where a
G
Z. Thus
3x  2 = 3(2a)  2 = 6a  2 = 2(3a  1). Since 3a — 1 is an integer, 3x — 2 is even. Next, suppose that x is odd. Then x = 2b + I for some integer b. So 3*  2 = 3(2b + 1)  2 = 6b + 1 = 2(3,/) + 1. Since 3a is an integer, 3x — 2 is odd. We can now give the following proof. P ro o f First, assume that 3x  2 is even. From the preceding result, x is even and so x = 2a, where a g Z. Thus 5.\ + 1 = 5(2 a) + 1 = 2(5a) + 1. Since 5a is an integer, 5x + 1 is odd. Next, assume that 3x  2 is odd. Again, by the preceding result, x is odd. Hence x = 2b + 1 for some integer b. Therefore, 5x + 1 = 5(2 b + 1) + 1 = W b + 6 = 2(5 b + 3). Since 5b + 3 is an integer, 5x + 1 is even.
4 More on Direct Proof and Proof by Contrapositive
he vast m ajority o f the exam ples illustrating direct p ro o f and p ro o f by contrapositive that w e have seen involve properties o f even and odd integers. In this chapter, we w ill give additional exam ples o f direct proofs and proofs by contrapositive concerning integers but in new surroundings. First, w e will see how even and odd integers can be studied in a m ore general setting, through divisibility o f integers. We w ill then explore som e properties o f real num bers and, finally, look at som e properties o f set operations.
T
4.1 Proofs Involving D ivisib ility of Integers We have now seen m any exam ples o f integers that can be w ritten as 2x for some integer x . These are precisely the even integers, o f course. However, som e integers can also be expressed as 3x or 4x, or as —5 x for som e integer x . In general, for integers a and b with a ^ 0, we say that a d ivides b if there is an integer c such that b = ac. In this case, we write a \ b. H ence if n is an even integer, then 2  n; m oreover, if 2 divides some integer n, then n is even. T hat is, an integer n is even if and only if 2  n. T heorem 3.17 (w hich states for integers a and b. that ab is even if and only if a or b is even) can therefore be restated for integers a and b as: 2  a b if and only if 2  a or 2  b. If a  b, then we also say that b is a m ultiple o f a and that a is a divisor o f b. Thus every even integer is a m ultiple o f 2. If a does not divide b, then w e w rite a / b. For exam ple, 4  48 since 48 = 4 • 12 and —3  57 since 57 = ( —3) • ( —19). On the other hand, 4 / 6 6 as there is no integer c such that 66 = 4c. We now apply the techniques w e ’ve learned to prove som e results concerning d i visibility properties o f integers. Result to
Prove Let a, b and c be integers w ith a ^ 0 and b ^ 0. If a  b and b \ c, then a  c.
PROOF STRATEGY
It seem s reasonable here to use a direct pro o f and to begin by assum ing that a \ b and b  c. T his m eans that b = a x and c = by for som e integers x and y. Since our goal is to show that a \ c, we need to show that c can be w ritten as the product of a and som e other integer. H ence it is logical to consider c and determ ine how w e can express it. ♦ 99
100
Chapter 4 Result 4.1 Proof
More oil Direct Proof and Proof by Contrapositive
L e t a , b and c be integers with a ^ 0 and b ^ 0. I f a \ b and b
\c, then a \ c.
A ssum e that a \ b and b \ c. Then b = a x and c = b y, w here x , y e Z. Therefore, c = by  (a x ) y = a {xy ). Since x y is an integer, a \c . m We now verify tw o other divisibility properties o f integers.
Result 4.2 Proof
L et a , b, c and d be integers with a / 0 and b / 0. I f a \ c and b \ d , then a b \ cd. L et a  c and b \ d. T hen e = a x and d = b y, w here x , y e Z. Then cd = (a x )(b y ) = (a b ) (x y ). Since x y is an integer, ab \ cd.
Result 4.3 P roof
m
L e t a. b, c, x , y e Z , where a / 0. I f a \ b a n d a \ c, then a \ (b x + cy). A ssum e that a  b and a  c. T hen b — a r and c. — as, w here r, s e Z. Then b x + cy = (ar).x + (as) y = a ( r x + sy). Since r x + s y is an integer, a  (bx + cy).
H
The exam ples that we have presented thus far concern general properties o f divisi bility o f integers. We now look at some specialized properties o f divisibility. Result 4.4 P roof
L e t x e Z. I f 2 \ ( x 2 — 1), then 4  ( x 2 — 1). A ssum e that 2  ( x 2  1). So x 2  1 = 2y for some integer y. Thus jt2 = 2 y + 1 is an odd integer. It then follow s by T heorem 3.12 that x too is odd. H ence x = 2z + 1 for som e integer 2. Then x" — 1 = (2 z + 1)” — 1 = (4z2 + 4z + 1) — 1 = 4 z2 + 4z = 4(z2 + z). Since z2 + z is an integer, 4  (x 2 — 1).
m
For each o f the Results 4 .1 4 .4 , a direct p ro o f w orked very well. F or the follow ing result, however, the situation is quite different. R e su lt to P ro v e PROOFSTRATEGY
L e t x , y e Z. If 3 / x j , then 3 / x and 3 / y . If w e let P: 3 / x y ,
O: 3 / x
and
R: 3 / v,
then w e w ish to prove that P =$■ Q a R for all integers x and y. (It should be clear that P , Q and R are open sentences in this case, but w e om it the variables here for sim plicity.) I f we use a direct proof, then w e w ould assum e that 3 / x y and attem pt to show that 3 / x and 3 / y . Thus we w ould know that x y cannot be expressed as 3 tim es an integer. O n the other hand, if we use a pro o f by contrapositive, then we are considering the im plication ( ~ ( Q A R) ) => ( ~ P ) , w hich, by De M o rg an ’s Law, is logically equivalent to
4.1
Proofs Involving Divisibility of Integers
101
( ( ~ G ) v ( ~ P ) ) =>• (~ P ) and w hich, in w ords, is: If 3  * or 3  y, then 3  x y . This m ethod looks m ore prom ising. ♦ Result 4.5
Proof
L e t x , y e Z. I f 3 / x y , then 3 / x an d 3 X y. A ssum e that 3  x o r 3  y . W ithout loss o f generality, assum e that 3 divides x . Then x = 3z for some integer z. H ence xy = (3z)y = 3(zy). Since zy is an integer, 3  x y. ■ We have already m entioned that if an integer n is not a m ultiple o f 2, then we can w rite n = 2q + 1 for som e integer q (that is, if an integer n is not even, then it is odd). T his is a consequence o f know ing that 0 and 1 are the only possible rem ainders w hen an integer is divided by 2. A long the same lines, if an integer n is n ot a m ultiple o f 3, then we can w rite n = 3q + 1 or n = 3q \2 for som e integer q; that is, every integer can be expressed as 3q, 3q + 1 or 3q + 2 for som e integer q since 0, 1 and 2 are the only rem ainders that can result w hen an integer is divided by 3. Similarly, if an integer n is not a m ultiple o f 4, then n can be expressed as Aq + 1, Aq + 2 or Aq + 3 for some integer q. This topic concerns a w ellknow n theorem called the D ivision A lgorithm , w hich will be explored in m ore detail in C hapter 11.
R esu lt to P rove : ROOF STRATEGY
Result 4.6 Proof
Let x e Z . If 3 / ( x 2 — 1), then 3 [ x . We have two options here, nam ely (1) use a direct pro o f and begin a pro o f by assum ing that 3 / ( x 2 — 1) or (2) use a p ro o f by contrapositive and begin a p ro o f by assum ing that 3 / x . Certainly, we cannot avoid assum ing that 3 does not divide som e integer. However, it appears far easier to know that 3 / x and attem pt to show that 3  (x 2 — 1) than to know that 3 / ( x 2 — 1) and show that 3  x. A lso, if 3 / x , then we now know that x = 3q + 1 or x = 3q + 2 for som e integer q , w hich suggests a p ro o f by cases. ♦ L e t x e Z . I f 3 / ( x 2 — 1), then 3 \ x . A ssum e that 3 / x . T hen either x = 3^ + 1 for som e integer q or x = 3q + 2 for som e integer q. We consider these tw o cases. Case 1. x = 3q
+ 1 fo r som e integer q . Then x2  1
= (3q + l )2  1= (9q 2 + 6q + 1)  1 = 9 q 2 + 6q = 3(3^ 2 + 2q).
Since 3q 2 + 2q is an integer, 3  (x 2 — 1). Case 2. x = 3q
+ 2 fo r som e integer q . Then x 2  1 = (3q + 2)2  1 = (9q 2 + \ 2 q + A )  \ = 9 q 2 + \ 2 q + 3 = 3(3 q 2 + Aq + 1).
Since 3q 2 + Aq + 1 is an integer, 3  (x 2 — 1). We now consider a biconditional involving divisibility.
■
102
Chapter 4 Result 4.7
Proof
More on Direct Proof and Proof by Contrapositive
L et x , y e Z.
Then
A ssum e first that x tw o cases.
4
 (x 2 — y 2) if and only i f x and y are o f the sam e parity.
and y
are o f the same parity. We show that 4  (x2  y 2). The
Case 1. x and y are both even. Thus x — 2a and y = 2b for some integers a and b. Then x 2  y ’ = (2a)2  (2b)2 = 4a 2  4 b2 — 4 (a 2  b2). Since a 2 — b 2 Case 2.x and d. Then
is an integer, 4  ( x 2 — y 2). y are both odd. So .v — 2c  1and y = 2 d + 1 for som e integers c and
x 2  v2 = (2c + l )2  (2d + l )2  (4c2 + 4c + 1)  (4d 2 + 4 d +
1)
= 4 c2 + 4c  4c/ 2  4 d = 4(c2 + c  d 2  cl). Since c2 + c — d 2 — d is an integer, 4
j
(x 2 — y 2).
For the converse, assum e that x and y are o f opposite parity. We show that 4 / (x 2 — y 2). We consider two cases. Case 1 .x is even and y is odd. Thus x = 2a and y = 2b + 1 for some integers a and b. T hen x 2 — y 2 = (2a)2  (2b + I )2 = 4 a 2 
[4b2 + 4b + l\
= 4 a 2  4 b 2  4b  1 = 4c/2  4 b 2  4b  4 +
3
= 4 (a 2  b2  b  1) + 3. Since a 2  b  b  1 is an integer, it follow s that there is a rem ainder o f 3 w hen x 2  y 2 is divided by 4 and so 4 / ( x 2 — y 2). C ase 2. x is odd and y is even. The pro o f o f this case is sim ilar to that o f C ase 1 and is therefore om itted. m We consider a result o f a som ew hat different nature. R e su lt to P ro v e PROOFSTRATEGY
Result 4.8
Proof
For every integer n > 7, there exist positive integers a and b such that n = 2a + 3b. First, notice that we can write 7 = 2 ■2 + 3 ■1, 8 = 2 • 1 + 3  t and 9 = 2 • 3 + 3 • 1. So the result is certainly true for n = 7, 8 , 9. On the other hand, there is no pair a , b of positive integers such that 6 — 2a + 3b. O f course, this observation only shows that we cannot replace n > 7 by n > 6 . Suppose that n is an integer such that n > 7. We could bring the integer 2 into the discussion by observing that we can write n = 2q or » = 2q + 1, w here q e Z. Actually, if n = 2 q, then q > 4 since n > 7; w hile if n = 2q + 1, then q > 3 since n > 7. This is a useful observation. 4 For every integer n > 7, there exist p ositive integers a and b such that n — 2a 4 3b. L et n be an integer such that it > 1. Then n = 2q or n = 2cy + 1 for som e integer q. We consider these tw o cases.
4.2
Proofs Involving Congruence of Integers
103
Case 1. n = 2q. Since n > 7, it follow s that q > 4. Thus n = 2 q — 2 {q — 3) + 6 = 2(q — 3) + 3 • 2. Since q > 4, it follow s that q — 3 e N. C ase 2. n — 2q + I. Since n > 7, it follow s that c/ > 3. Thus n = 2(? f 1 = 2(g — 1) — 2 — (—1 = 2{q — 1) + 3 • 1. Since q > 3, it follow s that q — 1 e N.
a
4.2 Proofs Involving C ongruence of Integers
We know that an integer x is even if x = 2q for som e integer q , w hile x is odd if x = 2q + 1 for som e integer q . Furtherm ore, two integers x and vare o f the same parit if they are both even or are both odd. From this, it follow s that x and y are o f the same parity if and only if 2  (x — y). Consequently, 2  (x — y) if and only if x and y have the sam e rem ainder when divided by 2. We also know that an integer x can be expressed as 3q, 3q + 1 or 3q + 2 for som e integer q , according w hether the rem ainder is 0, 1 or 2 w hen x is divided by 3. If integers * and >’ are both o f the form 3q + 1, then x = 3s + 1 and y = 3t + 1, w here s , t e Z, and so x — y — 3(s — t ). Since s — t is an integer, 3  (x — y). Similarly, if x and y are both of the form 3q or are both of the form 3q + 2, then 3  (x — y) as well. H ence if x and y have the sam e rem ainder w hen divided by 3, then 3  (x — y). The converse of this im plication is true as well. T his suggests a special interest in pairs x , y o f integers such that 2  (x — y) or 3  (x — y ) or, in fact, in pairs x , j o f integers such that n  (x  y) for som e integer n > 2. For integers a. b and n > 2, w e say that a is c o n g ru e n t to b m o d u lo n. w ritten a = b (m od n), if n  (a — b). For exam ple, 15 = 7 (m od 4) since 4  (15 — 7) and 3 = —15 (m od 9) since 9  (3 — (—15)). On the other hand, 14 is not congruent to 4 m odulo 6 , w ritten 14 ^ 4 (m od 6), since 6 /(1 4 — 4). Since w e know that every integer x can be expressed as x = 2q or as x = 2q + 1 for som e integer q, it follow s that either 2  (x — 0 ) or 2  (x — 1), that is, x s 0 (m od 2) or x = 1 (m od 2). A lso, since each integer x can be expressed as x = 3q, x = 3q + 1 or x = 3q + 2 for som e integer q, it follow s that 3  (x —0), 3  (x — 1) or 3  (x — 2). H ence x = 0 (m od 3),
x = 1 (m od 3)
or
x = 2 (m od 3).
M oreover, for each integer x , exactly one of x = 0 (m od 4),
x = 1 (m od 4),
x = 2 (m od 4),
x = 3 (m od 4)
holds, according to w hether the rem ainder is 0, 1, 2 or 3, respectively, w hen x is divided by 4. Sim ilar statem ents can also be m ade w hen x is divided by n for each integer n > 5. We now consider som e properties o f congruence o f integers. R esult to Prove PROOF STRATEGY
L et a, b , k and n be integers w here n > 2, If a = b (m od n ), then k a = k b (m od n). A direct p ro o f seems reasonable here. So, w e begin by assum ing that a = b (m od n). O ur goal is to show that ka = k b (m od n). B ecause we know that a = b (m od n), it follows
104
Chapter 4
More on Direct Proof and Proof by Contrapositive
from the definition that n  (a —b), w hich im plies that a —b — nx for som e integer x. We need to show that ka = kb (m od n), w hich m eans that w e need to show that n  (ka — kb). T hus, we m ust show that ka — kb = nl for som e integer t. This suggests considering the expression ka — kb. ♦ Result 4.9 Proof
Let a, b, k and n be integers where n > 2. If a = b (m od ri), then ka = kb (m od n). A ssum e that a = b (m od n). T hen n \ (a — b). H ence a — b = nx for som e integer x. Therefore,
ka —kb = k(a —b) = k(nx) — n(kx). Since kx is an integer, n \ (ka — kb) and so ka = kb (m od ri).
m
Result 4.10
Let a, b, c, d, n e Z where n > 2 . If a = b (m od n) and c = d (m od ri), then a + c = b + d (m od n).
Proof
A ssum e that a = b (m od n) and c = d (m od ri). T hen a — b = nx and c — d = ny for som e integers x and y. A dding these two equations, we obtain
(a —b) + (c —d) = nx + ny and so
(a + c)  (b + d) = n(x + y). Since x + y is an integer, n \ [(a + c) — (b + d)]. H ence a + c = b + d (m od ri).
m
The next result parallels that o f R esult 4.10 in term s o f m ultiplication. R esult to Prove
PROOFSTRATEGY
Result 4.11
P roof
L et a, b, c, d, n e Z w here n > 2. If a = b (m od ri) and c = d (m od «), then ac hd (m od n). This result and R esult 4.10 have the sam e hypothesis. In the p ro o f o f R esult 4.10, we arrived at the equations a —b = nx and c — d = ny and needed only to add them to com plete the proof. This suggests that in the current result, it w ould be reasonable to m ultiply these tw o equations. However, if w e m ultiply them , we obtain (a — b)(c — d) = (nx)(ny), w hich does not give us the desired conclusion that ac — bd is a m ultiple o f n. It is essential though that w e w ork ac — bd into the proof. By rew riting a — b = nx and c — d = ny as a = b + nx and c = d + ny, respectively, and then m ultiplying, we can accom plish this, however. ♦
Let a, b, c, d, n € Z where n > 2 . If a = b (m od ri) and c = d (m od n), then ac hd (m od n). A ssum e that a = b (m od ri) and c = d (m od ri). Then a — b = nx and c — d = ny, w here x , y e Z. Thus a = b + nx and c — d + ny . M ultiplying these tw o equations, we obtain
ac = (b + nx)(d + ny) = bd + dnx + bny + n2xy = bd + n(dx + by + nxy)
4.3
Proofs Involving Real Numbers
105
and so a c  bd = n (d x + b y + n x y ). Since d x + b y + n x y is an integer, a c = bd (m od n).
B
The proofs o f the preceding three results use a direct proof. This is n o t a convenient proof technique for the next result, however. R esult to P ro v e ;<>01 STRATEGY
L et n e Z, If n 2 # n (m od 3), then n # 0 (m od 3) and n # 1 (m od 3). Let P (n ) : n 2 # n (m od 3), Q (ri) : « # 0 (m od 3) and R (n ) : n # 1 (m od 3). O ur goal is then to show that P (n ) =>■ (Q (n ) A R (n )) is true for every integer n. A direct pro o f does not appear to be a good choice. However, a pro o f by contrapositive w ould lead us to the im plication ~ ( Q ( « ) A R (n)) =» ( ~ P ( « ) ) , w hich, by D e M organ s Law, is logically equivalent to
C eQ («))V (~ R (n))) =►(/>(*)). In w ords, we then have: If n = 0 (m od 3) or n = { (m od 3), then n 2 = n (m od 3). Result 4.12 Proof
L et n e Z. I f n 2 #
n(m od 3), then n # 0 (m od 3) and n # 1 (m od
L et n be an integer such that n = 0 (m od 3) or n = 1
♦
3).
(m od 3). We consider these two
cases. Case 1. «  • 0 (m od 3). T hen n = 3k for som e integer k. Hence n 2  n = (3k)2  (3k) = 9k 2  3k = 3(3k 2  k). Since 3k2  k is an integer, 3  (n2  n). Thus n 2 = n (m od 3). Case 2. n = 1 (m od 3). So n = 3£ + 1 for som e integer I and n   n = (31 + l )2  (31 + 1) = <9C + 6 1 + 1 )  (31 +
1)
= 9 £2 + 31 = 3(3£2 + 1 5 , Since 3£2 + £ is an integer, 3  («2  /;) and so n 2 =
11
(m od 3).
a
A s a consequence o f Result 4.12, if an integer n and its square n 2 have different rem ainders w hen divided by 3, then the rem ainder for n (w hen divided by 3) is 2.
4.3 P ro o fs Involving R eal N um bers We now apply the proof techniques we have introduced to verify som e m athem atical statem ents involving real num bers. To be certain that we are w orking under the same set o f rules, let us recall som e facts about real num bers w hose truth w e accept w ithout justification. We have already m entioned that a 2 > 0 for every real num ber a. Indeed, a" > 0 for every real num ber a if n is a positive even integer. If a < 0 and n is a positive odd integer, then a n < 0. O f course, the product of two real num bers is positive if and only if both num bers are positive or both are negative.
106
Chapter 4
More oil Direct Proof and Proof by Contrapositive
N ow let a, b, c e R. If a > b and c > 0, then the inequality a c > be holds. Indeed, if c > 0 , then a / c > b /c . If a > b and c > 0, then ac > be and a / c > b /c .
(4.1)
I f c < 0, then the inequalities in (4.1) are reversed; nam ely: If a > b and c < 0, then a c < be and a / c < b /c .
(4.2)
A nother im portant and w ellknow n property of real num bers is that if the product o f two real num bers is 0 , then at least one o f these num bers is 0 . Theorem to Prove PliOOF STRATEGY
Theorem 4.13
Proof
If x and y are real num bers such that x y = 0, then x = 0 or y = 0. If w e use a direct proof, then we begin by assum ing that x y = 0. If x = 0 , then w e already have the desired result. On the other hand, if x ^ 0, then w e are required to show that y = 0. However, if x ^ 0, then \ / x is a real number. This suggests m ultiplying x y = 0 by \ / x . ^ ♦ L e t x , y e R. I f x y = 0, then x = 0 or y = 0. A ssum e that x y — 0. We consider tw o cases, according to w hether x = 0 or x ^ 0. C ase 1. x = 0. Then we have the desired result. Case 2. x ± 0. M ultiplying x y = 0 by the num ber \ / x , we obtain —(xy) = —■0. x x Since
^(•xy) = Q * ) y = 1  y = y, it follow s that y = 0 .
■
We now use Theorem 4.13 to prove the next result. Result 4.14
Proof
L et x e R. I f x 3 — 5 x 2 + 3x = 15, then x = 5. A ssum e that x 3 — 5 x 2 + 3x = 15. Thus x 3 — 5 x 2 + 3x — 15 = 0. O bserve that x 3 — 5 x 2 + 3x — 15 = x 2(x — 5) + 3(x — 5) = (x2 + 3)(x — 5). Since x 3 — 5 x 2 + 3x — 15 = 0, it follow s that (x 2 + 3)(x — 5) = 0. By T heorem 4.13, x 2 + 3 = 0 or x — 5 = 0. Since x 2 + 3 > 0, it follow s that x — 5 = 0 and so x = 5. ■ N ext we consider an exam ple o f a p ro o f by contrapositive involving an inequality.
Result 4.15
Proof
L et x e R. I f x 5 — 3 x 4 + 2 x 3 — x 2 + 4x — 1 > 0 , then x > 0. A ssum e that x < 0. T hen x 5 < 0, 2 x 3 < 0 and 4x < 0. In addition, —3 x 4 < 0 and —x 2 < 0. Thus x 5 — 3 x 4 + 2 x 3  x 2 + 4x — 1 < 0  1 < 0, as desired.
■
4.3
Proofs Involving Real Numbers
107
On occasion we m ay encounter problem s that involve the verification o f a certain equality or inequality and w here it is convenient to find an equivalent form ulation of the equality or inequality w hose truth is clear. T his then becom es the starting point o f a proof. We now verify an inequality w hose p ro o f uses this com m on approach. R esu lt to P rove
If x, _y e R, then 1
x 3 PROOF STRATEGY
2
3
+ y 4J
2
> xy. ~ J
L e t’s first elim inate fractions from the expression. Showing that  x 2 + >>2 > x y is equivalent to show ing that
12 Q * 2 + l y2)  n x y ' that is, 4 x 2 + 9 y 2 > 12x y , w hich, in turn, is equivalent to 4 x 2 — 12x y + 9y 2 > 0. T hat is, if w e could show that 4 x 2 — 12xy + 9 y 2 > 0, then w e w ould be able to show that 1 , 3 x  +  y
> xy.
M aking a sim ple observation about 4 x 2 — 12 x y + 9y 2 leads to a proof. Result 4.16
♦
I f x , y e R, then 1
x
Proof
2
2
3
+  y 2 >xy.
2
Since (2x — 3 y )2 > 0, it follow s that 4 x 2 — 12xy + 9 y 2 > 0 and so 4 x 2 + 9 y 2 > 12x y . D ividing this inequality by 12, we obtain 1 2 3 x 2+  r
> xj
producing the desired inequality.
■
R ecall that for a real num ber x , its a b s o lu te v alu e x  is defined as
x
if x > 0
1*1 = i 1 x if x < n0 . A w ellknow n property o f absolute values is that \xy\ = x>> for every two real n um bers x and y (see Exercise 30). T he follow ing theorem gives another fam iliar property of absolute values o f real num bers (called the triangle inequality) that has num erous appli cations. Since the definition o f x  is essentially a definition by cases, proofs involving x  are often by cases. Theorem 4.17
(T h e T ria n g le In e q u a lity ) For every tw o real num bers x and y , \x + 7 1 < x  + IVI.
108
Chapter 4 Proof
More on Direct Proof and Proof b y Contrapositive
Since \x + y \ — x  + jy if either x or y is 0, we can assum e that x and y are nonzero. We proceed by cases. Case 1. x > 0 and y > 0. T hen x + y > 0 and x + y\ = x + y = \x\ +  v . Case 2. x < 0 and y < 0. Since x + y < 0, \x + y \ =  ( x + y ) = (  x ) + (—y) = \x] + y . Case 3. O ne o f x and y is positive and the other is negative. A ssum e, w ithout loss of generality, that x > 0 and y < 0. We consider two subcases. Subcase 3.1. x + y > 0. Then M + !?[ =
X
+ (y ) =
X
 y > x + y = x + y .
Subcase 3.2, x + y < 0. Here x + V = X + (—y) = x — y > —x — y = —(x + y) = x + y . Therefore, x + y f < x + jy [ for every two real num bers. Example 4.18 S o lu tio n
a
Show that i f x  11 < 1 a nd x  11 < r /4 , w here r 6 R + , then x 2 + x  2 < r . First, observe that x 2 + x  2  = \(x + 2)(x  1) = x + 2 11x  1. By Theorem 4.17, x + 2 =■ (x — 1) + 3 < \x — 11 + 3 < 1 + 3 = 4. Therefore, x 2 + x — 2 1 = x + 2   x  1 < 4 0
= r.
#
4 .4 P ro o fs In v o lv in g Sets We now turn our attention to proofs concerning properties o f sets. Recall, for sets A and B contained in som e universal set U, that the in te rse c tio n o f A and B is A n B — {x : x e A and x e B }. the u n io n o f A and B is A y B = {x : x e A or x e B } and the d ifferen ce o f A and B is A  B — {x : x e A a n d x ^ B}. The set A  B is also called the rela tiv e c o m p lem en t o f B in A and the relative co m plem ent o f A in U is called sim ply the c o m p lem en t o f A and is denoted by A. Thus, A = Li — A . In w hat follow s, we w ill always assum e that the sets under discussion are subsets o f some universal set U .
4.4
A —B Figure 4.1
Proofs Involving Sets
109
A nB Venn diagrams for A —B and A n B
Figure 4.1 shows Venn diagram s o f A  B and A n B for arbitrary sets A and B . The diagram s suggest that these two sets are equal. This is, in fact, the case. Recall that to show the equality o f two sets C and D , w e can verify the tw o set inclusions C C D and D C C . T o establish the inclusion C C £>, we show that every elem ent of C is also an elem ent o f D \ that is, i f X 6 C then x e D . This is accom plished w ith a direct pro o f by letting x be an (arbitrary) elem ent of C and show ing that x m ust belong to D as well. R ecall that w e need not be concerned if C contains no elem ents; for in this case x e C is false for every elem ent X and so the im plication “If x € C , then x e D .” is true for all x e U and in this case the statem ent follow s vacuously. As a consequence of this observation, if C = 0, then C contains no elem ents and it follow s that C c D . Result 4.19
For eveiy tw o sets A and B , A  B = AHB.
P roof
First w e show that A _ B c A n B . Let x e A  B .J h e n x e A and x £ B . Since x £ B , it follow s that x e B . T herefore, x e A and x e B ; so x e A O B . H ence A  K A n B. _ _ __ N ext w e show that A P S C A — B . L et y e A H B . T hen y € A and y e B . Since y € S , we see that y £ B . N ow because y 6 A and y £ B . we conclude that y e A  B. Thus, A H B c A  B .
PROOF A N A L Y SIS
■
In the second paragraph o f the pro o f o f R esult 4.19, we used y (rather than x ) to denote an arbitrary elem ent o f A n B . We did this only for variety. We could have used x twice. O nce we decided to use distinct sym bols, y w as the logical choice since x w as used in the first paragraph o f the proof. This keeps our use of sym bols consistent. A nother possibility w ould have been to use a in the first paragraph and b in the second. T his has some disadvantages, however. Since the sets are being called A and B, we m ight have a tendency to think that a e A and b e B, w hich m ay confuse the reader. For this reason, we chose x and y over a and b. B efore leaving the proof o f Result 4.19, we have one other rem ark. A t one point in the second paragraph, we learned that y e A and y £ B. From this w e could have concluded (correctly) that y £ A n B. but this is n ot w hat w e wanted. Instead, we wrote that y € A — B. It is always a good idea to keep our goal in sight. We w anted to show that y e A — B ; so it w as im portant to keep in m ind that it w as the set A — B in w hich we w ere interested, not A H B .
♦
110
Chapter 4
More oil Direct Proof and Proof by Contrapositive
(A U B )  (A n B ) Figure 4.2
Venn d ia g r a m s
{ A  B ) U ( B  A) fo r
( AU B ) — (A n B ) a n d (A — B ) U (B —A)
N ext, le t’s consider the Venn diagram s for ( A U B ) — ( A fi B) and {A — B ) U (5 — A ), w hich are show n in Figure 4.2. From these tw o diagram s, w e m ight conclude (correctly) that the two sets ( A U B) — (A C \ B ) and (A — B ) U (S — A ) are equal. Indeed, all that is lacking is a p r o o f that these two sets are equal. T hat is, Venn diagram s can be useful in suggesting certain results concerning sets, but they are only draw ings and do not constitute a proof. Result 4.20
For every tw o sets A an d B , (A U B }  (A fi B ) = (A  B ) U (B — A ).
Proof
First w e show that (A U B ) — (A n B ) C (A — B ) (A n B ). Then x e A U f i and x £ A n B . Since .v x e B . W ithout loss o f generality, we can assum e x £ X £ B . Therefore. X£ A — B and so ,v e (A (A
U (B — A). Let x £ (A U B ) — 4 U B . it follow s that x £ A or A. Since x £ A fl B , the elem ent — B ) U (B — A). Hence
U B )  (A n B ) C (A  B ) U (B  A).
N ext we show that (A  B ) U (B — A) C (A U B ) — fA fl B ). L et X 6 (A  B ) U (B — A ). Then .v £ A — B or .v £ B — A, say the former. So # £ A and x ^ B . Thus x € A U B and .v ^ A fl B . Consequently, x e (A U B ) — (A fl B ). Therefore, (A
 B ) U ( B  A) C ( A U B ) — ( A C B),
as desired. PROOF ANALYSIS
m
In the proof of R esult 4.20, when w e w ere verifying the set inclusion (A U B )  (A fl B) C (A  B ) U (B  A), we concluded that ,v c A or .v £ B . A t that point, we could have divided the pro o f into tw o cases (Case 1. x £ A and Case 2. x e B ); however, the proofs of the tw o cases w ould be identical, except that A and B w ould be interchanged. T herefore, w e decided to consider only one o f these. Since it really d id n ’t m atter w hich case we handled, we sim ply chose the case w here x £ A. This was accom plished by w riting: W ithout loss o f generality, assum e that x £ A. In the proof o f the reverse set containm ent, w e found ourselves in a sim ilar situation, nam ely, x £ A — B or x e B — A . A gain, these two situations w ere basically identical
4.5
Fundam ental Properties of Set Operations
111
and we sim ply chose to w ork w ith the first (form er) situation. (Had w e decided to assum e that x e B — A , we w ould have considered the latter case.) ♦ We now look at an exam ple o f a biconditional concerning sets. Result 4.21 P roof
L et A and B be sets. Then A U B = A i f and only if B 
First we prove that if A U B = A ,th e n # C A. W e u s e a p ro o f by contrapositive. A ssum e that B is not a subset o f A. Then there m ust be some elem ent x e B such that x <£ A. Since x e B. it follow s that x € A U B. However, since x £ A , w e have A U fi / A. N ext we verify the converse, nam ely, if B c A, then A U 5 = A. We use a direct proof here. A ssum e that B C A. To verify that A U B = A , we show that A C A U B and A U B C A , The set inclusion A c A U B is im m ediate (if x e A, then x 6 A U B). It rem ains only to show then that A U B C A. Let y e A U B . Thus y e A or y e B . If y e A, then w e already have the desired result. If y £ B , then since B C A, it follow s that y e A . Thus A U B c A.
PROOF ANALYSIS
■
In the first paragraph of the proof o f R esult 4 .2 1 we indicated that w e were using a proof by contrapositive, w hile in the second paragraph w e m entioned that we w ere using a direct proof. This really w asn ’t necessary as the assum ptions w e m ade w ould inform the reader w hat technique w e w ere applying. A lso, in the p ro o f o f Result 4.21, w e used a proof by contrapositive for one im plication and a direct pro o f for its converse. This w asn ’t necessary either. Indeed, it is quite possible to interchange the techniques we used (see Exercise 41).
4.5
A.
▼
F undam ental Properties of Set O perations M any results concerning sets follow from some very fundam ental properties o f sets, w hich, in turn, follow from corresponding results about logical statem ents that were described in C hapter 2. For exam ple, we know that if P and Q are two statem ents, then P V Q and Q V P are logically equivalent. Similarly, if A and B are tw o sets, then A U B = B U A. We list some o f the fundam ental properties o f set operations in the follow ing theorem .
Theorem 4.22
For sets A . B and C, (1) Com m utative Law s (a ) A U B = B U A (b ) A n B = B n A (2) A ssociative Law s (a) A U ( B U C ) = ( A U B ) U C
(/» A n ( s n c ) = ( A n B ) n c (3) D istributive Law s (a) A U (B n C) = (A U B ) n (A U C ) (b) A n (B u C) = (A n B) u (A n C)
112
Chapter 4
More
o ji
Direct Proof and Proof by Contrapositive
(4) D e M organ’s Law s (a) A U B = A n B (b) A H B = A U B . We present proofs o f only three parts o f T heorem 4.22, beginning w ith the com m u tative law of the union o f two sets.
P ro o f o f Theorem 4.22(la)
We show that A U B c B U A. A ssum e th a t* e A U B . T h e n * e A o r x e S . A pplying the com m utative law for disjunction o f statem ents, w e conclude that x e B or x e A; so x e B U A. Thus, A U B c B U A. T he pro o f o f the reverse set inclusion B U A c A U B is sim ilar and is therefore om itted. ■ N ext w e verify one o f the distributive laws.
P ro o f o f Theorem 4.22(3a)
First w e show that A U (B fl C ) c (A U B ) fl (A U C ). L et x e A U (B fl C ). T hen * e A or x e B PI C . If x e A , then x e A U B and l e A U C . Thus x e (A U B ) D (A U C ), as desired. O n the other hand, if x e B fl C , then x e B and x e C ; and again, x e A U f i and x e A U C . So x € (A U B ) fl (A U C ). Therefore, A U ( B f l C ) C (AUB)n(AUC). To verify the reverse set inclusion, let x e (A U B ) fl (A U C ). Then x e A U B and x e A U C . I f x e A , then x e A U (B fl C ). So w e m ay assum e that x £ A . Then the fact that x e A U B and x £ A im plies that x e B . By the sam e reasoning, x e C . Thus x 6 B ft C and so x e A U (B n C). T herefore, (A U B ) fl (A U C ) c A U (5 n C ). . As a final exam ple, we prove one o f D e M o rg an ’s laws.
P ro o f o f Theorem 4.22(4a)
First, we show that A U B c A n B . L et x e A U B . T hen x ^ A U 5 . H ence x ^ A and x f B. T herefore, x e A and x e B , so x e A D S ; Consequently, A U B c
Ans.
_
__
___
N ext we show that A f l S c A U B . Let x e A O B . T hen x e A and x e B . Thus, x ^ A and x ^ B ; s o x ^ A U B . Therefore, x e A U B , H ence A H B c A U B , ■
PROOF ANALYSIS
In the p roof o f the D e M organ law that we ju st presented, w e arrived at the step x £ A U B at one point and then next w rote x ^ A and x B. Since . r e A U B im plies that x e A o r x e B , you m ight have expected us to write that x £ A o r x £ B after w riting x ^ A U B , but this w ould not be the correct conclusion. W hen we say that x i A U B , this is equivalent to w riting ~ (x e A U B), w hich is logically equivalent to ~ ((x e A) or (x e B )). By the De M organ law for the negation o f the disjunction of tw o statem ents (or tw o open sentences), w e have that ~ ((x e A) or (x e B )) is logically equivalent to ~ (x e A) and ~ (x e B); that is, x ^ A and x £ B . ♦ Proofs o f som e other parts o f T heorem 4.22 are left as exercises.
4.6
Proofs Involving Cartesian Products of Sets
113
4.6 Proofs Involving C artesian P roducts of Sets R ecall that the C a rte sia n p ro d u c t (or sim ply the p ro d u c t) A is defined as A x B = {(a, b) : a
e
x
B o f tw o sets A and B
A and b e B }.
If A = 0 or B = 0, then A x B = 0. B efore looking at several exam ples of proofs concerning Cartesian products o f sets, it is im portant to keep in m ind that an arbitrary elem ent o f the C artesian product A x B o f tw o sets A and B is o f the form {a, b), w here a G A and b G B . Result 4.23 P roof
Result 4.24
L et A , B , C and D be sets. I f A C . C and B C D , then A x B c C x D. Let (x, y) e A x B. T hen x G A and y e B . Since A c C and B C D , it follow s that x e C and y g D . H ence (x, y) G C x D. m For sets A , B and C, A x (B U C ) = (A x B ) U (A x C).
Proof
We first show that A x ( £ U C ) C ( A x f i ) U ( A x C ). L et (x, y) e A x (B U C ). Then x e A and y e B U C . Thus y e B or y e C , say the former. T hen (x, y) G A x B and so (x, y) € (A x B ) U (A x C ). Consequently, A x ( 8 U C ) c ( A x B ) U (A x C). N ext w e show that (A x B ) U (A x C ) c A x ( 6 U C ). L et (x, y ) e ( A x B ) U (A x C ). T hen (x, y) g A x B or (x, y) e A x C , say the former. T hen x e A and y e 5 C B U C . H ence (x, y) e A x ( 5 U C ), im plying that ( A x f i ) U ( A x C ) c A x (BUC). • We give one additional exam ple of a p ro o f involving the C artesian products o f sets.
Result 4.25
For sets A , B and C , A x ( B  C ) = ( A x B )  ( A x C ).
Proof First w e show that A x ( B — C ) c (A x B) — (A x C ) .L e t( x , y) e A
x
( B — C ).T h e n x € A a n d y € 5 —C .S in c e y e B — C , it follow s that y e .B an d y ^ C .B e c a u se x e A and y & B, w e have (x, y) e A x B. Since y ^ C , however, (x, y) $£ A x C . Therefore, (x, y) € (A x B) — (A x C ). H ence A x ( 8  C ) C ( A x S )  ( A x C ). We now show that (A x B) — (A x C ) c A x ( £ — C ). L et (x, y ) e ( A x B ) — (A x C ). T hen { x , y ) e A x B and (x, y) £ A x C . Since (x, y) e A x B , it follow s th a tx g A and y e B. A lso, since x g A and (x, y ) £ A x C, it follow s that y $ C . So y e B — C. Thus (x, y) G A x ( B — C ) and ( A x f i )  ( A x C ) C A x ( B  C) . ■
PROOF ANALYSIS
We add one com m ent concerning the preceding proof. D uring the p ro o f o f (A x B) — (A x C) c A x ( B — C), w e needed to show that y <£ C. We learned that (x, y) ^ A x C. However, this inform ation alone did not allow us to conclude that y £ C. Indeed, if
114
Chapter 4
More on Direct Proof ancl Proof by Contrapositive
(x , y) ^ A x C , then x 4. .4 or y $ C. Since we knew, however, that x e A a n d ( x , y ) £ A x C, w e were able to conclude that y £ C. 4
E X E R C IS E S
FO R CH A PTER 4
S e c t i o n 4.1: P r o o f s I n v o l v i n g D i v i s i b i l i t y o f I n t e g e r s 4.1. Let a
and b be integers, where a ^ 0. Prove that
4.2. Let a, b 6 Z, where a # 0 and b / 0. Prove that
if a \b
if a \b.
then a2 \ b2.
and b a. then a = b or a =  b .
4.3. Let m e Z. (a) Give a direct proof of the following: If 3  m, then 3  m 2. (b) State the contrapositive of the implication in (a). (c) Give a direct proof of the following: If 3 /m , then 3 / m 2, (d) State the contrapositive of the implication in (c ). (e) State the conjunction of the implications in (a) and (c) using “if and only if.” 4.4. Let x, y e Z. Prove that if 3 / x and 3 / y , then 3  (x2 — y2). 4.5. Let a , b , c e Z, where a ^ 0. Prove that if a / b e , then a / b and a / c . 4.6.
Let a
e Z. Prove that if 3  2a. then 3  a.
4.7.
Let n
e Z. Prove that 3  (2n1 + 1) if and only if 3 / n .
4.8. In Result 4.4, it was proved for an integer x that if 2  (x2  1), then 4  (x2  1). Prove that if 2  (x2  1), then 8  (x2 — 1). 4.9. (a) Let x € Z. Prove that if 2  (x2 — 5), then 4 ] (x2  5). (b) Give an example of an integer x such that 2 ] (x2  5) but 8 / (x2  5). 4.10. Let n e Z. Prove that 2  («4 — 3) if and only if 4  (n2 + 3). 4.11. Prove that for every integer n > 8, there exist nonnegative integers a and b such that n = 3a + 5b. 4.12. In Result 4.7, it was proved for integers x and y that 4  (x2  y 2) if and only if x and y are of the same parity. In particular, this says that if x and y are both even, then 4  (x2  y 2); while if x and y are both odd, then 4  t.v2 — y2). Prove that if x and y are both odd, then 8  (x2 ~ y2). 4.13. Prove that if a, b, c e Z and a 2 + b1 = c2, then 3  ab. S e c t i o n 4.2: P r o o f s I n v o l v i n g C o n g r u e n c e o f I n t e g e r s 4.14. Let a , b ,n e Z, where n > 2. Prove that if a = b (mod n ), then a 2 = b2 (mod
id .
4.15. Let a , b, c, n g Z, where n > 2. Prove that if a = b (mod n) and a = c (mod n ), then b = c (mod n ). 4.16. Let a. b e Z. Prove that if a~ + 2b~ = 0 (mod 3), then either a and b are both congruent to 0 modulo 3 or neither is congruent to 0 modulo 3. 4.17. (a) Prove that if a is an integer such that a = 1 (mod 5), then a 2 = 1 (mod 5). (b) Given that b is an integer such that b = 1 (mod 5), what can we conclude from (a)? 4.18. Let m , n G N such that m > 2 and m \ n. Prove that if a and b are integers such that a = b (mod n). then a = b (mod m). 4.19. Let a, b e Z. Show that if a = 5 (mod 6) and b = 3 (mod 4), then 4a + 6b = 6 (mod 8).
Exercises for Chapter 4
115
4.20. (a) Result 4.12 states: Let ;/ e Z. If n2 =£ n (mod 3), then n # 0 (mod 3) and n # 1 (mod 3). State and prove the converse of this result. (b) State the conjunction of Result 4 .12 and its converse using “if and only if.” 4.21. Let a e Z. Prove that a 3 = a (mod 3). 4.22. Let n e Z. Prove each of the statements (a)(f).
(a) If n =
0 (mod 7), then n1= 0 (mod 7).
(b) If n =
1 (mod 7), then n2= 1 (mod 7).
(c) I f n =
2 (mod 7), then n2= 4 (mod 7).
(d) If n =
3 (mod 7), then n2= 2 (mod 7).
(e) For each integer n, n2 = (1 — n)2 (mod 7). (f) For every integer n, n2 is congruent to exactly one of 0, 1, 2 or 4 modulo 7.
4.23.
Prove for any set S = [a, a + 1.......a + 5} of six integers where 6  a that 24  (x 2  y 2) for distinct odd integers x and y in S if and only if one of x and v is congruent to 1 modulo 6 while the other is congruent to 5 modulo 6.
4.24. Let x and y be even integers. Prove that x 2 = y 2 (mod 16) if and only if either (1) x = 0 (mod 4) and v = 0 (mod 4) or (2) x ee 2 (mod 4) and y = 2 (mod 4).
S e c t i o n 4 .3 : P r o o f s I n v o l v i n g R e a l N u m b e r s .25. Let x , y e R. Prove that if x 2 — 4x = y 2 — 4y and x *£ y, then x + y = 4. .26. Let a , b and m be integers. Prove that if 2a + 3b > 12m + 1, then a > 3m + 1 or b> 2m + 1. .27. Let x e R. Prove that if 3x4 + 1 < x 7 + x 3, then x > 0. .28. Prove that if r is a real number such that 0 < r < 1. then r(.1l_ r) > 4. .29. Prove that if r is a real number such that r — 11 < 1, then
— 1•
.30. Let x, y G R. Prove that xv = x • yj. 4.31. 4.32.
Prove for every two real numbers x and y that x + y  > x — y.
(a) Recall that Jr > 0 for every positive real number r. Prove thatif a and b arepositive real numbers, then 0 < s/ab < (The number V a b is called the geometric mean of a and b, while (a + b )/2 is called the arithmetic mean or average of a and b .) (b) Under what conditions does sfab = (a + b )/2 for positive real numbers a and b l Justify your answe
.33. The geometric mean of three positive real numbers a, b and c is 4 abc and the arithmetic mean is ( a + b  1 c)/3. Prove that ija b c < (a + b + c ) /3. [Note: The numbers a, b and c can be expressed as a = r 3, b = s3 and c = r3 for positive numbers r, s and t.] 4.34. Prove for every three real numbers x, y and
z
that x — z < §£ « j  + [y:~ f  .
4.35. Prove that if x is a real number such that x(x + 1) > 2, then x < —2 or x > 1. 4.36. Prove for every positive real number x that 1 +
> j +
4.37. Prove f o rx , y, z e R that x 2 + y 2 + z2 > x y + xz + yz. 4.38. Let a. b, x, y € R and; e R + . Prove that if x  a\ < r /2 and y  b\< r / 2, then (x + y)  (a + h)\ < r. 4.39. Prove that if a, b , c , d e R, then (ab + cd)2 < (a2 + c2){b2 + d 2).
116
Chapter 4
More on Direct Proof and Proof by Contrapositive
Section 4.4: Proofs Involving Sets 4.40. Let A and fi be sets. Prove that A U B = (A — fi) U (B — A) U (A fl B). 4.41. In Result 4.21. it was proved for sets A and B that A U B = A if and only if fi C A. Provide another proof of this result by giving a direct proof of the implication “If A U f i = A, then B C A” and a proof by contrapositive of its converse. 4.42. Let A and B be sets. Prove that A fl B = A if and only if A C fi. 4 43
(a) Give an example of three sets A, B and C such that A fl fi = A fl C but fi / C. (b) Give an example of three sets A, fi and C such that A U fi = A U C but fi / C. (c) Let A, fi and C be sets. Prove that if A n fi = A f l C and A U fi = A U C ,then fi = C.
4.44. Prove that if A and fi are sets such that A U B ^ 0, then A ^ 0 or fi ^ 0. 4.45. Let A = {» 6 Z : /? = 1 (mod 2)} and B = {n e Z : n e 3 (mod 4)}. Prove that fic A. 4.46. Let A and fi be sets. Prove that A U f i = A H fi if and only if A = fi. 4.47. Let A = [n € Z : « = 2 (mod 3)} and fi = [n e Z : n = f (mod 2)}. (a) Describe the elements of the set A — fi. (b) Prove that if n e A fl 6 , then rr 4.48. Let A = {n € Z : 2 j «} and for some odd integer k.
= 1 (mod 12).
fi = {ne Z : 4  n}. Let « e Z. Prove that n e A — B if and only if n =
2*
4.49. Prove for every two sets A and fi that A = (A  fi) U (A fl fi). 4.50. Prove for every two sets A and B that A — B , B — A and A n fi are pairwise disjoint. 4.51. Let A and fi be subsets of a universal set. Which of the following is a necessary condition for A and fi to be disjoint? (a) Either A = 0 (b) W heneverx (c) Whenever x (d) Whenever x (e) Whenever x
or fi = 0. fi A , it must occur that x e fi. ^ A,it must occur that x fi B. e A,it must occur that x e f i . e A,it must occur that x fi B.
Section 4.5: Fundam ental Properties of Set Operations 4.52. Prove that A fl fi = fi n A for every two sets A and B (Theorem 4.22( lb)). 4.53. Prove that A n (fi U C ) = (A n fi) U (A fl C ) for every three sets A. B and C (Theorem 4.22(3b)). 4.54. Prove that A Pi fi = A U fi for every two sets A and fi (Theorem 4.22(4b)). 4.55. Let A, fi
and C
be
sets.Prove that(A — fi) fl (A — C) = A — (fi U C).
4.56. Let A, fi
and C
be
sets.Prove that(A — fi) U (A — C) = A — (B fl C).
4.57. Let A, fi and C be sets. Use Theorem 4.22 to prove that A U (fi n C ) = (A n fi) U (A — C). 4.58. Let A, B
and C
be
sets.ProvethatA fl (fi n C) = (A U B ) D (A U C).
4.59. Show for
every
three sets A, fi and C that A — (fi — C ) = (A fl C) U (A — fi).
Additional Exercises for Chapter 4
117
Section 4.6: Proofs Involving Cartesian Products of Sets 4.60. For A = [x, y}, determine A x V (A ). 4.61. For A = {1 j and fi = {2}, determine V (A x f i) and V( A ) x V( B) . 4.62. Let A and B be sets. Prove that A x B = 0 if and only if A = 0 or B = 0. 4.63. 4.64. 4.65. 4.66.
For For condition Let
sets A and 5 , k
find a necessary and sufficient condition for A x B = B x A.
sets A and 5 , find a necessary and sufficient condition for (A x fi) n (B x A) = 0.Verify that necessary and sufficient. A, B and C be nonempty sets. Prove that A x C c fi x C if and only if A c B.
Result 4.23 states
that if A, B, C and D are sets such that A c C and B c D, then A x S c C x i ) .
(a) Show that the converse of Result 4.23 is false. (b) Under what added hypothesis is the converse true? Prove your assertion. 4.67. Let A, B and C be sets. Prove that A x ( f i n C ) = ( A x J j ) n ( A x C).  . 68. Let A, B. C and D be sets. Prove that (A x fi) fl (C x D) = (A f ! C ) x (B n D). 4.69. Let A, fi, C and D be sets. Prove that (A x B) U (C x D) C (A U C ) x (fi U £>). 4.70. Let A and S be sets. Show, in general, that A x B / A x B.
ADDITIONAL EXERCISES FOR CHAPTER 4 .71. Let n e Z. Prove that 5  n2 if and only if 5 j n. 4.72. Prove for integers a and ft that 3  ab if and only if 3  a or 3 j ft. .73. Prove that if n is an odd integer, then 8  [n1 + (n + 6)2 + 6], .74. Prove that if n is an odd integer, then 8 j (n4 + 4n 2 + 11). .75. Let n, m e Z. Prove that if n = 1 (mod 2) and m = 3 (mod 4), then n2 + m
= 0 (mod
.76. Find two distinct positive integer values of a for which the following is true and give a
4). proof in each case:
For every integer n, a / (n2 + 1). 4.77. Prove for every two real numbers a and ft that ab <
\'h \
.78. Prove for every two positive real numbers a and ft that  +  > 2. .79. Prove the following: Let x e R. If x (x  5) =  4 , then v75.r2  4 = 1 implies that * +  =
2.
.80. Let x , j e R . Prove that if x < 0, then .v3 —,v2.v < x 2y  x y 2. .81. Prove that 3  («3 — 4ri) for every integer n. .82. Evaluate the proposed proof of the following result. R esult
Let x , y € Z. I f x = 2 (mod 3) and y = 2 (mod 3), then x y = 1 (mod
3).
P ro o f Let x = 2 (mod 3) and y = 2 (mod 3). Then .v = 3k + 2 and y = '3k + 2 for some integer k. Hence xy = (3k + 2)(3 k + 2) = 9 k 2 + 12fe + 4 = 9k 2 + 12A: + 3 + 1 = 3(3A2 + 4 k + 1 ) + 1. Since 3k 2 + 4 ^ + 1 is an integer, x y = 1 (mod 3).
■
118
Chapter 4
More 011 Direct Proof and Proof by Contrapositive:
4.83. Below is given a proof of a result. W hat result is proved? P ro o f Assume that x = 1 (mod 5 ) and y = 2 (mod 5). Then 5  (x  1) and 5 ] ( j — 2). Hence x  1 = 5a and y — 2 = 5b for some integers a and b. So x = 5a + 1 and y = 5b + 2. Therefore, X2 + y 2 = (5a + 1)2 + (5b + 2)2 = (25a2 + 10a + 1) + (25 b2 + 20 b + 4) = 25a2 + 10a + 25 b1 + 20 b + 5 = 5(5a2 + 2a + 5 b2 + 4 b + 1). Since 5a 2 + 2a + 5b2 + 4 6 + 1 is an integer, 5  (x2 + y2) and so x 2 + y2 = 0 (mod 5).
■
4.84. A proof of the following result is given. Result
Let n e Z. If n4 is even, then 3// + 1 is odd.
P ro o f Assume that n4 = (n2) is even. Since n4 is even, n 2 is even. Furthermore, since nr is even, n is even. Because n is even, n = 2k for some integer k. Then 3n + 1 = 3(2/c) + 1 = 6* + 1 = 2(3*) + 1. Since 3k is an integer, 3n + 1 is odd.
■
Answer the following questions. (1) Which proof technique is being used? (2) W hat is the starting assumption? (3) W hat must be shown to give a complete proof? (4) Give a reason for each of the following steps in the proof. (a) Since n4 is even, n 2 is even. (b) Furthermore, since n 2 is even, n is even. (c) Because n is even, n = 2k for some integer k. (d) Then 3n + 1 = 3(2*) + 1 = 6k + 1= 2(3*) + 1. (e) Since 3* is an integer, 3n + 1 is odd. 4.85. Given below is an attempted proof of a result. P ro o f First, we show that A C ( A U 8 )  5 . Let X e A. Since A n B = 0, it follows that x f B. Therefore, x e A U B and x f />’: so .v € (A U li) /». Thus A c (A U B )  B. Next, we show that ( A U f i )  B C A . Let x e (A U B )  B. Then x e A U £ and x <£B . From this, it follows that x 6 A. Hence (A U 5 ) — S c A. ■ (a) W hat result is being proved above? (b) W hat change (or changes) in this proof would make it better (from your point of view)? 4.86. Evaluate the proposed proof of the following result. Result
Let x, y e Z such that 3  x. If 3  (x + y), then 3  y.
P ro o f Since 3 j x, it follows that x = 3a, where a e Z. Assume that 3  (x + y). Then x + y = 3b for some integer b. Hence y = 3 b  x = 3b  3a = 3(b  a). Since b  a is an integer, 3 ( y. For the converse, assume that 3 j y. Therefore, y = 3c, where c e Z. Thus x + y = 3a + 3c = 3(a + cl. Since a + c is an integer, 3 [ (X + y). a 4.87. Evaluate the proposed proof of the following result. R esult Let x, y e Z. If x = 1 (mod 3) and y = 1 (mod 3),then x y = 1 (mod 3). P ro o f Assume that x = 1 (mod 3) and y = 1 (mod 3).Then 3 (x  1) and 3 (y  1). Hence x  1 = 3q and y  1 = 3q for some integer q and so x = 3q + 1 and y = 3q + 1. Thus xy = (3 q + 1)(3 q + 1) = 9 q 2 + 6q + 1 = 3(3 q 2 + 2 q) + 1 and so xy  1 = 3(3q 2 + 2q). Since 3q 2 + 2q is an integer, 3  (xy  1). Hence xy = 1 (mod
3).
n
Additional Exercises for Chapter 4
119
4.88. Evaluate the proposed proof of the following result. Result For every three sets A, B and C , (A x C ) — (B x C ) C ( A ~ f i ) x C . P ro o f Let (x, y) e (A x C) — (B x C). Then (x, y) 6 A x C and (*, y) ^ B x C. Since j ) e A x C, it follows that x e A and y e C . Since (x, y) ^ 5 x C, we have x £ B. Thus x e A — B. Hence (x, y) r (A  B ) x C. m 4.89. Prove that for every three integers a , b and c, the sum a — b\ + \a
— c\ +
\b — c\
is
an even intege
.90. Prove for every four real numbers a, b , c and d that ac + bd < \J a 2 + b2\/c 2 + d 2. .91. Prove that for every real number x , sin6 x + 3 sin2 x cos2 x + cos6 x = 1. .92. Let a e Z. Prove that if 6  a and 10  a, then 15  a. .93. Let A = {x}. Give an example of a set concerning the set A to which each of the following elements belong, (a) (x, {x}) (b) ({x}, x) (c) (x, x ) (d) ({a }, {.y}) (e) x CO [ v (g) {(x, x)} (h) ({x}, i(x}). 4.94. Let a, b € Z. Prove that if a = b (mod 2) and b = a (mod 3), then a = b (mod 6). 4.95. Let a, b, c e R. Prove that  ( a 2 + b2 + c2 + 1) > a(b + 1) + b(e + 1) + c(a + I). 4.96. Prove that if a ,b and c are positive real numbers, then (a ■ h r c) (~ . 
1) > 9 .
.97. Let T = { 1 , 2 , . . . , 8}. (a) Determine the elements of the set A = [a e T : 2m = a (mod 9) for some in e N}. (b) Determine the elements of the set B = {a e T : 5"' = a (mod 9) for some m e Nj. (c) W hat property do the sets A and B have in (a) and (b)? 4.98. Consider the open sentence P (m ) : 5m + 1 = a 1 for some a e Z, where m e N. That is, P (m ) is the open sentence: 5m + 1 is a perfect square. (a) Determine four distinct solutions t of t 2 = 4 (mod 5). For each solution t, determine m = 41f ST4 + 3 and show that P(in) is a true statement. (b) Show that the set S = {t e Z : t 2 = 4 (mod 5)} contains infinitely many elements. (c) Let t be an element of the set S in (b). Prove that if m =  ' ~4) + 3, then 5m + 1 is a perfect square, (d) As a consequence of the results established in (a)(c), what can be concluded about the set M = {m e N : 5m + 1is a perfect square}? 4.99. Let « i , a i , . . . , an (®s£ 3) be n integers such that (~ti{— 1  < integer that lies strictly between ci\ and a„, then there is an integer j with 1 < j < n such that Uj = k.
5 Existence and Proof by Contradiction
hus far, we have been prim arily concerned w ith quantified statem ents involving universal quantifiers, nam ely statem ents of the type Vx e S , R (x ). We now con sider problem s that involve, either directly or indirectly, quantified statem ents involving existential quantifiers, that is, statem ents o f the type 3 x e S , R {x).
T
5.1 C ounterexam ples It m ust certainly com e as no surprise that som e quantified statem ents o f the type Vx e S. R (x ) are false. We have seen that ~ (Vx e 5, R(x>) = 3 x e S, ~ R (x ), that is, if the statem ent Vx € S, R ( x ) is false, then there exists some elem ent x e S for w hich R (x ) is false. Such an elem ent x is called a c o u n te re x a m p le o f the (false) statem ent V x e S. R (x ). Finding a counterexam ple verifies that Vx e S, R (x ) is false. Example 5.1
C onsider the statem ent: I f x e R , then (x 2  l)2 > 0.
(5.1)
or, equivalently. For every real num ber x , (x 2  l )2 > 0. Show that the statem ent (5.1) is fa ls e by exhibiting a counterexam ple.
S o lu tio n
For x = 1, (x~ — 1) = ( 1 — 1) = 0. Thus x = 1 is a counterexam ple.
4
It m ight be noticed that the num ber x = —I is also a counterexam ple. In fact, * = 1 and x = ~ l are the only two counterexam ples o f the statem ent (5.1). T hat is, statem ent If x e R — {1, —1}, then (x 2 — l ) 2 > 0.
120
is true.
(5.2)
5.1
Counterexamples
121
If a statem ent P is show n to be false in som e m anner, then P is said to be d isp ro v ed . T he counterexam ple x = 1 therefore disproves the statem ent (5.1). Example 5.2
D isprove the statem ent: I f x is a real number, then tan2 x + 1 = sec2 x .
Solution
(5.3)
Since tan x and sec x are not defined w hen x = 7r / 2 , it follow s that tan2 x + 1 and sec2 x have no num erical value w hen x = jt / 2 and, consequently, tan2 x + 1 and sec2 x are not equal w hen x = n /2 . T hat is, x = n / 2 is a counterexam ple to the statem ent (5.3). ♦
A lthough tan2 x + 1 = sec2 x is a w ellknow n identity from trigonom etry, state m ent (5.3), as presented, is false. The follow ing is true, however:
If x is a real num ber for w hich tan x and sec x are defined, then tan2 x + 1 = sec2 x .
.
Indeed, it is probably statem ent (5.4) that was intended in Exam ple 5.2, rather than statem ent (5.3). Since ta n x and se c x are defined for precisely the same real num bers x (namely, those num bers x such that c o s x / 0), w e can restate (5.4) as I f x e R — [nrc + § : n e Z j , then tan2 x + 1 = sec2 x. Example 5.3
D isprove the statem ent: A'2 I X
X “I 1
I f x e Z , then —— — =  . X —X X — 1
Solution
(5.5)
X2 + X
If x — 0, then x 2 — x = 0 and so —r is not defined. O n the other hand, if x = 0,
x —x
1 . X2 + X X + 1 t h e n = —1; so the expressions — a n d  are certainly not equal w hen x —1 x  —x x —1 x = 0. Thus x = 0 is a counterexam ple to the statem ent (5.5). ♦ ,
X +
x2 X X 1 Since neither — n o r  is defined w hen x = 1, it follow s that x = 1 is also x 2 —X X — 1 a counterexam ple o f statem ent (5.5). Indeed, x = 0 and x = 1 are the only counterex am ples o f statem ent (5.5) and so the statem ent I f , € Z — (0, n , then
= XZ — X
iii. X — 1
is true. The three preceding exam ples illustrate the fact that an open sentence R ( x ) that is false over some dom ain S m ay very w ell be true over a subset o f S. T herefore, the truth (or falseness) o f a statem ent Vx e S, R ( x ) depends not only on the open sentence R( x ) but on its dom ain as well.
122
Chapter 5 Example 5.4
Existence and Proof by Contradiction
D isprove the statem ent: For every odd positive integer n. 3  Ur — 1).
S o lu tio n
Since 3 / ( 3
 1), it follow s that n = 3 is a counterexam ple.
(5.6) ♦
You m ight have noticed that even though 3 / ( 32  1), it is the case th at 3  (n 2  1) for odd positive integers. For exam ple, 3  («2 — l ) i f n ss 1 , 5 , 7 , 11, 13, 17, while 3 K (n — 1) if » = 3, 9, 15, 21. This should m ake you w onder for w hich odd positive integers n, the open sentence 3  (n 2  1) is true. (See Result 4.6.) We have seen that a quantified statem ent o f the type V.v e S , A’(.v) is false if 3 x e S, ~ R (x ) is true, that is, if there exists some elem ent x e S for w hich R (x ) is false. There will be m any instances w hen R {x) is an im plication P ( x ) =» Q (x). Therefore, the quantified statem ent V.v € S, P ( x ) ■> Q( x )
(5.7)
a.T e 5, ~ ( P{ x ) => Q( x) )
(5.8)
is false if
is true. B y T heorem 2.4(a), the statem ent (5.8) can be expressed as 3x 6 S, ( P( x )
A
( ~ Q(.x))).
T hat is, to show that the statem ent (5.7) is false, we need to exhibit a counterexam ple, w hich is then an elem ent ,v e S for w hich P ( x ) is true and Q (x ) is false. Example 5.5
D isprove the statem ent: L e t n e Z. I f n 2 + 3n is even, then n is odd.
S o lu tio n
If n = 2, then n 2 + 3n — 22 + 3 • 2 = 10 is even and 2 is even. Thus counterexam ple.
n — 2 is a ^
In the preceding exam ple, not only is 2 a counterexam ple, every even integer is a counterexam ple.
Example 5.6
D isprove the statem ent: I f n is an odd integer, then n 2 — n is odd.
Solution
(5 .9)
For the odd integer n — 1, the integer n 2  n = l 2 — 1 = 0 is even. Thus n ■ — I is a counterexam ple. 4
5.1
Counterexamples
123
Actually, it is not difficult to prove that the statem ent If n is an odd integer, then n 2 — n is even. is true. A lthough it m ay very w ell be of interest to know this, to show that statem ent (5.9) is false requires exhibiting only a single counterexam ple. It does not require proving some other result. O ne should know the difference betw een these two. Example 5.7
Show that the statem ent: L e t n 6 Z. I f 4  (n 2 — 1), then 4 \ (n — 1). is fa lse.
S o lu tio n
Example 5.8
Since 4  (32 — 1) but 4 / (3 — 1), it follow s that n = 3 is a counterexam ple.
♦
Show that the statem ent For positive integers a , b, c, a bC = (ah) c. is false.
S o lu tio n
Example 5.9
L et a = 2, b = 2 and c = 3. T hen a l,c = 223 = 2s = 256, w hile (a b) c = (22) 3 = 4 3 = 64. Since 256 ^ 64, the positive integers a = 2, b = 2 and c = 3 constitute a counterexam ple. +
Show that the statem ent: L et a and b be nonzero real num bers. I f x , y e R + , then (5.10) is false.
S o lu tio n
L et x = b2 and y = a 2. Then
2 b2
b2 2 a2b 2 a 2b2 = a 2b 2 = x y . rr = — + 2a 2 ' 2'2
Thus x = b2 and y = a 2 is a counterexam ple and so the inequality is false. A n a ly sis
♦
A fter reading the solution o f Exam ple 5.9, the only question that m ay occur to you is w here the counterexam ple x = b2 and y = a 2 cam e from . M ultiplying the inequal ity (5.10) by 2a 2b 2 (w hich elim inates all fractions) produces the equivalent inequality [,4 yz „2 > 2 a 21 a 4x 2 + b* zb.2, zx y
124
Chapter 5
Existence and Proof by Contradiction
and so a 4x 2  2 c rb 2x y + b4y 2 > 0 , w hich can be expressed as (a2.x ~ b 2y ) 2 > 0. O f course, (a 2x â€” b 2y)~ > 0. Thus any values o f x and y for w hich a 2x â€” b 2y = 0 produce a counterexam ple. A lthough there are m any choices for x and y, one such choice is x = b2 and y = a 2.
5.2 P roof by C ontradiction
j
Suppose, as usual, that we w ould like to show that a certain m athem atical statem ent R is true. If R is expressed as the quantified statem ent V x e S, P ( x ) => Q (x ), then we have already introduced tw o proof techniques, nam ely direct p ro o f and p ro o f by contrapositive, that could be used to establish the truth of R. We now introduce a third m ethod that can be used to establish the truth of R , regardless o f w hether R is expressed in term s o f an im plication. Suppose that we assum e R is a false statem ent and, from this assum ption, w e are able to arrive at or deduce a statem ent that contradicts som e assum ption w e m ade in the pro o f or som e know n fact. (The know n fact m ight be a definition, an axiom or a theorem .) If we denote this assum ption or know n fact by P , then w hat w e have deduced is ~ P and have thus produced the contradiction C : P A ( ~ P ). We have therefore established the truth o f the im plication ( ~ R ) =* C. However, because ( ~ R ) =$â– C is true and C is false, it follow s that ~ R is false and so R is true, as desired. This technique is called p ro o f by c o n tra d ic tio n . If R is the quantified statem ent V x g S', P( x ) =$ Q (x ). then a proof by contradiction o f this statem ent consists o f verifying the im plication ~ (V x e S, P l x ) => O( x) ) =>. C for some contradiction C . However, since ~~ (V x e S,
PLX) => Q( x) ) = 3,v e S, ~ ( P( x ) .> Q { x )) s
3x e S, ( P (x )
A
( ~ 0 (x ))),
it follow s that a pro o f by contradiction o f V x G 5, P ( x ) => Q( x ) w ould begin by assum ing the existence o f some elem ent x e S such that P( x ) is true and Q( x ) is false. T hat is, a proof by contradiction o f V x g S, P ( x ) =^> 0 ( x ) begins by assum ing the existence o f a counterexam ple o f this quantified statem ent. O ften the reader is alerted that a pro o f by contradiction is being used by saying (or w riting) Suppose that R is false. or A ssum e, to the contrary, that R is false.
5.2
Proof by Contradiction
1 25
Therefore, if R is the quantified statem ent V i e S, P ( x ) => 2 ( a ) , then a p ro o f by contradiction m ight begin with: A ssum e, to the contrary, that there exists som e elem ent x e S fo r which P (x) is true and Q ix ) is false. (or som ething along these lines). The rem ainder of the p ro o f then consists o f showing that this assum ption leads to a contradiction. L e t’s now look at som e exam ples o f pro o f by contradiction. We begin by establishing a fact about positive real num bers. R esu lt to P ro v e 'ROOF STRATEGY
T here is no sm allest positive real number. In a proof by contradiction, we begin by assum ing that the statem ent is false and attem pt to show that this leads us to a contradiction. H ence w e begin by assum ing that there is a sm allest positive real number. It is useful to represent this num ber by a sym bol, say r. O ur goal is to produce a contradiction. H ow do w e go about doing this? O f course, if w e could think o f a positive real num ber that is less then r , then this w ould give us a contradiction.
Result 5.10 Proof
^
There is no sm allest positive real number. A ssum e, to the contrary, that there is a sm allest positive real num ber, say r. Since 0 < r / 2 < r . it follow s that r / 2 is a positive real num ber that is sm aller than r. This, however, is a contradiction.
PROOF ANALYSIS
®
The contradiction referred to in the proof o f R esult 5 .10 is the statem ent: r is the sm allest positive real num ber and r / 2 is a positive real num ber that is less than r . T his statem ent is certainly false. We have assum ed that the reader understands w hat contradiction has been obtained. If we think that the reader m ay not see this, then, o f course, we should specifically state (in the proof) w hat the contradiction is. T here is another point concerning R esult 5.10 that should be m ade. T his result states that “there is no sm allest positive real num ber.” This is a negativesounding result. In the vast m ajority o f cases, proofs o f negativesounding results are given by contradiction. Thus the proof technique used in R esult 5.10 is not unexpected. ♦ L e t’s consider tw o additional exam ples.
Result 5.11 P roof
N o odd integer can be expressed as the sum o f three even integers. A ssum e, to the contrary, that there exists an odd integer n w hich can be expressed as the sum o f three even integers x , y and z. T hen x = 2 a , y = 2b and z = 2c w ith a , b , c e Z. T herefore, n — .v r >' +•  — 2a — 2h
2c
2(a + b + c).
Since a + b + c is an integer, n is even. T his is a contradiction.
PROOF ANALYSIS
C onsider the statem ent: R: N o odd integer can be expressed as the sum o f three even integers.
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126
Chapter 5
Existence and Proof by Contradiction
O bviously, Result 5.11 states that R is a true statem ent. In order to give a p ro o f by contradiction o f Result 5 .1 1, w e attem pted to prove an im plication o f the type ( ~ R ) =* C for som e contradiction C . The negation ~ R is ~ R: T here exists an odd integer that can be expressed as the sum o f three even integers. The proof w e gave o f Result 5.11 began by assum ing the truth o f ~ R . We introduced sym bols for the four integers involved to m ake it easier to explain the proof. Eventually, we w ere able to show that n is an even integer. On the other hand, we knew that n is odd. H ence n was both even and odd. This was our contradiction C . +
In the tw o exam ples o f proof by contradiction that w e have given, neither statem ent to be proved is expressed as an im plication. For our next exam ple, w e consider an im plication. Result 5.12
I f a is an even integer and b is an odd integer, then 4 / (a 2 + 2 b 1).
P roof A ssum e, to the contrary, that there exist an even integer a and an odd integer b such that 4 [ (a + 2b2% Thus a = 2x. b = 2y ■■ 1 and a 1 + 2b 2 = 4 z for som e integers x , y and z. H ence a 2 + 2 b 2 = (2 x)2 + 2(2y + l )2 = 4z. S im plifying, w e obtain 4.r2 + 8 y2 +  j + 2 = 4z or, equivalently, 2 = 4z  4 x 2  8y 2  8>> = 4 (z  x 2  2 y 2  2y ). Since z — x 2 — 2y 2 — 2y is an integer, 4  2, w hich is im possible.
PROOF ANALYSIS
■
L et S be the set o f even integers and T the set of odd integers. In Result 5.12, our goal was to prove that Va e S , V7? g f , P (a , b).
(5.11)
is true, where P( a, b)\ 4 K ( a 2 + 2b2). Since we w ere attem pting to prove (5.11) by contradiction, w e w anted to establish the truth of ~ (Vo e S, Wb € T, P (a . /?)) =» C. for som e contradiction C or, equivalently, the truth of Ba e S , 3 b e T , ( ~ P( a, b))) => C. H ence we began by assum ing that there exist an even integer a and an odd integer b such that 4  (a2 + 2 b 2). We eventually deduced that 4 j 2, w hich is a false statem ent and thereby produced a desired contradiction.
5.2
Proof by Contradiction
127
U sing som e facts we discussed earlier, we could have given a direct p ro o f o f R e sult 5.12. O nce we w rote a = 2 x and b = 2y + 1, we have a 2 + 2 b 2 = (2* )2 + 2(2y + 1)2 = 4 x 2 + 8y 2 + 8y + 2 = 4 (x 2 + 2 y 2 + 2y) + 2. H ence we have expressed a 2 + 2b 1 as Aq + 2, w here q — a 2 + 2y 2 + 2y. T hat is, di viding a 1 + 2b2 by 4 results in a rem ainder o f 2 and so 4 / (a 2 + 2b2).At this stage, however, a p ro o f by contradiction o f R esult 5.12 is probably preferred, in order to both practice and understand this proof technique. ♦
L e t’s consider two other negativesounding results. Result 5.13
The integer 100 cannot be w ritten as the sum o f three integers, an odd num ber o f which are odd.
Proof
A ssum e, to the contrary, that 100 can be w ritten as the sum o f three integers a , b and c, an odd num ber of w hich are odd. We consider tw o cases. Case 1. Exactly one o f a , b a nd c is odd, say a. Then a = 2 x + 1, b = 2 y and c — 2 z, w here x , y , z e Z. So 100 = a + b + c = Since
a
(2a
+ 1) + 2y +
2z
= 2(x + y + z) + 1.
+ y + z e Z, the integer 100 is odd, producing a contradiction.
C ase 2. A ll o f a, b and c are odd. T hen a = 2 x + 1, b = 2y + 1 and c = 2z + 1, w here x , y , z e Z. So 100 = a + b + c = (2x + 1) + (2 y + 1) + (2z + 1) = 2(x + y + Since x + y + z + 1 e Z, the integer 100 is odd, again a contradiction.
z +
1) + 1. ■
PROOF ANALYSIS
O bserve that the proof o f Result 5.13 begins by assum ing that 100 can be w ritten as the sum of three integers, an odd num ber o f w hich are odd (as expected). However, by introducing sym bols for these integers, nam ely a , b and c, this m ade for an easier and clearer proof. ♦
Result 5.14
For every integer m such that 2  m and 4 K m , there exist no integers x and y fo r which x 2 + 3 y 2  m.
Proof
A ssum e, to the contrary, that there exist an integer m such that 2  m and 4 / m and integers a and y for w hich x 2 + 3y 2 = m . Since 2  m , it follow s that m is even. By Theorem 3.16, a 2 and 3 y 2 are o f the same parity. We consider two cases.
128
Chapter 5
Existence and Proof by Contradiction
Case 1. x ^ and 3 y 2 are even. Since 3 y 2 is even and 3 is odd, it follow s by Theorem 3.17 that y 2 is even. B ecause .r2 and y 2 are both even, w e have by T heorem 3.12 that x and y are even. Thus x = 2a and y = 2b, w here a, b e Z. Therefore, x 2 + 3 y 2 = ( 2 a )2 + 3(2 b)2 = 4 a 2 + 12 b 2 — 4(a 2 + 3b2) = m . Since a~ + 3b2 e Z, it follow s that 4 j m , producing a contradiction. Case 2. x~ and 3 y 2 are odd. Since 3 y 2 is odd and 3 is odd, it follow s by (the contrapositive form ulation of) T heorem 3.17 that y 2 is odd. By (the contrapositive form ulation of) T heorem 3.12, x and y are both odd. T h e n x = 2 a + 1 and y = 2b + 1, where a , b e Z. Thus x 2 + 3 y 2  (2a + l )2 + 3(2b + l )2 = (4 a 2 + 4 a + 1) + 3(4b 2 + 4b + 1) = 4 a 2 + 4a + 12 b2 + 12b + 4 = 4 (a 2 + a + 3b2 + 3b + 1) = m. Since a~ + a + 3b + 3b + 1 e Z, it follow s that 4  m , producing a contradiction.
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The next result concerns irrational num bers. R ecall that a real num ber is rational if it can be expressed as m j n for som e /;?, n e Z, w here n / 0. Since “irrational” m eans "not rational,” it is not surprising that pro o f by contradiction is the pro o f technique we w ill use. Result 5.15 P roof
The sum o f a rational num ber and an irrational num ber is irrational. A ssum e, to the contrary, that there exist a rational num ber x and an irrational num ber y w hose sum is a rational num ber z. Thus, x + y = z, w here x = a / b and z = c / d for som e integers a , b, c, d e Z and b , d ^ 0. This im plies that c
a
be — a d
d
b
bd
Since be — a d and bd are integers and bd / 0, it follow s that y is rational, w hich is a u contradiction. Result 5.15 concerns the irrationality o f num bers. One o f the best know n irrational num bers is \f2 . A lthough we have never verified that this num ber is irrational, we establish this fact now. Theorem to Prove
The real num ber >J2 is irrational.
PROOFSTRATEGY
In the proof o f this result, we w ill use T heorem 3.12 w hich states that an integer x is even if and only if x~ is even. A lso, in the proof, it w ill be useful to express a rational num ber m / n , w here m . n e Z and n / 0, in low est term s, w hich m eans that m and n contain no com m on divisor greater than 1. ^
Theorem 5.16 Proof
The real num ber ~J2 is irrational. A ssum e, to the contrary, that s/2 is rational. Then s/2 — a / b , w here a, b e Z and b / 0. We m ay further assum e that a / b has been expressed in (or reduced to) low est terms.
5.2
Proof by Contradiction
129
T hen 2 = a 2/ b 2, so a 2 = 2b2. Since b2 is an integer, a 2 is even. By T heorem 3.12, a is even. So a — 2c, w here c e Z. Thus, (2c)2 = 2£>2 a n d s o 4 c 2 = 2b2. T herefore, b 2 = 2 c2. Because c2 is an integer, b 2 is even, w hich im plies by T heorem 3.12 that b is even. Since a and b are even, each has 2 as a divisor, w hich is a contradiction since a / b has been reduced to low est term s. ■
The Three Prisoners Problem
We now take a brief diversion from our discussion o f p ro o f by contradiction to present a “story” problem . Three prisoners (see Figure 5.1) have been sentenced to long term s in prison, but due to overcrow ded conditions, one prisoner m ust be released. The w arden devises a schem e to determ ine w hich prisoner is to be released. He tells the prisoners that he w ill blindfold them and then paint a red dot or a blue dot on each forehead. A fter he paints the dots, he will rem ove the blindfolds and a prisoner should raise his hand if he sees a red dot on at least one o f the other two prisoners. The first prisoner to identify the color of the dot on his ow n forehead w ill be released. O f course, the prisoners agree to this. (W hat do they have to lose?) The w arden blindfolds the prisoners, as prom ised, and then paints a red dot on the foreheads o f all three prisoners. He rem oves the blindfolds and, since each prisoner sees a red dot (in fact tw o red dots), each prisoner raises his hand. Som e tim e passes w hen one of the prisoners exclaim s, “I know w hat color my dot is! It's red!” This prisoner is then released. A lthough the story o f the three prisoners is over, there is a lingering question: How did this prisoner correctly identify the color o f the dot painted on his forehead? The solution is given next but try to determ ine the answ er for yourself before reading on.
Solution o f the Three Prisoners Problem
L e t’s assum e (w ithout loss o f generality) that it’s prisoner #1 (see Figure 5.1) who determ ined that he had a red dot painted on his forehead. How did he com e to this conclusion? Perhaps you think he ju st guessed since he had nothing to lose anyway. But this is not the answ er w e w ere looking for. P risoner #1 know s that the color of his dot is either red or blue. He thinks, “A ssum e, to the contrary, that m y dot is blue. Then, o f course, #2 know s this and he know s that #3 has a red dot. (T hat’s w hy #2 raised his hand.) B ut #2 also know s that #3 raised his
#1
# 2 Figure 5.1
The three prisoners
#3
130
Chapter 5
Existence and Proof by Contradiction
hand. So if m y dot is blue, #2 know s his dot is red. Similarly, if m y dot is blue, then #3 know s his dot is red. In other w ords, if m y dot is blue, then both #2 and #3 should be able to identify the colors o f their dots quite quickly. B ut tim e has passed and they h av en 't determ ined the colors o f their dots. So m y dot c a n 't be blue." T herefore, #1 exclaim s, “I know w hat color m y dot is! I t’s red !” W hat you probably noticed is that the reasoning #1 used to conclude that his dot is red is proof by contradiction. It seem s as if there is m ore to know about prisoner #1. But th a t’s another story. ^
5.3 A R eview of Three P roof T echniques We have seen that w e ’re often in the situation w here w e w ant to prove the truth o f a statem ent Vx e S , P ( x ) Q (x ). You have now been introduced to three p ro o f tech niques: direct proof, proof by contrapositive, p ro o f by contradiction. For each o f these three techniques, you should be aw are o f how to start a p ro o f and w hat your goal should be. You should also know w hat not to do. Figure 5.2 gives several ways that we m ight start a proof. O nly som e of these can lead to a proof, however. L e t’s now com pare the three p ro o f techniques w ith two exam ples.
1. 2. 3. 4. 5. ti. V. 8. y. 10,
F i r s t S te p o f “ P r o o f ”
R e m a rk s /G o a l
A s s u m e t h a t t h e r e e x is ts x £ S s u c h t h a t P ( x ) is t r u e . A ssum e th a t th e re exists x G S such th a t P ( x ) is false. Assum e th a t th e re exists x e S such th a t Q( x) is true. A s s u m e t h a t t h e r e e x is ts x € S s u c h t h a t Q{x) is false . A ssum e th a t th e re exists x e S such th a t P{ x ) and Q( x ) are true. A s s u m e t h a t t h e r e e x is ts x € S s u c h t h a t P (x ) is t r u e a n d Q( x) is false . A ssum e th a t th e re exists x e S such th a t P ( x ) is false and Q( x) is true. A ssum e th a t there exists x £ S such th a t P ( x ) and Q( x) are false. Assum e th a t th ere exists x € S such th a t P ( x ) => Q{x) is true. A s s u m e t h a t t h e r e e x is ts x e S s u c h t h a t P ( x ) =$• Q( x) is false .
A d ir e c t p r o o f is b e in g u s e d . S h o w t h a t Q( x) is t r u e fo r t h e e le m e n t x. \ A m istake has been m ade.
Figure 5.2
A m istake has been made. A p r o o f b y c o n t r a p o s iti v e is b e in g u se d . S h o w t h a t P ( x ) is fa lse fo r t h e e le m e n t x. A m istake has been made. A p r o o f b y c o n t r a d ic t io n is b e in g u s e d . P r o d u c e a c o n t r a d ic t io n . A m istake has been m ade. A m istake has been m ade. A m istake has been m ade. A p r o o f b y c o n t r a d ic t io n is b e in g u s e d . P r o d u c e a c o n t r a d ic t io n .
How to prove (and not to prove) that Vx e S, P(x) =4> Q(x) is true
0.3
Result 5.17
Direct Proof
A Review of Three Proof Techniques
131
I f n is an even integer, then 3n + 7 is odd. L et n be an even integer. T hen n = 2x for som e integer x . Therefore, 3 n + l = 3(2x) + 7 = 6x + 7 = 2(3x + 3) + 1. Since 3x + 3 is an integer, 3n + 7 is odd.
P ro o f by Contrapositive
A ssum e that 3n + 7 is even. T hen 3n + 7 = 2y for som e integer y . H ence n = (3n + 7) + (  2 n  l ) = 2 y  2n  1 = 2{y  n  4) + 1. Since y — n — 4 is an integer, n is odd.
P ro o f by Contradiction
■
■
A ssum e, to the contrary, that there exists an even integer n such that 3n + 7 is even. Since n is even, n = 2x for som e integer x . H ence 3n + 7 = 3(2x) + 7 = 6 x + 1 = 2(3x + 3) + 1. Since 3x + 3 is an integer, 3n + 1 is odd, w hich is a contradiction.
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A lthough a direct proof o f R esult 5.17 is certainly the preferred pro o f technique in this case, it is useful to com pare all three techniques. The follow ing exam ple is m ore intricate. Result 5.18
Direct P ro o f
L et x be a nonzero real number. I f x \— < 2 , then x < 0. x A ssum e that x — < 2. Since x ^ 0, w e know that x 2 > 0 . M ultiplying both sides o f the x 1 .. , inequality x H— < 2 by x , w e obtain x I x 4— < 2x . Sim plifying this inequality, x \ x) w e have x 3 + x — 2x 2 < 0 ; so x ( x 2 — 2x + 1) = x (x — l )2 < 0 . Since (x — l )2 > O an d x (x — l )2 ^ 0, we m ust have (x — l )2 > 0. Since x (x — l )2 < 0 and (x — l )2 > 0 , it follow s that x < 0 , as desired. ■
P roof Strategy fo r P ro o f by Contrapositive
For a p ro o f by contrapositive, w e will begin by assum ing that x > 0 and attem pt to show that x H— > 2 . This inequality can be sim plified by m ultiplying through by x , obtaining x x 2 + 1 > 2x. Subtracting 2x from both sides, w e have x 2 — 2x + 1 = (x — l )2 > 0, w hich, o f course, we know to be true. A p ro o f is suggested then by reversing the order o f these steps: 1 x 4" — > 2 x x 2 + 1 > 2x x 2  2x + 1 = (x  l )2 > 0 . This m ethod is com m on w hen dealing w ith inequalities.
♦
132
C h a p te r 5
P ro o f by Contrapositive
Existence and Proof by Contradiction
A ssum e that x > 0. Since x / 0, it follow s that x > 0. Since (x — l )2 > 0, we have (x — l )2 = x 2 — 2x + 1 > 0. A dding 2x to both sides o f this inequality, w e obtain x 2 + 1 > 2x. D ividing both sides o f the inequality x 2 + 1 > 2x by the positive num ber x , we obtain x \— > 2 , as desired. x
P ro o f by Contradiction
a
A ssum e, to the contrary, that there exists a nonzero real num ber x such that x H— < 2 x and x > 0. Since x / 0, it follow s that x > 0. M ultiplying both sides o f the inequality x — < 2 by x , w e obtain x 2 + 1 < 2x. Subtracting 2x from both sides, w e have x x 2 — 2x + 1 < 0 . It then follow s that (x — l )2 < 0, w hich is a contradiction. ■
M any m athem aticians feel that if a result can be verified by a direct proof, then this is the proof technique that should be used, as it is norm ally easier to understand. This is only a general guideline, how ever; it is n o t a hard and fast rule.
5.4 E xisten ce Proofs In an existence th e o re m the existence o f an object (or objects) possessing som e specified property or properties is asserted. Typically then, an existence theorem concerning an open sentence R ( x ) over a dom ain S can be expressed as a quantified statem ent 3x e S, R ( x ) : T here exists x 6 S such that R (x ).
(5.12)
We have seen that such a statem ent (5.12) is true provided that R (x) is true for som e x e S. A proof o f an existence theorem is called an ex isten ce p ro o f. A n existence p ro o f may then consist o f displaying or constructing an exam ple o f such an object or perhaps, with the aid o f know n results, verifying that such objects m ust exist w ithout ever producing a single exam ple o f the desired type. For exam ple, there are theorem s in m athem atics that tell us that every polynom ial of odd degree w ith real coefficients has at least one real num ber as a solution, but we d o n ’t know how to find a real num ber solution for every such polynom ial. Indeed, w e quote the great m athem atician D avid H ilbert, w ho used the follow ing exam ple in his lectures to illustrate the idea o f an existence proof: There is a t least one student in this class . .. let us nam e him 'X ' .. .fo r w hom the folloM’ing statem ent is true: N o other student in the class has more hairs on his head than X . W hich student is it? T hat w e shall never know ; but o f his existence we can be absolutely certain. L e t’s now see som e exam ples of existence proofs. R e su lt to P ro v e PROOF STRATEGY
T here exists an integer w hose cube equals its square. Since this result is only asserting the existence of an integer w hose cube equals its square, w e have a p ro o f once w e can think o f an exam ple. The integer 1 has this property. ♦
5.4
Result 5.19
Proof
Existence Proofs
133
There exists an integer w hose cube equals its square. Since l 3 = l 2 = 1, the integer 1 has the desired property.
■
Suppose that w e did n ’t notice that the integer 1 satisfied the required condition in the preceding theorem . T hen an alternate p ro o f m ay go som ething like this: L et x e Z s u c h th a tx 3 = x 2.T h e n x 3 — x 2 = 0 o r x 2(x — 1) = 0. Thus, there are only tw o possible integers w ith this property, nam ely 1 and 0 , and in fact both integers have the desired property. A com m on error in elem entary algebra is to w rite (a + b)2 = a 2 + b2. C an this ever be true? Result 5.20 P roof
There exist real num bers a and b such that (a + b)2 = a 2 + b 2. L et a , f t e R such that (a + b)2 = a 2 + b 2. T hen a 2 + l a b + b 2 = a 2 + b2, so la b = 0. Since a = 1, b = 0 is a solution to this equation, we have (a + b)2 = (1 + 0)2 = l 2 = l 2 + 02 = a 2 + b2.
m
T he pro o f presented o f R esult 5.20 is longer than necessary. We could have w ritten the follow ing proof:
Proof
L et a = 1 and b = 0. Then (a + b)2 = (1 + 0)2 = l 2 = l 2 + 02 = a 2 + b2.
•
In the first proof, we actually presented an argum ent for how we thought o f a = 1 and b = 0. In a proof, w e are not required to explain w here w e got the idea for the proof, although it m ay very w ell be interesting to know this. If we feel that such inform ation m ight be interesting or valuable, it m ay be w orthw hile to include this in a discussion preceding or follow ing the proof. The first p ro o f we gave o f Result 5.20 actually inform s us o f all real num bers a and b for w hich (a + b )2 = a 2 + b2, nam ely (a + b)2= a 2 + b 2 if and only if at least one o f a and b is 0. T his is m ore than w hat was requested of us but, nevertheless, it seems interesting. We saw in Section 5.2 that \ f l is irrational. Since \ f l = 2 1/2, it follow s that there exist rational num bers a and b such that a h is irrational, nam ely a — 1 and b = l / l have this property. L et’s reverse this question. T hat is, do there exist irrational num bers a and b such that a h is rational? A lthough there are m any irrational num bers (in fact, an infinite num ber), w e have only verified that  J l is irrational. (O n the other hand, w e know from the exercises for this section that r + \ f l is irrational for every rational num ber r and that both r  J l and r /  J l are irrational for every nonzero rational num ber r .) R esu lt to P rove PROOF STRATEGY
There exist irrational num bers a and b such that a b is rational. As w e m entioned, there are only certain num bers that w e know to be irrational, the \/2 sim plest being \ f l . This m ight suggest considering the (real) num ber s / l . If this
134
Chapter 5
Existence and Proof by Contradiction
fsl2 . num ber is rational, then this answ ers our question. B ut perhaps V 2 is irrational. Then w hat do we do? This discussion suggests tw o cases. ♦ Result 5.21 Proof
There exist irrational num bers a and b such that a h is rational. /—\/2 C onsider the num ber v 2 . O f course, this num ber is either rational or irrational. We consider these possibilities separately. Case 1.
^j2 ~ is rational. T hen we can take a = b = \ / 2 and w e have the desired result.
C ase 2. V 2 V is irrational. In this case, consider the num ber obtained by raising the r —Jl (irrational) num ber s/2 to the (irrational) pow er V 2 ; that is, consider a b, w here a = \/2 \ f l ~ and b = \/2 . O bserve that a
b
V2 = (V 2 ^ )
= V 2 ^
= V22 = 2,
w hich is rational. The p ro o f o f T heorem 5.21 m ay seem unsatisfactory to you since we still d o n ’t know tw o specific irrational num bers a and b such that a h is rational. We only know that two such num bers exist. We actually do know a bit m ore, nam ely either (1) V 2 or (2) 7 2
■Ji
(
J i\ ^
’ is irrational and I V 2 ~ I
is rational
is rational. (A ctually it has been proved that
•Ji
s/2 is an irrational number. H ence there are also irrational num bers o f the form o f a h, w here a and b are both irrational.) In the next result, w e w ant to show that the equation x 5 + 2 x — 5 = 0 has a real num ber solution betw een x = 1 and x = 2. It is not easy to find a num ber that satisfies this equation. Instead, w e use a w ellknow n theorem from calculus to show that such a solution exists. You m ay not rem em ber all the term s used in the follow ing theorem but this is not crucial.
Intermediate Value Theorem o f Calculus
If / is a function that is continuous on the closed interval [a , b\ and k is a num ber betw een f ( a ) and f { b ) , then there exists a num ber c e (a, b) such that / ( c ) = k. We now give an exam ple to show how this theorem can be used.
Result 5.22
Proof
The equation x 5 + 2x — 5 = 0 has a real num ber solution betw een x = 1 a nd x = 2. Let / ( x ) = x 5 + 2x — 5. Since / is a polynom ial function, it is continuous on the set all real num bers and so / is continuous on the interval f 1, 2]. N o w / ( l ) = —2 a n d / ( 2 ) 31. Since 0 is betw een / ( l ) and / ( 2 ) , it follow s by the Interm ediate Value T heorem Calculus that there is a num ber c betw een 1 and 2 such that / ( c ) = c5 + 2c — 5 = H ence c is a solution.
of = of 0. ■
A s we ju st saw, the equation x 5 + 2x — 5 = 0 has a real num ber solution betw een x = 1 and x = 2. A ctually, the equation x 5 + 2x — 5 = 0 has exactly one real num ber
5.4
Existence Proofs
135
solution betw een x = 1 and x = 2. This brings up the topic o f uniqueness. An elem ent belonging to som e prescribed set A and possessing a certain property P is unique if it is the only elem ent o f A having property P . Typically, to prove that only one elem ent of A has property P . we proceed in one o f two ways: (1) We assum e that a and b are elem ents o f /I possessing property P and show that a = b. (2) We assum e that a and b are distinct elem ents o f A possessing property P and show that a = b. A lthough (1) results in a direct proo f and (2) results in a p ro o f by contradiction, it is often the case that either proof technique can be used. As an illustration, we return to Result 5.22 and show, in fact, that the equation x 5 + 2x — 5 = 0 has a unique real num ber solution betw een x = 1 and x — 2. Result 5.23
The equation x 5 + 2x — 5 = 0 has a unique real num ber solution betw een x = 1 and x = 2.
Proof
A ssum e, to the contrary, that the equation x 5 + 2x — 5 = 0 has tw o distinct real num ber solutions a and b betw een x = 1 and x — 2. We m ay assum e that a < b. Since 1 < a < b < 2, it follow s that a 5 + 2a — 5 < b5 + 2b — 5. O n the other hand, a 5 + 2a — 5 = 0 and b 5 + 2b — 5 = 0. Thus 0
= a 5 + 2a  5 < b5 + 2b  5 = 0,
w hich produces a contradiction.
■
Actually, we could have om itted Result 5.22 altogether and replaced it by R esult 5.23 only (renum bering this result by Result 5.22), including the proofs o f both Results 5.22 and 5.23. We now present another result concerning uniqueness. R esu lt to P rove
For an irrational num ber r, let S = {sr + t : s, t e Q}. For every x e S, there exist unique rational num bers a and b such that x = a r + b.
PROOFSTRATEGY
To verify that a and b are unique, we assum e that x can be expressed in two w ays, say as ar + b and cr + d , w here a , b , c , d e Q , and then show that a — c and b = d. H ence a r + b — cr + d. If a ^ c, then we can show that r is a rational num ber, producing a contradiction. T hus a = c. Subtracting ar from both sides o f ar + b = cr + d, we obtain b = d as well. ♦ We now give a com plete proof.
Result 5.24
For an irrational num ber r , let S = {sr + 1 : s, t e Q}.
136
Chapter 5
Existence and Proof by Contradiction
For every x 6 S, there exist unique rational num bers a a n d b such that x = a r + b. P roof
L et x € S and suppose that x = a r + b and x = cr + d , w here a , b, c , d e Q . Then a r + b = cr + d. If a ^ c, then (a  c)r = d — b and so cl — b r =  . a —c Since is a rational num ber, this is im possible. So a = c. Subtracting a r = cr from both sides o f a r + b = cr + d , we obtain b = d . m
Example 5.25
S o lu tio n
Show that the equation 6a"1 + a 2 — 2 a = 0 has a root in the interval [—1, 1], (b) D oes this equation have a unique root in the interval [—1, 1]? W B inspection, w e can see that (b) O bserve that 6x3 + a2 
a
= 0 is a root o f the equation.
2 a = .v ( 6 a 2 + a 
2 ) = a (3 a + 2 ) (2 a 
1 ).
Thus a • — 2 / 3 and x — 1 / 2 are also roots o f the equation 6 a 3 + a 2 — 2 a = 0 and so this equation does not have a unique root in the interval [  1, 1]. $
5.5 D isproving E xisten ce Statem ents L et R i x ) be an open sentence w here the dom ain o f x is S. We have already seen that to disprove a quantified statem ent o f the type V a e S, R(x% it suffices to produce a counterexam ple (that is, an elem ent x in S' for w hich R ( x ) is false). However, disproving a quantified statem ent o f the type 3 a e S , R( x ) requires a totally different approach. Since ~
(3 a e
S, R( x ) )
=
Vx
e
S,
~
R (x ),
it follow s that the statem ent 3 a e 5 , R (x ) is false if R ( x ) is false for every x e S. L et's look at som e exam ples o f disproving existence statem ents. Example 5.26 S o lution
D isprove the statem ent: There exists an odd integer n such that n 2 + 2n + 3 is odd. We show that if n is an odd integer, then n 2 + 2n + 3 is even. Let n be an odd integer. T hen n = 2 k + 1 for som e integer k. Thus n 2 + 2n + 3 = (2 k + I )2 + 2(2 k + 1) + 3 = A k2 + Ak + 1 + 4 k + 2 + 3 = Ak2 + M + 6 = 2(2 k 2 + Ak + 3). Since 2k 2 + Ak + 3 is an integer, n 2 + 2n + 3 is even.
Example 5.27 S o lu tio n
#
D isprove the statem ent. There is a real num ber x such that x 6 + 2x4 + x 2 + 2 = 0. L et x 6 R. Since a 6, x 4 and x 2 are all even pow ers o f the real num ber a , it follow s that x 6 > 0, x 4 > 0 and x 2 > 0. Therefore, x 6 + 2x4 + x 2 + 2 > 0 + 0 + 0 + 2 = 2 and so
Exercises for Chapter 5
137
x 6 + 2.\'4 + a' 2 + 2 7^ 0. H ence the equation ,v6 + 2 x 4 + x 2 + 2 = 0 has no real num ber solution. ♦ Example 5.28 S o lu tio n
D isprove the statem ent: There exists an integer n such that n 3 — n + 1 is even. L et n e Z. We consider tw o cases. Case 1. n is even. T hen n = 2a, w here a e Z. So n 3 — n + 1 = (2 a )3 — (2a) + 1 = 8c/3 — 2a + 1 = 2 (4 a 3 — a) + 1. Since 4 a 3—a is an integer, n 3 — n + 1 is odd and so it is not even. C ase 2. n is odd. T hen n = 2b + 1, w here b e Z. H ence n 3  n + l = ( 2 b + l )3  (2b + 1) + 1 = 8fr3 + \2 b 2 + 6b + 1  2b  1 + 1 = 8 b 3 + 12 b2 + 4b + 1 = 2(4 b3 + 6 b 2 + 2 b ) + l . Since 4 b3 + 6b2 + 2b is an integer, n 3 — n + 1 is odd and so it is not even.
$
If w e had replaced E xam ple 5.26 by For every odd integer n, n 2 + 2n + 3 is even. replaced E xam ple 5.27 by For every real num ber x , x 6 + 2 x 4 + x 2 + 2 ^ 0. and replaced E xam ple 5.28 by For every integer n, n 3 — n + 1 is odd. then we would have a true statem ent in each case and the solutions o f Exam ples 5 .2 6 5 .2 8 w ould becom e proofs.
E X E R C ISE S FO R C H A PT E R 5 Section 5.1: Counterexamples 5.1. Disprove the statement: If a and b are any two real numbers, then log (ab) = log(a) + log(fo). 5.2. Disprove the statement: If n e {0, 1, 2, 3, 4}, then 2" + 3" + n(n — 1)(n — 2) is prime. 5.3. Disprove the statement: If n e {1, 2, 3, 4, 5}, then 3  (2n2 + 1). 5.4. Disprove the statement: Let n e N. If
is odd, then fe±l ?2(”±2> js
5.5. Disprove the statement: For every two positive integers a and b, (a + b)3 — a3 + 2a 2b + la b + la b 2 + b3. 5.6. Let a, b e Z. Disprove the statement: If ab and (a + b)2 are of opposite parity, then a 2b2 and a + ab + b are of opposite parity. 5.7. For positive real numbers a and b, it can be shown that (a + b) + £) > 4. If a = b, then this inequality is an equality. Consider the following statement: If a and b are positive real numbers such that (a + b) ( ^ +  ) = 4 , then a = b. Is there a counterexample to this statement?
138
C h a p te r 5
Existence and Proof by Contradiction
5.8. In Exercise 5.7, it is stated that (a + b) ( i +
> 4 for every two positive real numbers a and b. Does it
therefore follow that (c2 + d 2) [4j + J ,) > 42 for every two positive real numbers c and i i 5.9. Disprove the statement: For every positive integer x and every integer n > 2, the equation x" + (x + 1)" = (x + 2)" has no solution.
Section 5.2: Proof by Contradiction 5.10. Prove that there is no largest negative rational number. 5.11. Prove that there is no smallest positive irrational number. 5.12. Prove that 200 cannot be written as the sum of an odd integer and two even integers. 5.13. Use proof by contradiction to prove that if a and b are odd integers, then 4 / (a2 + ft2). 5.14. Prove that if a > 2 and b are integers, then a / ft or a £ ( b + 1). 5.15. Prove that 1000 cannot be written as the sum of three integers, an even number of which are even. 5.16. Prove that the product of an irrational number and a nonzero rational number is irrational. 5.17. Prove that when an irrational number is divided by a (nonzero) rational number, the resulting number is irrational. 5.18. Let a be an irrational number and r a nonzero rational number. Prove that if s is a real number, then either ar + j or ar — s is irrational. 5.19. Prove that s/3 is irrational. [Hint: First prove for an integer a that 3  a2 if and only if 3  a. Recall that every integer can be written as 3q, 3q  1 or 3</  2 for some integer q. Also, see Exercise 3 of Chapter 4.] 5.20. Prove that s/2 + s/3 is an irrational number. 5.21. (a) Prove that V 6 is an irrational number. (b) Prove that there are infinitely many positive integers n such that J n is irrational. 5.22. Let S = [p + q s /2 : p , q e Q} and T = [r + s V 3 : r, s e Q}. Prove that S n T = Q. 5.23. Prove that there is no integer a such that a = 5 (mod 14) and <7 = 3 (mod 21). 5.24. Prove that there exists no positive integer x such that 2x < x 2 < 3x. 5.25. Prove that there do not exist three distinct positive integers a , b and c such that each integer divides the difference of the other two. 5.26. Prove that the sum of the squares of two odd integers cannot be the square of an integer. 5.27. Prove that if x and y are positive real numbers, then J x + y /
+ Jy.
5.28. Prove that there do not exist positive integers m and n such that m 2 — n 2 = 1. 5.29. Let m be a positive integer of the form m = 2s, where s is an odd integer. Prove that there do not exist positive integers x and y such that x 2 — y 2 = m. 5.30. Prove that there do not exist three distinct real numbers a, b and c such that all of the numbers a + b + c, ab, ac, be, abc are equal. 5.31. Use a proof by contradiction to prove the following. Let m e Z. If 3 / ( m 2  1), then $ [ m . (A proof by contrapositive of this result is given in Result 4.6.) 5.32. 5.33.
Prove that there exist no positive integers m and n for which m 2 + m + 1 = n 2. (a) Prove that there is no rational number solution of the equation x 2 — 3x + 1 (b) The problem in (a) should suggest a more general problem. State and outline a proof of this.
=0.
Exercises for Chapter 5
139
Section 5.3: A Review of Three Proof Techniques 5.34. Prove that if n is an odd integer, then In — 5 is even by (a) a direct proof, (b) a proof by contrapositive and (c) a proof by contradiction. 5.35. Let x be a positive real number. Prove that if x  f > 1, then x > 2 by (a) a direct proof, (b) a proof by contrapositive and (c) a proof by contradiction. 5.36. Let a, b € R. Prove that if ab / 0, then a ^ 0 by using as many of the three proof techniques as possible. 5.37. Let x, y e R + . Prove that if x < y, then x 2 < y2 by (a) a direct proof, (b) a proof by contrapositive and (c) a proof by contradiction. 5.38. Prove the following statement using more than one method of proof. Let a, b e Z. If a is odd and a + b is even, then b is odd and ab is odd. 5.39. Prove the following statement using more than one method of proof. For every three integers a, b and c, exactly two of the integers ab, ac and bc cannot be odd.
Section 5.4: Existence Proofs 5.40. Show that there exist a rational number a and an irrational number b such
that ah
is
rational.
; .41. Show that there exist a rational number a and an irrational number b such
that ab
is
irrational.
5.42. Show that there exist two distinct irrational numbers a and b such that a h is rational. 5.43. Show that there exist no nonzero real numbers a and b such that V fl2 + b2 = 4 a 3 + b3'. ' .44. Prove that there exists a unique real number solution to the equation x 3 + x 2 — 1 = 0 between x = 2 /3 and x = 1. 5.45. Let R( x ) be an open sentence over a domain S. Suppose that Vx e S, R( x) is a false statement and that the set T of counterexamples is a proper subset of S. Show that there exists a subset W of S such that Vx € W , R{x) is true. ; 46. (a) Prove that there exist four distinct positive integers such that each integer divides the sum of the remaining integers. (b) The problem in (a) should suggest another problem to you. State and solve such a problem. 5 47. Let S be a set of three integers. For a nonempty subset A of S, let a A be the sum of the elements in A. Prove that there exist two distinct nonempty subsets B and C of S such that a B = a t (mod 6). 5.48. Prove that the equation cos2(x) — 4x + it = 0 has a real number solution in the interval [0, 4], (You may assume that cos2(x) is continuous on [0 ,4].)
'■ection 5.5: Disproving Existence Statements 5 49. Disprove the statement: There exist odd integers a and b such that 4  (3a2 + l b 2). 5.50. Disprove the statement: There is a real number x such that x 6 F x 4 + 1 = 2x2. 5.51. Disprove the statement: There is an integer n such that n4 + n3 + n2 + n is odd. 5.52. The integers 1, 2, 3 have the property that each divides the sum of the other two. Indeed, for each positive integer a, the integers a, 2a. 3a have the property that each divides the sum of the other two. Show that the following statement is false. There exists an example of three distinct positive integers different from a, 2a, 3a for some a e N having the property that each divides the sum of the other two.
140
Chapter 5
Existence and Proof by Contradiction
ADDITIONAL EXERCISES FOR CHAPTER 5 5.53. Show that the following statement is false. If A and B are two sets of positive integers with A = fi  = 3 such that whenever an integer s is the sum of the elements of some subset of A, then s is also the sum of the elements of some subset of B , then A = B. 5.54. (a) Prove that if a > 2 and n > 1 are integers such that a 2 + 1 = 2", then a is odd. (b) Prove that there are no integers a > 2 and n > 1 such that a 2 + 1 = 2". 5.55. Prove that there do not exist positive integers a and n such that a 2 + 3 = 3". 5.56. 5.57.
Let X, y < R + . Use a proof by contradiction to prove that if x < y , then «Jx < J y . Theking's daughter had three suitors and couldn’t decide which one to marry. So the king said, “I have three gold crowns and two silver ones. I will put either a gold or silver crown on each of your heads. The suitor who can tell me which crown he has will marry my daughter." The first suitor looked around and said he could not tell. The second did the same. The third suitor said: “I have a gold crown." He is correct, but the daughter was puzzled: This suitor was blind. How did he know? (Reference: Ask Marilyn, Parade M agazine, July 6, 2003).
5.58. Prove that if a , b , c , d are four real numbers, then at most four of the numbers ab, ac, a d, be, bd, cd are negative. 5.59. Evaluate the proposed proof of the following result. Result
The number 25 cannot be written as the sum of three integers, an even number of which are odd.
P ro o f Assume, to the contrary, that 25 can be written as the sum of three integers, an even number of which are odd. Then 25 = x + y + z, where i , y , z e Z . W e consider two cases. Case 1. x
and y are odd. Then x = 2a + 1, y = 2b + 1 and 2 = 2c, where a, b, c e Z.Therefore. 25 = x + y + u = (2a + 1) + (2b + 1) + 2c — 2a + 2b + 2c f 2 — 2(a + b + c + 1).
Since a + b + c + 1 is an integer. 25 is even, a contradiction. Case 2. x , y and z are even. Then x = 2a, y = 2b and z = 2c, where a , b , c e Z.Hence 25 = x + y + z = 2 a + 2b + 2c = 2 (a + b + c). Since a + b + c is an integer, 25 is even, again a contradiction.
m
5.60. (a) Let n be a positive integer. Show that every integer m with 1< m < 2n can be expressed as2ek, where £ is a nonnegative integer and k is an odd integer with 1 < k < 2n. S >= n + 1 that there exist (b) Prove for every positive integer n and every subset S of {1, 2 , _2n) with integers a , b e S such that a \ b. 5.61. Prove that the sum of the irrational numbers s/2, V3 and >/5 is also irrational. 5.62. Let a j, a2, . . ., ar be odd integers where fl, > 1 for j = 1 , 2 ,. .. ,/ * . Prove that if n = a \a 2 ■■ ar + 2, then fl, X n for each integer / (1 < i < r). 5.63. Below is given a proof of a result. W hat result is proved? P ro o f Let a, b, c e Z such that a2 + b2 = c2. Assume, to the contrary, that a. b and c are all odd. Then a = 2r + 1, b = 2$ + 1 and c = 2t + 1, where r, s , f € Z. Thus, a 2 + b2 = (4 r 2 + 4 r + 1) + (4s2 + 4s + 1) = 2(2r2 +
2r + 2s2 + 2 s + 1).
Additional Exercises for Chapter 5
141
Since 2r 2 + 2r + 2s2 + 2s + 1 is an integer, it follows that a 2 + b2 is even. On the other hand, c2 = (2t + l )2 = At2 + At + 1 = 2(2f2 + 2t) + 1. Since 212 + 21 is an integer, it follows that c2 is odd. Therefore, a 2 + b2 is even and c2 is odd, contradicting the fact that a 2 + b2 = c2. ■ .64. Evaluate the proposed proof of the following result. R esult
If x is an irrational number and y is a rational number, then z = x — y is irrational.
P ro o f Assume, to the contrary, that z = x — y is rational. Then z = a/ b, where a , b 6 Z and b ^ 0. Since \ f l is irrational, we let x — V z. Since y is rational, y = c/ d, where c , d e Z and d ^ 0. Therefore, r
c
a
a d + be
Since ad + bc and bd are integers, where bd ^ 0, it follows that V 2 is rational, producing a contradiction. ■ .65. Prove that there exist four distinct real numbers a, b , c, d such that exactly four of the numbers ab, ac, ad, bc, bd, cd are irrational. .66. Below is given a proof of a result. W hat result is proved? P ro o f Let a = 2 (m od 4) and b = 1 (m od 4) and assume, to the contrary, that 4  (a2 + 2b). Since a = 2 (m od A) and b = 1 (m od 4), it follows that a = Ar + 2 and b = As + 1, where r, s e Z. Therefore, a 2 + 26 = (4/ + 2)2 + 2(45 + 1) = (16r2 + 16r + 4) + (8s + 2) = 16; + 16/*
8s + 6.
Since 4  (a 2 + 2b), we have a 2 + 2b = 4f, where ( e Z. So 16r2 + 16r + 8s + 6 = At and 6 = 41 — 16r2 — 16r — 8s = A(t — 4 r 2 —Ar — 2s). Since t — 4 r 2 — Ar — 2s is an integer, 4  6, which is a contradiction.
■
6 M athem atical Induction
e have seen three pro o f techniques w hich could be used to prove that a quantified statem ent Vx e S. P ( x ) is true: direct proof, p ro o f by contrapositive, p ro o f by contradiction. For certain sets S, however, there is another possible m ethod o f proof: m athem atical induction.
W
6.1 The P rin cip le o f M athem atical Induction Let A be a nonem pty set o f real num bers. A num ber m e A is called a le a st elem e n t (or a m in im u m or sm allest elem en t) o f A if x > m for every x e A . Som e nonem pty sets of real num bers have a least elem ent; others do not. The set N has a sm allest elem ent, nam ely 1, w hile Z has no least elem ent. The closed interval [2, 5] has the m inim um elem ent 2, but the open interval (2, 5) has no m inim um elem ent. The set
also has no least elem ent. If a nonem pty set A o f real num bers has a least elem ent, then this elem ent is necessarily unique. We w ill verify this fact. Recall that w hen attem pting to prove that an elem ent possessing a certain property is unique, it is custom ary to assum e that there are two elem ents w ith this property. We then show that these elem ents are equal, im plying that there is exactly one such elem ent. Theorem 6.1
I f a set A o f real num bers has a least elem ent, then A has a unique least element.
P roof
L et m x a n d ÂŤ?2 be least elem ents o f A. Since n il is a least elem ent, ni2 > /wi. A lso, since m 2 is a least elem ent, m \ > m 2. Therefore, m \ â€” m 2. â–
The proof we gave o f T heorem 6.1 is a direct proof. Suppose that we had replaced the first sentence o f this proof by
142
Assum e, to the contrary, that A contains distinct least elem ents m \ and m 2.
6.1
The Principle of M athematical Induction
143
If the rem ainder o f the proof o f T heorem 6.1 were the sam e except for adding a concluding sentence that we have a contradiction, then this too w ould be a p ro o f o f T heorem 6.1. T hat is, w ith a sm all change, the pro o f technique used to verify T heorem 6 .1 can be transform ed from a direct proof to a p ro o f by contradiction. There is a property possessed by som e sets o f real num bers that w ill be o f great interest to us here. A nonem pty set S o f real num bers is said to be w e llo rd e re d if every nonem pty subset o f S has a least elem ent. Let S = {â€”7, â€”1, 2}. The nonem pty subsets o f Sâ€™ are {  7 ,  1 , 2}, {  7 ,  1 } , {  7 , 2}, {  1 ,2 } , { 7 } , { 1 } and {2}. Since each o f these subsets has a least elem ent, S is w ellordered. Indeed, it should be clear that every nonem pty finite set of real num bers is w ellordered. (See Exercise 6.20 in Exercises for Section 6.2.) T he open interval (0, 1) is not w ellordered, since, for exam ple, (0 ,1 ) itself has no least elem ent. The closed interval [0 ,1 ] has the least elem ent 0 ; however, [0 . 1] is not w ellordered since the open interval (0 , 1) is a (nonem pty) subset o f [0, 1] w ithout a least elem ent. B ecause none o f the sets Z , Q , and R have a least elem ent, none o f these sets are w ellordered. H ence, having a least elem ent is a necessary condition for a nonem pty set to be w ellordered, but it is not sufficient. A lthough it m ay appear evident that the set N o f positive integers is w ellordered, this statem ent cannot be proved from the properties o f positive integers that we have used and derived thus far. Consequently, this statem ent is accepted as an axiom , w hich we state below.
The W e llO rd e rin g P rin c ip le
The set N o f positive integers is w ellordered.
A consequence o f the W ellO rdering Principle is another principle, w hich serves as the foundation o f another im portant p ro o f technique. Theorem 6.2
(T h e P rin c ip le o f M a th e m a tic a l In d u c tio n ) For each p o sitive integer n, let P ( n ) be a statem ent. I f (1) P (1) is true and (2) the implication I f P(k) , then P (k + 1). is true fo r every p ositive integer k, then P ( n ) is true fo r every p ositive integer n.
Proof
A ssum e, to the contrary, that the theorem is false. T hen conditions (1) and (2) are satisfied, but there exist some positive integers n for w hich P{n) is a false statem ent. Let S = [n e N ; P i n ) is false}.
144
Chapter 6
M athematical Induction
Since 5 is a nonem pty subset o f N, it follow s by the W ellO rdering Principle that S con tains a least elem ent s. Since P ( 1) is true, 1 g S. Thus s > 2 and s  1 £ N. Therefore, 5 — 1 ^ S and so P( s — 1) is a true statem ent. By condition (2), P( s ) is also true and so s S. This, however, contradicts our assum ption that s e S. m The Principle o f M athem atical Induction is stated m ore sym bolically below. T h e P rin c ip le of M a th e m a tic a l In d u c tio n
For each p ositive integer n, let P( n ) be a statem ent. I f (1) P (1) is true and (2) V k e N, P ( k ) =» P ( k + 1) is true, then Vn £ N, P ( n ) is true.
As a consequence o f the Principle of M athem atical Induction, the quantified state m ent V« £ N , P( n ) can be proved to be true if (1) we can show that the statem ent P( \ ) is true and (2) we can establish the truth o f the im plication If P ( k ), then P ( k + 1). for every positive integer k. A proof using the Principle o f M athem atical Induction is called an induction p roof or a p roof by induction. The verification o f the truth o f P ( \ ) in an induction pro o f is called the base step, basis step or the anchor o f the induction. In the im plication If P( k) , then P ( k + 1 ) . for an arbitrary positive integer k, the statem ent P ( k ) is called the inductive (or in du c tion) hypothesis. O ften w e use a direct p ro o f to verify VA e N, P( k ) => P ( k + I).
(6.1)
although any proof technique is acceptable. T hat is, w e typically assum e that the inductive hypothesis P( k ) is true for an arbitrary positive integer k and attem pt to show that P ( k + 1) is tru e. Establishing the truth o f (6 .1) is called the inductive step in the induction proof. We illustrate this proof technique by show ing that the sum of the first n positive integers is given by n( n + l )/2 for every positive integer n, that is, . n(n + 1) l + 2 + 3 +   + « =  .
2
Result 6.3
Let Pi n) : 1 + 2 + 3 + . . . + » =
+ 1}
2
where n £ N. Then P ( n ) is true fo r every p o sitive integer n.
6.1 Proof
The Principle of M athematical Induction
145
We employ induction. Since 1 = (1 • 2)/2, the statement P ( l) is true. Assume that P(k) is true for an arbitrary positive integer k, that is, assume that I + 2 + 3 + . . . + t = *<*+i>.
2
We show that P(k + 1) is true, that is, we show that 1 + 2 + 3 + . . . + (t + 1 ) = £ ± M ± y .
Thus l + 2 + 3 +  " + ( £   l ) = (l + 2    3 4     + fc) + (fc + l) * (* + !> « i + l ) + 2 ( t + 1) = 2 + < * + ! > =  j 
(k + m
+ 2)
2
as desired. By the Principle of Mathematical Induction, P(n) is true for every positive integer n. m Typically, a statement to be proved by induction is not presented in terms of P(n) or some other symbols. In order to illustrate this, we give an alternative statement and proof of Result 6.3, as it is to be understood what P(n) would represent. Result 6.4
For every positive integer n, + !). l1+, 2o +, a3 +,   +, n — «( «2
Proof
We employ induction. Since 1 = (1 ■2)/2, the statement is true for n — 1. Assume that 1 + 2 + 3 + ...+ * = w ± i> ,
2
where k is a positive integer. We show that l + 2 + 3 + . . . + ( t + l ) = <t + 1 f
+ 2 >.
Thus l + 2 + 3 +   + ( £ + l ) = ( l + 2 + 3 +  " + & ) + ( f c + l )
k(k + 1) = — 5—
(k + m
k(k + 1) + 2{k + 1) + ( * + !> =  5
+ 2)
2 By the Principle of Mathematical Induction, 1 _l o i , n(n + l) 1 + 2 + 3Hb n =  
2
for every positive integer n .
m
146
C h a p te r 6
PROOF ANALYSIS
Mathematical Induction
The proof o f R esult 6.3 (or o f R esult 6.4) began by stating that induction was being used. This alerts the reader of w hat to expect in the proof. A lso, in the p ro o f o f the inductive step, it is assum ed that l + 2 + 3 +   + i =
k ( k + 1) 9....
for a positive integer k , that is, for an arbitrary positive integer k. We do n ot assum e that k (k + 1) 1 + 2 + 3 + • • • + A; = for every positive integer k as this w ould be assum ing w hat we are attem pting to prove in R esult 6.3 (and in Result 6.4). ♦
Carl Friedrich Gauss (17771855) is considered to be one o f the m ost brilliant m athem aticians o f all time. The story goes that w hen he was very young (in grade school in G erm any) his teacher gave him and his classm ates the supposedly unpleasant task of adding the integers from 1 to 100. He obtained the correct result o f 5050 quickly. It is believed that he considered both the sum 1 + 2 + • ■■+ 100 and its reverse sum 100 + 99 + ■ ■■+ 1 and added these to obtain the sum 101 + 101 + ■>• + 101,w hich has 100 term s and so equals 10,100. Since this is twice the required sum, 1 + 2 + • • ■+ 100 = 10100/2 = 5050. This, o f course, can be quite easily generalized to find a form ula for 1 + 2 + 3 + ••• + «, w here n e N. Let S = 1 + 2 + 3 + •• ■+ n .
(6.2)
If we reverse the order o f the term s on the right side o f (6.2), then we obtain S = n + (« — 1) + (n  2)HI+ 1.
(6.3)
A dding (6.2) and (6.3), w e have 2 S = (n + 1) + in + 1) + (« + 1) + • • • + (n + 1).
(6.4)
Since there are n term s on the right side o f (6.4), w e conclude that I S — n(n + 1) or S — n{n + l) /2 . Hence 11 +, o2 +, a3 Hih, n — n(n . + You m ight think that the pro o f o f Result 6.3 (and R esult 6.4) that we gave by m athem atical induction is longer (and m ore com plicated) than the one w e ju st gave and this m ay very w ell be true. But, in general, m athem atical induction is a technique that can be used to prove a w ide range of statem ents. In this chapter, we w ill see a variety o f statem ents w here m athem atical induction is a natural technique used in verifying their truth. We begin w ith an exam ple that leads to a problem involving m athem atical induction. Suppose that an n x n square S is com posed o f n 1 l x l squares. For all integers k w ith 1 < k < n, how m any different k x k squares does S contain? (See Figure 6.1 for the case w here h = 3.) For n — 3, the square S contains the 3 x 3 square S itself, four 2 x 2 squares and nine l x l squares (see Figure 6.1). T herefore, the num ber of different squares that I contains is 1 + 4 + 9 = I 2 + 22 + 32 = 14.
6.1
The Principle of M athematical Induction
147
u uu □ rn □□□ n
S :
1
Figure 6.1
The squares in a 3 x 3 square
V
In order to determ ine the num ber o f different k x k squares in an n x n square S, w e place S in the first quadrant o f the coordinate plane so that the low er left com er o f S is at the origin (0, 0). (See Figure 6.2.) Then the upper right co m er o f S is at the point («, ri). Consequently, the low er left com er o f a k x k square S*, w here 1 < k < n , is at som e point (x , y), w hile the upper right corner o f S* is at (x + k , y + k). N ecessarily, x and y are nonnegative integers w ith x + k < n and y + k < n (again, see Figure 6.2). Since 0 < x < n — k and 0 < y < n  k , the num ber o f choices for each o f x and y is n — k + 1 and so the num ber o f possibilities for (x, y) is (n — k + l )2. Because k is any o f the integers 1 ,2 .........n, the total num ber of different squares in S is  k + l )2 = n 2 + {n  l )2 + ■• • + 22 + l 2 k=\ = l 2 + 2 2 +    + n 2 = Y ^ k 2. k= 1 Is there a com pact form ula for the expression Y ^ k 2 = l 2 + 22 + • • • + /i2? k= 1
148
Chapter 6
M athematical Induction
For the problem we are describing, it w ould be very helpful to know the answ er to this question. Since w e brought up this question, you m ight have already guessed that the answ er is yes. A form ula is given next, along w ith a pro o f by induction. Result 6.5
For every positive integer n , 12 + 22 _(_ . . . + „ 2 _ «(» + 1X2.n + 1)
6 Proof
We proceed by induction. Since l 2 = (1 • 2 ■3)/6 = 1, the statem ent is true w hen n = 1. A ssum e that I 2 __ 22 h___ + *2 = — +
+ 12
6
for an arbitrary positive integer k. We show that l 2 + 22 + ••• + ( * + l ) 2 = £ ± . 1)(* + 2 )(2k + 3 ) .
6
O bserve that l 2 + 22 + • ■• + (* + l )2 = [ l 2 + 2r + ■• ■+ k 2] + (k + l )2 = w + i x 2 * + i) + ( t + 1 ) 2
6 k(k + 1)(2k + 1) 6(k + l) 2
6
“
+
6
(fc + l)[Jt(2it + 1) + 6(k + 1)] 6 _ (k + 1)(2k 2 + lk_ + 6)
6 (A + IX* + 2X2* + 3) 6 as desired. B y the Principle o f M athem atical Induction, l^ + 2^ + . . . + „ ^ ,l(” + 1)(2" + 1)
6 for every positive integer n.
a
Strictly speaking, the last sentence in the pro o f o f Result 6.5 is typical o f the last sentence o f every proof using m athem atical induction, for the idea is to show that the hypothesis o f the Principle o f M athem atical Induction is satisfied and so the conclusion follow s. Some therefore om it this final sentence since it is understood that once properties (1) and (2) o f T heorem 6.2 are satisfied, w e have a proof. For em phasis, w e w ill continue to include this concluding sentence, however. There is another question that m ight have occurred to you. We explained why 1 + 2 + ••• + « equals n(n + l ) / 2 , but how did we know that l 2 + 22 + ■■• + « 2 equals n(n + 1)(2« + l ) / 6 ? We can actually show that l 2 + 22 + ■• • + « 2 =
6.1
The Principle of Mathematical Induction
149
n(n + I)(2« + l ) / 6 by using the form ula 1 + 2 + • • • + « = n(n + l) /2 . We begin by solving (k + l ) 3 = k i + 3 k2 + 3 fc + 1 for k 2. Since 3k 2 = (k + l ) 3  k 3  3£  1, it follow s that k 2   [(k + l ) 3  k ' \  k  i and so ”
rc
1
> + d3 I > 3 S f, = 5 X k= 1 k=\ k= 1
E
*= i
1 w
^
E
*.=1
1
Therefore, V f c 2 = i [(« + l) 3  l 3]  \ n ( n + 1) 3 l' 1 " ‘ J 2 k= i 3 «3 + 3n2 + 3«
/72 + «
3
] n
n
2
3
2 « 3 + 6 « 2 + 6/2 — 3 n 2 — 3 n — 2 n
6
2 n 3 + 3 n2 + «
6
«(2m2 + 3 n + 1) n(n + 1)(2h
+
1)
T his is actually an alternative proof that
for every positive integer n but o f course this pro o f depends on know ing that n{n + 1) 1 + 2 + 3 + • ■• + 77 =  — for every positive integer n. We have now used m athem atical induction to establish the form ulas n(n + 1) 1 + 2 + ■•■ + « —  — 
(6.5)
and l 2 + 22 +
■n 2 —
n(n + l)(2 n + 1 )
(6 .6 )
for every positive integer n. We saw that (6.6) gives the num ber o f different squares in an n x n square com posed o f n 2 1 x 1 squares. Actually, (6.5) gives the num ber o f intervals in an interval o f length n com posed o f n intervals o f length 1. You can probably guess w hat l 3 + 23 + • • • + « 3 counts. E xercise 6.6 deals w ith this expression. We now present a form ula for 1
1
2^3 + F A + for every positive integer n .
1 (n + 1)(n + 2)
150
C h a p te r 6 Result 6.6
M athematical Induction
For every p ositive integer n , 1
1
1
n
2 ^ 3 + 3 ^ 4 + " ' + (n + i)(„ + 2) ~ 2;? + 4 ' P roof
We use induction. Since 1
1
2 3
21+4
the form ula holds for n — 1. A ssum e that 1
1
I
2^3 + 3^4 +
k
+ (A + 1)(A + 2) ~ 2A + 4
for a positive integer k. We show that 1
1
1
2^3 + 3^4 + ' '
k + 1
k + 1
(A + 2 )(A + 3) ~~ 2 (A’ + 1) + 4 “ 2A + 6 '
O bserve that 1
1
1
2^3 + 3^4
(k + 2)(A + 3) 1
1
1
1
2 ^ 3 + F " 4 + " ' + (A + 1 nk + 2) _ A
1
2k + 4
(A + 2)(A + 3)
A(A + 3) + 2
(A + 2)(A + 3) 1
A 2 (k + 2)
+
(A + 2)(A + 3)
A2 + 3 A + 2
2(A + 2)(A + 3) _ 2(A + 2)(A + 3) (A + 1)(A + 2)
k + 1
2(k + 2)(A + 3)
2(A + 3)
A+ 1 2A + 6 ’
giving us the desired result. B y the Principle o f M athem atical Induction, 1
1
2 3
i 3 4
_
for every positive integer n . PROOF ANALYSIS
n
(« + 1)(n + 2) 2n + 4 m
Each o f the exam ples of m athem atical induction proofs that we have seen involves a certain am ount o f algebra. W e’ll need to recall even m ore algebra soon. M any m istakes in these proofs are due to algebra errors. Therefore, care m ust be taken. For exam ple, in the p ro o f o f R esult 6.6, we encountered the sum A 1 2(A + 2) + (A + 2)(A + 3 )' To add these fractions, we needed to find a com m on denom inator (actually a least com m on denom inator), w hich is 2(A + 2)(k + 3). This w as used to obtain the next fraction, that is, A
1
2(k + 2) + (A + 2)(A + 3) '
A(A + 3) _2A(A+ 2(A + 2)(A + 3) + 2(A + 2)(A + 3) ~ 2(A + 2)(A + 3 )'
6.2
A More General Principle of Mathematical Induction
151
W hen we expanded and factored the num erator and then cancelled the term k + 2, this was actually expected since the final result w e w ere looking for was k + 1
k + I
2 k + 6 ~ 2(k + 3) ’ w hich does not contain k + 2 as a factor in the denom inator.
♦
6.2 A More G eneral P rin cip le of M athem atical Indu ction T he Principle of M athem atical Induction, described in the preceding section, gives us a technique for proving that a statem ent o f the type For every positive integer n , P (n ), is true. T here are situations, however, w hen the dom ain o f P ( n ) consists of those integers greater than or equal to som e fixed integer m different from 1. We now describe an analogous technique to verify the truth o f a statem ent of the follow ing type w here m denotes some fixed integer: For every integer n > m , P in ). A ccording to the W ellO rdering Principle, the set N o f natural num bers is wellordered; that is, every nonem pty subset o f N has a least elem ent. A s a consequence of the W ellO rdering Principle, other sets are also w ellordered. Theorem 6.7
For each integer m , the set S — {/ C Z : / > m ) is wellordered. The proof o f T heorem 6.7 is left as an exercise (see Exercise 6.17). The follow ing is a consequence o f T heorem 6.7. T his is a slightly m ore general form o f the Principle o f M athem atical Induction. Consequently, it is com m only referred to by the sam e nam e.
Theorem 6.8
(T he P rin c ip le o f M a th e m a tic a l In d u c tio n ) For a fix e d integer m , let S — {i e Z : i > m } . For each integer n e S, let P{ n) be a statem ent. I f (1) P ( m ) is true and (2) the im plication I f P( k) , then P (k + 1). is true fo r every integer k e S, then P ( n ) is true fo r every integer n e S.
152
Chapter 6
Mathematical Induction
The p ro o f o f T heorem 6.8 is sim ilar to the p ro o f o f T heorem 6.2. We also state T heorem 6.8 sym bolically.
T h e P rin c ip le o f M a th e m a tic a l I n d u c tio n
For a fix e d integer m , let S = {i e Z : i > m }. For each n e S , let P( n ) be a statem ent. I f D/ , , ., (1) P (m ) is true and (2) Wk € S, P ( k ) => P ( k + 1) is true, then V« e S,
P( n ) is
true.
This (m ore general) Principle o f M athem atical Induction can be used to prove that certain quantified statem ents o f the type Wn e S, Pi n ) are true w hen S — {i £ Z : i > m] for a prescribed integer m . O f course, if m = 1, then S = N. We now consider several exam ples. Result 6.9
For every nonnegative integer n,
2" > n. P roof
We proceed by induction. The inequality holds for n — 0 since 2° > 0. A ssum e that 2k > k, w here k is a nonnegative integer. We show that 2k+] > k + 1. W hen k  0, we have 2k+x = 2 > 1 = £ + 1. We therefore assum e that k > 1. Then 2k+i = 2 . 2k > 2 k = k + k > k + 1. B y the Principle o f M athem atical Induction, 2" > n for every nonnegative integer n. ■
PROOF ANALYSIS
L et's review the pro o f o f Result 6.9. First, since Result 6.9 concerns nonnegative integers, we are applying T heorem 6.8 w hen m = 0. We began by observing that 2" > n w hen n — 0. N ext w e assum ed that 2k > k, w here k is a nonnegative integer. O ur goal w as to show that 2k+1 > k + 1, It seems logical to observe that 2k+l = 2 ■2k. Since w e knew that 2k > k , we have 2^+1 = 2 ■2k > 2k. If we could show that 2k > k + 1, then we have a proof. However, w hen k = 0, the inequality 2 k > k + 1 d o esn ’t hold. T h at’s why we handled k = 0 separately in the proof. This allow ed us to assum e that k > 1 and then conclude that 2k > k + 1. We could have proved Result 6.9 a bit differently. We could have observed first that 2n > n w hen n = 0 and then proved that 2" > n for n > 1 by induction. $
O ur next exam ple is to show that 2" > n 2 if n is a sufficiently large integer. We begin by trying a few values o f n, as show n below. It appears that 2" > n 2 w henever n > 5 .
6.2
A More General Principle of M athematical Induction
n 0 1 2 3 4 5 6
2" 1 2 4 8 16 32 64
153
n1 0 1 4 9 16 25 36
Com paring 2" and n 2 R e su lt to P ro v e
For every integer n > 5, 2" > r r.
PROOFSTRATEGY
L e t’s see w hat an induction p roof o f this result m ight look like. O f course, 2” > n 2 w hen n = 5. We assum e that 2k > k~ w here k > 5 (and k is an integer) and we w ant to prove that 2k+x > (k + l ) 2. We start with 2k+i = 2  2 k > 2 k2. We w ould have a pro o f if w e could show that 2k 2 > (k + l ) 2 or that 2 k2 > k 2 + 2k + 1. T here are several convincing ways to show that 2k 2 > k 2 + 2k + I for integers k > 5 . H ere’s one way: O bserve that 2k 2 = k 2 + k 2  k 2 + k ■k > k 2 + 5k since k > 5. A lso k 2 + 5k k 2 + 2k + 3k > k 2 + 2k + 3 ■5 — k 2 + 2 k + 15, again since k > 5. Finally, k 2 + 2k 15 > k~ + 2k + 1. We now present a form al proof. (H ere we are using the Principle M athem atical Induction w ith m = 5.)
Result 6.10
= + of +
For every integer n > 5, 2" > n 2.
Proof
We proceed by induction. Since 25 > 52, the inequality holds for n = 5. A ssum e that 2k > k 2 w here k > 5. We show that 2k+l > (k + l ) 2. Observe that 2*+1 _ 2 . 2k > 2 k1 = k 2 + k 2 > k 2 + 5k = k 2 + 2k + 3k > k 2 + 2 k + \ 5 > k~ + 2k + 1 = (k + l ) 2. Therefore, 2A+I > (k + l ) 2. By the Principle of M athem atical Induction, 2" > n 2 for every integer n > 5. B
Result 6.11
For every nonnegative integer n, 3  (22"  1)
154
Chapter 6 Proof
Mathematical Induction
We proceed by induction. The result is true w hen n = 0 since in this case 22"  1 = 0 and 3  0. A ssum e that 3 j \22k — l ) , w here k is a nonnegative integer. We show that 3  (22**2  1). Since 3  (2lk  l ) , there exists an integer.v such that 2lk  1 = 3x and so 2lk = 3x + 1. Now 22k+2  1 = 4 • 22k  1 = 4(3* + 1)  1
= I2 x + 3 = 3(4x + 1).
Since 4x + 1 is an integer, 3  ('22k+2 — 1). By the Principle o f M athem atical Induction, 3  {2?n — 1) for every nonnegative integer n. B
PROOF ANALYSIS
L et's review the preceding proof. A s expected, to establish the inductive step, w e assum ed that 3  (2~k — 1) for an arbitrary nonnegative integer k and attem pted to show that 3  ^22*4 2 — l) . To verify that 3  (22k+2 — l ) , it was necessary to show that 22*+2 — 1 is a m ultiple o f 3; that is, we needed to show that 22k+1 — 1 can be expressed as 3z for som e integer z. Since our goal was to show that 22k+2  1 can be expressed in a certain form , it is natural to consider 2~k+2 — 1 and see how w e m ight w rite it. Since w e knew that 2~k — 1 = 3 * , w here x e Z, it was logical to rew rite 22k+1 — 1 so that 2lk appears. Actually, this is quite easy since 22£+2 _ 22 . 2lk — 4 . 22k Therefore, 22k+~  1 = 4  2^k — 1. A t this point, we need to be a b it careful because the expression w e are currently considering is 4 • 22*  1, not 4(2®  1). T hat is, it w ould be incorrect to say that 4 • 22k  1 = 4(3*), H ence we need to substitute for 22k in this case, not for 2~k — 1. This is the reason that in the p ro o f w e rew rote 22k — 1 = 3 r as 22k = 3 x + 1. $
We reinforce this kind o f pro o f w ith another exam ple. Result 6.12
For every nonnegative integer n , 9  (43"  1 ).
Proof
We proceed by induction. W hen n = 0, 4 3"  1 = 0 . Since 9 j 0, the statem ent is true w hen n = 0. A ssum e that 9  (43i I), w here k is a nonnegative integer. We now show that 9  (43i+3 — l) . Since 9  (4 y — 1), it follow s that 43,c — 1 = 9 x for some integer x. H ence 43* = 9x j 1. N ow observe that 4^+3 _
j =
4 3 . 4 3i _
j = 6 4 (9 x +
1} _
j
= 64 • 9x + 64  1 = 64 • 9x + 63 = 9(64x + 7). Since 64x + 7 is an integer, 9  (43A+3 — l) . B y the Principle o f M athem atical Induction, 9  (43” — l) for every nonnegative integer n. H
6.2
A More General Principle of M athematical Induction
155
A s a final com m ent regarding the preceding proof, notice that we did not m ultiply 64 and 9 since w e w ere about to factor 9 from the expression in the next step in any case. For a positive integer n, the integer n fa c to ria l, denoted by «!, is defined as n\ = n(n — 1) • • • 3 • 2 ■1. In particular, 1! = 1,2! = 2 • 1 = 2 and 3! = 3 • 2 • 1 = 6. A lso, 0! is defined as 0! = 1. A m ong the m any equalities and inequalities involving n \ is the follow ing.
R e su lt to P rove
For every positive integer n, ,1 • 3o • 5* • ■■n(2n — 1) n = —(2«)! 2" ■n !
PROOFSTRATEGY
First, observe that (2 n)\ = 2n ■(2 n  1) • (2n  2) • • • 3 • 2 ■1.
(6.7)
O f the 2« term s in the right side o f expression (6.7), n o f them are even, nam ely 2, 4, 6, . . . , 2/7. If we w ere to factor 2 from each of these num bers, w e obtain 2" • 1 • 2 • 3 • • • n = 2n ■n \, w hile the rem aining 77 integers, nam ely 1, 3, 5 , . . . , 2n — 1, are all odd. Thus (2/7.)! — 2"  «!  1 • 3 • 5    (2n — I). Therefore, 1 ■3 • 5 ■• ■(In — 1) =
{2n)[
2n ■n !
This equality can also be established by induction. Result 6.13
For every positive integer n, ( 2 /7)!
1 • 3 • 5■• ■(2«  1) = Proof
2" ■n !
We proceed by induction. Since 1=
( 2 1) ! , 2 1 • 1!
2
.
the statem ent is true for n = 1. A ssum e, for a positive integer k , that 1.3. 5..(2*—! ) = < * ” . 2 k ■k\ We show that (2k + 2)! 2k+] (k + 1)!' O bserve that (2k + 2)!
(2k + 2)(2k + 1)
(2*)!
2k+1 • (k + 1)!
2 • (k + 1)
2* • (k)\
= 1 • 3 • 5 • • • (2k + 1).
= (2k + 1)[1  3  5    (2Jt — 1)]
156
C h a p te r 6
M athematical Induction
B y the Principle o f M athem atical Induction, (2«)! 2'1 • n ! for every positive integer n.
u
We saw in T heorem 3.12 that for an integer x , its square X2 is even if and only if x is even. This is actually a consequence of T heorem 3.17, w hich states that for integers a and b, their product ab is even if and only if a or b is even. We now present a generalizatio o f Theorem 3.12. Result 6.14
L e t x e Z. For every integer n > 2, x" is even i f and only i f x is even.
P roof
A ssum e, first, that x is even. Then x = 2y for some integer y . H ence x n = x ■x n~ ] = (2y ).v" 1 = 2 (vx" : ) . Since v x ”1 is an integer, x " is even. We now verify the converse, nam ely if x n is even, w here n > 2, then x is even. We proceed by induction. If x 2 is even, then w e have already seen that x is even. H ence the statem ent is true for n = 2. A ssum e that if x is even for som e integer k > 2, then x is even. We show that if x k+x is even, then x is even. L et x k+l be an even integer. T hen x ■x k is even. By T heorem 3.17, x is even or x k is even. I f x is even, then the result is proved. On the other hand, if x k is even, then, by the induction hypothesis, x is even as well. By the P rinciple o f M athem atical Induction, it follow s, for every integer n > 2, that if x n is even, then x is even. m
A lthough it is im possible to illustrate every type o f result w here induction can be used, w e give tw o exam ples that are considerably different than those we have seen. O ne o f De M organ’s laws (see T heorem 4.22) states that
AUB = I n s for every tw o sets A and B . It is possible to use this law to show that a u b u c
= I n s n c
for every three sets A, B and C . We show how induction can be used to prove De M organ’s law for any finite num ber o f sets. Theorem 6.15
I f A \, A,2, . . . , A n are n > 2 sets, then
Ai u a 2 u ■• ■u An = a 7 n A/?n • • •n ~a ~„. Proof
We proceed by induction. For n = 2, the result is De M o rg an ’s law and is therefore true. A ssum e that the result is true for any k sets, w here k > 2; that is, assum e that if B \ , B 2, ■. ■, B[: are any k sets, then Bx u
b
2 u ■• ■u B k =
n s T n ■■• n W k.
6.2
A More General Principle of M athem atical Induction
157
We prove that the result is true for any k + 1 sets. Let S i, S2, . . . , S^+i be k + 1 sets. We show that
S', u s2u •••u sk+1 = s[ ns^n •••ns^T. L et T = S] U S2 U • • • U Sk. Then Si U S 2 U • • • U S/fc+i
— (Si U S 2 'J ■■■U S*)U Sk+i
= T U Sk+i
Now, by De M organ’s law,
t us*+i = r n sk+l. By the definition o f T and by the inductive hypothesis, we have
T = Si u s2u ■••u sk = s7 n s^n •■■n s~k. Therefore,
Si u s 2 u • • • u s * + i = t u Sk+i = r n s^+i = s7ns^nns^n s ^ . By the P rinciple o f M athem atical Induction, for every n > 2 sets A x, A2, . . ■, A n, A i u a 2 u    u A n = M n A ^ n • ■• n Aii, as desired.
PROOF ANALYSIS
■
A few com m ents m ay be useful concerning the notation used in the statem ent and the p ro o f o f T heorem 6.15. First, the sets A ], A2, . . . , A„ were used in the statem ent o f Theorem 6.15 only as an aid to describe the result. Theorem 6.15 could have also been stated as: For every integer n > 2 , the com plem ent o f the union o f any n sets equals the intersection o f the com plem ents o f these sets. To verify the inductive step in the proof o f T heorem 6.15, w e assum ed that the statem ent is true fo r any k > 2 sets, w hich w e denoted by S i , S 2, . . . , Bk. The fact that w e used A j, A2, . . . , A„ to describe the statem ent of T heorem 6.15 did not m ean that we should use A ], A 2, ■■■, A k for the k sets in the inductive hypothesis. In fact, it is probably better that we do not use this notation. In the inductive step, w e now need to show that the result is true for any k + 1 sets. W e used S i, S2, . . . , S*+i for these sets. It w ould have been a bad idea to denote the k + 1 sets by 6 1 , S 2, . . ., B k+\ because that w ould have (im properly) suggested that k o f the k + 1 sets m ust specifically be the sets m entioned in the inductive hypothesis. ♦
We are now able to prove another w ellknow n theorem concerning sets, to w hich w e earlier referred. Theorem 6.16
I f A is a fin ite set o f cardinality n > 0, then the cardinality o f its p o w er set V ( A ) is 2".
160
Chapter 6
M athematical Induction
As it stands, we can factor 3 from 6 x + 3k (k + 1) but cannot factor 6 unless we can prove that k (k + 1) is even. This is the same difficulty we encountered w hen we w ere considering an induction proof. In any case, no contradiction is obtained. $
If a result can be proved by induction, then it can also be proved by m inim um coun terexam ple. It is not difficult to use induction to prove that 3  (22"  l) for every non negative integer n. We also give a pro o f by m inim um counterexam ple o f this statem ent. Result 6.18
For every nonnegative integer n, 3  (22”  1 ).
Proof
A ssum e, to the contrary, that there are nonnegative integers n for w hich 3 / (22"  1 ). By T heorem 6.7, there is a sm allest nonnegative integer n such that 3 / ( 2 2"  l) . D enote this integer by m . T hus 3 K (21,h — 1) and 3  (22" — l) for all nonnegative integers n for w hich 0 < n < m . Since 3  (22/! — l) w hen n = 0, it follow s that m > 1. H ence m can be expressed by m = k + 1, w here 0 < k < m . T hus 3  (2lk  1), w hich im plies that 22k  1 = 3x for som e in te g e r* . Consequently, 2lk — 3 x + 1. O bserve that
_i — 22(*+1) __j __ 2 ^ + 2 __j __22 . 2 ^ __1 = 4(3.v + 1)  1 = 12.v + 3 = 3(4.v + 1). Since 4 x + 1 is an integer, 3  (22m  l) , w hich produces a contradiction.
m
We give one additional exam ple using a p ro o f by m inim um counterexam ple. Result 6.19
For every positive integer n ,
. n(n + 1) 1 + 2 + 3 + ■ • ■ + « =  . 2 Proof
A ssum e, to the contrary, that n(n + 1) 1 + 2 + 3 +   +77 ^ — —
2
for som e positive integers n. By the W ellO rdering Principle, there is a sm allest positive integer n such that 1+2 + 3+ •••+ « /
n{n + 1)
2
D enote this integer by m . Therefore, n , . , , , , m ( m + 1) 1 + 2 + 3 + ■• • + /;? dz 2 w hile 11 + 2 + 3Hq i n(~n + !) V, n — 
.
6.4
The Strong Principle of M athematical Induction
161
for every integer n w ith 1 < n < m . Since 1 = 1(1 + l ) / 2 , it follow s that m > 2. H ence w e can w rite m — k + 1, w here 1 < k < m . Consequently, 1+2 + 3 + .+£=
k(k + 1)
O bserve that 1 + 2 + 3 + • • • + m = 1 + 2 + 3 + • • • + (/: + 1) = (1 + 2 + 3 + ■• • + A)+ (^' + = — 2—
*(* + 1) + 2(* + 1) + <* + ! » =  y 
(.k + \) (k + 2) =
1)
2
m (m + 1) "
2
’
w hich produces a contradiction.
*
6.4 The Strong P rin cip le o f M athem atical Induction We close w ith one last form o f m athem atical induction. This principle goes by m any nam es: the Strong Principle o f M athem atical Induction, the Strong Form o f Induction, the A lternate Form o f M athem atical Induction and the Second Principle o f M athem atical Induction are com m on nam es. Theorem 6.20
(T h e S tro n g P rin c ip le o f M a th e m a tic a l In d u c tio n ) For each p ositive integer n, let P{n) be a statem ent. I f (ia) P (1) is true and (b) the im plication I f P ( i ) f o r every integer i with I < i < k, then P (k + 1). is true fo r every p ositive integer k, then P{n) is true fo r every positive integer n. As w ith the Principle o f M athem atical Induction (Theorem 6.2), the Strong P rinciple o f M athem atical Induction is also a consequence o f the W ellO rdering Principle. The Strong Principle o f M athem atical Induction is now stated m ore sym bolically below.
T h e S tro n g P rin c ip le of M a th e m a tic a l I n d u c tio n
For each p ositive integer n, let P( n ) be a statem ent. I f (1) P (1) is true and (2) V& e N, P ( l ) a P ( 2) A • • • A P( k ) =4> P( k
+ 1) is true,
then Wn e N, P( n) is true.
T he difference in the statem ents o f the Principle o f M athem atical Induction and the Strong Principle of M athem atical Induction lies in the inductive step (condition 2).
162
C h a p te r 6
M athematical Induction
To prove that V/; e N, P ( n ) is true by the Principle o f M athem atical Induction, we are required to show that P ( l ) is true and to verify the im plication: If P( k) , then P ( k + 1).
(6.9)
is true for every positive integer k. On the other hand, to prove Wn e N, P (n ) is true by the Strong Principle o f M athem atical Induction, w e are required to show that P ( l ) is true and to verify the implication: If P( i ) for every i w ith I < / < k, then P{ k + 1).
(6.10)
is true for every positive integer k. If we w ere to give direct proofs o f the im plications (6.9) and (6.10), then w e are perm itted to assum e m ore in the inductive step (6.10) o f the Strong P rinciple o f M athem atical Induction than in the induction step (6.9) o f the P rinciple of M athem atical Induction and yet obtain the same conclusion. If the assum ption that P( k) is true is insufficient to verify the truth o f P ( k + 1) for an arbitrary positive integer k, but the assum ption that all of the statem ents P ( \ ) , P ( 2 ) , . . . , P( k ) are true is sufficient to verify the truth o f P (k + 1), then this suggests that we should use the Strong Principle o f M athem atical Induction. Indeed, any result that can be proved by the Principle of M athem atical Induction can also be proved by the Strong Principle o f M athem atical Induction. Just as there is a m ore general version o f the Principle o f M athem atical Induction (namely, Theorem 6.8), there is a m ore general version o f the Strong Principle o f M ath em atical Induction. We shall also refer to this as the Strong Principle o f M athem atical Induction. Theorem 6.21
(The Strong Principle o f M athem atical Induction) For a fix e d integer m , let S {/ e Z : i > m }. For each n e S , let P{n) be a statem ent. I f (1) P (m ) is true and (2) the im plication I f P( i ) fo r every integer i with m < i < k, then P (k + 1). is true fo r every integer k e S, then P {n ) is true fo r every integer n e S. We now consider a class o f m athem atical statem ents w here the Strong Principle of M athem atical Induction is com m only the appropriate pro o f technique. Suppose that w e are considering a sequence U j. c/%, . . . o f num bers. O ne w ay of defining a sequence {a„} is to specify explicitly the 7 2 th term a n (as a function o f n). 1 (1 )" F or exam ple, we m ight have a„ —  , a„ = — — or a„ = n 3 + n for each n e N . A n nsequence can also be defined recursively. In a recursively defined sequence {an}, only the first term or perhaps the first few term s are defined specifically, say a x, a2, ■■■, ak for som e fixed k e N. T hese are called the initial values. T hen ak+x is expressed in term s of d \ , a%,, . . , af, and m ore generally, for n > k, an is expressed in term s o f a \ , <22, ■■■ , a n 1• This is called the recurrence relation.
6.4
The Strong Principle of M athematical Induction
163
A specific exam ple o f this is the sequence {an} defined by a\ = \ , a 2 = 3 and a„ — 2a„ \ — a „  2 for n > 3. In this case, there are tw o initial values, nam ely a \ = 1 and c>2 = 3. The recurrence relation here is an = 2a„_i — a „ ~ 2 f ° r n > 3 . L etting n = 3, w e find that a3 — 2a2 — a \ = 5; w hile letting n = 4, we have a 4 = 2a 3 a 2 = 7. Similarly, a 5 = 9 and 06 = 11 From this inform ation, one m ight w ell conjecture (guess) that an — 2n — 1 for every n e N. (C onjectures w ill be discussed in m ore detail in Section 7.1.) U sing the Strong Principle o f M athem atical Induction, w e can, in fact, prove that this conjecture is true. Result 6.22
A sequence [a„} is defined recursively by a, = 1, a 2 = 3 and a„ = 2an \ — a n 2fo r n > 3. Then an — 2n — I fo r all n e N.
Proof
We proceed by induction. Since a i — 2  1 — 1 = 1, the form ula holds for n = 1. A ssum e for an arbitrary positive integer k that a, = 2i — 1 for all integers i w ith 1 < i < k . We show that = 2 (k + 1) — 1 = 2k + 1. If k = 1, then ak+1 = a 2 = 2  l + l = 3 . Since a2 = 3, it follow s that ak+\ = 2k + 1 w hen k = 1. H ence w e m ay assum e that k > 2 . Since k + 1 > 3, it follow s that ak+ 1 = 2a t — ak \ = 2(2k — 1)  (2k — 3) = 2 k + 1, w hich is the desired result. By the Strong Principle o f M athem atical Induction, a„ = 2n — 1 for all n e N. ■
PROOF ANALYSIS
Problem 6.23
A few com m ents about the proof o f R esult 6.22 are in order. At one point, w e assum ed for an arbitrary positive integer k that a t = 2/ — 1 for all integers i w ith 1 < i < k. O ur goal was to show that a^+i = 2k + 1. Since k is a positive integer, it m ay occur that k = 1 or k > 2. If k = 1, then w e need to show that ak+\ = a 2 = 2  1 + 1 = 3 . T hat a 2 = 3 is know n because this is one of the initial values. If k > 2, then k + 1 > 3 and ak+l can be expressed as 2ak  ak^ \ by the recurrence relation. In order to show that ak+\ — 2 k + \ w hen k > 2, it was necessary to know that ak = 2 k — 1 and that ak1 = 2(k — 1) — 1 = 2k — 3. Because w e w ere using the Strong Principle o f M athem atical Induction, w e knew both pieces of inform ation. If w e had used the P rinciple of M athem atical Induction, we w ould have assum ed (and therefore knew) that ak = 2k — 1 but w e w ould not have know n that ak \ = 2k — 3, and so w e w ould have been unable to establish the desired expression for ak+\ . ♦ A sequence [an] is defined recursively by a\ — 1, a 2 — 4 a nd a„ = 2a„_i — a „ ^ 2 + 2 fo r n > 3. C onjecture a fo rm u la fo r a„ and verify that your conjecture is correct.
164
Chapter 6 S o lu tio n
Result 6.24
M athematical Induction
We begin by finding a few m ore term s o f the sequence. O bserve that <33 = 2a2  &i + 2 = 9, w hile <74 — 2o3 — a 2 + 2 = 16 and 05 = 2 a4 — a 3 + 2 = 25. The obvious conjecture is that a„ = n 2 for every positive integer n. We verify that this conjecture is correct in the next result. ^
A sequence {an} is defined recursively by ci\ = 1, a2 = 4 and an = 2an \ — an . 2 + 2 fo r n > 3. Then a„ = n 2 fo r all n e N.
Proof
We proceed by induction. Since a \ — l = l 2, the form ula holds for n = 1. an arbitrary positive integer k that a, — i 2 for every integer / w ith 1 < / < k. We sho that #£+! = (/: + 1). Since a 2 = 4, it follow s that ak+\ — (k + l ) 2 w hen k = 1. Thus w e m ay assum e that k > 2. H ence £ i 1 > 3 and so a k+1 — 2ak  cik~i + 2 = 2k 2  (k  l ) 2 + 2 = 2k 2  (k2  2 k \ I ) i 2 — £ 2 + 2k + 1 = (A + l ) 2. By the Strong Principle o f M athem atical Induction, a n = « 2 for all n e N.
■
A lthough we m entioned that problem s involving recurrence relations are com m only solved w ith the aid o f the Strong Principle o f M athem atical Induction, it is by no m eans the only kind o f problem w here the Strong Principle o f M athem atical Induction can be applied. A lthough the best exam ples o f this require a know ledge of m athem atics beyond w hat we have covered thus far, w e do present another type of exam ple. Result 6.25
P roof
For each integer n > 8, there are nonnegative integers a a n d b such that n ~ 3a + 5b. We proceed by induction. Since 8 = 3 • 1 + 5 • 1, the statem ent is true for n = 8. A ssum e for each integer i w ith 8 < / < k , w here k > 8 is an arbitrary integer, that there are nonnegative integers s and t such that i = 3s + 5t. C onsider the integer k + 1. We show that there are nonnegative integers x and j such that k + 1 = 3x + 5_y. Since 9 = 3 • 3 + 5 • 0 and 10 = 3  0 + 5  2 , this is true if k + 1 = 9 and k + 1 = 10. H ence w e m ay assum e th at/: + 1 > ll .T h u s 8 < (k + 1) — 3 < k. By the induction hypothesis, there are nonnegative integers a and b such that (k ! 1 ) 3 — 3a 1 5b and so k + 1 = 3{a +
1) + 5b.
Letting x = a + 1 and _v = b, we have the desired conclusion. By the Strong Principle of M athem atical Induction, for every integer n > 8, there are nonnegative integers a and b such that n = 3a + 5b. ■
Exercises for Chapter 6
165
EXERCISES FOR CHAPTER 6 Section 6.1: The Principle of M athem atical Induction 6.1. Which of the following sets are wellordered? (a) (b) (c) (d)
S = {x e Q : x > 1 0 } S = { 2 ,  1 , 0 , 1, 2} S = {x e Q :  1 < x < 1} S = {p : p is a prime} = {2, 3, 5, 7, 11,13, 1 7 ,...} .
6.2. Prove that if A is any wellordered set of real numbers and B is a nonempty subset of A , then B is also wellordered. 6.3. Prove
that every nonempty set of negative integers has a largest element.
6.4. Prove
that 1 + 3 + 5 + ■■• + (2n — 1) = n 2 for every positive integer n
( 1) by mathematical induction and (2) by adding 1 + 3 + 5 +  h (2/7 — 1) and (2n — 1) + (2n — 3) + ■■• + 1. 6.5. Use mathematical induction to prove that 1 4 5 + 9 + • ■• + (4n — 3) = 2n 2 — n for every positive integer n. 6.6. (a) We have seen that l 2 + 22 + ■• • + n 2 is the number of squares in an n x n square composed of n 2 l x l squares. W hat does l 3 + 23 + 33 + • • • + n 3 represent geometrically? o 3 q •> n2(n + 1)2 (b) Use mathematical induction to prove that 1 + 2 + 3 ' + • • • + « =   for every positive integer n. 6.7. Find another formula suggested by Exercises 6.4 and 6.5, and verify your formula by mathematical induction. 6.8. Find a formula for 1 + 4 + 7 + • • • + (3« — 2) for positive integers n, and then verify your formula by mathematical induction. 6.9. Prove
that
1  3 + 2 4 + 3  5 +   + n(n + 2) =
1)6(2"7> for every positive integer n.
6. 10. Let r / 1 be a real number. Use induction to prove that a + ar + a r 2 + • • ■+ o r "1 positive integer n. 6 .1 1. Prove that 55 +
+ •• ■+ (n+2)(n+3) =
7^+ 9
for every positive integer n.
6.12. Consider the open sentence P(n): 9 + 13 +  b (4n + 5) = 4,r+^4n+ 1; where n e N. (a) Verify the implication P(k) =)• P{k + 1) for an arbitrary positive integer k. (b) Is Sn e N, P in) true? 6.13. Prove
that
1  1! + 2  2! + •• + « n! = ( « + 1)! — 1 for every positive integer n.
6.14. Prove
that
2! • 4! ■6 ! ...............(2«)! > [(« + 1)!]" for every positive integer n.
6.15. Prove that ^y + ~^ +  ^ + "  +  ^ < 2 s/n — 1 for every positive integer n. 6.16. Prove that 7  [34"+1 — 52"1 ] for every positive integer n.
= for eve
166
Chapter 6
Mathematical Induction
Section 6.2: A More General Principle of M athem atical Induction 6.17. Prove Theorem 6.7: For each integer m, the set S = [i e Z : i > m ] is wellordered. [Hint: For every subset T of S, either T C N or T  N i s a finite nonempty set.] 6.18. Prove that 2" > n 3 for every integer n > 10. 6.19. Prove the following implication for every integer n > 2: If A ] , a 2 , . . ., x n are any n real numbers such that x i ■X2 .........x n = 0, then at least one of the numbers x i , x 2. ■■■, x„ is 0. (Use the fact that if the product of two real numbers is 0, then at least one of the numbers is 0.) 6.20. (a) Use mathematical induction to prove that every finite nonempty set of real numbers has a largest element. (b) Use (a) to prove that every finite nonempty set of real numbers has a smallest element. 6 .2 f. Prove that 4  (5” — f) for every nonnegative integer n. 6.22. Prove that 3" > n 2 for every positive integer n. 6.23. Prove that 7  (32'1 — 2") for every nonnegative integer n. 6.24. Prove Bernoulli's Identity: For every real number x > —1 and every positive integer n,
6.25. Prove that n\ > 2 " for every integer n > 4. 6.26. Prove that 81  (f0 ”+1 —9n — f 0) for every nonnegative integer n. 6.27. Prove that f + 3 +  H f
i 2
for every positive integer n.
6.28. In Exercise 6 of Chapter 4, you were asked to prove that if 3  l a , where a e Z, then 3 j a. Assume that this result is true. Prove the following generalization: Let a e Z. For every positive integer n, if 3  2/!a ,th e n 3  a. 6.29. Prove that if A \, A 2 ........ A n are any n > 2 sets, then
a 1 n a 2 n •*
An — Ai
• n
u
a 2 u • • ■ An. u
6.30. Recall for integers n > 2, a. b. c. d, that if a = b (m od n) and c = d (m od n), then both a + c = b + d (m od n) and ac = bd (m od ri). Use these results and mathematical induction to prove the following: For any 2m integers a {, a2, . . . , am and b\, b2 ........ bm for which a, = bt (mod ri) for 1 < i < m , (a) a\ + a2 + ■■■+ am = bi + b 2 + ■■■+ bm (m od n ) and (b) a\a 2 ■■■am = bib 2 ■■■bm (m od n). 6.31. Prove for every n > 1 positive real numbers a\, a 2 ........ an that
6.32. Prove for every n > 2 positive real numbers a \ , a 2 ........ a„ that n
(6.11) <i<j<n
[Note: W hen n = 4, for example, (6.11) states that 3(cZj + #
1
+ ^
3 + a )
> 2 ( ^ i <^2 +
CH&3
+
ci\a4
+
ct2&3
+ ^ 2^4 + ^ 3^ 4)]
Section 6.3: Proof by Minimum Counterexample 6.33. Use proof by minimum counterexample to prove that 6 j I n ( / r  l) for every positive integer n. 6.34. Use the method of minimum counterexample to prove that 3 j (22" — l) for every positive integer n.
Additional Exercises for Chapter 6
167
6.35. Give a proof by minimum counterexample that 1 + 3 + 5 + • • • + (2n — 1) = n 2 for every positive integer n. 6.36. Prove that 5  (n 5 — n ) for every integer n. 6.37. Use proof by minimum counterexample to prove that 3 (2" + 2"+1) for every nonnegative integer n. 6.38. Give a proof by minimum counterexample that 2" > n 2
for every integer n > 5.
6.39. Prove that 12  (n 4 — n2) for every positive integer n. 6.40. Let S = {2'' : r e Z, r > 0}. Use proof by minimum counterexample to prove that for every n there exists a subset S„ of S such that Yl,rs„ ' — n ■
e N,
Section 6.4: The Strong Principle of M athematical Induction 6.41. A sequence {a„} is defined recursively by a\ = 1 and a„ = 2an \ for n >2. Conjecture a formula for a„ and verify that your conjecture is correct. 6.42. A sequence [a,,] is defined recursively by «i = 1, «2 = 2 and an = <7„_i+ 2a „_2 for n > 3. Conjecture a formula for an and verify that your conjecture is correct. 6.43. A sequence {a,,} is defined recursively by a\ = 1,
= 4, <23 = 9 and
a„ = an 1  a „_2 + fln3 + 2(2«  3) for n > 4. Conjecture a formula for a„ and prove that your conjecture is correct. 6.44. Consider the sequence F\, F2, F3, . . . , where F i = 1, F 2 = 1, F 3 = 2, F4 = 3, F 5 = 5 and Fg = 8. The terms of this sequence are called F ibonacci num bers. (a) Define the sequence of Fibonacci numbers by means of a recurrence relation. (b) Prove that 2  F„ if and only if 3  n. 6.45. Use the Strong Principle of Mathematical Induction to prove that for each integer /? > 12, there are nonnegative integers a and b such that n = 3a + lb. 6.46. Use the Strong Principle of M athematical Induction to prove the following. Let S = {/' e Z : i > 2} and let P be a subset of S with the properties that 2, 3 e P and if n e S, then either n e P or n = ab, where a, b e S. Then every element of S either belongs to P or can be expressed as a product of elements of P. [Note: You might recognize the set P of primes. This is an important theorem in mathematics, which appears as Theorem 11.17 in Chapter 11.] 6.47. Prove that there exists an odd integer m such that every odd integer n with n > m can be expressed either as 3a + 11 b or as 5c + I d for nonnegative integers a, b, c and d.
ADDITIONAL EXERCISES FOR CHAPTER 6 6.48.
Prove that 1 ■2 + 2  3 + 3  4 +   + /?(« + 1) = /('l+1X"+2> for every positive integer n.
6.49. Prove that 4" > « 3 for every positive integer n. 6.50. Prove that 24  (52" —1) for every positive integer n. 6.51. By Result 6.5,
168
Chapter 6
Mathematical Induction
for every positive integer n . (a) Use ( 6 . 1 2 ) to determine a formula for 2 2 + 42 + 6 2 H b ( 2 k ) 2 for every positive integer n. (b) Use (6.12) and (a) to determine a formula for l 2 + 32 + 52 H f ( 2 n  l ) 2 for every positive integer/;. (c) Use (a) and (b) to determine a formula for l 2 — 2 2 + 32 — 42 H b (  i ) " +1B2 for every positive integer n. (d) Use mathematical induction to verify the formulas in (b) and (c). 6.52. Use the Strong Principle of M athematical Induction to prove that for each integer n > 28, there are nonnegative integers x and y such that n = 5x + 8y. 6.53.
Find a positive integer m such that for each integer n > m, there are positive integers x and y such that n = 3x + 5y. Use the Principle of M athematical Induction to prove this.
6.54. Find a positive integer m such that for each integer n > m, there are integers x , y > 2 such that n = 2x + 3y. Use the Principle of M athematical Induction to prove this. 6.55. A sequence {a,,} of real numbers is defined recursively by <•/, = 1. «2 ~ 2 and a„ = £ " = i (*’  l)^r for n > 3. Prove that a„ = (n  1)! for every integer n > 3. 6.56. Consider the sequence a\ = 2, a 2 = 5, a 3 = 9,
= 14, etc.
(a) Find a recurrence relation that expresses a„ in terms of for every integer n > (b) Conjecture an explicit formula for a„, and then prove that your conjecture is correct.
2.
6.57. The following theorem allows one to prove certain quantified statements over some finite sets. The Principle of F inite Induction For a fixed positive integer m, let S = {1,2 ........ let P(n) be a statement. I f (a) P ( l ) is true and (b) the implication I f P(k), then P (k + 1). is true fo r every integer k with 1 < k < m, then P(n) is true fo r every integer n e S. Use the Principle of Finite Induction to prove the following result. Let 5 = {1, 2 , . . ., 24}. For every integer t with 1 < t < 300, there exists a subset S, c S such that
YieS, i = f6.58. Evaluate the proposed proof of the following result. R esult For every positive integer n , 1 + 3 + 5 + • • ■+ (2n  1) = ir . P ro o f We proceed by induction. Since 2 • 1 — 1 = l 2, the formula holds for n = 1. Assume that 1 + 3 + 5 + • ■■+ (2k — 1) = k 2 for a positive integer k. We prove that 1 + 3 + 5 + ■• • + (2k + 1) = (k + l) 2. Observe that 1 + 3 + 5 + • ■• + (2k + 1) = (k + i f l + 3 + 5 +   + ( 2 *  l ) + ( 2 * + 1) = (k +
l)2
k 2 + (2 k + 1) = (k + I)2
0. + 1)2 = ( * +
l)2.
6.59. Below is given a proof of a result. Which result is being proved and which proof technique is being used?
Additional Exercises for Chapter 6
169
P roof Assume, to the contrary, that there is some positive integer n such that 8 / ( 32,1 — 1). Let m be the smallest positive integer such that 8 / (32m — 1). For n = 1, 32" — 1 = 8. Since 8  8, it follows that m > 2. Let m = * + 1. Since 1 < * < m, it follows that 8  (32k — 1). Therefore, 32* — 1 = 8x for some integer x and so 32k = 8x + 1. Hence 32"'  1 = 32(ir+1)  1 = 32*+2  1 = 9 • 32*  1 = 9(8* + 1)  1 = 72x + 8 = 8(9x + 1). Since 9x + 1 is an integer, 8  (32m — 1), which produces a contradiction.
■
6.60. Below is given a proof of a result. Which result is being proved and which proof technique is being used? P roo f First observe that a\ = 8 = 3 • 1 + 5 and «2 — 1 1 = 3  2 + 5. Thus a„ = 3n + 5 for n = 1 and n = 2. Assume that a, = 3/ + 5 for all integers i with 1 < i < k, where k > 2 . Since k + 1 > 3, it follows that cik+\ — 5ctk •—4<7£_i —9 = 5(3k + 5) — 4(3k + 2) — 9 = 15* + 25  12*  8  9 = 3k + 8 = 3(* + 1) + 5.
■
6.61. By an «gon, we mean an wsided polygon. So a 3gon is a triangle and a 4gon is a quadrilateral. It is well known that the sum of the interior angles of a triangle is 180°. Use induction to prove that for every integer n > 3, the sum of the interior angles of an /?gon is (n — 2) • 180°. ■?.62. Suppose that {an} is a sequence of real numbers defined recursively by a\ = 1, <22 = 2, #3 = 3 and an = 2an \ — a„  3 for n > 4. Prove that a„ = a„^\ + a„  2 for every integer n > 3. 5.63. Suppose that {«„} is a sequence of real numbers defined recursively by a\ = 1, a.2 = 2 and a„ = a „ _ i/a „_2 for n > 3. (a) Prove that 1 2 1/2
if n= 1 ,4 (mod 6) if n= 2, 3 (mod 6) if n= 0 ,5 (mod 6)
for every positive integer n. (b) Prove, for each nonnegative integer j , that Y^i= 1 ai+> =
^■
6.64. Let x e R where x > 3. Prove that (1 + x)n > [ ”("~ 1g('1~2:>] x 3 for every integer n > .65. Prove that
3.
^X )/= i!) — '’("+l^ ni:2 for every positive integer n.
6.66. Prove that YTj = 1 ( S / = 1(2/ — 1)^) = »(»+iX2«+i) for every positive integer n. 6.67. Prove that there exists an odd integer m such that every odd integer n with n > m can be expressed as 3a + 5b + 7c for positive integers a, b and c.
Prove or Disprove
n every m athem atical statem ent that you have seen so far, you have been inform ed of its truth value. If the statem ent was true, then we have either provided a p ro o f or asked you to provide one o f your own. W hat you d id n ’t know (perhaps) was how we or you were to verify its truth. If the statem ent was false, then here too w e either verified this or asked you to verify that it was false. A s you proceed further into the w orld o f m athem atics, you w ill m ore and m ore often encounter statem ents w hose truth is in question. Consequently, each such statem ent presents tw o problem s for you: (1 ) D eterm ine the truth or falseness o f the statem ent. (2) Verify the correctness o f your belief.
I
7.1 C onjectures in M athem atics In m athem atics, w hen w e d o n 't know w hether a certain statem ent is true but there is good reason to believe that it is, then w e refer to the statem ent as a conjecture. So the w ord conjecture is used in m athem atics as a sophisticated synonym for an intelligent guess (or perhaps ju st a guess). O nce a conjecture is proved, then the conjecture becom es a theorem . If, on the other hand, the conjecture is show n to be false, then we m ade an incorrect guess. This is the w ay m athem atics develops— by guessing and showing that our guess is correct or w rong and then possibly m aking a new conjecture and then repeating the process (possibly often). As we learn w h a t’s true and w h a t’s false about the m athem atics w e ’re studying, this influences the questions w e ask and the conjectures we make. L e t’s consider an exam ple o f a conjecture (although there is always the possibility that som eone has settled the conjecture betw een the tim e it was w ritten here and the m om ent you read it). A w ord is called a palindrom e if it reads the sam e forw ard and backw ard ( such as deed, noon and radar). Indeed, a sentence is a palindrom e if it reads the same forw ard and backw ard, ignoring spaces (Name no one man). A positive integer is called a palindrom e if it is the same num ber w hen its digits are reversed. (It is considerably easier to give an exam ple o f a num ber that is a p alindrom e than a w ord that is a palindrom e.) F or exam ple, 1221 and 47374 are palindrom es. C onsider the integer 27.
170
7.1
Conjectures in M athematics
171
It is not a palindrom e. Reverse its digits and w e obtain 72. N eedless to say, 72 is not a palindrom e either. A dding 27 and 72, w e have: 27 + 72 99 A palindrom e results. C onsider another positive integer, say 59. It is not a palindrom e. Reverse its digits and add: 59 + 95 154 The result is not a palindrom e either. Reverse its digits and add: 154 + 451 605 O nce again w e arrive at a num ber that is not a palindrom e. B ut reverse its digits and add: 605 + 506
1111 This tim e the result is a palindrom e. It has been conjectured that if we begin w ith any positive integer and apply the technique described above to it, then we w ill eventually arrive at a palindrom e. However, no one know s if this is true. (It is know n to be true for all tw odigit num bers.) Som e conjectures have becom e fam ous because it has taken years, decades or even centuries to establish their truth or falseness. O ther conjectures rem ain undecided still today. We now consider four conjectures in m athem atics, each o f w hich has a long history. In 1852, a question occurred to the British student Francis G uthrie w hen he was coloring a m ap o f the counties o f England. Suppose that some country (real or im aginary) has been divided into counties in some manner. Is it possible to color the counties in this m ap w ith four or few er colors such that one color is used for each county and tw o counties that share a com m on boundary (not sim ply a single point) are colored differently? For exam ple, the m ap of the “country” show n in Figure 7.1 has eight “counties,” w hich are colored w ith the four colors red (R), blue (B), green (G) and yellow (Y), according to the rules described above. This m ap can also be colored w ith m ore than four colors but not less than four. W ithin a few years, som e o f the best know n m athem aticians o f the tim e had becom e aw are o f Francis G uthrie’s question, and eventually a fam ous conjecture developed from this.
The F our C o lo r C o n je c tu re
Every m ap can be colored w ith four or few er colors.
M any attem pted to settle this conjecture. In fact, in 1879 an article was published containing a reported proof o f the conjecture. However, in 1890, an error w as discovered in the p roof and the “theorem ” returned to its conjecture status. It was not until 1976 w hen
172
Chapter 7
Prove or Disprove
an actual pro o f by K enneth A ppel and W olfgang H aken, com bining both m athem atics and com puters, was presented. The period betw een the origin o f the problem and its solution covered som e 124 years. This is now a theorem . The Four C olor Theorem
Every map can be colored w ith four or few er colors.
We now describe a conjecture w ith an even longer history. One o f the fam ous m athem aticians o f the 17th century was Pierre Ferm at. He is undoubtedly best know n for one particular assertion he m ade. He w rote that for each integer n > 3, there are no nonzero integers x , y and z such that x " + y n = z " . O f course, there are m any nonzero integer solutions to the equation x 2 + y 2 = z 2. For exam ple, 32 + 42 = 52, 5“ + 12~ = 13" and 8" + 152 = 172. A triple (x, y , z) o f positive integers such that x 2 + y 2 = z 2 is often called a P ythagorean triple. T herefore, (3, 4, 5), (5, 12, 13) and (8, 15, 17) are P ythagorean triples. Indeed, if (a, b, c) is a P ythagorean triple and k e N, then (ka, kb, k c ) is also a Pythagorean triple. F erm at’s assertion w as discovered, un proved, in a m argin o f a book o f F erm at’s after his death. In the m argin it was w ritten that there was insufficient space to contain his “truly rem arkable dem onstration.” C on sequently, this statem ent becam e know n as F erm at’s L ast Theorem . It w ould have been m ore appropriate, however, to have referred to this statem ent as F erm at’s L ast Conjecture, as the truth or falseness o f this statem ent rem ained in question for approxim ately 350 years. However, in 1993, the B ritish m athem atician A ndrew W iles settled the conjecture by giving a truly rem arkable pro o f o f it. H ence F erm at’s L ast T heorem is at last a theorem . F erm at’s Last Theorem
For each integer n > 3, there are no nonzero integers x , y and z such that x" + y n — z n.
The final tw o conjectures we m ention concern prim es. A lthough w e have m entioned prim es from tim e to tim e, we have not yet presented a form al definition. We do this now. A n integer p > 2 is a prim e if its only positive integer divisors are 1 and p. A Ferm at num ber (Yes, the sam e Ferm at!) is an integer o f the form F, = 22' + 1,w here t is a nonnegative integer. The first five Ferm at num bers are F 0 = 3, F l = 5, F 2 = 17, F 3 = 257, FA = 65, 537, all o f w hich happen to be prim es.
7.2
Revisiting Quantified Statements
173
In 1640, F erm at w rote to m any (including to the fam ous m athem atician B laise Pascal) that he believed every such num ber (he d id n ’t call them F erm at num bers) w as a prim e, but he was unable to prove this. H ence w e have the follow ing. F e r m a t's C o n je c tu re
Every F erm at num ber is a prim e. N early one century later (in 1739), the fam ous m athem atician L eonhard E uler proved that F$ = 4, 294, 967, 297 is divisible by 641, thereby disproving F erm at’s Conjecture. M ore specifically, E uler proved the follow ing.
E u le r ’s T h e o re m
If p is a prim e factor of Ft , then p = 2r+1k + 1 for som e positive integer k. Letting t = 5 in E u le r’s Theorem , w e see that each prim e factor o f F$ is o f the form 64* + 1. The first five prim es o f this form are 193, 257, 449, 577 and 641, the last of w hich divides F 5 . In recent decades, other F erm at num bers have been studied and have been show n not to be prim e. Indeed, m any students o f this topic now lean tow ard the opposing view point (and conjecture): Except for the F erm at num bers Fq, F i, • • ■, F 4 (all o f w hich were observed to be prim e by Ferm at), no Ferm at num ber is prim e. The last conjecture we describe here has its origins around 1742. The G erm an m athem atician C hristian G oldbach conjectured that every even integer exceeding 2 is the sum o f tw o prim es. O f course, this is easy to see for sm all even integers. For exam ple, 4 = 2 + 2, 6 = 3 + 3, 8 = 5 + 3 and 10 = 7 + 3 = 5 + 5 . The m ajor difference betw een this conjecture and the three preceding conjectures is that this conjecture has never been resolved. H ence w e conclude w ith the follow ing.
G o ld b a c h ’s C o n je c tu re
Every even integer at least 4 is the sum o f tw o prim es.
7.2 R evisiting Q uantified Statem ents M any (in fact, m ost) o f the statem ents w e have encountered are quantified statem ents. Indeed, for an open sentence P ( x ) over a dom ain S, we have often considered a quantified statem ent w ith a universal quantifier, nam ely Vx e S, P(x): For every x e S, P(x). or If x e S, then P(x). or a quantified statem ent w ith an existential quantifier, nam ely 3x e S , P(x): T here exists x e S such that P(x). Recall that Vx 6 S, P( x) is a true statem ent if P(x) is true for every x e S', w hile 3x € S, P(x) is a true statem ent if P( x) is true for at least one x e S. L et’s review these again. Example 7.1
L e t S = {1, 3, 5, 7} and consider P (n ): n 2 + n + 1 is prim e.
174
Chapter 7
Prove or Disprove
f o r each n e S. Then both Vb € S. P ( n ): For every n e S, n 2 + ;? + 1 is prime. and 3n e S, P u t ): There exists n e S such that n 2 + n + 1 is prime. are quantified statements. Since P ( 1): P(3): P(5): P(7):
l 2 + 1 + 1 = 3 zs prime. 32 + 3 j 1 = 13 is prime, 52 + 5 + 1 = 31 is prime. 1 + 1 + 1 = 5 7 is prime,
is true, is true, is true, is false,
it follow s that V« e S. P ( n ) is fa ls e and 3/; e S, P (n ) is true. On the other hand, the statement Q : 323 is prime. is not a quantified statement but Q is fa ls e (as 323 = 17 • 19 is not prime).
#
L et P (x) be a statem ent for each x in som e dom ain S. R ecall that the negation of V.v e S, P i x ) is ~ (Vx e S, P ( x ) ) = 3 x e S , ~ PQc). and the negation o f 3x e S. P ( x ) is ~ (3x e S. P {x)) = Vx e S , ~ F (x ). A gain, consider F (« ): /?“ + n + 1 is prim e. from Exam ple 7.1, w hich is a statem ent for each n in S = {1, 3, 5, 7}. The negation of 'in 6 S. P (n ) is Bn e S t ~ P in): T here exists n e S such that n 2 + n + 1 is not prim e. is true as 7 e S but 72 + 7 + 1 = 57 is not prim e. O n the other hand, the negation of 3n 6 S. P (n ) is V// e S, ~ BC/O: If « e S , then n 2 + n + 1 is not prime. is false since, for exam ple. 1 e S and l 2 + 1 + 1 = 3 is prim e. In Section 2.10 w e began a discussion o f quantified statem ents containing two quantifiers. We continue this discussion here. Example 7.2
C onsider P { s , t): 2s + 3‘ is prim e.
7.2
Revisiting Quantified Statements
175
where s is a positive even integer and t is a positive odd integer. I f we let S denote the set o f positive even integers and T the set o f positive odd in tegers, then the quan tified statement 3s e S, 3r e T , P (s , t). can be expressed in words as There exist a positive even integer s and a positive odd integer t such that 2s + 3' is prime.
(7.1)
The statement (7.1) is true since P { 2, 1): 22 + 3 1 — 1 is prime, is true. On the other hand, the quantified statement Vs e S , V t e / . P (s, t ). can be expressed in words as For every positive even integer s and every positive odd integer t, 2 s + 3 ' is prime.
^ .2 )
The statement (7.2) is fa ls e since P ( 6, 3): 26 + 33 = 91 is a prime, is false, ® 91 = 7 • 13 is not a prime.
♦
L et P ( s , t ) be an open sentence, w here the dom ain of the variable 5 is S and the dom ain o f the variable t is T . R ecall that the negations o f the quantified statem ents 3s e S, 3t e f , P{s, t) and Vs e S, V? e T , P ( s , t) are  (3 i e 5, I t g T . P (s, /)) = Vs G S, v r e T, ~ P ( s , t). and
~ (Vs e S , V / e
. /J(.v. r)) = 3s e S, 3r g T, ~ /’(.v. 0
Therefore, the negation o f the statem ent (7.1) is F or every positive even integer s and every positive odd integer t , 2s + 3' is not prime. w hich is a false statem ent. O n the other hand, the negation of the statem ent (7.2) is There exist a positive even integer s and a positive odd integer t such that 2s + 3‘ is not prim e. w hich is a true statem ent. We have seen that quantified statem ents m ay also contain different kinds of quan tifiers. For exam ple, it follow s by the definition o f an even integer that for every even
176
Chapter 7
Prove or Disprove
integer n, there exists an integer k such that n = 2k. There is another m athem atical sym bol w ith w hich you should be familiar. The sym bol 3 denotes the phrase such that (although som e m athem aticians sim ply w rite s.t. for “such that”). For exam ple, let S denote the set o f even integers again. Then Vra e S, 3 k Z 3 / i = 2k.
(7.3)
states: For every even integer n, there exists an integer k such that n ~ 2k. T his statem ent can be rew orded as: I f n is an even integer, then n = 2k f o r some integer k. If w e interchange the tw o quantifiers in (7.3), we obtain, in words: There exists an even integer n such that f o r every integer k, n = 2k. This statem ent can also be rew orded as There exists an even integer n such that n = 2k f o r every integer k. This statem ent can be expressed in sym bols as 3n
e S,Vfc€ Z , n
= 2k.
(7.4)
Certainly, the statem ents (7.3) and (7.4) say som ething totally different. Indeed, (7.3) is true and (7.4) is false. A nother such exam ple of this is For every real num ber x , there exists an integer n such that \x — n\ < 1.
(7.5)
T his statem ent can also be expressed as I f x is a real num ber, then there exists an integer n such that x — n\ < 1. In order to state (7.5) in sym bols, let P ( x , n): \x — n \ < 1. w here the dom ain o f the variable x is R and the dom ain o f the variable n is Z. Thus, (7.5) can be expressed in sym bols as Vx e R , 3n e Z, P ( x , n). The statem ent (7.5) is true, as w e now verify. In the pro o f o f the follow ing result, we w ill refer to the ceiling [a ] o f a real num ber, w hich is the sm allest integer greater than or equal to x . Result 7.3 Proof
For every real num ber x , there exists an integer n such that \x — n\ < 1. L et x be a real num ber. If we let n =
X;T>then
x — n\ = \x — fx ]  = fx]  x < 1. ■
7.2
Revisiting Quantified Statements
177
A nother exam ple o f a quantified statem ent containing two different quantifiers is T here exists a positive even integer m such that for every positive integer
n , \ ~ —^  <
5.
(7.6)
L et S denote the set o f positive even integers and let P (vm ,’n )7: II1 m  n±1I< — ±2 w here the dom ain o f the variable m is 5 and the dom ain o f the variable n is N. Thus, (7.6) can be expressed in sym bols as Bm e S, V/i e N, P (m , n). T he truth o f the statem ent (7.6) is now verified.
1
1
m
n
C onsider m = 2 . Let n be a positive integer. We consider three cases. 1J_ —  1 I 1 1I _ 1m n 1“ 12 11 — 11 —  1 1I 21 1m n1 IJ_ —  1 _ I 1 —  1 _ n 1~ 1m n 1 “ 12
1 2
II O
Proof
There exists a positive even integer m such that f o r every positive integer n,
11 N >1—
Result 7.4
1 2
Thus 15 —  1 < I for every n e N.
■
L et P(s, t) be an open sentence, w here the dom ain o f the variable s is S and the dom ain o f the variable t is T . The negation o f the quantified statem ent Vse S, Bt e T, P(s, t) is
~ (Vs <= S, Bt e T, P(s, t)) = Bs e S, ~ (Bt e T, P(s, t))
= 3s 6 S, Vf 6 T, ~ P(s, t); w hile the negation o f the quantified statem ent 3s e S, Vf e T, P(s, t) is
~ (3s e S, Vr e T, P(s , t)) = Vs e S, ~ (Vt e T, P(s, t)) = Vs e S, Bt e T, ~ P(s, t). Consequently, the negation o f the statem ent (7.5) is There exists a real num ber x such that for every integer n, \x — n\
> 1.
T his statem ent is therefore false. The negation o f the statem ent (7.6) is For every positive even integer m , there exists a positive integer n such that
I i _ II > 1
Im
T his too is false.
nI
2‘
178
Chapter 7
Prove or Disprove
L e t’s consider the follow ing statem ent, w hich has m ore than two quantifiers. For every positive real num ber e, there exists a positive real num ber d such that for every real n u m b e rx , [x < d im plies that 2 x  < e.
(7.7)
If w e let P ( x , d ): \x\ < d . and Q ( x ,e ) : \2x\ < e. w here the dom ain o f the variables e and d is R + and the dom ain of the variable x is R, then (7.7) can be expressed in sym bols as Ve e R + . 3d e R + , Vx e R, P ( x , d) =»Q (x , e). The statem ent (7.7) is in fact true, w hich we now verify. Result 7.5
Proof
For every positive real number e, there exists a positive real num ber d such that i f x is a real num ber with x  < d, then \2 x\ < e. Let e be a positive real number. N ow choose d = e /2 . L et x be x  < d = e /2 . Then
a real num ber with
[2x1 = 2x < 2 ( ^ ) = e , as desired.
h
7.3 Testing Statem ents We now turn our attention to the m ain topic o f this chapter. For a given statem ent w hose truth value is not provided to us, our task is to determ ine the truth or falseness o f the statem ent and, in addition, show that our conclusion is correct by proving or disproving the statem ent, as appropriate. Example 7.6
Prove or disprove: There is a real num ber solution o f the equation x 6 + 2x2 + 1 = 0 .
Strategy
O bserve that x 6 and x 2 are even pow ers o f x . Thus if x is any real num ber, then x 6 > 0 and x 2 > 0, so 2 x 2 > 0. A dding 1 to x 6 + 2 x 2 shows that x 6 + 2 x 2 + 1 > 1. H ence it is im possible for x 6 + 2.i2 + 1 to be 0. These thoughts lead us to our solution. We begin by inform ing the reader that the statem ent is false, so the reader know s w hat w e w ill be trying to do. f
Solution of Example 7.6
The statem ent is false. L et x e R. Since x 6 > 0 and x 2 > 0, it follow s that x 6 + 2 x 2 + 1 > 1 and so x 6 + 2 x 2 + 1 ^ 0 . t For the preceding exam ple, we w rote “ Strategy” rather than “P roof Strategy” for two reasons: (1) Since the statem ent m ay be false, there m ay be no pro o f in this case. (2) We
7.3
Testing Statements
179
are essentially “thinking out loud,” trying to convince ourselves w hether the statem ent is true or false. O f course, if the statem ent turns out to be true, then our strategy may very well turn into a proof strategy. Example 7.7
Prove or disprove: Let x , y , z e Z. Then two of the integers x, y and z are of the same parity.
Strategy
For any three given integers, either two are even or two are odd. So it certainly seems as if the statem ent is true. The only question appears to be w hether w hat w e said in the preceding sentence is convincing enough to all readers. We try another approach. ♦
S o lution o f E x a m p le 7.7 Proof
The statem ent is true. C onsider x and y. If x and y are o f the same parity, then the p ro o f is com plete. Thus we m ay assum e that x and y are of opposite parity, say x is even and y is odd. If z is even, then x and z are o f the sam e parity; w hile if z is odd, then >> and z are o f the same parity. ■ O f course, the preceding p ro o f could have been done by cases as well.
Example 7.8
Prove or disprove: Let A , B and C be sets. If A x C — B x C , then A = B.
Strategy
The elem ents o f the set A x C are ordered pairs of elem ents, nam ely they are of the form (x. y), w here x e A and y 6 C. L et (x, y ) e A x C. If A x C = B x C, then it follow s that (x, y) m ust be an elem ent o f B x C as well. This says that x e B and y e C . Conversely, if (x, y ) e B x C , then (x, y ) e A x C , w hich im plies that x e A as well. These observations certainly seem to suggest that it should be possible to show that A = B under these conditions. However, this argum ent depends on A x C containing an elem ent (x, y). C ould it happen that A x C contains no elem ents? If A or C is empty, this w ould happen. However, if C / 0 and A x C = 0, then A m ust be empty. But B x C = A x C = & w ould m ean that B m ust also be em pty and so A = B. This suggests a different response. ♦
Solution o f E x a m p le 7.8
The statem ent is false. L et A = {1}, B = {2} and C = 0. T hen A x C = B x C = 0, but A ^ B . H ence these sets A, B and C form a counterexam ple. ♦ In some instances, w e m ight consider m odifying a false statem ent so that the revised statem ent is true. O ur preceding discussion seem s to suggest that if the set C were required to be nonem pty, then the statem ent w ould have been true.
Result 7.9 Proof
Let A , B and C be sets, where C ^ 0. If A x C — B x C , then A — B . A ssum e that A x C = B x C. Since C / 0, the set C contains som e elem ent c. Let x e A. Then (x, c) e A x C. Since A x C = B x C , it follow s that (x, c) e B x C. H ence x € B and so A c B. By a sim ilar argum ent, it follow s that B c A. Thus A = B. m
180
Chapter 7 Example 7.10
Strategy
Solution of Example 7.10 Proof
Prove or Disprove
Prove or disprove: There exists a real num ber % such that x 3 < x < x 2. If there is a real num ber x such that ,v3 < x < x 2, then this num ber is certainly not 0. Consequently, any real num ber x w ith this property is either positive or negative. If x > 0, then we can divide x 3 < x < x 2 by x , obtaining x 2 < 1 < x . However, if x > 1, th e n x 2 > 1. T herefore, there is no positive real num ber x for w hich x 3 < x < x 2. Hence any real num ber x satisfying x 3 < x < x 2 m ust be negative. D ividing x 3 < x < x 2 by x gives us x 2 > 1 > x or x < 1 < x 2. E xperim enting w ith some negative num bers tells us that any num ber less than —1 has the desired property. ♦
The statem ent is true.
C onsider x = —2. T hen x 3 = —8 and x 2 = 4. T hus x 3 < x < x 2.
a
Example 7.11
Prove or disprove: For every positive irrational number b, there exists an irrational num ber a such that 0 < a < b.
Strategy
We begin w ith a positive irrational num ber b. If this statem ent is true, then we m ust show that there is an irrational num ber a such that 0 < a < b. If w e let a = b j 2, then certainly 0 < a < b. The only question is w hether b / 2 is necessarily irrational. We have seen, however, that b / 2 is irrational (in Exercise 5.17 in Section 5.2). ♦
Solution of Example 7.11 Proof
Example 7.12
Strategy
Solution of Example 7.12 P roof
The statem ent is true.
Let b be a positive irrational num ber. Now let a = b /2 . T hen 0 < a < b and a is irrational by Exercise 5.17 in Section 5.2. m Prove or disprove: Every even integer is the sum o f three distinct even integers. T his statem ent can be rew orded in a variety o f w ays. O ne rew ording o f this statem ent is: If n is an even integer, then there exist three distinct even integers a, b and c such that n = a + b + c. W hat this statem ent does not say is that the sum o f three distinct even integers is even; that is, w e do not begin w ith three distinct even integers and show that their sum is even. We begin w ith an even integer n and ask w hether w e can find three distinct even integers a, b and c such that r — a + b + c. This is certainly true for n = 0 since 0 = ( —2) + 0 + 2. It is also true for n — 2 since 2 = ( —2) + 0 + 4. If n = 4, then 4 = ( —2) + 2 + 4. T his last exam ple m ay suggest a pro o f in general. For every even integer n, w e can write n = 2 + (  2 ) + n. Certainly, n, 2 and  2 are even. B ut are they distinct? They are not distinct if n = 2 or n — —2. This provides a p lan for a proof. 4 T he statem ent is true.
Let n be an even integer. We show that n is the sum o f three distinct even integers by considering the follow ing three cases.
7.3
Testing Statements
181
Case 1. n — 2. O bserve that 2 = (—2) + 0 + 4. Case 2. n — —2. O bserve that —2 = ( —4) + 0 + 2. Case 3. n ^ 2, —2. Then n — 2 + (—2) + n. Example 7.13 Strategy
e
Prove o r disprove: L et k e N. I f k 2 + 5k is odd, then (k + l ) 2 + 5(k + 1) is odd. O ne idea that m ight occur to us is to assum e that k2 + 5k is an odd integer, w here k e N, and see if we can show that (k + l ) 2 + 5 (k + 1) is also odd. If k2 + 5k is odd, then we can w rite k2 + 5k — 21 + 1 for som e integer t . T hen
(k + l) 2 + 5(k + l) = k2 + 2k + l + 5 k + 5 = (k2 + 5k) + (2k + 6) = (21 + 1) + (2k + 6) = (21 + 2k + 6) + 1 = 2(l + k + 3) + 1, w hich is an odd integer and we have a proof. S olution o f E x a m p le 7.13
Proof
♦
T he statem ent is true.
A ssum e that k 2 + 5k is an odd integer, w here k e N. T hen k2 + 5k = 21 + 1 for some integer i. H ence
(k + l ) 2 + 5(k + 1) = k2 + 2k + 1 + 5k + 5 = (k2 + 5k) + (2k + 6) = (21 + 1) + (2k + 6) = (2l + 2k + 6 ) + \ = 2(1 + k + 3) + 1. Since I + k + 3 is an integer, (k + l ) 2 + 5 (k + 1) is an odd integer. Example 7.14 Strategy
Solution o f E x a m p le 7.14
■
Prove or disprove: For every positive integer n, n2 + 5 n is an odd integer. It seem s like the reasonable thing to do is to investigate n 2 + 5n for a few values of n. For n = 1, w e have n2 + 5n = 1 + 5 • 1 = 6. We have already solved the problem ! For n = 1, n2 + 5 n is not an odd integer. We have discovered a counterexam ple. ♦ The statem ent is false. For n is a counterexam ple.
= 1, n2 + 5n = 1 + 5 • 1 = 6, w hich is even. Thus n = ♦
L ooking at Exam ples 7.13 and 7.14 again, w e m ight be w ondering w hat exactly is going on. Certainly, these two exam ples seem to be related. Perhaps the follow ing thought m ay occur to us. For each positive integer n, let
P(n): The integer n2 + 5 n is odd. and consider the (quantified) statem ent For every positive integer n, n 2 + 5n is odd. or in sym bols, Vn e N, P(n).
(7.8)
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Chapter 7
Prove or Disprove
We m ight ask w hether (7.8) is true. B ecause o f the dom ain, a p ro o f by induction seem s appropriate. In fact, the statem ent in Exam ple 7.13 is the inductive step in an induction pro o f o f (7.8). By E xam ple 7.13, the inductive step is true. On the other hand, the statem ent (7.8) is false as n = 1 is a counterexam ple. This em phasizes the im portance o f verifying both the basis step and the inductive step in an induction proof. R eturning to Exam ple 7.13 once again, we can show (using a p ro o f by cases) that k 2 + 5k is even for every k e N, w hich w ould provide a vacuous pro o f o f the statem ent in Exam ple 7.13. In this chapter, we have discussed analyzing statem ents, particularly understanding statem ents, determ ining w hether they are true or false and proving or disproving them. A ll o f the statem ents that we have analyzed were provided to us. B ut how do w e obtain statem ents to analyze for ourselves? This is an im portant question and concerns the creative aspect of m athem atics— how new m athem atics is discovered. O bviously, there is no rule or form ula for creativity, but creating new statem ents often com es from studying old statem ents. L e t’s illustrate how we m ight create statem ents to analyze. In Exercise 4.6, you w ere asked to prove the follow ing: Let a e Z. If 3  2a, then 3  a.
(7.9)
W hat other statem ents does this suggest? For exam ple, is its converse true? (The answ er is yes, but the converse is not very interesting.) Is (7.9) true if 3 and 2 are interchanged? W hat integers can we replace 2 by in (7.9) and obtain a true statem ent? T hat is, for w hich positive integers k is it true that if 3  ka , then 3  a ? O f course, this is true for k = 1. A nd we know that it’s true for k = 2. It is not true for k = 3; that is, it is not true that if 3  3a, then 3  a. The integer a = 1 is a counterexam ple. O n the other hand, it is possible to prove that if 3  4a, w here a e Z, then 3  a. W hat we are attem pting to do is to extend the result in (7.9) so that w e have a result o f the type: Let a 6 Z. If 3  ka, then 3  a.
(7.10)
for a fixed integer k greater than 2. We w ould like to find a set S o f positive integers such that the follow ing is true: L et a e Z. If 3  k a , w here k e S, then 3  a.
(7.11)
Surely 2 e S. R esult (7.9) then becom es a special case and a corollary o f (7.11). For this reason, (7.11) is called a g e n e ra liz a tio n o f (7.9). Ideally, w e w ould like S to have the added property that (7.11) is true if k e S, w hile (7.11) is false if k £ S. We are thus seeking a set S o f integers such that the follow ing is true: L et a e Z. T hen 3  ka im plies that 3  a if and only if k e S. If w e w ere successful in finding this set S, then w e m ight start all over again by replacing 3 in (7.9) by some other positive integer. In m athem atics it is often the case that a new result is obtained by looking at an old result in a new w ay and extending it to obtain a generalization o f the old result. H ence it frequently happens that: Today’s theorem becomes tomorrow's corollary. We conclude this chapter w ith a quiz. Solutions are given follow ing the quiz.
7.3
Testing Statements
183
Q uiz Prove or disprove each of the following statements. 1. If n is a positive integer and s is an irrational number, then n /s is an irrational number. 2. For every integer b, there exists a positive integer a such that \a  \b\\ < 1. 3. If x
and y are integers of the same parity, then x y and (x + y Y are of the same parity.
4. Let a, b e Z. If 6 / a b , then either (1) 2 I a and 3 l b or (2) 3 / a and 2 Kb. 5. For every positive integer n. 22” > A"'. 6. If A, B and C are sets, then (A — B ) U (A — C ) = A — (B U C). 7. Let n e N. If (n + 1)0? + 4) is odd, then (n + 1)(/? + 4) + 3" is odd. 8. (a) There exist distinct rational numbers a and b such that (a — \ )(h — 1) = (b) There exist distinct rational numbers a and b such that 
1. \ = 1.
9. Let a . b . c e Z. If every two of a, ft and c are of the same parity, then a + b + c iseven. 10. If n is a nonnegative integer, then 5 divides 2 ■4" + 3 • 9”.
S o lu tio n s fo r Q uiz 1. The statement is true. P roo f Assume, to the contrary, that there exist a positive integer nand an irrational number ssuch that n /s is a rational number. Then n / s = a /b , where a . b e Z and a, b # 0. Therefore, s = n b /a , where nb, a e Z and a / 0. Thus s is rational, producing a contradiction. ■ 2. The statement is true. P roof
Let b e Z. Now let a = h\ + 1. Thus a e N and a  fc[ = (6  + 1)  Z? = 1.
b
3. The statement is false. Observe that  = 1 and v = 3 are of the same parity. Then ,vy = 3 and (x + y ) 2 = 16 are of opposite parity. Hence X = 1 and y = 3 produce a counterexample. ♦ 4. The statement is false. Let a = b = 2. So ab = 4. Hence 6 J a b . Since 2 ' a and 2 ' b. false. Thus a = b = 2 constitute a counterexample.
both (1) and (2) are t
5. The statement is false. For n = 3, 22 = 28 = 256 while 4"' = 43, = 46 = 4096. Thus 2~ < 43' and so n = 3 is a counterexample. ^ 6. The statement is false. Let A = {1, 2, 3}, B = {2} and C = {3}. Thus B U C = {2, 3}. Hence A — B = {1. 3}, A  C = {1, 2} and A  (B U C) = {1}. Therefore, (A  B) U (A  C) = {1, 2, 3} / A  (S U C). So A = {1, 2, 3}, B = {2} and C = {3} constitute a counterexample. ♦ 7. The statement is true. P roof Let n e N and consider (n + l)(n + 4). We show that (n + 1)0? + 4) is even, thereby giving a vacuous proof. There are two cases. Case 1. n is even. Then n = 2k for some integer k. Thus (n + 1)(7? + 4) = (2k + 1 )(2k + 4) = 2(2 k + 1)(k + 2). Since (2k + l)(k + 2) e Z, it follows that (n + 1)(« + 4) is even.
184
Chapter 7
Prove or Disprove
Case 2. n is odd. Then n = 2 1 + 1 for some integer I. Thus (n + 1)(« + 4) = (21 + 2)(2€ + 5) = 2(1 + l)(Z f + 5). Since (I + 1)(21 + 5) e Z, it follows that (n + l)(n + 4) is even. 8. (a) The statement is true. P ro of
Let a = 3 and b =  . Then (a — 1)(b — 1) = 2 ( i ) = 1.
(b) The statement is true. P roof
Let a = \ and b = —1. Then
P roof Analysis
Observe that if a and b are two (distinct) rational numbers that satisfy  + g = 1, then
= 1 and so a + b = ab. Thus ab — a — b = 0, which is equivalent to ab — a — 6 + 1 = 1 and so (a — 1)(b — 1) = 1. Therefore, two distinct rational numbers a and b satisfy (<a  1)(b  1) = 1 if and only if a and b satisfy
if and only if a and b satisfy a + b = ab.
^
9. The statement is false. Let a = 1 ,6 = 3 and c = 5. Then every two of a, b and c are of the same parity; yet a + b + c. is odd. Hence a = 1 ,6 = 3 and c = 5 produce a counterexample. # 10. The statement is true. P roo f We proceed by induction. For n = 0, 2 • 4" + 3 • 9" = 2 • 1 + 3 • 1 = 5. Thus 5  (2 ■4° + 3 ■9°) and the statement is true for n = 0. Assume that 5  (2 • 4* + 3 ■9k) for a nonnegative integer k. We show that 5  (2 • Ak+l + 3 • 9A+1). Since 5  (2 ■4k + 3 • 9k), it follows that 2 • # + 3 • 9k = 5x for some integer x. Thus 2 • 4fe s= 5x — 3 ■9k . Hence 2 . 4k+' + 3 • 9i+1 = 4(2 ■4k) + 3 ■9^+' = 4(5.v  3 • 9k) + 3 • 9k + 1 = 20x  12 • 9k + 27 • 9k = 20.v + 15 • 9k = 5(4.v + 3 • 9*). Since 4x + 3 • 9k e Z, it follows that 5  (2 • 4k+l + 3 ■9i+1). By the Principle of M athematical Induction, 5 divides 2 ■4" + 3 ■9” for every nonnegative integer n . ■
Exercises for Chapter 7
185
EXERCISES FOR CHAPTER 7 Section 7.1: Conjectures in Mathematics 7.1. Consider the following sequence of equalities: 1 = 0+1 2 + 3 + 4 = 1 +8 5 + 6 + 7 + 8 + 9 = 8 + 27 1 0 + 1 1 + 1 2 + 1 3 + 1 4 + 1 5 + 16 = 27 + 64 (a) W hat is the next equality in this sequence? (b) W hat conjecture is suggested by these equalities? (c) Prove the conjecture in (b) by induction. 7.2. Consider the following statements: (1 + 2 )2  l 2 = 2 3
(1 + 2 + 3)2  (1 + 2 )2 = 33 (1 + 2 + 3 + 4)2  (1 + 2 + 3)2 = 43 (a) Based on the three statements given above, what is the next statement suggested by these? (b) W hat conjecture is suggested by these statements? (c) Verify the conjecture in (b). 7.3. A sequence {a,,} of real numbers is defined recursively by ai = 2 and for n > 2, an =
2 + 1 ■a\ + 2 ■a\ + 1 (n  1)a2_l n
.
(a) Determine <23 and <24. (b) Clearly, an is a rational number for each n e N. Based on the information in (a), however, what conjecture does this suggest? 7.4. It has been stated that the German mathematician Christian Goldbach is known for a conjecture he made concerning primes. We refer to this conjecture as Conjecture A. C o n jecture A Every even integer at least 4 is the sum o f two primes. Goldbach made two other conjectures concerning primes. C o n jecture B Every integer at least 6 is the sum o f three primes. C o n jecture C Every odd integer at least 9 is the sum o f three odd primes. Prove that the truth of one or more of these three conjectures implies the truth of one or two of the other conjectures. 7.5. By an ordered partition of an integer n > 2 is meant a sequence of positive integers whose sum is n. For example, the ordered partitions of 3 are 3,1 + 2, 2 + 1, 1 + 1 + 1. (a) Determine the ordered partitions of 4. (b) Make a conjecture concerning the number of ordered partitions of an integer n > 2. 7.6. Two recursively defined sequences {a„} and {bn} of positive integers have the same recurrence relation, namely a„ = 2a„_i + a „ _ 2 and bn = 2bn \ + bn 2 for n > 3. The initial values for {a„} are a\ — 1 and a 2 = 3, while the initial values for {b„} are b\ = 1 and b 2 = 2. (a) Determine a 3 and <24. (b) Determine whether the following is true or false: C onjecture: an  2”" 2 • n + 1 for every integer n > 2 .
186
Chapter 7
Prove or Disprove
(c) Determine h3 and %, (d) Determine whether the following is true or false: C onjecture: bn =
for every integer n > 2.
7.7. We know that 1 + 2 + 3 = 1  2  3 ; that is, there exist three positive integers whose sum equals their product. Prove or disprove (a) and (b). (a) There exist four positive integers whose sum equals their product. (b) There exist five positive integers whose sum equals their product. (c) What conjecture does this suggest to you? 7.8. Observe that 3 = 1 + 2 , 6 = 1 + 2 + 3, 9 = 4 + 5 and 12 = 3 + 4 + 5. (a) Show that 13 and 14 can be expressed as the sum of two or more consecutive positive integers. (b) Make a conjecture as for which integers n > 3 can be expressed as the sum of two or more consecutive positive integers. (c) Prove your conjecture in (b). [Note that every positive integer n can be expressed as n = 2rs, where r is a nonnegative integer and 5 is a positive odd integer.] 7.9. An m x n checkerboard has m > 2 rows, n > 2 columns and mn squares in all. Two squares are adjacent if they belong to the same row or column and there is no square strictly between them.Conjecture for which integers m and n it is possible to number the squares from 1 to mn such that consecutively numbered squares are adjacent as are the squares numbered 1 and mn.
Section 7.2: Revisiting Quantified Statements 7.10. (a) Express the following quantified statement in symbols: For every odd integer n, the integer 3n + 1 is even. (b) Prove that the statement in (a) is true. 7.11. (a) Express the following quantified statement in symbols: There exists a positive even integer n such that 3n + 2” ~ 2 is odd. (b) Prove that the statement in (a) is true, 7.12. (a) Express the following quantified statement in symbols: For every positive integer n, the integer n"~l is even. (b) Show that the statement in (a) is false. 7.13. (a) Express the following quantified statement in symbols: There exists an integer n such that 3n 2 — 5n + 1 is an even integer. (b) Show that the statement in (a) is false. 7.14. (a) Express the following quantified statement in symbols: For every integer n > 2, there exists an integer m such that n < m < 2n. (b) Prove that the statement in (a) is true. 7.15. (a) Express the following quantified statement in symbols: There exists an integer n such that in in — 3) < 1 fo r every integer m . (b) Prove that the statement in (a) is true. 7.16. (a) Express the following quantified statement in symbols: For every integer n, there exists an integer m such that (n — 2)(m — 2) > 0. (b) Express in symbols the negation of the statement in (a). (c) Show that the statement in (a) is false. 7.17. (a) Express the following quantified statement in symbols: There exists a positive integer n such that —nm < 0 fo r every integer m.
Exercises for Chapter 7
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(b) Express in symbols the negation of the statement in (a). (c) Show that the statement in (a) is false. 7.18. (a) Express the following quantified statement in symbols: For every positive integer a, there exists an integer b with \b\ < a such that \bx\ < a fo r every real number x. (b) Prove that the statement in (a) is true. 7.19. (a) Express the following quantified statement in symbols: For every real number x, there exist integers a and b such that a < x < b and b —a = 1. (b) Prove that the statement in (a) is true. 7.20. (a) Express the following quantified statement in symbols: There exists an integer n such that fo r every two real numbers x and y , x 2 + y 2 > n. (b) Prove that the statement in (a) is true. 7.21.
(a) Express the following quantified statement in symbols: For every even integer a and odd integer b, there exists arational number c such that either a < c < b or b < c < a. (b) Prove that the statement in (a) is true.
7.22. (a) Express the following quantified statement in symbols: There exist two integers a and b such that fo r every positive integer n, a < ~ < b. (b) Prove that the statement in (a) is true. 7.23. (a) Express the following quantified statement in symbols: There exist odd integers a, b and c such that such that a + b + c= (b) Prove that the statement in (a) is true. 7.24.
1.
(a) Express the following quantified statement in symbols: For every three odd integers a , b and c, their product abc is odd. (b) Prove that the statement in (a) is true.
7.25. Consider the following statement. R : There exists a real number L such that for every positive real number e, there exists a positive real number d such that if x is a real number with .r < d, then 3x — L\ < e. (a) Use P (x , d): \x\ < d and Q ( x , L , e): \3x — L\ < e to express the statement R in symbols. (b) Prove that the statement R is true. 7.26. Prove the following statement. For every positive real number a and positive rational number b, there exist a real number c and irrational number d such that ac + bd = 1. 7.27. Prove the following statement. For every integer a, there exist integers b and c such that \a — b\ > cd for every integer d.
Section 7.3: Testing Statements 7.28. For the set 5 = {1, 2, 3, 4}, let P(n): 2"+1 + (—1)"+1 (2" + 2'!_1) is prime, and Q{n): 2n + 3 is prime. Prove or disprove: V« e S, P(n) => Q{n). 7.29. Let P(n): n 2 + 3/; + 1 is even. Prove or disprove: (a) V* e N, P(k) =$■ P (k + 1). (b) V« e N, P(n).
188
Chapter 7
Prove or Disprove
F o r each of the exercises 7.307.81: Prove or disprove. 7.30. Let x € Z. If 4x + 7 is odd, then x is even. 7.31. For every nonnegative integer n, there exists a nonnegative integer k such that k < n. 7.32. Every even integer can be expressed as the sum of two odd integers. 7.33. If x , y , z 6 Z such that x + y + z = 101, then two of the integers x , y and z are of opposite parity. 7.34. For every two sets A and B , (A
U
B) — B = A.
7.35. Let A be a set. If A n B = 0 for every set B, then A = 0. 7.36. There exists an odd integer, the sum of whose digits is even and the product of whose digits is odd. 7.37. For every nonempty set A, there exists a set B such that A U B = 0 . 7.38. If x and y are real numbers, then x + y \ = x + y . 7.39. Let S be a nonempty set. For every proper subset A of S, there exists a nonempty subset B of S such that A U B = S and A n B = 0. 7.40. There is a real number solution of the equation x 4 + x 2 + 1 = 0 . 7.41. There exists an integer a such that a ■c > 0 for every integer c. 7.42. There exist real numbers a, b and c such that
= .
7.43. If x, y e R a n d x 2 < y 2, th e n x < J . 7.44. Let x, y, z e Z. If z = x — y and z is even, then x and y are odd. 7.45. Every odd integer can be expressed as the sum of three odd integers. 7.46. Let x, y , z e Z. If z = x + y and x is odd, then y is even and z is odd. 7.47. For every two integers a and c, there exists an integer b such that a + b = c. 7.48. Every even integer can be expressed as the sum of two even integers. 7.49. For every two rational numbers a and b with a < b, there exists a rational number r such that a < r < b. 7.50. Let A. B , C and D be sets with A
C
C and B
C
D. If A and B are disjoint, then C and D are disjoint.
7.51. Let A and B be sets. If A U B ^ 0, then both A and B are nonempty. 7.52. For every two positive integers a and c, there exists a positive integer b such that a + b = c. 7.53. For every odd integer a, there exist integers b and c of opposite parity such that a + b = c. 7.54. For every rational number a /b , where a, b e N, there exists a rational number e /d , where c and d are positive odd integers, such that 0 < ^ <  . 7.55. The equation x 3 + x 2 
1 = 0 has a real number solution between x = 0 and x = 1.
7.56. There exists a real number x such that x 2 < x < x 3. 7.57. Let A and B be sets. If A — B = B — A, then A — B = 0. 7.58. If x e Z, then
== j f c j .
7.59. For every positive rational number b, there exists an irrational number a with 0 < a < b. 7.60. Let A be a set. If A  B = 0 for every set B , then A = 0. 7.61. Let A, B and C be sets. If A n B = A fl C, then B = C. 7.62. For every nonempty set A, there exists a set B such that \A — B\ = \B — A . 7.63. Let A be a set. If A U B / 0 for every set £ , then A ^ 0. 7.64. There exist an irrational number a and a rational number bsuch that ab isirrational.
Additional Exercises for Chapter 7
189
7.65. There exists a real number solution of the equation x 2 + x + 1 = 0. 7.66. For every two sets A and B , V ( A
U
B) = V {A )
U
V (B ).
7.67. Every nonzero rational number can be expressed as the product o f two irrational numbers. 7.68. If A and B are disjoint sets, then V ( A ) and V ( B ) are disjoint. 7.69. Let S be a nonempty set and let T be a collection of subsets of S. If A n B ^ 0 for all pairs A , B of elements of T , then there exists an element x € S such that x e C for all C e T . 7.70. Let A, B and C be sets. Then A U ( B  C ) = ( A U 5 )  ( A U C). 7.71. Let a, b, c e Z. If ab, ac and be are even, then a, b and c are even. 7.72. Let n € Z. If »3 + n is even, then n is even. 7.73. There exist three distinct integers a, b and c such that ab = bc. 1.14. Let a, b , c e Z. Then at least one of the numbers a + b, a + c and b + c is even. 7.75. Every integer can be expressed as the sum of two unequal integers. 7.76. There exist positive integers x and y such that x 2 — y 2 = 101. 7.77. For every positive integer n, n 2 — n + 11 is a prime. 7.78. For every odd prime p, there exist positive integers a and b such that a 2 — b 2 = p. 7.79. If the product of two consecutive integers is not divisible by 3, then their sum is. 7.80. The sum of every five consecutive integers is divisible by 5 and the sum of no six consecutive integers is divisible by 6. 7.81. There exist three distinct positive real numbers a , b , c, none of which are integers, such that all of ab, ac, bc and abc are integers.
ADDITIONAL EXERCISES FOR CHAPTER 7 7.82. (a) Show that the following statement is false: For every natural number x, there exists a natural number y such that x < y < x 2. (b) Make a small addition to the statement in (a) so that the new statement is true. Prove the new statement. 7.83. (a) Show that the following statement is false: Every positive integer is the sum of two distinct positive odd integers. (b) Make a small addition to the statement in (a) so that the new statement is true. Prove the new statement. 7.84. (a) Prove or disprove: There exist two distinct positive integers whose sum exceeds their product. (b) Your solution to (a) should suggest another problem to you. State and solve this new problem. 7.85. (a) Prove or disprove: If a and b are positive integers, then + b = J a + ~Jb. (b) Prove or disprove: There exist positive real numbers a and b such that y/a + b = ~fa + \[b. (c) Complete the following statement so that it’s true and provide a proof: Let a, b e R + U {0}. Then + b = yfa + \ f b if and only i f ____________________. 7.86. Consider the open sentence P(n) : nl > ( ) " for n e N. Prove or disprove:
e N, P(n).
7.87. Evaluate the proof of the following statement. R esult Every even integer can be expressed as the sum of three distinct even integers. P roo f Let n be an even integer. Since n + 2, n — 2 and —n are distinct even integers and n = (n + 2) + (n  2) + (  n ), the desired result follows.
■
Chapter 7
190
Prove Or Disprove
7.88. It is known (although challenging to prove) that for every nonnegative integer m, the integer 8m + 3 can be expressed as a 2 + b 2 + c2 for positive integers a, b and c. (a) For every integer m with 0 < m < 10, find positive integers a, b and c such that 8m + 3 = a 2 + b 2 + c2. (b) Prove or disprove: If a , b and c are positive integers such that a 2 + b 2 + c 2 = 8 m + 3 for some integer m, then all of a, b and c are odd. 7.89. In Exercise 6 in Chapter 4, you were asked to prove the statement P: Let a e Z. If 3 ] 2a, then 3  a. (a) (b) (c)
Prove that the converse of P is true. Now state P and its converse in Is the statement obtained by interchanging 2 and 3 in P true? Find a set S of positive integers with 2 e S and S > 3 such that the
a
more familiar manner.
following is true:
Let a 6 Z. If 3  ka, where k e S, then 3  a. Prove this generalization of the statement P . (d) In Exercise 72 in Chapter 4, it was shown for integers a and b that 3  ab if and only if 3  a or 3  b. How can this be used to answer (c)? 7.90. In Exercise 20 in Chapter 5, you were asked to prove that \/2 + V3 is irrational. (a) Prove that J 2 + \f5 is irrational. (b) Determine, with proof, another positive integer a such that ~Jl + (c) State and prove a generalization of the result in (a).
is irrational.
7.91. In Exercise 27 in Chapter 3, you were asked to prove the statement: P: If n € Z, then n 3 — n is even. This can be restated as: P: If n S Z, then 2  (n — n). (a) Find a positive integer a ^ 2 such that If n e Z. then a \ («3 —n). is true and prove this statement. (b) Find a positive integer k / 3 such that If n e Z, then 2 j (nk — n). is true and prove this statement. (c) Ask a question of your own dealing with P and provide an answer. 7.92. Let A denote the set of odd integers. Investigate the truth (or falseness) of the following statements. (a) (b) (c) (d) (e) (f)
For all x, y e A , 2 \ (x 2 + 3y 2). There exist x, y 6 A such that 4  (x 2 + For all x , y c. .1. 4  (.v2 : 3 y2). There exist x , y e A such that 8  (x 2 + There exist x , y e A such that 6  (x2 + Provide a related statement of your own
3y 2). 3y2). 3y2). and determine whether it is true or false.
Additional Exercises for Chapter 7
191
7.93. (a) Prove or disprove the following: There exist four positive integers a, b. c and d such that a 2 + b 2 + c 2 = d 2. (b) Prove or disprove the following: There exist four distinct positive integers a, b , c and cl such that a 2 + b 2 + c 2 = d 2. (c) The problems in (a) or (b) above should suggest another problem that you can solve. State and solve such a problem. (d) The problems in (a) or (b) above should suggest a conjecture to you (that you probably cannot solve). State such a conjecture.
8
E quivalence Relations
T
here are m any com m on exam ples o f relations in m athem atics. For instance, three different w ays that a real num ber x can be related to a real num ber v are: (1) x < v, (2) y = x 2 + 1 or (3) x = y.
T hree different w ays that an integer a can be related to an integer b are: (1) a  b , (2) a and b are of opposite parity or (3) a = b (m od 3). In the area o f geom etry, three different w ays that a line i in 3space can be related to a plane n in 3space are: (1) I lies on n , (2) t is parallel to n or (3) I intersects FI in exactly one point. Three different w ays that a triangle T can be related to a triangle T are: (1) T is congruent to T ', (2) T is sim ilar to T or (3) T has the sam e area as T . All of the preceding exam ples concern two sets A and B (where possibly A = B ) such that elem ents o f A are related to elem ents o f B in some manner. We now study this idea in a m ore general setting.
8.1 R elations Let A and B be tw o sets. By a re la tio n R fro m ,4 to B we m ean a subset of A x B . That is, R is a set of ordered pairs, w here the first coordinate o f the pair belongs to A and the second coordinate belongs to B . If (a, b) e R , then w e say that a is re la te d to b by R and w rite a R b. If (a. b) R. then a is not related to b by R and we w rite a b. For the sets A = {x, y. z} and B â€” {1, 2}, the set R = {(x, 2), (y, 1), (y, 2)}
192
(8.1)
is a subset o f A x S and is therefore a relation from A t o B . Thus, x R 2 (x is related to 2) and x If. 1 (x is not related to 1). For tw o given sets A and B . it is always the case that 0 and A x B are subsets o f A x B . Therefore, 0 and A x B are both exam ples of relations from A to B. (Indeed, these are the extrem e exam ples.) For the relation 0, no elem ent o f A is related to any elem ent of B \ w hile for the relation A x B, each elem ent o f A is related to every elem ent o f B . Sim ply said then, a relation from a set A to a set
8.2
Properties of Relations
193
B tells us w hich elem ents o f A are related to w hich elem ents o f B . A lthough this m ay seem like a fairly sim ple idea, it is very im portant that w e have a thorough understanding of it. Let R be a relation from A to S . The d o m a in o f R , denoted by dom ( R ), is the subset of A defined by dom (i?) â€” [a e A : ( a , b ) e R for some b e B}\ w hile the ra n g e o f R , denoted by range(ft), is the subset of B defined by range(W) = [b e B : (a, b) e R for some a e A}. Hence dom (/?) is that set o f elem ents o f A that occur as first coordinates am ong the orÂ dered pairs in R and range(i?) is the set of elem ents o f B that occur as second coordinates am ong the ordered pairs in R. The dom ain and range o f the relation R given in (8.1) are dom (R ) = {x, y } and range(7?) = {1, 2}. The reason that z ^ dom (/?) is because there is no ordered pair in R w hose first coordinate is z. Let R be a relation from A to B . By the in v erse re la tio n o f R is m eant the relation R 1 from B to A defined by / T 1 = {(b , a ) : (a , b ) e R). For exam ple, the inverse relation o f the relation R = {(x, 2), (y, 1), (y, 2)} from A = {x, y, z} to B = {1,2} is the relation / ?  ' = { ( l , y ) , ( 2 , x ) , ( 2 ,y )} from 6 to A. By a re la tio n on a set A, w e m ean a relation from A to A. T hat is, a relation on a single set A is a collection of ordered pairs w hose first and second coordinates belong to A. Therefore, {(1, 2), (1, 3), (2, 2), (2, 3)} is an exam ple o f a relation on the set A = {1, 2, 3,4}. If A = {1,2}, then A x A = {(1,1), (1, 2), (2, 1), (2, 2)}. Since  A x A = 4, the num ber o f subsets o f A x A is 24 = 16. However, a relation on A is, by definition, a subset o f A x A. Consequently, there are 16 relations on A. Six of these 16 relations are 0, {(1, 2)}, {(1, 1), (1,2)}, {(1, 2), (2, 1)}, {(1, 1), (1, 2), (2, 2)}, A x A.
8.2 Properties o f R elations For a relation defined on a single set, there are three properties that a relation may possess and w hich w ill be o f special interest to us. A relation R defined on a set A is called reflexive if x R x for every x e A. T hat is, R is reflexive if ( x , x ) e R for every x e A. L et S = {a, b, c} and consider the follow ing six relations defined on S: Ri = {(a , b), (b , a), (c , a)} R 2 = {(a, b), ib, b ), (fo, c), (c, b), (c , c)j R 3 = {(a, a ), (a , c), (b, b), (c, a), (c, c)}
194
Chapter 8
Equivalence Relations R 4 = {(fl, a), (a , b), (b , b), (b , c), (a, c)}
7?5 = {(a, a), (a, &)} = {(a , fe), (a, c)}. The relation is not reflexive since (a, a) e A’;, for exam ple. Since (a , a ) ^ i?2, it follow s that is not reflexive either. Because (a, a), (b , b), (c, c) e # 3, the relation /s’3 is reflexive. N one o f the relations R 4 , i?5, are reflexive. A relation R defined on a set A is called sy m m e tric if w henever x R y , then y R x for all x , y e A. H ence for a relation R on A to be “n ot sym m etric,” there m ust be some ordered pair (m\ z ) in R for w hich (z, w ) £ R. Certainly, if such an ordered pair ( w , z ) exists, then w / z. The relation R { is not sym m etric since (c, a) e R\ but (a , c) 0 R \. N otice that (a, b) e R \ and (/?, a) e but this does not m ean that R \ is sym m etric. Recall that the definition of a sym m etric relation R on a set A says that w henever x R y, then y R x f o r all x , y 6 A. The relation R 3 is sym m etric, however, since both [a, c) and (c, a) belong to R 3 . N one of the ordered pairs (a, a), (fo, b), (c, c) in R 3 are relevant as to w hether R 3 is sym m etric. N one o f the relations R 2 , R 4 , R$, R& are sym m etric. A relation R defined on a set A is called tra n sitiv e if w henever x R y and y R z, then x R z for all x , y, z £ A . N otice that in this definition, it is not required that x , y and z be distinct. H ence for a relation R on A to be “not transitive,” there m ust exist two ordered pairs  u, v) and (v, w ) in R such that (u, w ) £ i? . If this should occur, then necessarily u ^ v and v ^ w (although perhaps u w ). F or exam ple, the relation R i is not transitive since (a, b ), (b , c) e R 2 but (a , c) ^ Ri Actually, R \ is not transitive either because (a, b ), (b, a) e R\ but (a , a) £ R \. The exam ple (counterexam ple) that shows that R \ is not transitive illustrates the fact that show ing a relation is not transitive m ay not be easy. A ll o f the relations R$, R 4 , R 5, R 6 are transitive. It is not always easy to convince oneself that a relation is transitive either. L et's give a careful argum ent as to w hy the relations R$ and R(, are transitive. For R$ to be transitive, it is required that (x, z) belongs to R$ w henever (x, y) and (y, z) belong to R 5 for all x , y , z e A. To verify that the transitive property holds in R 5, w e m ust consider all possible pairs o f ordered pairs o f the type (x, y) and ( y, z). We have two choices for (,\. >•) in R 5, nam ely (a, a) and (a, b), that is, x = a and y = a, o r x = a and y = b. If (x, y) = (a, a), then y = a and so either (y, z) = (a, a) or (y, z) = (a, b). In the first case, w e have (a, a) e R$ and (a, a) € R 5 , and (x, z) = (a , a) belongs to R$. In the second case, (a, a) e Rs and (a, b) e R$, and (x, z) = (fl, b) belongs to R$. This exam ple suggests (correctly!) that if (x, y) and (y, z) belong to som e relation R and x = y, then certainly (x, z) e R. The sam e could be said if y — z. Thus, w hen checking transitivity, w e need only consider ordered pairs (x, y) and (y\ z) for w hich x / y and y / z. Suppose next that (x, y) = («, b), so that y = b. H ere there is no ordered pair of the type (y, z); that is, there is no ordered pair of Rs w hose first coordinate is b. Thus, there is nothing to check w hen (x, y) = (a, b). For R$, there are only tw o possibilities for two ordered pairs of the type (x, y), (y, z) and in each case, (x, z) e R5. Thus R$ is transitive.
8.2
Properties of Relations
195
L e t’s turn to R 6 now. The relation R(, does not contain two ordered pairs of the type (x, y), (y , z) since if (x, y ) = (a, b), no ordered pair has b as its first coordinate; w hile if (x, v) = (a, c), no ordered pair has c as its first coordinate. Consequently, the hypothesis o f the transitive property is false and the im plication “If (x, y ) e R& and (y, z ) e R(,, then (x, z) e Rq.” is satisfied vacuously. H ence, R(, is transitive. A nother w ay to convince yourself that R 6 is transitive is to think of w hat m ust happen if R$ is not transitive; namely, there m ust be two ordered pairs (x, y ), (y, z) in R 6 such that (x, z) £ R(,. B ut there are no such ordered pairs (x, y ) and ( y , z ) \ In the preceding discussions, w e have m ade use o f an im portant point w hen testing a relation for transitivity. It bears repeating here. W hen w e are attempting to determine whether a relation R is transitive and, consequently, checking all pairs o f the type (x, y ) and ( j , z), we need not consider the situation where x = y or y — z. In this case, the ordered pair (x, z) will alw ays be present in R. If a relation R is not transitive, then there m ust exist ordered pairs (x, y ) and ( y , z ) in R , w here x ^ y and y z, such that (x, z) is not in R. T hat is, (x, y) and (>', z) constitute a counterexam ple to the im plication “If (x, y ) e R and (y, z) e R , then (x, z) e R ,” w hich is a requirem ent o f R being transitive. We already m entioned that relations occur frequently in m athem atics. L et R be the relation defined on the set Z o f integers by a R h if a < b; that is, R is the relation < . Since x < x for every integer x , it follow s that x R x for every x e Z; that is, R is reflexive. Certainly, 2 R 3 since 2 < 3. However, 3 > 2; so 3 If 2. T herefore, R is not sym m etric. On the other hand, it is a w ellknow n property o f integers that if a < b and b < c, then a < c. T herefore, if a R b and b R c, then a R c. So R is transitive. A nother relation R we could consider on the set Z is defined by a R b if a ^ b. However, then 1 If 1 since 1 = 1. Consequently, this relation is not reflexive. If a and b are integers such that a ^ b, then w e also have b ^ a. So if a R b, then b R a. This says that this relation is sym m etric. N otice that 2 ^ 3 and 3 ^ 2 but 2 = 2. T hat is, 2 R 3 and 3 R 2 but 2 If 2. T herefore, R is not transitive. T he d ista n c e betw een two real num bers a and b is \a — b\. So the distance betw een 2 and 4.5 is 2 — 4 .5 1 =  — 2 .5 1 = 2.5. Thus if the real num bers (points) a and b are plotted on the real num ber line (xaxis), then the length o f the segm ent betw een them is the distance. T his is illustrated in Figure 8.1 for the real num bers a = 3 and b = —2, w here the distance betw een them is thus \a — b\ = 3 — (—2)] = 5 = ( —2) — 3 = \b — a\. Define a relation R on the set R o f real num bers by a R b if \a — b\ < 1; that is, a is related to b if the distance betw een a and b is at m ost 1. Certainly, the distance from a real num ber to itself is 0; that is, \a — a\ — 0 < 1 for every x e R. So a R a and R is reflexive. If the distance betw een tw o real num bers a and b is at m ost 1, then the distance
Figure 8.1
The distance between 3 and —2
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Chapter 8
Equivalence Relations
betw een b and a is at m ost 1. In sym bols, if a — b\ < 1, then \b — a\ = \a — b\ < 1; that is, if a R b, then b R a. Therefore, R is sym m etric. N ow to the transitive property. If a R b and /> R c. is a R c? T hat is, if the distance betw een a and b is at m ost 1 and the distance betw een b and c is at m ost 1, does it follow that the distance betw een a and c is at m ost 1? The answ er is no. For exam ple, 3 R 2 and 2 R 1 since 3 — 2 < 1 and 2 — 11 < 1. However, 3 — 11 = 2. So 3 1 and R is not transitive.
8.3 E quivalen ce R elations Perhaps the m ost fam iliar relation in m athem atics is the equals relation. For exam ple, let R be the relation defined on Z by a R b if a = b. F or every integer a, w e have a — a and so a R a. If a = b, then b = a. H ence if a R b, then b R a. A lso, if a = b and b = c, then a ~ c. So if a R b and b R c, then a R c. These observations tell us that the equals relation on the set o f integers possesses all three o f the properties reflexive, sym m etric and transitive. T his suggests the question o f asking w hat other relations (on the set Z or indeed on any set) have these sam e three properties possessed by the equals relation. These are the relations that w ill be our prim ary focus in this chapter. A relation R on a set A is called an eq u iv alen ce re la tio n if R is reflexive, sym m etric and transitive. O f course then, the equals relation R defined on Z by a R b if a — b is an equivalence relation on Z. For another exam ple, consider the set A = {1, 2, 3, 4, 5, 6} and the relation R = {( 1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (1, 3), (1, 6), (6, 1), (6, 3),
(8.2)
(3, 1), (3, 6), (2 ,4 ), (4, 2)}
defined on A. T his relation has all three o f the properties reflexive, sym m etric and transitive and is consequently an equivalence relation. Suppose that R is an equivalence relation on som e set A. I f a 6 A, then a is related to a since R is reflexive. Q uite possibly, other elem ents o f A are related to a as well. The set o f all elem ents that are related to a given elem ent o f A w ill turn out to be im portant and, for this reason, these sets are given special nam es. For an equivalence relation R defined on a set A and for a e A, the set [a] ~ (x e A : x R a] consisting o f all elem ents in A that are related to a, is called an eq u iv alen ce class, in fact, the equivalence class containing a since a e [a ] (because R is reflexive). L oosely speaking, then, [a] consists of the “relatives” o f a. F or the equivalence relation R defined in (8.2), the resulting equivalence classes are [1] = { 1 ,3 ,6 } ,
[2] = {2, 4},
[3] = {1 ,3 , 6},
[4] = {2,4},
[5] = {5},
[6] = { 1 ,3 ,6 } .
(8.3)
Since [1] = [3]= [6] and [2] = [4], there are only three distinct equivalence classes in this case, nam ely [1{. {2} and [5]. L e t’s return to the equals relation R defined on Z by a R b if a = b and determ ine the equivalence classes for this equivalence relation. For a e Z, [a] = {x € Z : x R a] = {x e Z : x = a] = [a}\ that is, every integer is in an equivalence class by itself.
8.3
Equivalence Relations
197
A s another illustration, define a relation R on the set L o f straight lines in a plane by l \ R I 2 if either l \ — l 2 (the lines coincide) or i \ is parallel to £2 Since every line coincides w ith itself, R is reflexive. If a line l \ is parallel to a line £ 2 (or they coincide), then £2 is parallel to l \ (or they coincide). Thus R is sym m etric. Finally, if £ 1 is parallel to £2 and £ 2 is parallel to £ 3 (including the possibility that such pairs o f lines m ay coincide), then either £j is parallel to £ 3 or they coincide. Indeed, it m ay very w ell occur that i \ and £ 2 are distinct parallel lines, as are l 2 and £ 3 but i \ and £ 3 coincide. In any case, though, this relation is transitive. T herefore, R is an equivalence relation. H ence for £ e L , the equivalence class [£] — {x e L : x R £} = {x e L : x = £ or x is parallel to £}; that is, the equivalence class [£] consists o f £ and all lines in the plane parallel to £. To describe additional exam ples o f relations from geom etry, let T be the set o f all triangles in a plane. For tw o triangles T and T ' in T , define relations R\ and R i on T by T R\ T if T is congruent to T and T AN T ' if T is sim ilar to T ' . T hen both R \ and A 2 are equivalence relations. For a triangle T and the relation R 1, [T] is the set o f triangles in T that are congruent to T ; w hile for R 2 , T ] is the set o f triangles in T that are sim ilar to T . The relation R defined on Z by x R y if \x\ = y  is also an equivalence relation. In this case, for a e Z, the equivalence class [a] consists o f the two integers a and —a, unless a = 0, in w hich case [0] = {0}. We now consider an exam ple that requires m ore thought and explanation. Result 8.1
A relation R is defined on Z by x R y i f x + 3y is even. Then R is an equivalence relation. Before proving this result, le t’s be certain that w e understand this relation. First, notice that 5 R I since 5 + 3 • 7 = 26 is even. However, 8 ^ 9 since 8 + 3 • 9 = 35 is not even. On the other hand, A R A because 4 + 3  4 = 16 is even.
P r o o f o f R e s u lt 8.1
First w e show that R is reflexive. L et a e Z. Then a + 3a = 4a = 2 (2 a) is even since 2a e Z. Therefore a R a and R is reflexive. N ext w e show that R is sym m etric. A ssum e that a R b . T hus a + 3b is even. H ence a + 3b = 2k for som e integer k. So a — 2k — 3b. T herefore, b + 3a = b + 3(2k  3b) = b +
6
k 9 b =
6
k 
8
b = 2(3k  4b).
Since 3k — Ab is an integer, b + 3a is even. T herefore, b R a and R is sym m etric. Finally, we show that R is transitive. A ssum e that a R b and b R c. H ence a + 3b and b + 3 ca reev e n ; s o a + 3 b — 2 k a n d b + 3c = 2£ for som e integers £ a n d £ . A dding these tw o equations, w e obtain (a + 3b) + {b + 3c) = 2k + 21. So a + Ab + 3c = 2k + 2£ and a + 3c = 2k + 2£ — Ab = 2(k + £ — 2b). Since k + £ — 2b is an integer, a + 3c is even. H ence a R c and so R is transitive. T herefore, R is an equivalence relation. ■ PROOF ANALYSIS
A few rem arks concerning the preceding p ro o f are in order. R ecall that a relation R defined on a set A is reflexive if x R x for every x e A. T he reflexive property m ay also be rew orded to read: For every x e A , x R x or: If x e A , then x R x . H ence w hen we proved that R is reflexive in R esult 8.1, w e began by assum ing that a was an arbitrary
198
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Equivalence Relations
elem ent of Z. (W e’re giving a direct proof.) We w ere then required to show that a + 3a is even, w hich w e did. It w ould be incorrect, however, to assum e that a + 3a is even or that a R a. This, in fact, is w hat w e w ant to prove. ♦ Since the relation defined in Result 8.1 is an equivalence relation, there are equiva lence classes, nam ely an equivalence class [a] for each a e Z. L e t’s start w ith 0, say. The equivalence class [0] is the set o f all integers related to 0. In sym bols, this equivalence class is [0] = {.v € Z : x R 0] — .v e Z : x + 3 • 0 is even} = [x e Z : x is even}
{0, ± 2 , ± 4 , . . . } ;
that is, [0] is the set of even integers. It sh o u ld n 't be difficult to see that if a is an even integer, say a = 2k, w here k e Z, then [a] = {.v e = [x
Z
: x R a } = {x e Z : x + 3a is even} : x + 3(2k) is even} = [x e Z : x + 6 k is even}
e Z
is also the set o f even integers. O n the other hand, the equivalence class consisting of those integers related to 1 is [1] = j.v e Z : x R l } = { x e Z : x + 3 • 1 is even} = [x e Z : x + 3 is even} = {±1, ± 3 , ± 5 , . . .}, w hich is the set o f odd integers. In fact, if a is an odd integer, then a — 2 1 + 1 for some integer i and [,a]
= j.v e Z = {x e
Z
:
x  3a is even} = }.v e Z : x + 3(2£ + 11 iseven} :
x + 61 + 3 is even}
is the set o f odd integers. T herefore, if a and b are even, then [a J = [fo] is the set o f even integers; w hile if a and b are odd, then [a] ~ [ft] is the set o f odd integers. H ence there are only tw o distinct equivalence classes, nam ely [0] and [1], the sets o f even and odd integers, respectively. We w ill soon see that there is a good reason for this observation.
8.4 Properties of E quivalen ce C lasses You m ay have noticed that in the preceding exam ples o f equivalence relations, we have seen several situations w here tw o equivalence classes are equal. It is possible to determ ine exactly w hen this happens. Theorem 8.2
L et R be an equivalence relation on a nonempty set A and let a and b be elements o f A. Then [a] = [£>] i f a nd only if a R b.
Proof
A ssum e that a R b. We show that the sets [a] and [b ] are equal by verifying that [a] c [b ] and [ft] C [a ]. First, we show that [a] c [/?]. L e tx e [a], T h e n x R a. Since a R b and R is transitive, x R b. Therefore, x e [b] and so [a] C [b]. N ext, let y e \ h ]. Thus, y R b. Since a R b and R is sym m etric, b R a. A gain, by the transitivity of R , we have y R a. Therefore, y e [a] and so [/;] C [a]. H ence [a] = [/?].
8.4
Properties of Equivalence Classes
199
For the converse, assum e that [«] = [b]. B ecause R is reflexive, a e [a]. But, since [a] = [b], it follow s that a e [b]. Consequently, a R b . ■
A ccording to T heorem 8.2 then, if R is an equivalence relation on a set A and a is related to b, then the set [a] o f elem ents o f A related to a and the set [b\ o f elem ents o f A related to b are equal, that is, [ a ] = [b ]. B ecause the theorem characterizes w hen [a] — [b], w e know that if a f i b , then [a] / [ H L e t’s return once m ore to the equivalence relation defined in (8.2) on the set A = { 1, 2, 3, 4, 5, 6} and the equivalence classes given in (8.3). We observed earlier that [1] = [3] = [6], Since every two o f the integers 1, 3, 6 are related to each other (according to the definition o f R ), T heorem 8.2 tells us that the equality of [1], [3] and [6] is expected. The sam e can be said of [2] and [4], However, sin ce(5 , 6) i R , for exam ple, T heorem 8.2 tells us that [5] / [6], w hich is the case. T herefore, as we also observed earlier, there are only three distinct equivalence classes, namely, [1] = [3] = [6] = { 1 , 3 , 6},
[2] = {4] = {2, 4},
[5] = {5}.
(8.4)
Now, you m ight have noticed one other thing. E very elem ent o f A belongs to exactly one equivalence class. This observation m ight rem ind you of a concept we discussed earlier. R ecall that a p a rtitio n P of a nonem pty set S is a collection o f nonem pty subsets o f S w ith the property that every elem ent o f S belongs to exactly one o f these subsets; that is, P is a collection of pairw ise disjoint, nonem pty subsets o f S w hose union is S. H ence the set of the distinct equivalence classes in (8.4) is a partition of the set A = {1, 2, 3, 4, 5, 6}. We now show that this too is expected. Theorem 8.3
Let R be an equivalence relation defined on a nonempty set A. Then the set P — {[a] : a e A} o f equivalence classes resulting from R is a partition o f A.
Proof
Certainly, each equivalence class [a] is nonem pty since a e [a] and so each elem ent of A belongs to at least one equivalence class. We show that every elem ent o f A belongs to exactly one equivalence class. A ssum e that som e elem ent x o f A belongs to two equivalence classes, say [a] and [/>]. Since .v e [a] and .v e \b\. it follow s that x R a and x R b. B ecause R is sym m etric, a R x. T hus a R x and x R b. Since R is transitive, a R b . By T heorem 8.2, it follow s that [a] = [b\. So any two equivalence classes to w hich x belongs are equal. H ence x belongs to a unique equivalence class. ■
In the pro o f o f T heorem 8.3, w e were required to show that each elem ent x e A belongs to a unique equivalence class. D uring this proof, w e assum ed that x belongs to tw o equivalence classes [a] and [b\. O bserve that we m ade no assum ption w hether [ia] and [b ] are distinct. L ater w e learned that [a ] = [&]; so x can only belong to one equivalence class. W ith a very sm all change, w e could have reached the same conclusion by a different proof technique. We could have said: A ssum e, to the contrary, that x belongs to tw o distinct equivalence classes [ c / ] and [ft]. By the same argum ent as above, w e can show that [a] = [6]. However, n o w , this produces a contradiction, and w e have ju st given a proof by contradiction.
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Chapter 8
Equivalence Relations
A ccording to T heorem 8.3 then, w henever w e have an equivalence relation R defined on a nonem pty set A, a partition o f A into the associated equivalence classes o f R results. Perhaps unexpectedly, the converse is also true. T hat is, if we are given a partition of A , then there is a corresponding equivalence relation that can be defined on A, w hose resulting equivalence classes are the elem ents o f the given partition. For exam ple, let P = ({1> 3 ,4 } , {2, 7}, {5, 6}} be a given partition o f the set A = {1, 2, 3, 4, 5, 6, 7}. (N otice that every elem ent of A belongs to exactly one subset in P .) Then R = {(1, 1), (1, 3), (1, 4), (2, 2), (2, 7), (3, 1), (3, 3), (3. 4), (4, 1), (4, 3), (4, 4), (5, 5), (5, 6), (6, 5), (6, 6), (7, 2), (7, 7)} is an equivalence relation on A, w hose distinct equivalence classes are [1] = { 1 ,3 ,4 } ,
[2] = {2, 7}
and
[5] = {5. 6},
and so P — {[1], [2], [5]}. We now establish this result in general; that is, if we have a nonem pty set A and a partition P o f A, then it is possible to create an equivalence relation S o n A such that the distinct equivalence classes o f R are precisely the subsets in P . Since w e are trying to verify this in general (and not for a specific exam ple), we need to describe the subsets in P w ith the aid o f an index set. Since we w ill w ant every subset in P to be an equivalence class, we w ill need every tw o elem ents in the same subset to be related. On the other hand, since we will w ant tw o different subsets in P to be different equivalence classes, we w ill need elem ents in distinct subsets not to be related. Theorem 8.4
L et P = {A a : a e 1} be a p a r titio n o fa nonempty set A .T h e n there existsan equivalence relation R on A such that P is the set o f equivalence classes determined by R, that is, P = {[a ] : a e A}.
Proof
Define a relation R on A by x R y if x and y belong to the sam e subset in P \ that is, x R y if x , y c A a for some a e / . We now show that R is an equivalence relation. Let a € A. Since P is a partition o f A, it follow s that a e Ap for som e fi e I . Trivially, a and a belong to Ap\ so a R a and R is reflexive. N ext, let a, b e A and assum e that a R b. Then a and b belong to A y for some y e I . H ence b and a belong to A r ; so b R a and R is sym m etric. Finally, le t a, b and c be elem ents o f A such that a R b and b R c. Therefore, a, b e Ap and b, c e A y for som e f}, y e / . Since P is a partition o f A, the elem ent b belongs to only one set in P. Hence Ap — A y and so a, c e A p. Thus a R c and R is transitive. T herefore, R is an equivalence relation on A. We now consider the equivalence classes resulting from R . L et a e A. T hen a e A a for some a e / . T he equivalence class [a] consists o f all elem ents o f A related to a. From the w ay that R is defined, however, the only elem ents related to a are those elem ents belonging to the sam e subset in P to w hich a belongs, that is, [a] = A a . H ence {[«} : a e A} = {A„ : a e 1} = P.
8.4
Properties of Equivalence Classes
201
We now give an additional exam ple o f an equivalence relation. A lthough this ex am ple is sim ilar to the one described in R esult 8.1, it is different enough to require som e thought. R esult to Prove
PROOFSTRATEGY
A relation R is defined on Z by relation.
X
R y if 1 Lv — 5 v is even. T hen R is an equivalence
Since w e w ant to verify that R is an equivalence relation, w e need to show that R is reflexive, sym m etric and transitive. L et's start w ith the first o f these. We begin with an integer a. To show that a R a, we need to show that 11a — 5 a is even. However, 11a — 5a = 6a = 2(3a), so this should n ’t cause any difficulties. To verify that R is sym m etric, w e begin w ith a R b (where a, b e Z, o f course) and attem pt to show that b R a. Since a R b , it follow s that 11a — 5b is even. To show that b R a, w e need to show that 1 l b — 5a is even. Since 1 l a  5b is even, w e can write 1 l a — 5b = 2k for some integer k. A t first though, it m ight seem like a good idea to solve for a in term s o f b or solve for b in term s o f a . However, because neither the coefficient o f a nor the coefficient o f b is 1 or —1 in 1 l a — 5b — 2k, fractions w ould be introduced. We need another approach. N otice that if we write 1 l b  5a = (1 l a  5b) + (? a + ? b). then we have l i b — 5a = (11a — 5b) + ( —16a + 16 b) — 2 k — 16a + 16 b =
2
(k — 8a + 8 b).
This w ill work. To verify that R is transitive, We begin by assum ing that a R b and b R c (and attem pt to show that a R c). T hus 1 l a — 5b and 1 l b  5c are even and so 11a — 5b = 2k
and
l i b — 5c = 21
(8.5)
for integers k and t . To show that a R c, we m ust verify that 1 l a  5c is even. We need to w ork the expression 1 l a — 5c into the discussion. However, this can be done by adding the expressions in (8.5). W e’re ready to give a p ro o f now. ♦
Result 8.5
A relation R is defined on Z by x R y if 1 Lr — 5;y is even. Then R is an equivalence relation.
Proof
First, w e show that R is reflexive. Let a e Z. T hen 1 l a — 5a = 6 a = 2(3a). Since 3a is an integer, 1 l a  5a is even. Thus a R a and R is reflexive. N ext w e show that R is sym m etric. A ssum e that a R b, w here a , b e Z. Thus 1 l a — 5b is even. Therefore, 1 l a — 5b = 2k, w here k e L. O bserve that 11 b — 5a = (11a — 5b) + (—16a + 16 b) =
2
k — 16 a + 16 b — 2 (k — 8a + 8 b).
Since k  8a + 8b is an integer, 1 lb  5a is even. H ence b R a and R is sym m etric. Finally, w e show that R is transitive. A ssum e that a R b and b R c. H ence 1 l a — 5b and 1 l b — 5c are even. Therefore, 1 l a — 5b = 2k and 1 lb — 5c = 21, w here k, I e Z.
202
Chapter 8
Equivalence Relations
A dding these equations, we obtain (11a — 5b) + (11 b — 5c) = 2k + 21. Solving for 1 l a — 5c, w e have 1 l a  5c = 2k + 21 
6
b = 2(k + l ~ 3b).
Since k + 1 — 3b is an integer, 11a — 5c is even. H ence a R c and R is transitive. T herefore, R is an equivalence relation. ■ We now determ ine the equivalence classes for the equivalence relation ju st discussed. L e t’s begin w ith the equivalence class containing 0, say. Then [0] = {x e Z : x R 0} = [x e Z : 11* is even} = {x e Z : x is even} = {0, ± 2 , ± 4 , . . .}. R ecall that the distinct equivalence classes alw ays produce a partition o f the set involved (in this case Z). Since the class [0} does not consist o f all integers, there is at least one other equivalence class. To determ ine another equivalence class, w e look for an ele m ent that does not belong to [0]. Since 1 ^ [0], the equivalence class [1] is distinct (and disjoint) from [0], Thus [1] = {x e Z : x R 1} = [x e Z : 1 l x — 5 is even} = {x e Z : x is odd} = ( ± 1 , ± 3 , ± 5 , . ..}. Since [0] and [1] produce a partition o f Z (that is, every integer belongs to exactly one o f [0] and [1]), these are the only equivalence classes in this case.
8.5 C ongruence M odulo n N ext w e describe one o f the m ost im portant equivalence relations. If you have m ore m athem atics in your future, it is likely that you will see the equivalence relation we are about to describe again— indeed often. R ecall again that for integers a and b, w here a ^ O , the integer a is said to d iv id e b, w ritten as a \ b, if there exists an integer c such that b = ac. A lso, for integers a, b and n > 2, a is said to be c o n g ru e n t to b m o d u lo n, w ritten a = b (m od n), if n \ (a — b). For exam ple, 24 = 6 (m od 9) since 9  (24 — 6); w hile l s 5 (m od 2) since 2  ( 1 —5). A lso, 4 s 4 (m od 5) since 5  (4 — 4). However, 8 ^ 2 (m od 4) since 4 / ( 8 — 2). These concepts w ere introduced in C hapter 4. L e t’s consider a few exam ples o f pairs a , b o f integers such that a = b (m od 5). N otice that 7 = 7 (m od 5), —1 = —1 (m od 5) andO = 0 (m o d 5). A lso, 2 =  8 (m od 5) and  8 s 2 (m od 5). N otice also that 2 s 17 (m od 5). T herefore, both  8 = 2 (m od 5) and 2 = 17 (m od 5). F urtherm ore, —8 = 17 (m od 5). These exam ples m ight suggest that the reflexive, sym m etric and transitive properties are satisfied here, a fact w hich w e are about to verify. This is the im portant equivalence relation w e referred to at the beginning o f this section, not ju st for n = 5 but for any integer n > 2 . Theorem 8.6
L et n € Z, where n > 2 . Then congruence modulo n (that is, the relation R defined on Z by a R b if a = b (m od n)) is an equivalence relation on Z.
P roof
L et a e Z. Since n \ 0, it follow s that n \ (a — a) and so a = a (m od n). Thus, a R a, im plying that R is reflexive.
8.5
Congruence Modulo n
203
N ext, w e show that R is sym m etric. A ssum e that a R b , w here a , b e Z. Since a R b , it follow s that a = b (m od ri) and so n \ (a — b). H ence, there exists k e Z such that a — b = nk. Thus, b — a = —(a — b) = —(nk) = n ( —k). Since —k e Z , it follow s that n \ ( b — a) and so b = a (m od n). T herefore, b R a and R is sym m etric. Finally, w e show that R is transitive. A ssum e that a R b and b R c, w here a , b, c e Z. We show that a R c. Since a R b and b R c, we know that a = b (m od n) and b = c (m od n). Thus, n \ (a — b) and n \ ( b — c). Consequently, a — b = nk
and
b —c — ni
( 8 .6 )
for som e integers k and t. A dding the equations in (8.6), w e obtain (a — b) + (b — c) = n k + n l = n(k + £); so a — c = n (k + 1). Since k + i e Z, we have n \ (a — c) and so a = c (m od n). T here fore, a R c and R is transitive. m PROOF ANALYSIS
Theorem 8.6 describes a w ellknow n equivalence relation. L e t’s review how w e verified this. The proof w e gave to show that congruence m odulo n is an equivalence relation is a com m on proof technique for this kind of result and we need to be fam iliar w ith it. To prove that R is reflexive, we began with an arbitrary elem ent o f Z. We called this elem ent a. O ur goal w as to show that a R a. By definition, a R a if and only if a = a (m od n). However, a = a (m od n) if and only if n \ (a — a), w hich is the sam e as the statem ent n  0. Clearly, n \ 0 and this is w here w e decided to start. To prove that R is sym m etric, we started (as alw ays) by assum ing that a R b. O ur goal was to show that b R a. Since a R b, the definition o f the relation R tells us that a = b (m od n). From this, w e knew that n \ (a — b) and a — b = n k for som e integer k. However, to show that b R a, w e needed to verify that b = a (m od n). B ut this can only be done if we can show that n \ (b — a) or, equivalently, that b — a = n i for some integer I. H ence w e needed to verify that b — a can be expressed as the product o f n and som e other integer. Since b — a is the negative o f a — b and we have a convenient expression for a — b, this provided us w ith a key step. Finally, to prove that R is transitive, w e began by assum ing that a R b and b R c, w hich led us to the expressions a — b = n k and b — c = n i , w here k , l € Z. Since our goal was to show that a R c, we w ere required to show that a — c is a m ultiple of n. Som ehow then, we needed to w ork the term a — c into the problem , know ing that a — b = n k and b — c — n i . The key step here was to observe that a —c = (a— b) + (b  c). ♦ A ccording to T heorem 8.6 then, congruence m odulo 3 is an equivalence relation. In other w ords, if we define a relation R on Z by a R b if a = b (m od 3), then it follows that R is an equivalence relation. L et’s determ ine the distinct equivalence classes in this case. First, select an integer, say 0. T hen [0] is an equivalence class. Indeed, [0] = {x e Z : x R 0} = [x e Z : x = 0 (m od 3)} = [x e Z : 3  x ] = {0, ± 3 , ± 6 , ± 9 , . . .}.
'204
Chapter B Equivalence Relations H ence the class [0] consists o f the m ultiples o f 3. This class could be denoted by [3] or [6] or even [—300]. Since there is an integer that is not in [0], there m ust be at least one equivalence class distinct from [0]. In particular, since 1 f [0], it follow s that [1] ^ [0]; in fact, necessarily, [1] fl [0] = 0. The equivalence class [1] = {x e Z : x R 1} = {x e Z : x = 1 (m od 3)} 
{
v C Z : 3 j (.v  1)} = {1,  2 , 4,  5 , 7,  8 , . ..}.
Since 2 ^ [0] and 2 ^ [1], the equivalence class [2] is different from both [0] and [1]. By definition, [2] = {x e Z : x R 2} = [x e Z : x = 2 (m od 3)} = {.v e Z : 3  (,v  2)} = {2,  1 , 5,  4 , 8,  7 , . . . } . Since every integer belongs to (exactly) one o f these classes, w e have exactly three distinct equivalence classes in this case, nam ely: [0] = {0, ± 3 , ± 6 , ± 9 , . . . } , [1] = {1,  2 , 4 ,  5 , 7 ,  8 , . . . } , [2] = {2,  1 , 5 ,  4 , 8,  7 , . ..}. T hese equivalence classes have a connection w ith some very fam iliar m athem atical concepts, division and rem ainders, w hich w e encountered in Section 4.1 and w hich is useful to review here. If m and n > 2 are integers and m is divided by n, then we can express this division as m = n q + r, w here q is the quotient and r is the rem ainder. The rem ainder;has the requirem ent that 0 < r < n. W ith this requirem ent, q and r are unique and the result that w e have ju st referred to is the D ivision A lg o rith m . (The D ivision A lgorithm w ill be studied in considerably m ore detail in C hapter 11.) Consequently, every integer m can be expressed as 3q + r , w here 0 < r < 3 ; that is, r has one o f the values 0, 1 or 2. H ence, every integer can be expressed as 3q, 3q + 1 or 3q + 2 for some integer q. In this case, the equivalence class [0] consists o f the m ultiples o f 3, and so every integer having a rem ainder o f 0 w hen divided by 3 belongs to [0]. Furtherm ore, every integer having a rem ainder o f 1 w hen divided by 3 belongs to [1], w hile every integer having rem ainder o f 2 w hen divided by 3 belongs to [2], Since 73 =
24 •
3+ 1
and
 2 2 = (  8 ) 3 +
2,
for exam ple, it follow s that 73 e [1] a n d  2 2 e [2]. In fact, [73] = [1] and [  2 2 ] = [2], In general, for n > 2, the equivalence relation congruence m odulo n results in n distinct equivalence classes. In other w ords, if w e define a R b b y a = b (m od «), then there are n distinct equivalence classes: [0], [ 1 ] , » . . , [ « — 1], In fact, for an integer v w ith 0 < ;• < n, an integer m belongs to the set [/•] if and only if there is an integer q (the quotient) such that m = nq + r. In fact, the equivalence class [r] consists o f all integers having a rem ainder o f r w hen divided by n. L e t’s consider another equivalence relation defined on Z involving congruence but w hich is seem ingly different from the class o f exam ples we have ju st described. R e su lt to P ro v e
L et R be the relation defined on Z by a R b if 2a + b = 0 (m od 3). Then R is an equivalence relation.
8.5 PROOF STRATKCY
Congruence Modulo n
205
To prove that R is reflexive, w e m ust show that x R x for every x e Z. This m eans that w e m ust show that 2 x + x = 0 (m od 3 ) or that 3 j s 0 (m od 3 ) . This is equivalent to show ing that 3 \ 3 x , w hich is obvious. This tells us w here to begin the pro o f o f the reflexive property. P roving that R is sym m etric is som ew hat m ore subtle. O f course, we know w here to begin. W e assum e that x R y. From this, w e have 2 x + y s 0 (m od 3 ) . So 3  (2x + y) or 2x + y — 3r for som e integer r. O ur goal is to show that y R x or, equivalently, that 2y + x = 0 (m o d 3 ) . Eventually then, w e m ust show that 2 v + x = 3s for some integer s. We cannot assum e this o f course. Since 2x + y = 3r, it follow s that y = 3r — 2x. So 2 y + x = 2 ( 3 r — 2x) + x =
6
r — 3x = 3 ( 2 r — x).
Since 2r — x € Z, we have 3  (2y + x ) and the verification o f sym m etry is nearly com plete. Proving that R is transitive should be as expected. ♦ Result 8.7
L et R be the relation defined on Z by a R b i f 2a + b = 0 (m od 3). Then R is an equivalence relation.
Proof
L et x e Z. Since 3  3x, it follow s that 3x = 0 (m od 3). So 2x + x = 0 (m od 3). Thus, x R x and R is reflexive. N ext we verify that R is sym m etric. A ssum e that x R y, w here x , y e Z. Thus 2 x + y = 0 (m od 3) and so 3  (2x + y). T herefore, 2x + y = 3r for som e integer r . H ence y = 3r — 2x. So 2y + x — 2(3 r — 2 x ) + x = 6 r — 3x = 3 (2 r — x ). Since 2r — x is an integer, 3  (2v + x). So 2y + x = 0 (m od 3). T herefore, y R x and R is sym m etric. Finally, we show that R is transitive. A ssum e that x R y and}) R z, w here x , y , z 6 Z. T hen 2x + 3/ = 0 (m od 3) and 2 y + z = 0 (m od 3). Thus, 3  (2x + y) and 3  (2y + z). From this, it follow s that 2x + y — 3r and 2y + z = 3s for som e integers r and s. A dding these tw o equations, we obtain 2x + 3 y + z = 3r + 3 s ,4 so 2x + z = 3r + 3s — 3y — 3 (r + s — y). Since r + s — y is an integer, 3 (2x + z); so 2x + z = 0 (m od 3). H ence x R z and R is transitive. ■
PROOF ANALYSIS
A few additional com m ents about the p ro o f o f the sym m etric property in R esult 8.7 m ight be helpful. A t one point in the pro o f w e knew that 2x + y = 3 r for som e integer r and w e w anted to show that 2 y + x = 3s for som e integer s. If w e added these two equations, then w e w ould obtain 3x + 3y — 3 r + 3s. O f course, we c a n ’t add these because we d o n ’t know that 2 y + x = 3s. B ut this does suggest another idea.
206
Chapter 8
Equivalence Relations
A ssum e that a R y. Thus 2a + y = 0 (m od 3). H ence 3 [ (2.v + y); so 2 x + y = 3r for som e integer r . O bserve that 3 x + 3 y = ( 2 a + y ) + ( 2 y + .v i = 3 r + ( 2 y + x ) .
Therefore, 2 y + x —: 3 a + 3 y — 3 r
B ecause a + y — r e Z , it follow s that 3 y R x and R is sym m etric.
j
(2y
= 3 (a
+ y —
r).
+ a ). Consequently, 2 y + a
ee
0 (m o d 3), ♦
The distinct equivalence classes for the equivalence relation described in R esult 8.7 are [0] = {a e Z : # R 0} = pc e Z : 2x = 0 (m od 3)} =
[1} = 
{x
eZ:
=
{ 0 , ± 3 , ± 6 , ± 9 , . . .},
{ae Z : x R 1} = {x € Z : 2a  1  0 (m od x c
[2] =
3  2a} =
Z :3 
(2 a
+ 1)}  {1,
2 ,
3)}
4,  5 , 7,  8 , . ..},
{ae Z : a R 2} = {a e Z : 2x + 2 = 0 (m od
3)}
{a€ Z : 3 1(2a + 2)} = {2,  1 , 5,  4 , 8,  7 , . ..}.
L e t’s discuss how we obtained these equivalence classes. We started w ith the integer 0 and saw that [0] = {a e Z : 3  2 a }. By trying various values o f x (namely, 0, 1, 2, 3, 4, 5, etc. and —1, —2, —3, —4, etc.), we see that w e are obtaining the m ultiples o f 3. (Exercise 6 o f C hapter 4 asks you to show that if 3  2 a , then a is a m ultiple o f 3.) The contents o f [1] and [2] can be justified, if necessary, in a sim ilar manner. We have seen that if w e define a relation R { on Z by a R \ b if a = b (m od 3), then we have three distinct equivalence classes; w hile if w e define a relation R i on Z by a Ro b i f 2 a + b = 0 (m od 3), then w e also have three distinct classes— in fact, the sam e equivalence classes. L e t’s see w hy this is true. Result 8.8
Let a, b e Z. Then a = b (mod 3) i f a n d only if'2a + b = 0 (m od 3).
Proof First, assum e that a = £>(mod3 ) . T hen Thus a = 3 a + b. Now 2a + b —
2(3a
3
 (a — b) and so a — b =
+ b) + b — 6 x +
3b
3a
for som e
integer a .
= 3(2x + b).
Since 2 a + b is an integer, 3  (2a + b) and so 2a + b = 0 (m od 3 ) . For the converse, assum e that 2a + b = 0 (m od 3). H ence 3  (2a + £>), w hich im plies that 2a + b = 3y for some integer y. Thus b = 3y — 2a. O bserve that a — b — a — (3y — 2a) = 3 a — 3 y = 3 (a — y). Since a — y is an integer, 3  (a — b) and so a = b (m od 3).
s
We sho u ld n 't jum p to the conclusion that just because w e are dealing w ith an equivalence relation defined in term s o f the integers m odulo 3, w e w ill necessarily have three distinct equivalence classes. For exam ple, suppose that we define a relation R on
8.6
The Integers Modulo n
207
Z by a R b if a 2 = b2 (m od 3). Then, here too, R is an equivalence relation. In this case, however, there are only two distinct equivalence classes, nam ely: [0]  {0, ± 3 , ± 6 , ± 9 , . . . } and [1] = {±1. ± 2 , ± 4 , ± 5 , . . since w henever an integer n has a rem ainder 1 or 2 w hen it is divided by 3, then n 2 has a rem ainder of 1 w hen it is divided by 3.
8.6 The Integers M odulo n We have already seen that for each integer n > 2, the relation R defined on Z by a R b if a = b (m od n) is an equivalence relation. Furtherm ore, this equivalence relation results in the n distinct equivalence classes [0], [1], . . . , [n  1]. We denote the set of these equivalence classes by L„ and refer to this set as the integers m odulo n . Thus, Z 3 = {[0]. [1], [2]} and, in general, Z ,  0 , [I  .........[n — 1]}. H ence each elem ent [r] o f Z„, w here 0 < r < iu is a set that contains infinitely m any integers; indeed, as w e have noted, [/■] consists o f all those integers having the rem ainder r w hen divided by n. For this reason, the elem ents o f Z„ are som etim es called residue classes. A lthough it m akes perfectly good sense to take the union and intersection o f two elem ents of Z„ since these elem ents are sets (in fact, subsets o f Z), it d o esn ’t m ake sense atthis point to add or m ultiply tw o elem ents o f Z „ . However, since the elem ents o f Z„ have the appearance of integers, say [a] and [ft], w here a , b e Z, it does suggest the possibility o f defining addition and m ultiplication in Z„. We now discuss how these operations can be defined on the set Z„. O f course, w e have seen addition and m ultiplication defined m any tim es before. W hen w e speak o f addition and m ultiplication being operations on a set S, we m ean that for x , y e S , the sum x + y and the product x y should both belong to S. For exam ple, in the set Q o f rational num bers, the sum and product o f two rational num bers a / b and c / d (so a, b , c, d e Z and b , d ^ 0) are defined by a c a d + be a c — I—  and b d bd
ac  •—= — , b d bd
both o f w hich are rational num bers and so belong to Q. A s we m entioned, if addition and m ultiplication are operations on a set S, then x + y e S and x y e S for all x , y e S. T herefore, if T is a nonem pty subset of S and x , y e T , then x + y e S and x y e S. The set T is closed under addition if * + y e T w henever x , y e T . Similarly, T is closed under m ultiplication if x y e T w henever x , j e T , N ecessarily, if addition and m ultiplication are operations on a set S, then S is closed under addition and m ultiplication. For exam ple, addition and m ultiplication are operations on Z. If A and B denote the sets o f even integers and odd integers, respectively, then A is closed u nder both addition and m ultiplication but B is closed under m ultiplication only. H ow ever addition and m ultiplication m ight be defined in Z„, w e w ould certainly expect that the sum and product o f tw o elem ents o f Z„ to be an elem ent o f L n. There
208
Chapter 8
Equivalence Relations
appears to be a natural definition o f addition and m ultiplication in Z„; nam ely, for two equivalence classes [a] and [b] in Z„, we define [a] + [b ] = [a + b]
and
[a] • [b ] = [ab],
(8.7)
L e t’s suppose that w e are considering Zg, for exam ple, w here then Z(, = {[0], [ 1 ] , . . . , [5]}. From the definitions o f addition and m ultiplication that we ju st gave, [1] — 13] — 11 1 3 1  [4] and [1] ■[3] = [1 • 3] = [3]. This certainly seems harm less enough, but le t’s consider adding and m ultiplying two other equivalence classes, say [2] and [3]. A gain, according to the definitions in (8.7), [2] + [3] = [2 + 3] = [5] and [2] ■[3] = [2 • 3] = [6]. However, w e have been expressing the elem ents o f Z(, by [0], [1], [2], [3], [4] and [5] and w e d o n ’t explicitly see [2 • 3] = [6] am ong these ele m ents. Since 6 = 0 (m od 6), it follow s that 6 e [0], that is, [6] = [0]. A lso, the rem ainder is 0 w hen 6 is divided by 6 and so [6] = [0]. T herefore, [2] • [3] = [0]. By sim ilar reason ing, [3] + [5] = [2] and [3] • [5] = [3], In fact, the com plete addition and m ultiplication tables for Z(, are given in Figure 8.2. If w e add [1] to [0], add [1] to [1] and continue in this m anner, then we obtain [0] + [1] = [1], [1] + [1] = [2], [2] + [ lj = [ 3 ] , . . . . [5] + [1] = [6] = [0], [6] + [1] = [0] + [1] = [1], etc.; that is, we return to [0] and cycle through all the classes o f Zg again (and again). If, instead o f Z^, w e w ere dealing w ith Z 12, we w ould have [0] + [1] = [1], [1] + [1] = 12 1. [2] + [1]  [ 3 ] , . . . , [11] + [1] = [12] = [0], [12] + [1] = [0] + [1] ' I . etc. and this should rem ind you of w hat occurs w hen a certain num ber o f hours is added to a tim e (in hours), w here, o f course, 12 o ’clock is represented here as 0 o ’clock. (For exam ple, if it is 11 o ’clock now, w hat tim e w ill it be 45 hours from now?) A lthough the definitions o f addition and m ultiplication in Z„ that we gave in (8.7) should seem quite reasonable and expected, there is a possible point o f concern here that needs to be addressed. A ccording to the definition o f addition in Z&, [4] + [5] = [3]. However, the class [4], w hich consists o f all integers x such that x = 4 (m od 6), need not be represented this way. Since 10 e [4], it follow s that [10] = [4], A lso, [16] = [4] and [—2] = [4], for exam ple. M oreover, [11] = [5], [17] = [5] and [—25] = [5], H ence adding the equivalence classes [4] and [5] is the same as adding [10] and [—25], say, since [10] = [4] and [—25] = [5], But. according to the definition we have given, [10] + [—25] = [—15], Luckily, [—15] = [3] and so w e obtain the same sum as before. B ut will this happen every tim e? T hat is, does the definition o f the sum of the equivalence classes [a] and [b ] that w e gave in (8.7) depend on the representatives a and b o f these classes? If the sum (or product) o f two equivalence classes does not depend on the representatives, then we say that this sum (or product) is w elldefined. We certainly
+ [0] [1] [2] [3] [4] [5]
[0] [0] [1] [2] [3] [4] [5]
[1] [1] [2] [3] [4] [5] [0]
[2] [2] [3] [4] [5] [0] [1]
[3] [3] [4] [5] [0] [1] [2]
Figure 8.2
[4] [4] [5] [0] [1] [2] [3]
[5] [5] [0] [1] [2] [3] [4]
[0] [1] [2] [3] [4] [5]
[0] [0] [0] [0] [0] [0] [0]
[1] [0] [1] [2] [3] [4] [5]
[2] [0] [2] [4] [0] [2] [4]
The Addition and Multiplication Tables for Z,
[3] [0] [3] [0] [3] [0] [3]
[4] [0] [4] [2] [0] [4] [2]
[5] [0] [5] [4] [3] [2] [1]
8.6
209
The Integers Modulo n
w ould w ant this to be the case, w hich, fortunately, it is. M ore precisely, addition and m ultiplication in Z„ are w elldefined if w henever [<7 ] = [ft] and [c] = [d] in Z„, then [a + c] = \b + d] and [ac] = [bd]. Theorem 8.9 Proof
Addition in Z„, n > 2, is welldefined. The set Z„ is the set o f equivalence classes resulting from the equivalence relation R defined on Z by a R b if a = ft (m od n). L et [<7], [ft], [c], [d] e Z„, w here [a] = [b ] and [c] = [d]. We prove that \a + c] = [ft + d]. Since [a] = [ft], it follow s by T heorem 8.2 that a R b. Similarly, c R d. T herefore, a = b (m od n) and c = d (m od n). Thus, n  (a — ft) and n j (c — d). H ence, there exist integers x and y so that a — b — n x and c — d — ny.
(8.8)
A dding the equations in (8.8), w e obtain (a — b) + (c — d) = n x + ny = n (x + y); so (a + c) — (ft + d) — n (x + y). This im plies that n \ [(<7 + c) — (b + d)]. Thus, (a + c) = (b + d) (m od n). From this, w e conclude that (a + c) R (b + d), w hich im plies that [a + c] = [b + d\. m
If the proof o f Theorem 8.9 looks a bit fam iliar, review R esult 4.10 and its proof. As an exam ple, in Z 7, [118] + [26] = [144]. Since the rem ainder is 4 w hen 144 is divided by 7, it follow s that [118] + [26] = [4], Furtherm ore, [118] = [6] and [26] = [5]; so [118] + [26] = [6] + [5] = [11] = [4], A s we have m entioned, the m ultiplication in Z„ that w e described in (8.7) is also w elldefined. The verification o f this fact has been left as an exercise (Exercise 8.58). A ddition and m ultiplication in Zn satisfy m any fam iliar properties. A m ong these are: Com m utative Properties [a] + [ft] = [ft] + [a] and [a] ■[ft] = [ft] • [a]
for all a , ft e Z;
A ssociative Properties ([a] + M ) + [c] = [a] + ([ft] + [c]) and ([a] • [ft]) • [c] = [a] • ([ft] • [c])
for all a , ft, c e Z;
D istributive Property [a] ■([ft] + c ) = \a\ ■[ft] + [a] ■[c]
for all a , b , c e Z.
A lthough w e defined m ultiplication in Z„ in a m anner that was probably expected, this is not the only w ay it could have been defined. For exam ple, suppose that we are considering the set Z 3 o f integers m odulo 3. For equivalence classes [a] and [ft] in Z 3, define the “product’' [a] • [ft] to equal [q], w here [<7] is the quotient w hen ab is divided by 3. Since the “product” o f every tw o elem ents o f Z 3 is an elem ent o f Z 3, this operation is closed. In particular, [2] • [2] = [1] since the quotient is 1 w hen 2 • 2 = 4 is divided
210
C h a p te r 8
Equivalence Relations
by 3. However. [2] = [5] but [5] â€˘ [5] = [8] = [2], N otice also that [5] â€˘ [2] = [3] = [0], H ence this m ultiplication is not welldefined.
EXERCISES FOR CHAPTER 8 Section 8.1: Relations 8.1. Let A = {a , ft, c} and B = {r, s , t. u). Furthermore, let R = {(a, s), (a, t), (ft, t)} be a relation from A to B. Determine dom[R) and range(ft)8.2. Let A be a nonempty set and B C V (A ). Define a relation R from A to B by x R Y if x e Y . Give example of two sets A and B that illustrate this. W hat is R for these two sets?
an
8.3. Let A = {0. 1}. Determine all the relations on A. 8.4. Let A = { a . b , c } m d B = {1, 2, 3, 4}. Then Ri = {(a, 2), (a, 3), (ft, 1), (ft, 3), ( c , 4)} is a relation from A to 5 , while R 2 = {(1, ft), (1, c), (2, a), (2, ft), (3, c), (4. a), (4, c)} is a relation from B to A. A relation R is defined on .4 by x R y if there exists z e B such that x R \ z and z R 2 y Express R by listing its elements. 8.5. For the relation R = {(1, 1), (1. 2), (1, 3), (2, 2), (2, 3), (3. 3)} defined on the set{1,2, 3}, what is 8.6. A relation R is defined on N by a R b if a / b e N. For
c,
J?1 ?
d e N, under whatconditions is c R ~' d l
8.7. For the relation R = {(,v. y) : x + 4y is odd} defined on N. what is R ~ l ? 8.8. For the relation R = {(x, y) : x < }>} defined on N, what is tf â€œ !? 8.9. Let A and B be sets with A = \B\ = 4 . (a) Prove or disprove: If R is a relation from A to B where R .*.9 and R = R ^ 1, then A = B. (b) Show that by making a small change in the statement in (a), a different response to the resulting statement can be obtained. 8.10. Let A be a set with  A = 4 . W hat is the maximum number of elements that a relation R on A can contain so that R fl R ~ l = 0?
Section 8.2: Properties of Relations 8.11. Let A = {a, ft, c, d] and let R = {(a, a), (a , ft), {a, c), (a , d), (ft, ft), (ft, c), (ft, d), (c, c), B, d), ( d ,d ) } be a relation on A. W hich of the properties reflexive, symmetric and transitive does the relation R possess? Justify your answers. 8.12. Let S = {a, ft, c). Then R = {(a , a), (a, ft), (a , c)} is a relation on S. W hich of the properties reflexive, symmetric and transitive does the relation R possess? Justify your answers. 8.13. Let S = {a , ft, r}. Then R = {(a, ft)} is a relation on S. W hich of the properties reflexive, symmetric and transitive does the relation R possess? Justify your answers. 8.14. Let A = {a , ft, c, d}. Give an example (with justification) of a relation R on Athat has none of the following properties: reflexive, symmetric, transitive. 8.15. A relation R is defined on Z by a R b if \a  ft < 2 . W hich of the properties reflexive, symmetric and transitive does the relation R possess? Justify your answers. 8.16. Let A = [a. ft, c, d}. How many relations defined on A are reflexive, symmetric and transitive and contain the ordered pairs (a, ft), (ft, 6 % (c, d i? 8.17. Let R = 0 be the empty relation on a nonempty set A. W hich of the properties reflexive, symmetric and transitive does R possess?
Exercises for C hapter 8
211
8.18. Let A = {1, 2, 3. 4j. Give an example of a relation on A that is: (a) reflexive and symmetric but not transitive. (b) reflexive and transitive but not symmetric. (c) symmetric and transitive but not reflexive. (d) reflexive but neither symmetric nor transitive. (e) symmetric but neither reflexive nor transitive. (f) transitive but neither reflexive nor symmetric. 8.19. A relation R is defined on Z by x R y if x ■y > 0. Prove or disprove the following: (a) R is reflexive, (b) R is symmetric, (c) R is transitive. 8.20. Determine the maximum number of elements in a relation R on a 3element set such that R has none of the properties reflexive, symmetric and transitive. 8.21. Prove or disprove: If there exists a relation R\ on the set \a.. <i.~} that is not reflexive, not symmetric and not transitive, then there exists a relation R 2 on the set [b\, b 2 .b^,} that is not reflexive, not symmetric and not transitive. 8.22. Let S be the set of all polynomials of degree at most 3. An element s(x) of S' can then be expressed as s(x) = a x 3 + b x 2 + cx + d, where a, b, c, d s R . A relation R is defined on S by p (x) R q(x) if p(x) and q(x) have a real root in common. (For example, p = (x — 1)2 and q = x 2 — 1 have the root 1 in common so that p R q.) Determine which of the properties reflexive, symmetric, and transitive are possessed by R. 8.23. A relation R is defined on N by a R b if either a  b or b \ a. Determine which of the properties reflexive, symmetric and transitive are possessed by R .
Section 8.3: Equivalence Relations 8.24. Let R be an equivalence relation o n / l = {a, b, c, d, e, f , g} such that a R c , c R d , d R g and b R f .I f there are three distinct equivalence classes resulting from R, then determine these equivalence classes and determine all elements of R . '.25. Let A = {1, 2, 3, 4, 5, 6}. The relation R = {(1, 1), (1, 5), (2. 2). (2, 3), (2, 6), (3, 2), (3, 3), (3, 6), (4, 4), (5, 1), (5, 5), (6, 2), (6, 3), (6, 6)} is an equivalence relation on A. Determine the distinct equivalence classes. 8.26. Let A = {1 , 2, 3, 4. 5, 6}. The distinct equivalence classes resulting from an equivalence relation R on A are {1, 4. 5), {2, 6} and {3}. W hat is R7 8.27. Let R be a relation defined on Z by a R b if a 3 = b . Show that R is an equivalence relation on Z and determine the distinct equivalence classes. 8.28. (a) Let R be the relation defined on Z by a R b if a + b is even. Show that R is an equivalence relation and determine the distinct equivalence classes. (b) Suppose that “even” is replaced by “odd” in (a). W hich of the properties reflexive, symmetric and transitive does R possess? 8.29. Let R be an equivalence relation defined on a set A containing the elements a, b, c and d. Prove that if a R b, c R d and a R d, then b R c. 8.30. Let H = {2m : m e Z}. A relation R is defined on the set Q+ of positive rational numbers by a R b if a / b G II. (a) Show that R is an equivalence relation. (b) Describe the elements in the equivalence class [3].
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Equivalence Relations
8.31. A relation R on a nonempty set A is defined to be circ u lar if whenever x R y and y R z, then z R x for all x , y , z e A. Prove that a relation R on A isan equivalence relation if and only if R is circular and reflexive. 8.32. A relation R is defined on the set A = {a + b s / l :a , b e Q , a + b  J l ^ 0} by x R y if x / y e that R is an equivalence relation and determine the distinct equivalence classes.
Q. Show
8.33. Let H = {4k : k e Z}. A relation R is defined on Z by a R b if a — b e H . (a) Show that R is an equivalence relation. (b) Determine the distinct equivalence classes. 8.34. Let H be a nonempty subset of Z. Suppose that the relation R defined on Z by a R b if a — b e H is an equivalence relation. Verify the following (a) 0 e H. (b) If a 6 //.th e n  a e H. (c) If a, b e H , then a + b e H . 8.35. Prove or disprove: There exist equivalence relations R 2 2 R 1 and R t U R 2 = S x S.
and R 2 on the setS — {a , b, c} such that R 1 % R 2,
S e c t i o n 8 .4 : P r o p e r t i e s o f E q u i v a l e n c e C l a s s e s 8.36. Give an example of an equivalence relation R on the set A = {v, w , x , y, z } such that there are exactly three distinct equivalence classes. W hat are the equivalence classes for your example? 8.37. A relation R is defined on N by a R b if a 2 + b 2 is even. Prove that R is an equivalence relation. Determine the distinct equivalence classes. 8.38. Let R be a relation defined on the set N by a R b if either a \ 2b or b \ 2a. Prove or disprove: R is an equivalence relation. 8.39. Let 5 be a nonempty subset o f Z and let R be a relation defined on S by x R y if 3  (x + 2y). (a) Prove that R is an equivalence relation. (b) If S = {—7, —6, —2, 0, 1,4, 5, 7}, then what are the distinct equivalence classes in this case? 8.40. A relation R is defined on Z by x R y if 3x — l y is even. Prove that R is an equivalence relation. Determine the distinct equivalence classes. 8.41. (a) Prove that the intersection of two equivalence relations on a nonempty set is an equivalence relation. (b) Consider the equivalence relations Rn and Rt, defined on Z by a R 2 b if a = b (mod 2) and a R $ b if a = b (mod 3). By (a), R\ = R 2 fl R 3 is an equivalence relation on Z. Determine the distinct equivalence classes in R 1. 8.42. Prove or disprove: The union of two equivalence relations on a nonempty set is an equivalence relation. 8.43. Let A = {u, v, w, x , y, z}. The relation R = {(u , u ), (m, v), (u , w), (v, m), (v, v), (v, w), (w, m),
(w, v),
(w, w), cX, x), (x, y), (y, x), (y, y), (z, z)} defined on A is an equivalence relation. In particular, [«] = [v] = [w] = {m, v, w ], [x ] = [y] = {x, y) and [z] = {z}; so [m] = [v] = \[w]\ = 3 and [x] = [v] = 2, while [z] = 1. Therefore, [w] + IM I + I M I + IM l + IM I + ltz]l = 14.Let A = {ai, a2, .. ■, an} be an nelement set and let R be an equivalence relation defined on A. Prove thatI M I is even if and only if n is even.
Exercises for Chapter 8
213
Section 8.5: Congruence Modulo n 8.44. Classify each of the following statements as true or false. (a) 25 = 9 (mod 8), (b) —17 = 9 (mod 8), (c) —14 = —14 (mod 4), (d) 25 = —3 (mod 11). 8.45. A relation R is defined on Z by a R b if 3a + 5b = 0 (mod 8). Prove that R is an equivalence relation. 8.46. Let R be the relation defined on Z b y a R b if a + b = 0 (mod 3). Show that R is not an equivalence relation. 8.47. The relation R on Z defined by a R b if a 2 = b 2 (mod 4) is known to be an equivalence relation. Determine the distinct equivalence classes. 8.48. The relation R defined on Z by x R y if x 3 == y 3 (mod 4) is known to be an equivalence relation. Determine the distinct equivalence classes. 8.49. A relation R is defined on Z by a R b if 5a = 2b (mod 3). Prove that R is an equivalence relation. Determine the distinct equivalence classes. 8.50. A relation R is defined on Z by a R b if 2a + 2b = 0 (mod 4). Prove that R is an equivalence relation. Determine the distinct equivalence classes. 8.51. Let R be the relation defined on Z by a R h if 2 a + 3b = 0 (mod 5). Prove that R is an equivalence relation and determine the distinct equivalence classes. 8.52. Let R be the relation defined on Z by a R b if a 2 = b 2 (mod 5). Prove that R is an equivalence relation and determine the distinct equivalence classes. 8.53. For an integer n > 2, the relation R defined on Z by a R b if a = b (mod n) is an equivalence relation. Equivalently, a R b if a — b = kn for some k € Z. Define a relation R on the set R of real numbers by a R b if a — b = k n for some k e Z. Is this relation R on R an equivalence relation? If not, explain why. If yes, prove this and determine [0], [re] and [ \ / 2].
Section 8.6: The Integers Modulo n 8.54. Construct the addition and multiplication tables for Z4 and Z 5. 8.55. In Zg, express the following sums and products as [/•], where 0 < r < 8. (a) [ 2 ] + [6] (b) [2] • [6] (c) [1 3 ] + [138] (d) [1 3 ]  [138] 8.56. In Z] 1, express the following sums and products as [r], where 0 < r < 11. (a) [7] + [5] (b) [7] • [5] (c) [8 2 ] + [207] (d) [ 8 2 ] • [207] 8.57. Let S = Z and T = [4k : k s Z}. Thus T is a nonempty subset of S. (a) (b) (c) (d) (e)
Prove that T is closed under addition and multiplication. If a e S — T and b e T , is ab e T1 If a e S — T and b e T , is a + b e T1 If a, b e S — T, is it possible that ab € T1 If a, b € S — T , is it possible that a + b e T?
8.58. Prove that the multiplication in Z„, n > 2, defined by [<?][£>] = [ab] is welldefined. (See Result 4.11.) 8.59. (a) Let [a], [b] e Z 8. If [a] • [b] = [0], does it follow that [a] = [0] or [b ] = [0]? (b) How is the question in (a) answered if Zs is replaced by Z9? by Z 10? by Z n ? (c) For which integers n > 2 is the following statement true? (You are only asked to make a conjecture, not to provide a proof.) Let [a], [£>] e Z„, n > 2. If [a] • [b ] = [0], then [a] = [0] or [b] = [0]. 8.60. For integers m, n > 2 consider Z,„ and Z„. Let [a] e Z„, where 0 < a < m — 1. Then a, a + m e [a] in Z„,. If a, a + m e [b\ for some [b] e Z„, then what can be said of m and n ?
214
Chapter 8
Equivalence Relations
8.61. (a) For integers m, n > 2 consider Z,„ and Z„. If some element of Z„, also belongs to Z„, then what can be said of Z„, and Z„? (b) Are there examples of integers m, n > 2 for which Z„, n Z„ = 0?
ADDITIONAL EXERCISES FOR CHAPTER 8 8.62. Prove or disprove: (a) There exists an integer a such that ab = 0 (mod 3) for every integer b. (b) If a € Z, then ab = 0 (mod 3) for every b € Z. (c) For every integer a , there exists an integer b such that ab = 0 (mod 3). 8.63. A relation R is defined o n R b y a R b if a — b e Z. Prove that R is an equivalence relation and determine the equivalence classes [1/2] and [\/2], 8.64. A relation R is defined o n Z b y a R b if \a  2\ = \b  2 \ . Prove that R is an equivalence relation and determine the distinct equivalence classes. 8.65. Let k and I be integers such that k + I = 0 (mod 3) and let a , b e Z. Prove that if a = b (mod 3). then ka + t b = 0 (mod 3). 8.66. State and prove a generalization of Exercise 8.65. 8.67. A relation R is defined on Z by a R b if 3  (a3  b). Prove or disprove the following: (a) R is reflexive, (b) R is transitive. 8.68. A relation R is defined on Z by a R b if a = b (mod 2) and a = b (mod 3). Prove or disprove: R is an equivalence relation on Z. 8.69. A relation R is defined on Z by a R b if a = b (mod 2) or a = b (mod 3). Prove or disprove: R is an equivalence relation on Z. 8.70. Determine each of the following. (a) [4]3 = [4] [4] [4] in Z5 (b) [7]5 in Z 10 8.71. Let S = {(a , b ) : a , b e R, a
0}.
(a) Show that the relation R defined on S by {a. b) R (c, d) if a d = be is an equivalence relation. (b) Describe geometrically the elements of the equivalence classes [(1, 2)] and [(3, 0)]. 8.72. In Exercise 8.19. (of this chapter), a relation R was defined on Z by x R y if x ■y > 0, and we were asked to determine which of the properties reflexive, symmetric and transitive are satisfied. (a) How would our answers have changed if x ■y > 0 was replaced by: (i) x  j S 0. (h) x ■y > 0, (iii) x ■y # 0. (iv) x ■y > 1, <v) ,v • v is odd, (vi) x ■y is even, (vii) x y # 2 (mod 3)? (b) What are some additional questions you could ask? 8.73. For the following statement S and proposed proof, either (1) S is true and the proof is correct, (2) S is true and the proof is incorrect or (3) S is false (and the proof is incorrect). Explain which of these occurs. S: Every symmetric and transitive relation on a nonempty set is an equivalence relation. P roof Let R be a symmetric and transitive relation defined on a nonempty set A. We need only show that R is reflexive. Let ,r e A. We show that x R x. Let y € A such that x R y, Since R is symmetric, y R x. Now x R y and y R x. Since R is transitive, x R x. Thus R is reflexive. ■ 8.74. Evaluate the proposed proof of the following result. R esult
A relation R is defined on Z by a R b if 3  £& + 2b). Then R is an equivalence relation.
215
Additional Exercises for Chapter 8
Proof 3
Assume that a R a. Then 3  (a t 2a). Since a + 2a = 3a and a e Z, it follows that 3  3a or  {a + 2a). Therefore, a R a and R is reflexive.
Next, we show that R is symmetric. Assume that a R b. Then 3 ( (a + 2b). So a + 2b = 3.v. where x £ Z. Hence a = 3x — 2b. Therefore, b + 2a — b + 2(3.r — 2b) = b + 6 .x — 4 b = 6 .x — 3 b = 3(2.v
— b).
Since 2x — b is an integer, 3  (b + 2a). S o b R a and R is symmetric. Finally, we show that R is transitive. Assume that a R b and b R c. Then 3  (a + 2b) and 3  (b + 2c). So a + 2b = 3.v and b + 2c = 3y, where a , y £ Z. Adding, we have (a + 2b) + (b + 2c) = 3 a + 3 y . So a + 2c — 3a + 3_y — 3 b = 3(.v + y — b). Since x + y — b is an integer, 3 j (a + 2c). Hence a R c and R is transitive. 8.75. (a) Show that the relation R defined on R x R by {a. b) R (c, d) if \a\ + \b\ = c + \d\ is an relation. (b) Describe geometrically the elements of the equivalence classes [(1, 2)] and [(3, 0)].
* equivalence
8.76. Let x e Z m and y e Z„, where m, n > 2. If a c y, then what can be said of m and n? 8.77. Let A be a nonempty set and let B be a fixed subset of A. A relation R is defined on V ( A ) by X R Y if X H B = Y CB. (a) Prove that R is an equivalence relation. (b) Let A = {1, 2, 3, 4} and B = {1,3,4}. For X = {2, 3, 4}, determine [X]. 8.78. Let Ri and R 2 be equivalence relations on a nonempty set A. Prove or disprove each of the following. (a) If /?j fl R 2 is reflexive, then so are R\ and R 2. (b) If R\ fl R 2 is symmetric, then so are R\ and R 2. (c) If R\ fl R 2 is transitive, then so are R\ and R 2. 8.79. Prove that if R is an equivalence relation on a set A, then the inverse relation R _1is an equivalence relation on A. 8.80. Let Ri and R 2 be equivalence relations on a nonempty set A. A relation R = R \ R 2 is defined on A as follows: For a , b e A, a R b if there exists c £ A such that a R \ c and c R 2 b. Prove or disprove: R is an equivalence relation on A. 8.81. A relation R on a nonempty set S is called sequential if for every sequence x, y , z of elements of S (distinct or not), at least one of the ordered pairs ( a , y) and (y, z) belongs to R. Prove or disprove: Every symmetric, sequential relation on a nonempty set is an equivalence relation. 8.82. Consider the subset H = {[3&] : k £ Z} of Z l2, (a) Determine the distinct elements of H and construct an addition table for H . (b) A relation R on Z \ 2 is defined by [a] R [b] if [a — b] e H . Show that R is an equivalence relation and determine the distinct equivalence classes. 8.83. For elements a, b e Z„, n > 2, a = [c] and b = [rf] for some integers c and d. Define a — b = [r — [d] as the equivalence class  r — d]. Let H = {.vj, x 2, . . . , x j] be a subset of Z n, n > 2, such that a relation R defined on Z„ by a R b if a — b £ H is an equivalence relation. (a) For each a e Z n, determine the equivalence class [a] and show that [a] consists of d elements. (b) Prove that d  n.
9
J
Functions
I
f R is a relation from a set A to a set B and x is an elem ent o f A, then either x is related to no elem ents o f B or x is related to at least one elem ent o f B . In the latter case, it m ay occur that x is related to all elem ents o f B or perhaps to exactly one elem ent o f B . If every elem ent o f A is related to no elem ents o f B , then R is the em pty set 0. If every elem ent o f A is related to all elem ents o f S , then R is the C artesian product A x B. However, if every elem ent o f A is related to exactly one elem ent o f B , then w e have the m ost studied relation of all: a function. Surely, you have encountered functions before, at least in calculus and precalculus. B ut it is likely that you have not studied functions in the m anner we are about to describe here.
9.1 The D efinition of F u nction Let A and B be nonem pty sets. By a function / from A to B , w ritten / : A > 5 , we m ean a relation from A to B w ith the property that every elem ent a in A is the first coordinate o f exactly one ordered pair in / . Since / is a relation, the set A in this case is the dom ain o f f , denoted by dom ( / ) . The set B is called the codom ain o f / . For a function / : A —> B , let (a, b) e f . Since / contains only one ordered pair w hose first coordinate is a, it follow s that b is the unique second coordinate o f an ordered pair w hose first coordinate is a\ that is, if (a, b) e / and (a , c) e / , then b — c. If (a , /;) e / , then we write b — f ( a ) and refer to b as the im age o f a. Som etim es / is said to m ap a into b. Indeed, / itself is som etim es called a m apping. The set r a n g e ( /) — \b e B : b is an im age under / o f som e elem ent of A} = { / ( x ) : x e A }
216
is the range o f f and consists o f the second coordinates o f the elem ents o f / . If A is a finite set, then the function / is a finite set and the num ber o f elem ents in / is  A since there is exactly one ordered pair in / corresponding to each elem ent o f A. T hroughout this chapter, as w ith earlier chapters, w henever w e refer to cardinalities o f sets, we are concerned w ith finite sets only. Suppose that / : A —> B and g : A * B are tw o functions from A to B and a e A. T hen / and g contain exactly one ordered p air having a as its first coordinate, say (a, x ) e / and (a, y) e g. If the sets / and g are equal, then (a, x) belongs to g as well. Since g contains only one ordered pair w hose first coordinate is a, it follow s that (a, x ) = (a, y). B ut this im plies that x — y, that is, f ( a ) = g(a). H ence it is natural to define two
9.1
The Definition of Function
217
functions f : A B and g : A —> B to be eq u al, w ritten / = g, if / ( a ) = g(a) for all a e A. L et A = {1. 2, 3} and B = { w , x , y , z}. Then f \ = {(1, y), (2, w ), (3, y)} is a func tion from A to B and so w e m ay w rite f \ : A B . On the other hand, f 2 = {(1, x ), (2, z), (3, y), (2, x)} is not a function since there are two ordered pairs w hose first coordinate is 2. In addition, fo = {(1, z), (3 ,x )} is not a function from A to B either because d o m ( /3) ^ A. O n the other hand, fo is a function from A — {2} to B. It is often convenient to “visualize” a function / : A  » Z? by representing the two sets A and B by diagram s and draw ing an arrow (a directed line segm ent) from an elem ent x e A to its im age f ( x ) e B. This is illustrated for the function f \ described above in Figure 9.1. Therefore, in order to represent a function in this way, exactly one directed line segm ent m ust leave each elem ent o f A and proceed to an elem ent o f B . In calculus, functions such as f ( x ) = x 1 are considered. This function / is from R to R , that is, A — R and B = R . A lthough f ( x ) = x 2 is com m only referred to as a function in calculus and elsew here, strictly speaking f (x) is the im age o f a real num ber x under / . The function / itself is actually the set f = { ( x , x 2) : x e R ). So (2, 4) and (—3, 9), for exam ple, belong to / . The set {(x, x 2) : x e R} o f points in the plane is the graph o f / . In this case, the graph is a parabola. H ere the function / : R —> R defined by / ( x ) = x 2 can also be thought o f as defined by a rule, nam ely the rule that associates the num ber x 2 w ith each real num ber x. Example 9.1
A nother function encountered in calculus is g (x ) — ex . A s w e mentioned above, this function is actually the set g = { ( x , e x) : x e R ). More precisely, this is the function g : R —*■ R d e fin e d by g ( x ) = ex f o r all x e ~R. In gen eral, w e will fo llo w this latter convention f o r defining functions that are often described by some rule or form ula. Consequently, the function h (x) — from calculus is the x —1
Figure 9.1
A function
218
Chapter 9
Functions
function h : R — {1} > R defined by h ( x ) = f o r all x e R with x ^ 1 and the x —1 fu nction cp(x) = In x is the function <fi : R + —> R defined by (j>(x) = \ n x f o r all x e R + , where, recall, R + is the set o f all positive real numbers. ♦ For a function / : A —» B and a subset C of A, the im ag e / ( C ) o f C is defined as / ( C ) = [ f ( x ) : x 6 C}. Therefore, / ( C ) c B for each subset C o f A. If C = A, then / ( A ) is the range o f / . Example 9.2
For A = {a , b, c, d. e] and B = {1, 2 , . . . , 6 }, / = {(a, 3), (6 , 5), (c, 2), (d , 3), (e, 6)} is a function fro m A to B . F o r C \ — {a, b, c], C 2 — {a, d ), C 3 = {e} an d C 4=
A,
/ ( C , ) = {2, 3, 5), / ( C 2) = {3}, / (C 3) = {6 }, / ( C 4) = ran< ?e(/) = {2, 3, 5, 6 }.
#
For a function / : A —»• S and a subset D of B . the in v erse im ag e f ~ l (D ) o f D is defined as f  \ D ) = {a e A : f ( a ) e D ). T herefore, / ~ ' ( D ) C A for each subset D o f B. N ecessarily then, f ~ l (B ) = A. In particular, for an elem ent b e B, f\{b }) = { a e A : Example 9.3
f ( a ) = b}.
For the function f : A = {a, b, c, d, e} —»■ B — {1, 2, . . . , 6 } defined in Exam ple 9.2 by / = {(a,3),(&,5),(c,2),W ,3),(e,6)},
it fo llow s that r\B )= A .
/ “ '({3}) = [a, d],
/ ~ ' ( { 1 , 3}) = {a, d},
/ “ '({4}) — 0
and ♦
Am ong the m any classes o f functions encountered in calculus are the polynom ial functions, rational functions and exponential functions. The function / : R —»■ R defined earlier by / ( x ) = x 2 for x e R is a polynom ial function. The function h in E xam ple 9.1 is a rational function and g is an exponential function. O ther im portant classes o f functions encountered often in calculus are continuous functions and differentiable functions. T he definition o f function that we have given is m ost likely not the definition you recall from calculus; in fact, you m ay not recall the definition o f function given in calculus at all. If this is the case, then it is not surprising. The evolution of w hat is m eant by a function has spanned hundreds o f years. It w as in the developm ent o f calculus that the necessity o f a form al definition o f function becam e apparent. Early in the 18th century, the Swiss m athem atician Johann B ernoulli wrote: I call a function o f a variable magnitude a quantity composed in any manner whatsoever fro m this variable magnitude and fro m constants.
9.2
The Set of All Functions from A to H
219
L ater in the 18th century, the fam ous Sw iss m athem atician L eonhard E uler studied calculus as a theory o f functions and did not appeal to diagram s and geom etric interpre tations, as m any o f his predecessors had done. T he definition o f function that E uler gave in his w ork on calculus is: A function o f a variable quantity is an analytic expression com posed in any way whatsoever o f the variable quantity and numbers or constant quan tities. Early in the 19th century, the G erm an m athem atician P eter D irichlet developed a m ore m odern definition o f function: y is a function o f x when to each value o f x in a given interval there corresponds a unique value o f y. D irichlet said that it d id n ’t m atter w hether y depends on.r according to som e form ula, law, or m athem atical operation. H e em phasized this by considering the function / : R —» R defined by r
_
1 if x is rational 0 if x is irrational.
L ater in the 19th century, the G erm an m athem atician R ichard D edekind wrote: A fun ction (j>on a set S is a law according to which to every determinate element s o f S there belongs a determinate thing which is called the transform o f s and is denoted by 4 >(s). So, by this tim e, the m odern definition o f function had nearly arrived.
9.2 The Set of All F unctions from A to B For nonem pty sets A and B , we denote the set o f all functions from A to B by B A. T hat is, B a = { / : / is a function from A to B] or, m ore simply, BA = {f:
f :A
> B \.
A lthough this m ay seem like peculiar notation, it is actually quite logical. In particular, le t’s determ ine B A for A = {a, b } and B = {x, y , z j. Each function / from A to B is necessarily o f the form / = { ( a , c t ) , ( b . 0 )}, w here a, fi e B . Since there are three choices for a and three choices for fi, the total num ber o f such functions / is 3 ■3 = 32 = 9. These nine functions are listed below: f i = {{a, x), (b, x)}, U = {{a ,y ),(b ,x )}, f i = {(a ,z),(b ,x )},
f 2 = {(a ,x ),(b ,y )}, h = {(a ,y ),(b ,y )}, fs = {(a ,z),(b ,y )} ,
/ 3 = {(a, x ) , (b, z)}, f 6 = {(a, y ) , ( b , z)}, f 9 = {(a ,z),(b ,z )}.
Hence the num ber o f elem ents in B A is 32. In general, for finite sets A and B , the num ber o f functions from A to B is \ B a \ =  5  /41. If B = {0, 1}, then it is com m on to represent the set o f all functions from A to B by 2 A.
220
Chapter 9
9.3
Functions
O netoO ne and Onto F u nctions We now consider tw o im portant properties that a function m ay possess. A function / from a set A to a set B is called onetoone or injective if every tw o distinct elem ents of A have distinct im ages in S . In sym bols, a function / : A >■ B is onetoone if w henever .v, y € A and .v / v, then f i x ) ± ./'(>’). Thus, if a function f :A B is not onetoone, then there exist distinct elem ents w and z in A such that f { w ) = / (z). Let A = [a, b , c, d }, B = {r, f,z/, v} and C = {x, y , z}. Then /] = {(a, s ), (fr, m),
(c ,
vj, ( J , ?•)}
is a onetoone function from A to B since distinct elem ents o f A have distinct im ages in B : w hile the function f 2 = {(a , s ) , ( b , t ) , ( c , s ) , ( d , u )} from A to B is not onetoone since a and c have the same im age, nam ely s. T here is no onetoone function from A to C , however. In general, for a function / : A B to be onetoone, w here A and B are finite sets, every tw o elem ents o f A m ust have distinct im ages in B and so there m ust be at least as m any elem ents in B as in A, that is, A < \B\. A t tim es, the definition o f a onetoone function is difficult to w ork w ith since it deals w ith unequal elem ents. However, there is a useful equivalent form ulation o f the definition using the contrapositive: A function / : A > B is onetoone if w henever f i x ) = f ( y ) , w here x , y e A, then x = y. We show how this form ulation can be applied to functions defined by form ulas. Result 9.4
L et the functio n f : R
R be defined by f i x ) = 3x  5. Then f is onetoone.
Proof A ssum e that f ( a ) = f i b ) , w here a , b e R. T hen 3a  5 = 3b  5. A dding 5 to both sides, w e obtain 3a = 3b. D ividing by 3, w e have a = b and so / is onetoone. ■ Example 9.5
L et the function f : R  » R be defined by f i x ) — x 2  3x  2. Determine whether f is onetoone.
Solution
Since / ( 0 ) =  2 and / ( 3 ) =  2 , it follow s that / is not onetoone.
♦
A naly sis
To show that the function / defined in E xam ple 9.5 is not onetoone, w e m ust show that there exist two distinct real num bers having the sam e im age under / . T his was accom plished by show ing that / ( 0 ) = /( 3 ) . B ut w hat if w e ca n ’t find tw o real num bers w ith this property? N aturally, if w e ca n ’t find two such num bers, then we m ight think that / is onetoone. In that case, w e should be trying to prove that / is onetoone. We w ould probably begin such a pro o f by assum ing that f ( a ) = f i b ) , that is a 2 — 3a — 2 = b 1 — 3b  2. We w ould then try to show that a — b. We can sim plify a 2  3a — 2 = b 2  3 b — 2 by adding 2 to both sides, producing a 2 — 3a — b l — 3b. W hen attem pting to solve an equation, it is often convenient to collect all term s on one side o f the equation
9.3
OnetoOne and Onto Functions
221
w ith 0 on the other side. R ew riting this equation, we obtain a 2 — 3a — b 2 + 3b — 0. R earranging some term s and factoring, we have a 2 — 3a  b 2 + 3b = (a 2  b 2)  3(a  b) = (a — b)(a + b) — 3 (a — b) = (a — b)(a + b — 3) = 0. H ence if f ( a ) = f ( b ) , then (a — b)(a + b — 3) = 0. Since (a — b)(a + b — 3) = 0, it follow s that either a — b = 0 (and so a — b) or a + b — 3 — 0. Therefore, f i a ) — f (b ) does not im ply that a = b. It only im plies that a = b or a + b = 3. Since 0 + 3 = 3, w e now see w hy / ( 0 ) = / ( 3 ) . In fact, if a and b are any tw o real num bers w here a + b = 3, then J'(a) = f ( b ) . This tells us how to find all possible counterexam ples to the statem ent: / is onetoone. L ooking at f i x ) = x 2 — 3.v — 2 once again, w e see that f ( x ) = x(x  3)  2. Since x ( x  3) = 0 if x = 0 or x = 3, it is now m ore apparent why 0 and 3 are num bers for w hich / ( 0 ) = / ( 3 ) . ♦ A function / : A » B is called o n to or su rje c tiv e if every elem ent o f the codom ain B is the im age o f some elem ent of A. Equivalently, / is onto if / (A) = B . A function we considered earlier w a s / i : A —> B , w here A = {1, 2, 3) , B = {x, y, z, vi’} and f i  {(1, y), (2, w ), (3, y )}. This function f i is not onto since neither x n o rz is an im age o f som e elem ent o f A. You m ight notice that for these two sets A and B , there is no function from A to B that is onto since any such function has exactly three ordered pairs but B has four elem ents. Thus for finite sets A and B , if / : A —»■ B is a surjective func tion, then \B\ <  A. The function g : B > A w here g = {(.v. 3), (v , 1), (z, 3), (»•. 2)} is a surjective function, however, since each o f the elem ents 1, 2 and 3 is an im age of some elem ent o f S . Next, w e determ ine w hich o f the functions defined in Result 9.4 and Exam ple 9.5 are onto. R e su lt to P ro v e PROOFSTRATEGY
The function / : R
R defined by /( J t) = 3x — 5 is onto.
L et’s m ake a few observations before we begin the proof. To show that / is onto, we m ust show that every elem ent in the codom ain B = R is the im age o f som e elem ent in the dom ain A = R. Since / ( 0 ) =  5 and / ( l ) =  2 , certainly  5 and  2 are im ages of elem ents o f R. The real num ber 10 is an im age as w ell since / ( 5 ) = 10. Is n an im age o f som e real num ber? To answ er this question, we need to determ ine w hether there is a real num ber x such that f i x ) = n . Since f i x ) = 3x — 5, w e need only find a solution for x to the equation 3x — 5 = n . Solving this equation for x , we find that X = i n + 5 )/3 , w hich, o f course, is a real num ber. Finally, observe that / 7T + 5 \
/W = / ( —
/ JT + 5 \
) = 3(—
)  5 = *.
This discussion, however, gives us the inform ation we need to prove that / is onto since for an arbitrary real num ber r , say, w e need to find a real num ber x such that / (,v) = r. However, then, 3x — 5 = r and x = (r + 5 )/3 . ♦ Result 9.6 Proof
The function /' : R — R defined by f i x ) = 3x — 5 is onto. L et r e R. We show that there exists l e R
such that f i x ) = r . C hoose x = (r + 5 )/3 .
222
Chapter 9
[■’unctions
T hen x e R and
PROOF ANALYSIS
N otice that the proof itself o f Result 9.6 does not include consideration of the equation 3x — 5 — r. O ur goal was to show' that som e real num ber x exists such that f ( x ) = r. How we obtain this num ber, though possibly interesting, is not part o f the proof. On the other hand, it m ay be a good idea to accom pany the proof w ith this inform ation. ♦ L et A = {1, 2, 3}, B = {x, y, z, w } and C = {a, b, c}. Four functions gi : A g 2 : B —*■ C , gi : A —> C and gn : A —» C are defined as follows:
B,
g\ = {(1, y), (2, w ), (3, jc)}, g i = {(jc, b), (y, a), (z, c), (w, b)}, 83
= { ( l , a ) , ( 2 , c ) , (3 ,6 )},
«4 = { d , 6 ) , ( 2 , b), (3, b)}. The functions gi and g3 are onetoone; w hile gi and g4 are not onetoone since g i( x ) = g2(vt’) = b and #4( 1) = gt(2) = b. Both g 2 and gs are onto. The function g 1 is not onto because z is not an im age of any elem ent o f A; w hile $4 is not onto since neither a nor c is an im age o f an elem ent o f A.
9.4 Bijective Functions We have already m entioned, for finite sets A and B, that if / : A —>• B is a surjective function, then A > \B\. A lso, w e m entioned that if / : A —»■ B is onetoone, then  A <  B . H ence if A and B are finite sets and there is a function f : A > B that is both onetoone and onto, then  A [ = \B . W hat happens w hen A and B are infinite sets w ill be dealt w ith in detail in C hapter 10. A function f : A *■ B is called bijectiv e or a o n eto o n e c o rre sp o n d e n c e if it is both onetoone and onto. From w hat we m entioned earlier, if a function f : A ^ B is bijective and A and B are finite sets, then  A  = \B\. Perhaps it is also clear that if A and B are finite sets w ith A =  B , then there exists a bijective function / : A —» B. A bijective function from a set A to a set B creates a pairing o f the elem ents o f A with the elem ents o f B . In the case w here A and B are sets w ith A = B  = 3 , say A = {a, b, c} and B = [x, y, z], the bijective functions from A to B are f \ = {{a ,x ),(b ,y ),(c ,z)}
h = {(a, y), (b , z), (c, a)} / 3 = {(a, z), (b, x), (c, j)} / 4 = {( a, y) , (b ,x ), ( c, z )}
fs = i (a ,z ),( b,y) , (c, jc)} f 6 = {(a,x), (b , z), (c, j)}.
9.4
Bijective Functions
223
T hat is, there are six bijective functions from A to B ; indeed there are six bijective functions from any 3elem ent set to any 3elem ent set. M ore generally, we have the follow ing. (See Exercise 9.34.) Theorem 9.7
I f A a n d B are finite sets with A  = \B\ = n, then there are n\ bijective functions from A to B.
Proof
Suppose that A = {ax, a 2, ■■■, an}. T hen any bijective function / : A »■ B can be ex pressed as / = {(at,  ) , (a2,  ) , . . . , (an,  ) } , w here the second coordinate o f each ordered pair o f f belongs to B. There are n pos sible im ages for a\ in / . O nce an im age for a \ has been determ ined, then there are n — 1 possible im ages for a2. Since / is onetoone, no elem ent o f B can be the im age of tw o elem ents o f A. B ecause neither o f the im ages o f a\ and a 2 can be an im age o f 03, there are n — 2 possibilities for a 3. C ontinuing in this manner, w e see that there is only one possibility for the im age o f an. It turns out that the total num ber o f p o s sible bijective functions / is obtained by m ultiplying these num bers and so there are n(n — 1)(n — 2) • • ■1 = nl bijective functions from A to S . ■ T here is another interesting fact concerning the existence o f bijective functions / : A —> B for finite sets A and B with A  = B .
Theorem 9.8
L et A and B be finite nonempty sets such that A  = \B\ and let f be a function fro m A to B . Then f is onetoone i f and only i f f is onto.
Proof
L et A  = fi = n. A ssum e first that / is onetoone. Since the n elem ents o f A have distinct im ages, there are n distinct im ages. Thus r a n g e (/) = B and so / is onto. For the converse, assum e that / is onto. T hus each o f the n elem ents o f B is an im age o f som e elem ent o f A. Consequently, the n elem ents o f A have n distinct im ages in B , w hich im plies that no two distinct elem ents o f A can have the sam e im age and so / is onetoone. ■ T heorem 9.8 concerns finite sets A and B w ith ]A = \B\. Even though we have not defined cardinality for infinite sets, w e w ould certainly expect that  A =  A for every infinite set A. W ith this understanding, T heorem 9.8 is false for infinite sets A and B , even w hen A = B. For exam ple, the function / : Z —» Z defined by f ( n ) = 2n is onetoone; yet its range is the set o f all even integers. T hat is, / is not onto, even though / is a onetoone function from Z to Z. The function / ; N > N defined by g(n) — n — 1 w hen n > 2 and g ( l ) = 1 is onto but not onetoone since g ( l ) = g ( 2) = 1. For the sets A = (1 , 2, 3}. B — {x, y, z, w} and C = {a, b, c) described at the end o f Section 9.3, no function from A to B or from B to C can be bijective. It is possible to have a bijective function from A to C , however, since A  = C . In fact, £3 is such a function, although other bijective functions from A to C exist. Certainly, not every function from A to C is bijective, as g 4 illustrates.
224
Chapter 9
Functions
For a nonem pty set A, the function iA : A —»• A defined by i A(a) = a for each a e A is called the identity function on A . If the set A under discussion is clear, w e write the identity function iA by i. For S = {1, 2, 3}, the identity function is is — i — {(1> 1). (2 ,2 ), (3,3)}. N ot only is this identity function bijective, the identity function iA is bijective for every nonem pty set A. Identity functions are im portant and we will see them again soon. We give one additional exam ple o f a bijective function. Result 9.9
The function f : R — {2} —» R — {3} defined by
fix)
«
3x x  2
is bijective. Proof
H ere it is necessary to show that / is both onetoone and onto. We begin w ith the first o f these. A ssum e that f ( a ) = f ( b ) , w here a, b 6 R — {2}. T h e n  — . M ultia —2 b —2 plying both sides by (a  2){b  2), we obtain 3a(b  2) = 3b(a  2). Sim plifying, we have 3ab — 6a = 3ab — 6 b. A dding —3a b to both sides and dividing by —6, w e obtain a — b. Thus / is onetoone. To show that / is onto, let r g R — {3}. We show that there exists j c e R — {2} such 2T that f (x) — r. Choose x —  . T hen ,n _
r ( 2r \ _ 3 ( r b ) _ 6;__ 6r _ _ \r —3/ —2 2r — 2(r — 3) 6
im plying that / is onto. T herefore / is bijective. PROOF ANALYSIS
■
Som e rem arks concerning the proof that the function / in R esult 9.9 is onto m ay be useful. F or a given real num ber r in R  {3}, we need to find a real num ber x in R  {2} 3x such that f ( x ) = r. Since w e w anted f i x ) = — :— — r , it was required to solve this x —2 equation for x . This can be done by rew riting this equation as 3,v = r ( x — 2) and then sim plifying it to obtain r x  3x = 2r . Now, factoring x from r x — 3 x and dividing by r — 3, w e have the desired choice o f x, nam ely x — 2r / ( r  3). Incidentally, it was perfectly perm issible to divide by r  3 since r e R  {3} and so r ^ 3 . N otice also that X e R — {2} for if x = 2r / ( r — 3) = 2, then 2/' = 2r — 6, w hich is im possible. A lthough 3x solving   = ;• for * is not part o f the proof, again it m ay be useful to include this w ork in addition to the proof.
$
Suppose that / i s a function from A to B , th at is, / : A B. If /'(.v) = / ( j ) im plies that x = y for all x , y e A. then / is onetoone. It m ay seem obvious that if x — y, then f ( x ) — f ( y ) for all x , y e A since this is sim ply a requirem ent o f a function. In order for a relation / from a set A to a set B to be a function from A to B , the follow ing two conditions m ust be satisfied: (1) For each elem ent a e A, there is an elem ent b e B such that (a. b) e f . (2) If (a , b), (a , c) € / , then b — c.
9.5
Composition of Functions
225
C ondition (1) states that the dom ain of / is A , that is, every elem ent o f A has an im age in B ; w hile condition (2) says that if an elem ent o f A has an im age in B , then this im age is unique. Occasionally, a function / that satisfies condition (2) is called w elldefined. Since (2 ) is a requirem ent o f every function however, it follow s that every function m ust be w elldefined. There are situations though w hen the definition o f a function / m ay m ake it unclear w hether / is w elldefined. This can often occur w hen a function is defined on the set o f equivalence classes o f an equivalence relation. The next result illustrates this w ith the equivalence classes for the relation congruence m odulo 4 on the set o f integers. R esult to Prove
T he function / : Z 4 —»■ Z 4 defined by / (.v) — [3jc + 1] is a w elldefined bijective function.
PROOF STRATEGY
To prove that this function is w elldefined, w e are required to prove that if [a ] = [b], then / ([a]) = / ([&]), that is, [3a + 1] = [3b + 1], It seem s reasonable to use a direct proof, so we assum e that [a] = [ft]. Since [a] and [b ] are elem ents o f Z 4, to say that [a] = \h \ m eans that a = b (m od 4). Since a = b (m od 4), it follow s that 4  («  b) and so a  b = 4k for som e integer k. To verify that [3a + 1] = [3b + 1], w e are required to show that 3a + 1 = 3b + 1 (m od 4) or, equivalently, that (3a + 1)  (3b + 1) = 3a — 3b = 3(a — b) is a m ultiple o f 4. Since Z 4 consists only o f four elem ents, nam ely, [0], [1], [2], [3], to prove that / is bijective, we need only observe that the elem ents / ([0]), / ([1]), / ([2]), / ([3]) are distinct. §
Result 9.10
Proof
The function f function.
:
Z 4 —>• Z 4 defined by f ([.*]) = [3* + 1] is a welldefined bijective
First, w e verify that this function is w elldefined; that is, if [a] = [b ], then / ([a]) = /([& ]). A ssum e then that [a] = [b], Thus a = b (m od 4) and so 4  (a — b). H ence a — b = 4k for som e integer k. Therefore, (3a + 1)  (3b + 1) = 3(a  b) = 3(4*') = 4(3*). Since 3k is an integer, 4  [(3a + 1)  (3b + 1)]. T hus 3a + 1 = 3b + 1 (m od 4) and [3a + 1] = [3b + 1]; so / ([a]) = / ([6]). H ence / is w elldefined. Since / ([0]) = [1], / ([1]) = [0], / ([2]) = [3] and / ([3]) = [2], it follow s that / is both onetoone and onto; that is, / is bijective. m
________ 9.5 C om position o f Functions

A s it is com m on to define operations on certain sets o f num bers (and on the set Z„ of equivalence classes, as w e described in C hapter 8), it is possible to define operations on certain sets o f functions, under suitable circum stances. For exam ple, for functions / : R —> R and g : R —> R, you m ight recall from calculus that the sum f + g and product f g o f / and g are defined by (/ +
= f i x ) + g( x) and ( f g ) ( x ) = f ( x ) ■g( x)
(9.1)
226
Chapter 9
Functions
for all x € R. So if / is defined by f i x ) = x 2 and g is defined by g ( x ) = sin a , then ( / + g)(x) — x 2 + siiiX and ( f g ) ( x ) = x 2 sin a for all x e R. In calculus we are espe cially interested in these operations because once w e have learned how to determ ine the derivatives o f / and g, we w ant to know how to use this inform ation to find the deriva tives o f f + g and f g . The derivative o f f g , for exam ple, gives rise to the w ellknow n product rule for derivatives: ( f g ) ' M = f ( x ) ■g '(x) + g ( x ) ■f { x ) . This later led us to study the quotient rule for derivatives. The definitions in (9.1) o f the sum f + g and product f g of the functions / : R R and g : R —> R depend on the fact that the codom ain o f these tw o functions is R, w hose elem ents can be added and m ultiplied, and so / ( a ) + g ( x ) and f i x ) • g (x ) m ake sense. O n the other hand, if / : A  * B a n d g : A > B, w here B = {a, b, c}, say, then f i x ) * ■g(x) and / ( a ) • g ( x ) have no meaning. There is an operation that can be defined on pairs o f functions satisfying appropriate conditions that has no connection w ith num bers. For nonem pty sets A, B and C and functions / : A —> B and g : />’ > C . it is possible to create a new function from / and g, called their com position. The co m p o sitio n g o / o f / and g is the function from A to C defined by (g ° f ) ( a ) = g ( f { a ) )
for all a e A.
To illustrate this definition, let A = { 1 ,2 ,3 , 4}, B — [a, b, c, d) and C — {r, s, t, u, v} and define the functions / : A —> B and g : B  » C by / = { (l,& )f ( 2 ,rf)>( 3 ,f l) ,( 4 ,a ) } , g = {(a, u), (b, r), (c, r), (d , s )}. We now have the correct arrangem ent o f sets and functions to consider the com position g o f . Since g o: / is a function from A to C , it follow s that g o f has the follow ing appearance: # ° / = {(!> “ )> (2, P), (3, y ) , (4, <5)}, w here a , f i , y , 8 e C . It rem ains only to determ ine the im age o f each elem ent o f A. First, w e find the im age o f 1. A ccording to the definition o f g o / , (g ° / ) ( 1 ) = g ( f ( 1)) = g(b) = r, so (1, r) e g o f . Similarly, (g o / ) ( 2 ) = g ( f ( 2)) = g(d) = s and so (2, s) e g o / . Continuing in this m anner, we obtain g ° f = {(!> r), (2, s), (3, « ), (4, «)}. A diagram that illustrates how g o f is determ ined is show n in Figure 9.2. To find the im age o f 1 under g q f , w e follow the arrow from 1 to b and then from b to r . The function g o / is basically found by rem oving the set B . The fact that g o f is defined does not necessarily im ply that / o g is also defined. Since g is a function from B to C and / is a function from A to B , the only way that f o g w ould be defined is if range(g) c A. In the exam ple we have ju st seen, f c g is not defined since range(g) = {/% s, u } 2 A.
9.5
Composition of Functions
227
g °f
/l cx\ /
2
a
i
30
/ \
/
a
g
r \ >o s \
■ ,4 h \ C s' \ d
V 4
A
yj
O t
\ X) u / \ o V•/
B Figure 9.2
c
The composition function g o f
C om position o f functions was also encountered in calculus. L e t’s consider an ex am ple o f com position that you m ight have seen in calculus. A gain, suppose that the functions f : R —»■ R and g : R —> R are defined by f ( x ) — x 2 and g ( x ) = sin x . In this case, w e can determ ine both g o f and f o g ; namely, ig ° / ) ( * ) = g i f M ) = g ( x 2) = sin ( x 2) if °
0 0 0
= f i g ( x ) ) = / ( s i n x ) = (s in x )2 = sin2 x.
Second, this exam ple also serves to illustrate that even w hen g o f and / o g are both defined, they need not be equal. The study o f com position o f functions in calculus led us to the w ellknow n chain rule for differentiation: C? ° / ) '( x )  g' ( f ( x ) ) ■f \ x ) . There are tw o facts concerning properties o f com position o f functions that w ill be especially useful to us. First, if / and g are injective functions such that g o f is defined, then g o f is injective. The corresponding statem ent is also true for surjective functions. R e su lt to P ro v e
L et / : .4  .» />' and g : B (a) (b)
PROOFSTRATEGY
C be tw o functions.
If / and g are injective, then so is g o f . If / and g are surjective, then so is g a f .
To verify (a), we use a direct pro o f and begin by assum ing that / and g are onetoone. To show that g © f is onetoone, we prove that w henever (g o /')(«,) = (g o f ) ( a 2). then a : = a2. However, (g q /)(<%) = ( g o /)(<72) m eans that gi f (a\)) = g ( f ( a 2)). But g is onetoone; so g ( x ) = g ( y ) im plies that x = y. The form g ( x ) = g ( y ) is exactly w hat we have, w here x — f (« ;) and _v = / (a2). This leads us to f U i \ ) = f U i i ) . B ut we also know that / is onetoone.
228
Chapter 9
Functions
To verify (b), w e need to prove that if / and g are onto, then g o / is onto. To show that g o f is onto, it is necessary to show that every elem ent o f C is an im age o f some elem ent o f A under the function g o f . So w e begin w ith an elem ent c e C . Since g is onto, there is an elem ent b e B such that gib) — c. B ut / is onto; so there is an elem ent a e A such that f ( a ) = b. This suggests considering (g o /')(«). # Theorem 9.11
Let f : A —» B and g : B —> C be two functions. {a) I f f and g are injective, then so is g o f . (b) I f f and g are surjective, then so is g o f .
Proof
L et / : A —> B and g : B —> C be injective functions. A ssum e that (g o f ) ( a \ ) = (g o f ) ( a 2). w here a \ , a2 e A. By definition, g i f ( a 0 ) = g i f (0 2 )) Since g is injective, it follow s that f i a \ ) = f ( a 2 ) Flowever, since / is injective, it follow s that a\ = a 2This im plies that g o / is injective. N ext let / : A  » B and g : B —> C be surjective functions and let c e C. Since g is surjective, there exists b e B such that gib) — c. On the other hand, since / is surjective, it follow s that there exists a e A such that f { a ) = b. H ence ig o f ) i a ) = g i f (a)) = gib) = c, im plying that g o / is also surjective. ■ C om bining the tw o parts o f T heorem 9.11 produces an im m ediate corollary.
Corollary 9.12
I f f : A  » B and g : B — C are bijective functions, then g o f is bijective. F or nonem pty sets A . B , C and Z), let / : A —►B , g : B ►C and h : C —> D be functions. T hen the com positions f o / : A > C and h o g : B *> D are defined, as are the com positions h o (g o f ) : A D and ih q g ) o f : A D . C om position o f the functions / , g and /? is associative if the functions h o ( g o f ) and (// o g) o / are equal. T his is, in fact, the case.
Theorem 9.13
Proof
F or nonempty sets A, C Z3, let f : A —>■ B , g : B > C and h : C functions. Then ih q g) o f = h o (g o / ) .
D be
L et a e A and suppose that f ( a ) = b, gib) = c and h(c) = d. Then
i(h o g ) o J)(a) = {h o g ) i f i a ) ) = <7i o g)(b) = % (Z>)) = hie) = </; w hile Qi o i g o f ) ) ( a ) = /;((.? o / ) ( « ) ) = h i g i f i a ))) = A(g(fe).} = /?(c) = d. T hus ih
f = h o (g
0
/).
h
A s w e have m entioned, it is com m on, w hen considering the com position o f func tions, to begin w ith tw o functions / and g, w here / : A —►B and g : B —*■ C and arrive at the function g o f : A —* C . Strictly speaking, however, all that is needed is for the dom ain o f g to be a set B ' w here r a n g e ( /) is a subset o f B'. In other w ords, if / and g are functions w ith / : A —>• B a n d g : B ' —> C , w here r a n g e (/) c B ' , then the com position g o f : A —> C is defined.
9.6 Example 9.14
229
For the sets A = {—3, —2, . . . , 3} an d B = {0, 1 , . . . , 10}, B' = {0, 1, 4, 5, 8, 9}and C = { 1 , 2 , . . . , 10}, let f : A —> B and g : B '  » C be functions defined by f ( n ) = n 2 f o r all n e A and g{n) = n + 1 f o r all n e S '. (a) (.b)
Solution
Inverse Functions
Show that the composition g o / : A * Cis defined. For n e A , determine (g o / ) ( « ) .
(a)
Since r a n g e ( /) = {0, 1, 4, 9} and r a n g e / c 5 ', it follow s that the com position g o / : A —> C is defined. (b) For n e A , { g o / ) ( « ) = g ( f ( n ) ) = g (n 2) = n 2 + 1.
♦
9.6 Inverse F u nctions N ext w e describe a property possessed by all bijective functions. In preparation for doing this, w e return to relations to recall a concept introduced in C hapter 8. F or a relation R from a set A to a set B, the inv erse re la tio n R ~ x from B to A is defined as R
1
= {(b , a ) : (a, b) e R}.
F or exam ple, if A — [a, b , c , d ] , B = {1, 2, 3} and R = {(a, 1), (a, 3), (c, 2), (c, 3), (d, 1)} is a relation from A to 6 , then / T 1 = { ( l , a ) , ( 3 , a ) f (2 ,c ), ( 3 ,c ) , ( l ,d ) } is the inverse relation o f R . O f course, every function / : A  » B is also a relation from A to B and so there is an inverse relation f ~ 1 from B to A . This brings up a natural question: U nder w hat conditions is the inverse relation / 1 from B to A also a function from B to A? If the inverse relation f ~ l is a function from B to A, then certainly d o m ( / _1) = B. This im plies that / m ust be onto. If / is not onetoone, then f { a \ ) = f ( a 2) = b for someai,a2 € A andfo e B ,w h e r e a \ / a 2B ut then (b, a \), (b , a 2 ) € / _1, w hich cannot occur if / _l is a function. This leads us to the follow ing theorem . In the proof, tw o basic facts are used repeatedly, nam ely (1) f ( a ) = b i f and only i f (a, b) e / and (2) i f f ~ x is a function and f ( a ) = b. then (b, a) e / _1. Theorem 9.15
Proof
L et f : A —»■ B be a function. Then the inverse relation / “ ’ is a fu nc tion fro m B to A i f and only i f f is bijective. Furthermore, i f f is bijective, then f ~ x is also bijective. First, assum e that / ~ ! is a function from B to A. T hen w e show that / is both onetoone and onto. A ssum e that f { a \ ) — f { a 2) = y, w here y € B. T hen ( a \ , y), (a2, y) e / , im plying that ( j , a \), (y, a2) e / _1. Since f ~ l is a function from B to A, every elem ent of B has a unique im age under f ~ [. Thus, in particular, y has a unique im age under f ~ l . Since / _1(>0 = a\ and / 1(>>) = a 2, it now follow s th a ta i = a 2 and so / is onetoone. To show that / is onto, Set b e B . Since f ~ l is a function from B to A, there exists a unique elem ent a e A such that f ~ l (b) = a. H ence (b, a) e / _1, im plying that (a, b) e / , that is, f ( a ) = b. Therefore, / is onto.
230
Chapter 9
Functions
For the converse, assum e that the function / ' : A —> B is bijective. We show that f ~ l is a function from B to A . Let ft e B . Since / is onto, there exists a e A such that (a , b) e / . H ence (b, a ) e f . It rem ains to show that (b, a) is the unique elem ent of f ~ x w hose first coordinate is b. A ssum e that (ft, a) and (ft, a') are both in f ~ ]. T hen (a, b), (a', b) e / , w hich im plies that / ( « ) = f ( a ' ) = b. Since / is onetoone, a = a'. T herefore, w e have show n that for every b € B there exists a unique elem ent a e A such that (b , a) e f 1: that is, f ~ x is a function from B to A. Finally, w e show that if / is bijective, then / 1 is bijective. A ssum e that / is bijective. We have ju st seen that f ~ x is a function from B to A. First, w e show that f ~ x is onetoone. A ssum e that f ~ x(b\) = / _1(ft2) = a. T hen {b\, a), (b 2 , a) e / _l and so (a , b\), (a, b2) e / • Since / is a function, and / _1 is onetoone. To show that / “ ' is o n to ,le ta e A. Since / is a function, there is an elem ent b e 6 such that (a, b) e f . C on sequently, (ft, a ) e / _1 s o t h a t / _1(ft) — a a n d / _1 is onto. T h e re fo re ,/ 1 is bijective. ■ Let f : A B be a bijective function. By T heorem 9.15 then, / _1 : B 4* A is a bijective function, w hich is referred to as the in v erse fu n ctio n or sim ply the inverse o f / . H ence both com position functions f ~ l o f and f o f are defined. In fact, / _1 o / is a function from A to A and / o is a function from fi to B. A s we are about to learn, f ~ l o / and / o / _l are functions w e'v e visited earlier. L et a € A and suppose that f ( a ) = b. So (a, b) e / and therefore (b , a ) e / _1, that is, f ~ ](b) = a. Thus ( Z ” 1 o / ) (a) = (f ( c i )) = / M/» = a and ( / o /  1) (ft) = / ( / _1(^)) = / ( a ) = ft. So it follow s that / “' 0 / = U
and
f o f ~ l = iB
are the identity functions on the sets A and B. (See Figure 9.3.) In fact, if / : A  » B and g : B  4* A are functions for w hich g o f = iA and f o g = i B, then / and g have som e im portant properties. Theorem 9.16
I f f : A —»■ 6 a n d g : B > A are two fun ctions such that g o f — iA a n d f o g = i s , then f and g are bijective an d g —■f 1.
Proof
First, w e show that / is onetoone. A ssum e that / ( a j ) = / ( 02X w here a ; , </2 € A. Then .?(/(</•,)) = g ( f ( a 2)) Since g o f = iA, it follow s that
a\ = (g o./)(«!) = g(/(tfi)) = g {f(a 2)) = (go /)(a2) = and so / is onetoone.
A
Figure 9.3
A bijective function and its inverse
9.6
Inverse Functions
231
N ext, w e show that / is onto. L et b e B and suppose that gib) = a. Since / o g = i B, it follow s that ( / o g)(b) = b. Therefore, ( / o g)ib) — f i g i b )) = / ( a ) = b and so / is onto. H ence / is bijective and so f ~ l exists. Similarly, g is bijective. L et a e A and suppose that f i a ) = b e B . T hen f ~ l (b) = a. Since g o f — i&, it follow s that a = ig o f ) ( a ) = g { f ( a ) ) = gib). Therefore, g = f ~ l . • If a bijective function / has a relatively sm all num ber o f ordered pairs, then it is easy to find f ~ l . But w hat if / is a bijective function that one m ight encounter in calculus, say? W e illustrate this next w ith a function described in R esult 9.9. Example 9.17
The function f : R — {2} — R — {3} defined by 3x /( * ) =
x 2
is known to be bijective. Determine f ~ \ x ) , where x e R — {3}. Solution
Since ( / o / “ ') (x) = x for all x e R — {3}, it follow s that ( / o /  ■ ) (x) = / ( /  ' (*)) =
3f~ H x ) f~ K x )  2
T hus 3 / “ 1(x) = x ( / _1(x) — 2) and 3 f ~ \ x ) = x / _1(x) — 2x. Collecting the term s in volving /  I (x) on the sam e side o f the equation and then factoring out the term / ~ '( x ) , w e have x f ~ \ x )  3 /  1 (x) = 2x so / ” (x)(x — 3) = 2x. Solving for /
(x), w e obtain r \x ) =
A nalysis
X
2x  3
You m ight very w ell have dealt w ith the problem o f finding the inverse o f a function before and m ight recall a som ew hat different approach than the one w e ju st gave. L et’s look at this exam ple again, but from a different perspective. W hen we consider functions from calculus, rather than w riting f i x ) = x 2, g ix ) = 1 , 1 5x + 1 or n ix ) = x H— , w e som etim es w rite these as y = x , y = 5x + 1 o r y = x 1— . x x 3x , 2x In Exam ple 9.17, w e w ere given f i x ) =  and found that / (x) =  . L et’s x —2 x  3 2x 2jc w rite the inverse as y =  instead. T hat is, (x, y ) e f ~ l , w here v = . O f x —3 ' x —3 course, initially, w e d o n 't know w hat y is. B ut if (x, y) e f ~ l , then (y, x ) e / and we 3y know that x = f i y ) = — — Solving this equation for v, w e have x (y — 2) = 3y, so y  2 x y — 2x = 3y. C ollecting the term s w ith v on the same side o f the equation and factoring out the term y, w e obtain
232
Chapter 9
Functions x y — 3)’ = 2x
and
y ( x  3) — 2x.
2 x. Solving for y , w e obtain 1 = ; that is, x —3
In short, to find /
1 if / ( x ) =  —, we replace / (x) by x and x by y and then solve for
y. The result is / _ 1(x). O f course, the procedure w e have described for finding / _1(*) is exactly the same as before. The only difference is the notation. You m ight have also noticed that the algebra perform ed to determ ine f ~ ' ( x ) in Exam ple 9.17 is exactly the sam e as the algebra perform ed in proving / is onto in Result 9.9. ♦ Finding the inverse o f a bijective function is not always possible by algebraic m anip ulation. For exam ple, the function / : R •> {(). oc) defined by / ( x ) — ex is bijective but ./' '(.v.) = ln x . Indeed, the function g : R R defined by g (x )  3 x 7 + 5 x 3 + 4x  1 is bijective, but there is no w ay to find an expression for « 1(.v). If / : A —>■ B is a onetoone function from A to B that is not onto, then, o f course, / is not bijective, and, according to T heorem 9 . 1 5 , / does not have an inverse (from B to A). On the other hand, if w e define a new function g : A »• r a n g e ( /) by g (x ) = f ( x ) for all x e A , then g is a bijective function and so its inverse function g '' : r a n g e (/)  j A exists. For exam ple, let E denote the set o f all even integers and consider the function / :Z Z by / ( « ) = 2n. T hen this function / is injective but not surjective and so there is no inverse function o f / from Z to Z. O bserve that r a n g e ( /) = E. If w e define g : Z  * E by g ( n ) = f ( n ) for all n G Z, then g is bijective and g 1 : E Z is a (bijective) function. In fact, g ~ l (n) = n j 2 for all n e E .
9.7 P erm u tatio n s We have already m entioned that the identity function iA defined on a nonem pty set A is bijective. N orm ally, there are m any bijective functions that can be defined on nonem pty sets. Indeed, the num ber o f bijections on an //elem ent set is n ! according to T heorem 9.7. These types o f functions occur often in m athem atics, especially in the area o f m athem atics called abstract (or m odern) algebra. A p e rm u ta tio n o f (or on) a nonem pty set A is a bijective function on A, that is, a function from A to A that is both onetoone and onto. By Results 9.4 and 9.6, the function / : R — R defined by / ( x ) = 3x — 5 is a perm utation o f R. L et's consider an even sim pler exam ple. For A = {1. 2, 3}, let / be a perm utation o f A. Then / is com pletely determ ined once w e know the im ages o f 1, 2 and 3 under / . We saw that there are three possible choices for / ( l ) , tw o choices for / ( 2) once / ( l ) has been specified and one choice for / ( 3 ) once / ( l ) and / ( 2 ) have been specified. From this, it follow s that there are 3 • 2 • 1 = 3! = 6 different perm utations / 'o f the set A = {1, 2, 3}. This agrees w ith T heorem 9.7. O ne o f these functions is the identity function defined on {1, 2, 3}, w hich w e denote by « i; that is,
on = {(1, 1), (2,2), (3,3)}.
9.7
Perm utations
233
A nother perm utation o f {1, 2, 3} is a 2 = { (1,1 ), (2, 3), (3, 2)}. There are other com m on w ays to represent these perm utations. A perm utation o f {1, 2, 3} is also w ritten as 1 2 3
w here the num bers im m ediately below 1, 2 and 3 are their im ages. H ence a \ , a 2 and the other four perm utations o f {1, 2, 3} can be expressed as:
= (1 2 3 )
"2 =
12 3\ 1 3 2J
(123 “3= 1 ^3 2 1
“4 = ( 2 1 3 )
“5 =
1 2 3\ 2 3 1/
a 6
(12 3 = 1 i 3 12
Since each perm utation a, (1 < i < 6) is a bijective function from {1, 2, 3} to {1,2, 3}, it follow s that from C orollary 9.12 that the com position o f any two perm u tations o f {1, 2, 3} is again a perm utation o f {1, 2, 3}. For exam ple, let’s consider 12 3 \ (\ 2 3\ < 2 2 °a5 = i 1 3 2 ; ° V 2 3 1
(\
2 3
Since ( a 2 o 0:5) (1) = ( ^ ( “ sO )) = “ 2(2) = 3, (a 2 o a$) (2) = 2 and (a 2 ° a s ) (3) = 1, it follow s that 12 3\
a2 o a5 =

12
2
/1 2 3 \
J ° \
(\ 2 3
2 3 1J =
\ 3 2 1 ' = a 3 ‘
By Theorem 9 . 13, it follow s that com position o f perm utations on the sam e nonem pty set A is associative. H ence for all integers i, j , k 6 {1, 2, • ■•, 6 },
(a; o a j) oau = a/ 0 (a; o a*). Also, by T heorem 9.15, since a perm utation is a bijective function, each perm utation has an inverse, w hich is also a perm utation. Thus for each i (1 < i < 6), a ~ = a j for som e j (1 < j < 6). T he inverse o f a perm utation can be found by interchanging the tw o rows and then reordering the colum ns so that the top row is in the natural order 1 ,2 , 3, ....T h u s 2 3 1\
(I 2 3
12 3 /
\3
1 2 ' ~ a6’
The set o f all n \ perm utations o f the set {1, 2 ,
is denoted by S n. Thus
(S3 = {a,,a2, . .., a6}. As we have seen w ith S 3 , the elem ents o f S n satisfy the properties o f closure, associativity and the existence o f inverses for every positive integer n. This w ill be revisited in C hapter 13.
234
Chapter 9
Functions
E X E R C IS E S FOR C H A PT E R 9 Section 9.1: The D efinition of Function 9.1. Let A = {a, b, c, cl) and B = {x, y , z}. Then / = {(a, y), (b, z), (c, y), (d , 2)} is a function from A to B. Determine d o m (/) and ran g e (/). 9.2. Let A = {1. 2, 3} and B = [a. b , c, d}. Give an example of arelation R from A to B containing exactly three elements such that R is not a function from A to B. Explain why R is not a function. 9.3. Let A be a nonempty set. If R is a relation from A to A that is both an equivalence relation and a function, then what familiar function is R1 Justify your answer. 9.4. For the given subset A; of R and the relation R , (1 < i < 3) from A, to R, determine whether Rj is a function from A, to R. (a) Ai = R, R i =ss {(x, y) : x e A i , j = 4x — 3} (b) A2 = [0, 00), R 2 = {(.v. y) : x e A2, (y + 2)2 = x} (c) A3 = R. R 3 = {(.v. y) : x e A3, (x + y ) 2 = 4} 9.5. Let A and B be nonempty sets and let R be a nonempty relation from A to B. Show that there exists a subset A' of A and a subset / of R such that / is a function from A' to B . 9.6. In each of the following, a function / : A; >• R (1 < i < 5) is defined, where the domain A, consists of all real numbers x for which f:(x ) is defined. In each case, determine the domain A, and the range of fi. (a) /;(,v) = 1 : x 2 (b) / 2(x) = 1  1 (c) / 3(x) = a /3 7  1 (d) / 4(x) = x 3  8 (e) f s(x) = ^ 3. 9.7. Let A = {3, 17, 29, 45} and B = {4. 6 , 22, 60}. A relation R from A to B is defined by a R b if a + b is a prime. Is R a function from A to B1 9.8. Let A = {5, 6}. B = {5, 7, 8} and S = {h : n > 3 is an odd integer}. A relation R from Ax B to S is defined as (a. b) R s if s  (a + b). Is R a function from A x B to S I 9.9. Determine which of the following five relations R t {i = 1 , 2 , . . . , 5) are functions. (a) lb) (c) (d) (e)
R\ is defined on R by x R\ y Ro is defined on R by x R 2 y R 3 is defined from N to Q by R 4 is defined on R by x R 4 y R 5is defined on R by x R 5 y
if x 2 + y2 = 1. if 4x2 + 3y2 = 1. a R 3 b if 3a + 5b — 1. if y = 4 — x — 2. if \x + 7 } = 1.
9.10. A function g : Q —»• Q is defined by g{r) = 4r + 1 for each r
6
Q.
(a) Determine g(Z) and g(E), where E is the set of even integers. (b) Determine g  1(N) and g ~ l(D), where D is the set of odd integers. 9 .1 1. Let C = {x 6 R : x > 1} and D = R + . For each function / defined below, determine /( C ) , / _1(C), and /  1({1}). (a) (b) (c) (d) (e)
/ / / / /
: R > R is defined by / Cx) = x 2. : R + > R is defined by f ( x ) = ln x . : R — R is defined by / ( x )  e x. : R —> R is defined by / ( x ) = sinx. : R —> R is defined by / ( x ) = 2x —x 2.
235
Exercises for Chapter 9
9.12. For a function /' : A ■•> B and subsets C and D of A and E and F of B. prove the following. (a) (b) (c) (d) (e) (f)
f( C ' u D) = n o U f ( D ) f(C ’ n D )C n c ) n f( D ) / ( C )  ■ f ( D ) c f ( C — D) n (E U F) = r (E) U f ~ \ F ) r 1 (E n F) = r ' ( E } n f  \ F ) r 1(E  F ) = r \ E )  f  '(F )
Section 9.2: The Set of All Functions from A to B 9.13. Let A = {1. 2, 3} and B — j.v. y>. Determine B A. 9.14. For sets A = {1, 2, 3, 4} and B = {a. y, r j. give an example of a function g e B A and a function h e B b . 9.15. For A = [a, b, c}, determine 2A. 9.16.
(a) Give an example of two sets A and B such that = 8. (b) Give an example of an element in B A for the sets A and B given in (a).
9.17. (a) For nonempty sets A, B and C, what is a possible interpretation of the notation C BA ? (b) According to the definition given in (a), determine C bA for A = {0, 1}. B = {a , b} and C =
{a ,
y).
Section 9.3: OnetoOne and Onto Functions 9.18. Let A = {w, x, y, z } and B = {r, s, t). Give an example of a function / nor onto. Explain why / fails to have these properties.
: A —!►B that is neither oneto
9.19. Give an example of two finite sets A and B and two functions / : A —►B and g : B —>• A such that / is onetoone but not onto and g is onto but not onetoone. 9.20. A function /
:
Z» Z is defined by / ( « )
= 2n + 1. Determine
9.21. A function /
:
Z 4Z is defined by f ( n )
= n — 3. Determine whether / is (a) injective, (b)
whether / is (a) injec
9.22. A function /
:
Z> Z is defined by f ( n )
=■ 5n + 2. Determine
whether / is (a) injec
9.23. Prove or disprove: For every nonempty set A, there exists an injective function / : T • * V (A ). 9.24. Determine whether the function / : R —» R defined by J '(x ) = 9.25.
a2
+ 4x + 9 is (a) onetoone, (b) onto.
Is there a function / : R —>■ R that is onto but not onetoone? Explain your answer.
9.26. Give an example of a function / : N —>■N that is (a) onetoone and onto (b) onetoone but not onto (c) onto but not onetoone (d) neither onetoone nor onto. 9.27. Let A = {2, 3, 4, 5} and B = {6, is a prime.
8, 10}. A relation R is defined from A to B by a R b if a \ band b/a +
(a) Is R a function from A to B ? (b) If R is a function from A to B , then determine whether this function is onetoone and/or 9.28. Let A = {2, 4, 6} and B = {1, 3, Is / a onetoone function?
onto.
4, 7, 9}. A relation / is defined from A to B by a f b if 5 divides ab
9.29. Let / be a function with d o m (/) = A and let C and D be subsets of A. Prove that if / is onetoone, then f(C C D )= f(C )n f(D ).
236
Chapter 9
Functions
Section 9.4: Bijective Functions 9.30. Prove that the function / : R —>■ R defined by f i x ) = I x — 2 is bijective. 9.31. Let / : Z 5 —> Z 5 be a function defined by /( [ a ] ) = [2a + 3], (a) Show that / is welldefined. (b) Determine whether / bijective. 9.32. Prove that the function / : R — {2} —s R — [5) defined by f i x ) —
js bijective.
9.33. Let A = [0, 1] denote the closed interval of real numbers between 0 and 1. Give an example of two different bijective functions f \ and /> from A to A, neither of which is the identity function. 9.34. Give a proof of Theorem 9.7 using mathematical induction. 9.35. For two finite nonempty sets A and B, let R be a relation from A to B such that range{R) = B. Define the domination number y i R ) of R as the smallest cardinality of a subset S c A such that for every element y of B , there is an element x e S such that x is related to y. (a) Let A = (1, 2, 3, 4, 5, 6 , 7} and B = [ a , b , c , d , e , f , g ] and let R = {(1, c), (1, e), (2, c), (2, / ) , (2, g), (3, b), (3, / ) , (4, a), (4, c), (4, g), (5, a), (5, b), (5, c), (6 , d), (6 , e), (7, a), (7,g)}. Determine y (R ). (b) If R is an equivalence relation defined on a finite nonempty set A (and so B = A), then what is y iR )7 (c) If / is a bijective function from A to B, then what is y ( / ) ? 9.36. Let A — [a, b , c, d, e, / } and B = [u, v, w , x , y, z}. With each element r e A, there is associated a list or subset L (r) c B. The goal is to define a “list function” cp : A —*■ B with the property that cp(r) e L (r) for each r € A. (a) For L{a) — [w, x , y], L(b) = {u, z], L(c) = {u, v}, L id ) = {u, w}, L(e) = {u , x , y}, L { f ) = {v, y}, does there exist a bijective list function cp : A B for these lists? (b) For Lia) = {«, v, x, y}, L(b) = [v, w , y ] , L ie) = [v, y}, L id ) = {u, w , x , z], L(e) = [v, w], L i f ) = [h1, y], does there exist a bijective list function cp : A —»• B for these lists?
Section 9.5: Composition of Functions 9.37. Let A = [1, 2, 3, 4}, B = [a, b , c) and C = { w , x , y, z}. Consider the functions / : A » B and g : B —> C, where / = {(1, b), (2, c), (3, c), (4, a)} and g = {(a, x), i b , >>), (c, x)}. Determine g o / . 9.38. Two functions / : R —»• R and g : R —>■ R are defined by f i x ) = 3x2 + 1 and g{x) — 5x — 3 for all x Determine (g o /) ( 1 ) and ( / o g )(l).
e R.
9.39. Two functions / : Z 10 >■ Z i0 and g : Z i0 —►Z !0 are defined by f ( [ a ]) = [3a] and g([a]) = [7a]. (a) Determine g o / and f o g . (b) W hat can be concluded as a result of (a)? 9.40. Let A and B be nonempty sets. Prove that if / : A * B, then / o iA = f and iB o / = / . 9.41. Let A be a nonempty set and let / : A > A be a function. Prove that if / o / = iA, then / is bijective. 9.42. Prove or disprove the following: (a) If two functions / : A » B and g : B > C are both bijective, then g o / : A *■ C is bijective. (b) Let / : A — B and g : B —> C be two functions. If g is onto, then g o / : A > C is onto. (c) Let / : A > B and g \ B *■ C be two functions. If g is onetoone, then g o / : A —»■ C is onetoone. (d) There exist functions / : A —>■ 5 and g : B C such that / is not onto and g o / : A » C is onto. (e) There exist functions / : A —> B and g : B —> C such that / is not onetoone and g o / : A C is onetoone.
237
Exercises for Chapter 9 9.43. For nonempty sets A, B and C , let / : A —s B and g : B —>■ C be functions. (a) Prove: I f g o f is onetoone, then / is onetoone. using as many of the following proof techniques as possible: direct proof, proof by contrapositive, proof by contradiction. (b) Disprove: If g o / is onetoone, then g is onetoone.
9.44. Let A denote the set of integers that are multiples of 4, let B denote the set of integers that are multiples of 8 and let B ' denote the set of even integers. Thus A = {4k : k e Z}, B = {8£ : k e Z} and B' = [2 k : k e Z}. Let / : A x A  > 8 and g : B' —* Z be functions defined by / ( ( x , y)) = x y for x , y e A and g(n) = n /2 for n e B '. (a) Show that the composition function g o / : A x A —> Zi s defined. (b ) For k, I e Z, determine (g o f) (( 4 k , 41)). 9.45. Let A be the set of even integers and B the set of odd integers. A function / : A x B  > £ x A is defined by / ( ( a , b)) = / ( a , b) = (a + b , a) and a function g : B x A  ^ B x B is defined by g(c, c/) = (c + d, c). (a) Determine (g o /)(1 8 , 11). (b) Determine whether the function g o f : A x B —> B x B i s onetoone. (c) Determine whether g o / is onto. 9.46. Let A be the set of odd integers and B the set of even integers. A function f : A x B  > A x A is defined by f ( a , b) = (3 a  b, a + b) and a function g : A x A  ^ B x A is defined by g(c, d) = (c — d, 2c + d). (a) Determine (g o /) ( 3 , 8). (b) Determine whether the function g o f  . A x B ^  B x A i s onetoone. (c) Determine whether g o / is onto. 9.47. For functions / , g and h with domain and codomain R, prove or disprove the following. (a) (g + h) a / = (g a f ) + (h o / ) (b) / o (g + h) = ( / o g) + ( / o /?). 9.48. The composition g o f : (0, 1) —*■ R of two functions / and g is given by (g o f ) ( x ) = o^=i_L, where / : (0, 1)
(—1, 1) is defined by / ( x ) = 2x — 1 fo rx e (0, 1). Determine the function g.
Section 9.6: Inverse Functions 9.49. Let A = {a, b, c}. Give an example of a function / : A —»■ A such that the inverse (relation) / _1 is not a function. 9.50. Show that the function / : R —»• R defined by / ( x ) = 4x — 3 is bijective and determine / ~ ’(x) for x 6 R. 9.51. Show that the function / : R — {3} > R — {5} defined by / ( x ) = f o rx e R  {5}.
is bijective and determine / 1 0 0
9.52. The functions / : R * R and g : R —>■ R defined by f ( x ) = 2x + 1 and g(x) = 3x — 5 for x e R are bijective. Determine the inverse function of g o f ~ l . 9.53. Let A and B be sets with A = £  = 3.How many functions from A to 5 9.54. Let the functions / : R > R and g : R » R be defined (a) Show that / is onetoone and onto. (b) Show that g is onetoone and onto. (c) Determine the composition function g o f .
have inverse functions?
by f [ x ) = 2x + 3 and g(x) = —3x +
5.
Chapter 9
238
Functions
(d) Determine the inverse functions / (e) Determine the inverse function (g
o
1 and g 1. f ) ~ l of g o f and the composition f ' x ° X
9.55. Let A = R — {1} and define / : A — A by f i x ) =   for all .v e A. x — 1
(a) Prove that / is bijective. (b) Determine Z ^ 1. (c) Determine f o / o f . 9.56. Let A, B and C be nonempty sets and let / , g and h be functions such that / : A > B , g : B » C and h : B »• C. For each of the following, prove or disprove: (a) If g o / = h o / , then g = h. (b) If / is onetoone and g o / = h 9.57. The function / : R
o
/ , then g = h.
R is defined by m
=
—■— r 1
X —
if x < 1
•Jx — 1 if v > 1.
(a) Show that / is a bijection. (b) Determine the inverse / _1 of f . 9.58. Suppose, for a function / : A is surjective, then g o f = .
B, that there is a function g : B
9.59. Let f <A B, g : B —»• C and h :B — C g o f = h o / , then g = /i.
A such that /
o
g = iB. Prove that if g
be functions where / is a bijection. Prove that if
Section 9.7: Perm utations /1 ^ 3 4 5 \ /1 2 3 4 5 \ 9.60. L et a = I 9 g 4 5 i ) and ^ = ( 3 5 2 4 1 j be permutations in <S5. Determine a
_i
o and p
.
12 34 5 6 j and p = ; 1 , g 2 1 4 J be elements of S 6. 2 6 4 15 3 (a) Determine a 1 and p (b) Determine a o p and p o a .
9.61. Let a =
9.62. Prove for every integer n > 3, there exist a, ft G Sn such that a
o
p / p
o
a.
ADDITIONAL EXERCISES FOR CHAPTER 9 9.63. Let / : R — R be the function defined by f i x ) = x 2 + 3x + 4. (a) (b) (c) (d) (e)
Show that f is not injective. Find all pairs n , r% of real numbers such that / ( r j ) = / ( r 2). Show that / is not surjective. Find the set S of all real numbers such that if s e S, then there is no real number x such that / (.v) = 5. W hat wellknown set is the set S in (d) related to?
9.64. Let / : R —> R he the function defined by /(.v ) = x 2 + a x + b, where a, h e R. Show that / is not onetoone. [Hint: It might be useful to consider the cases a ± 0 and a — 0 separately.]
Additional Exercises for Chapter 9
239
9.65. In Result 9.4, we saw that the (linear) function / : R * R defined by f i x ) = 3x —5 is onetoone. In fact, we have seen that other linear functions are onetoone. Prove the following generalization of this result: The function / : R R defined by f ( x ) = ax + b, where a, b e R and a ^ 0, is onetoone. 9.66. Evaluate the proposed proof of the following result. Result
3x
The function / : R — {1} > R — {3} defined by f ( x ) = — ;— is bijective. x —1
First, we show that f is onetoone. Assume that f ( a ) = f( b ) , where a, b e R — {1}. Then 3b  j =  — Crossmultiplying, we obtain 3a(b  1) = 3b(a  1). Simplifying, we have
Proof 3a
3ab — 3a — 3ab — 3b. Subtracting 3ab from both sides and dividing by —3, we have a = b. Thus / is onetoone.
3x
Next, we show that / is onto. Let f ( x ) = r. T h e n   = r ; so 3x = r(x  1). Simplifying, we have
3x — rx — r and so 3x  rx =  r . Therefore, x{3  r) =  r . Since r e R  {3}, we can divide by 3  r —r r
and obtain x =  = . Therefore, m
= f ( ^ — ) =
\r3J
7=5  1
r —(r —3)
= \= r .
3
Thus / is onto.
■
9.67. For each of the following functions, determine, with explanation, whether the function is onetoone and whether it is onto. (a) (b) (c) (d) (e)
/ : R x R ^ R x R , where / ( x , y ) = (3x — 2, 5_y + 7) g :Z x Z Z x Z, where g(m, n ) = (n + 6, 2 —m ) h : Z x Z —>■ Z x Z, where /?(/•, s) = (2r + 1, 4s + 3) <p : Z x Z —► S = {a + b  J l : a, b e Z), where <p(a, b) = a + b \ f l t c R ^ R x R , where a ( x ) = (x2, 2x + 1).
9.68. Let S be a nonempty set. Show that there exists an injective function from V (S ) to V (V (S)). 9.69. Let A = {a, b, c, d, e). Then / = {(a , c), (b, e), (c, d), (d, b), (,e, a)} is a bijective function from A to A. (a) Show that it is possible to list the five elements of A in such a way that the image of each of the first four elements on the list is to the immediate right of the element and that the image of the last element on the list is the first element on the list. (b) Show that it is not possible to list elements of A as in (a) for every bijective function from A to A. 9.70. Let A = R — {0} and let / : A —*■ A be defined by f { x ) = 1 — j for all x e R. (a) Show that / o / o / = iA. (b) Determine / 1 . 9.71. Give an example of a finite nonempty set A and a bijective function / : A —>■ A such that (1) / / iA, (2) / o / ^ iA and (3) / o / o / = iA. 9.72. For nonempty sets A and B and functions / : A > B and g : B >■ A, suppose that g o f = iA, the identity function on A. (a) (b) (c) (d) (e) (f)
Prove that / is onetoone and g is onto. Show that / need not be onto. Show that g need not be onetoone. Prove that if / is onto, then g is onetoone. Prove that if g is onetoone, then / is onto. Combine the results in (d) and (e) into a single statement.
240
Chapter 9
Functions
9.73. Let A = {1, 2}, B = {1,  I. 2,  2 } and C  L 2 .3 .4 ) . Then / = {(1, 1), (1. 11. (2, 2), (2,  2 ) } is a relation from A to B while g = {(1, 1), (  1 . 1), (2, 4), (  2 . 4)} is a relation from B to C. Furthermore, g f = i(.v, z) : (.v. y) G / and (y. z) e g for some y e B] is a relation from A to C. Observe that even though the relation / is not a function from Ato S . the relation g f is a function from A to C. Explain why. 9.74. A relation / on R is defined by / = {(x, y) : x GR and y = x or y =  x ] and a function g : R > R is defined by g(x) = x 2. Then g f — { u , z) : (x, y)
e f and (y, z) g g for some y e
(a) Explain why / is not a function from R to R. (b) Show that g f is a function from R to R and explicitly determine it. (c) Even though the relation / is not a function from R to R, the relation g f is a function from R Explain why.
R}.
to R.
9.75. Let A = {1, 2}, B = {1, 2, 3, 4} and C = {1, 2, 3, 4, 5, 6}. Give an example of a function / from A to 5 and a relation g from B to C that is not a function from B to C such that g f = {'(x, z) : (x, y ) e / and (y, z) G g for some y e B j is a function from A to C. 9.76. Let T be the set of all functions with domain and codomain R. Define a relation R on T by f R g if there exists a constant C such that / ( x ) = g(x) + C for all x G R. (a) Show that R is an equivalence relation. (b) Let / G T . If the derivative of / is defined for all x G R.then use this information to describe the elements in the equivalence class [ / ] . 9.77. Let S be the set of odd positive integers. A function F : N » S is defined by F (n) = k for each n G N, where k is that odd positive integer for which 3n + 1 = 2 mk for some nonnegative integer m. Prove or disprove the following: (a) F is onetoone. (b) F is onto. 9.78. A function F : N » N U {0j is defined by F (n) = m for each n G N. where m is that nonnegative integer for which 3n + 1 = 2mk and k is an odd integer. Prove or disprove the following: (a) F is onetoone. (b) F is onto. 9.79. Recall that the derivative of ln x is 1/x and that the derivative of x" is n x * " 1 for every integer n. In symbols, A 0 n x) = X1 and O4.X (.Xn) = n x n^ i . Let / : R+ R be defined by f ( x ) = ln x for every x £ R+ . Prove &X that the /7th derivative of / ( x ) is given by / fw)(x) = 1 1
for every positive integer n.
9.80. Let / : R —>■ R be defined by f i x ) = xe~x for every x GR. Prove that the «th derivative of f ( x ) is given by f {n\ x ) = (—l)"e~*(x — n) for every positive integer n. 9.81. The function h : Z 16 > Z 24 is defined by h([a]) = [3fl] for a g Z. (a) Prove that the function h is welldefined; that is, prove that if [a] = [ft] in Z i6, then h([a]) = h([b]) in Z24(b) For the subsets A = {[0], [3], [6], [9], [12], [15]} and B = {[0], [8]} of Z 16, determine the subsets h(A) and h(B ) of Z 24. (c) For the subsets C = {[0], [6], [16], [18]} and D = {[4], [8], [16]} of Z 24, determine h ~ l {C) and
Additional Exercises for Chapter 9
241
9.82. Let U be some universal set and A a subset of U. A function g^ : U —* {0, 1} is defined by g A & ) '■
1 if x e A 0 il'.v £ A.
Verify each of the following. (a) (b) (c) (d) (e)
gu(x) = 1 for all x e U. ga(x) = 0 for all x e U. For U = R and A = [0, oo), o gA)(x) = 1 for x € R. For subsets A and B of U and C — A D B , gc = (g,\) ■(gs), where ((g^) • (,?/!))(.v) = g i(.v) • g s(x ). For A Q U , g j ( x ) = 1  gA(x) for each ,v e U.
9.83. (a) Let S = {a , b, c, d } and let T be the set of all six 2element subsets of S. Show that there exists an injective function f : S  » {0,1, 2 , . . . , \T\} such that the function g : T  * { 1 , 2 ,  T} defined by ,?({/. /'!) = 1 /( 0 ~ / ( / ) ! is bijective. (b) Let S = {a, b, c, d, e} and let T be the set of all ten 2element subsets of S. Show that there exists no injective function / : S {0,1, 2 , . . ., \T\] such that the function g : T  * { 1 , 2 ........ ]T} defined by g({i, j }) = 1 /( 0  /G ')i is bijective. (c) For the sets S and T in (b), show that there exists an injective function / : S  * { 0 ,1 ,2 , \T\ + 2 } such that the function g : T —* { 1 ,2 ........  + 2} defined by g({i, j} ) = \ f ( i ) — / (j)\ is injective. (d) The results in (b) and (c) should suggest a question to you. Ask and answer such a question.
IDj Cardinalities of Sets
any consider the Italian m athem atician and scientist G alileo G alilei to be the founder o f m odern physics. A m ong his m ajor contributions was his m athem atical view o f the laws o f m otion. Early in the 17th century, G alileo applied m athem atics to study the m otion o f the Earth. H e was convinced that the E arth revolved about the sun, an opinion not shared by the C atholic C hurch at that tim e. This led him to be im prisoned for the last nine years o f his life. G alileo â€™s tw o m ain scientific w ritings w ere Dialogue Concerning the Two C hief World Systems and Discourses and M athematical Demonstrations Concerning Two N ew Sciences, the first published before he w ent to prison and the second published (in the N etherlands) w hile he was in prison. In these tw o w orks, he would often discuss scienÂ tific theories by m eans o f a dialogue am ong fictional characters. It is in this m anner that he could state his positions on various theories. One topic that intrigued G alileo was infinite sets. G alileo observed that there is a onetoone correspondence (that is, a bijective function) betw een the set N o f positive integers and the subset S o f N consisting o f the squares o f positive integers. This led G alileo to observe that even though there are m any positive integers that are n ot squares, there are as m any squares as there are positive integers. This led G alileo to be faced with a property of an infinite set that he found bothersom e: T here can be a onetoone correspondence betw een a set and a proper subset o f the set. W hile G alileo concluded correctly that the num ber o f squares o f positive integers is not less than the num ber of positive integers, he could not bring him self to say that these sets have the sam e num ber o f elem ents. Bernhard Bolzano was a B ohem ian priest, philosopher and m athem atician. A lthough best know n for his w ork in calculus during the first h alf o f the 19th century, he too was interested in infinite sets. H is Paradoxes o f the Infinite, published tw o years after his death and unnoticed for tw enty years, contained m any ideas o f the m o d em theory o f sets. H e noted that onetoone correspondences betw een an infinite set and a proper subset o f itself are com m on and was com fortable w ith this fact, contrary to G alileo â€™s feelings. The G erm an m athem atician R ichard D edekind studied under the brilliant Carl F riedrich G auss. D edekind had a long and productive career in m athem atics and m ade m any contributions to the study o f irrational num bers. W hat had confused G alileo and interested B olzano gave rise to a definition of an infinite set by D edekind during the last part o f the 19th century: A set S is infinite if it contains a proper subset that can be put in
M
242
10.1
Numerically Equivalent Sets
243
onetoone correspondence w ith S. Certainly, then, understanding infinite sets w as not an easy task, even am ong w ellknow n m athem aticians o f the past. We m entioned in C hapter 1 that the cardinality S  o f a set S is the num ber of elem ents in S and. for the present, we w ould use the notation S only w hen S is a finite set. A set S is finite if either S — 0 or  S \ = n for som e n e N; w hile a set is in fin ite if it is not finite. It m ay seem that w e should w rite 5  = oo if 5 is infinite but we w ill soon see that this is not particularly inform ative. Indeed, it is considerably m ore difficult to give a m eaning to  S  if S is an infinite set; however, it is precisely this topic that w e are about to explore.
10.1 N um erically E q u ivalent Sets It is rather obvious that the sets A = {a, b, c } and B = {x, y, z] have the same cardinality since each has exactly three elem ents. T hat is, if w e count the num ber o f elem ents in two sets and arrive at the same value, then these tw o sets have the sam e cardinality. T here is, however, another w ay to see that the sets A and B described above have the same cardinality w ithout counting the elem ents in each set. O bserve that we can pair off the elem ents of A and B , say as (a, x ) , (b , y) and (c, z). This im plies that A and B have the sam e num ber of elem ents, that is, A] = B . W hat w e have actually done is describe a bijective function f : A > B. nam ely / = {(a, .v). (b , r ) . (c, z)}. A lthough it is m uch easier to see that JA  \B  by observing that each set has three elem ents than by constructing a bijective function from A to B , it is this latter m ethod o f show ing that A = B\ that can be generalized to the situation w here A and B are infinite sets. Two sets A and B (finite or infinite) are said to have the sam e c a rd in a lity , w ritten  A  =  B , if either A and B are both em pty or there is a bijective function / from A to B . Two sets having the sam e cardinality are also referred to as n u m e ric a lly eq u iv ale n t sets. Two finite sets are therefore num erically equivalent if they are both em pty or if both have n elem ents for some positive integer n . Consequently, tw o nonem pty sets A and B are not num erically equivalent, w ritten  A  ^ \B , if there is no bijective function / from one set to the other. T he study o f num erically equivalent infinite sets is m ore challenging but considerably m ore interesting than the study of num erically equivalent finite sets. The justification for the term “num erically equivalent sets” lies in the follow ing theorem , w hich com bines the m ajor concepts o f C hapters 8 and 9. Theorem 10.1
Proof
Let S be a nonempty collection o f nonempty sets. A relation R is defined on S by A R B if there exists a bijective function fro m A t o B . Then R is an equivalence relation. Let A € S. Since the identity function i& ■ A —> A is bijective, it follow s that A R A. Thus R is reflexive. N ext, assum e that A R B , w here A, B e S. Then there is a bijective function f : A + B . By T heorem 9.15, / has an inverse function f ~ l : B —^ A and, furtherm ore, / “ ' is bijective. Therefore, B R A and R is sym m etric. Finally, assum e that A R B and B R C , w here A, B . C e S. T hen there are bijective functions / : A  > B a n d a : B C . It follow s by C orollary 9.12 that the com position g o f : A —* C is bijective as w ell and so A R C . T herefore, R is transitive. C onsequently, R is an equivalence relation. ■
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Chapter 10
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According to the equivalence relation defined in Theorem 10.1, if A is a nonempty set, then the equivalence class [A] consists o f all those elements o f S having the same cardinality as A; hence the term “numerically equivalent sets” is natural for two sets having the same cardinality. Example 10.2
Let S = {A i , A 2, A 3 , A 4 , A 5 , A(,} where Aj = { 1 , 2 , 3}, A 2 — {a, ft, c, J ). A 3 = {.v. v. z}, A 4 = {/', s, t}, A 5 = {m, n }, A 6 = {7, 8 , 9, 10}.
Then every W o o f the sets A \, A 3 and A 4 are numerically equivalent, while A 2 and A 6 are numerically equivalent.This says that \ A X\ =  A 3 1 = \A 4 \ a n d \ A 2 \ = \A6\. The only set in S numerically equivalent to A 5 is A 5 itself. Thus, [Ai] = {A i , A 3 , A4}, [Aj] = {A t , A$} and [A5] = {A5 }
are the distinct equivalence classes o f S.
♦
W hile Example 10.2 deals only with finite sets, it is infinite sets in which we w ill be primarily interested in this chapter. In particular, w e w ill have a special interest in sets that are numerically equivalent to N or to R.
10.2 D en u m erab le Sets In order to start gaining an understanding o f the cardinality o f an infinite set, we begin with a particular class o f infinite sets. A set A is called d enum erable if A = N , that is, if A has the same cardinality as the set o f natural numbers. Certainly, if A is denu merable, then A is infinite. B y definition, if A is a denumerable set, then there is a bijec tive function / : N > A and so / = {(1, / ( l ) ) , (2, / ( 2)), (3, / ( 3 ) ) , . . . }. Consequently, A = { / ( l ) , / ( 2 ) , / ( 3 ) , .. ,};thatis, w eca n listth eelem en tso f A a s / ( l ) , f ( 2) . / ( 3 ) ........ Equivalently, w e can list the elements o f A as a \ , a2, <2 3 , • • •, where then a, = f ( i ) for i 6 N. Conversely, if the elem ents o f A can be listed as a\, a2, <%,, . • • , where at ^ a , for i ^ j , then A is denumerable since the function g : N —* A defined by g(n) = an for each n € N is certainly bijective. Therefore, A is a denumerable set if and only if it is possible to list the elements o f A as a x, a 2, a3, . . . and so A = {«i, a 2, 0 3 , . . .}. A set is countable if it is either finite or denumerable. C ountably infinite sets are then precisely the denumerable sets. Hence, if A is a nonempty countable set, then we can either write A = {%, a2, as, a,,} for some n e N or A = {a\, a2, <2 3 , . . .}. A set that is not countable is called uncountable. An uncountable set is necessarily infinite. It may not be clear whether any set is uncountable but w e w ill soon see that such sets do exist. Let’s look at a few examples o f denumerable sets. Certainly, N itself is denumerable since the identity function % : N > N is bijective. However, not only is the set o f positive integers denumerable, the set o f all integers is denumerable. The proof o f this fact that w e give illustrates a common technique for showing that a set is denumerable; namely, if w e can list the elements o f a set A as a x, a 2, 0 3 , . . . such that every element of A appears exactly once in the list, then A is denumerable.
i 0.2
/: Figure 10.1
Result 10.3 Proof
1
2
3
4
1 0
I 1
1 1
1 2
Denumerable Sets
245
5 1
I • ••
 2
A bijective function / : N > Z
The set Z o f integers is denumerable. O bserve that the elem ents o f Z can be listed as 0, 1, —1, 2, —2 .........T hus the function / : N Z described in Figure 10.1 is bijective and so Z is denum erable. m The function / : N —> Z given in Figure 10.1 can be also defined by /( „ )  '
'*
^
".
(10.1)
A lthough w e have already observed that this function / is bijective. E xercise 10.8. asks for a form al proof o f this fact. The fact that Z is denum erable illustrates w hat G alileo had observed centuries ago: It is possible for two sets to have the same cardinality w here one is a proper subset of the other. (Such a situation could never occur w ith finite sets, however.) For exam ple, N C Z and N = Z . This fact serves as an illustration o f a result, the p ro o f o f w hich is a bit intricate. T h e o re m to P rove PROOFSTRATEGY
Every infinite subset o f a denum erable set is denum erable. In the proof, we begin w ith two sets, w hich w e’ll call A and B , w here A is denum erable, g c A and B is infinite. B ecause A is denum erable, w e can w rite A = {<it, a 2, a%,...} . Because our goal is to show that B is denum erable, we need to show that we can write B = [bi, b2, h i , . ..}. The question, o f course, is how to do it. B ecause B is an infinite subset o f A, som e o f the elem ents o f A belong to B (in fact, infinitely m any elem ents o f A belong to S ); while, m ost likely, som e elem ents of A do not belong to B . We can keep track of the elem ents o f A that belong to B by m eans o f a set, w hich w e ’ll denote by S. If a\ e B , then 1 e S; if a\ fi B , then 1 fi S. In general, n e S if and only if a„ e B . Certainly, S C N . Since N is a w ellordered set (by the W ellO rdering Principle), S contains a sm allest elem ent, say s. T hat is, as € B. Furtherm ore, if r is an integer such that 1 < r < s, then ar fi B . It is the elem ent as that we w ill call b \ . It is now logical to look at the (infinite) set S — {s} and consider its sm allest elem ent, say t. Thus t > s. The elem ent a, will becom e b2. A nd so on. Since we w ant to give a precise and careful proof, w e are already faced w ith two problem s. First, denoting the sm allest elem ent o f S by s and denoting the sm allest elem ent o f S — {s} by t will present difficulties to us. We need to use better notation. So let us denote the sm allest elem ent o f S' by i > (so b\ = a Q and the sm allest elem ent of S — {ii} by h (so b 2 — aj2). This is m uch better notation. The other problem we have is w hen we w rote “A nd so on.” O nce w e have the positive integers z'i and i2, it will follow that the positive integer i 3 is the least elem ent o f S — {/‘i , i2} In general, once we have determ ined the positive integers i\. i2. , it, w here k e N, the positive integer
246
Chapter 10
Cardinalities of Sets
z*+ 1 is the sm allest elem ent o f S — {ii , ij, • • ■, ifc}. In fact, this suggests that the elem ents b \.b n , b%,. .. can be located in A using induction. A fter using induction to construct the set {b\, bo, 63, . . .}. w hich we w ill denote by S ' , say, then we still have one m ore concern. A re we certain that S ' = S ? B ecause each elem ent o f B ' belongs to B , we know that S ' c g . To show that B ' — B , we m ust also be sure that 5 c fi'. As w e know, the standard way to show that B c B ' is to take a typical elem ent b e B and show that b e B ' . L e t’s now w rite a com plete proof. $ Theorem 10.4 Proof
Every infinite subset o f a denumerable set is denum erable. L et A be a denum erable set and let B be an infinite subset o f A . Since A is denum erable, we can w rite A = {a\, 0 2 , 0 3 , . . . } . Let S — {i e N : a, e S}; that is, S consists o f all those positive integers that are subscripts o f the elem ents in A that also belong to B . Since B is infinite, S is infinite. First we use induction to show that B contains a denum erable subset. Since S is a nonem pty subset of N, it follow s from the W ellO rdering Principle that S has a least elem ent, s a y /[.L e t ft] — </;.. Let .S'i = S — { ii} .S in ceS i / 0 (indeed, Si is infinite), Si has a least elem ent, say %. Let bn = tf;,, w hich, of course, is distinct from b \ . A ssum e that for an arbitrary integer k > 2, the (distinct) elem ents b \, bn, . . . , ^ have been defined by bj = a, , for each integer j w ith 1 < j < k, w here i\ is the sm allest elem ent in S and ij is the m inim um elem ent in S j  i = S — {i\, i% , , . , /y_i} for 2 < j < k. N ow let % .) be the m inim um elem ent o f Sn = S — { /j, % . . . ,    and let bk+ 1 = a !f+1. H ence it follow s that for each integer n > 2, an elem ent b„ belongs to B that is distinct from b i, 62.........bn 1. Thus we have exhibited the elem ents b\, bn, b%t . . . in B. Let B ' = [by, bn. b% . .. } . Certainly B ' C B . We claim , in fact, that B = S '. It re m ains only to show that S c S '. L et b e B . Since B c A, it follow s that b — an for som e n e N and so n e S. If n = i\, then b — b\ = a„ and so b e S '. Thus w e m ay assum e that n > i \ . Let S' consist o f those positive integers less than n that belong to S. Since n > i\ and i\ e S, it follow s that S' / 0. Certainly, 1 < S' < n — 1; so S' is finite. Thus S' = m for som e m e N. T he set S' therefore consists o f the m sm allest integers o f S, that is, S' = {/1, i n ........... im }. The sm allest integer that belongs to S and is greater than im m ust be ;m+i, o f course, and im+\ > n. B ut n e S, so n = im+\ and b = a„ = ajm+l e S '. H ence S = S ' = {b\, bn, ft3, . . .}, w hich is denum erable. ■ In order to use T heorem 10.4 to describe other denum erable sets, it is convenient to introduce some additional notation. L et k e N. T hen the set k L is defined by k Z = {kn : n e Z}. Similarly, A'N = {kn : « 6 N), Thus 1Z = Z and IN = N, w hile 2Z is the set o f even integers. A n im m ediate conse quence o f T heorem 10.4 is stated next.
Result 10.5 Proof
The set 2Z o f even integers is denumerable. Since 2Z is infinite and 2Z C Z, it follow s by Theorem 10.4 that 2Z is denum erable. ■
10.2
b\
b2
b3
a2
(a2, b i )
«3
(0 3,
(a2 . b2) (ia2, b3)
a\  • •
!>■) (a3. b2) (a3, b3)
(a y t0
a2 a3
(a3 ,b£r
247
t3
...
:'/)(')
• ••
(«2v
(a2, b3)
• ••
( a 3 , b2)
(03, b3)
• • •
b\
...
(ah b : ) (a 1, b2) i a u h )
Denumerable Sets b2 (ay. 2)
(a)
(b)
Figure 10.2
Constructing a bijective function
/ :
N
/I x B
O f course, k Z is denum erable for every nonzero integer k. We now describe a denum erable set that can be obtained from two given sets. Recall, for sets A and B , that the C artesian product A x B = {(a, b) : a B A , b e B}. Result 10.6 Proof
I f A and B are denumerable sets, then A x B is denumerable. Since A and B are denum erable sets, we can w rite A = {au a 2, a3, . . .} and B = (/>; ■b2, b3, ...} . C onsider the table show n in Figure 10.2(a), w hich has an infinite (de num erable) num ber o f rows and colum ns, w here the elem ents a\ , a2, a 3, . . . are w ritten along the side and />;. b2. b3. . . . are w ritten across the top. In row /, colum n j of the table, w e place the ordered pair (a,, bj). Certainly, every elem ent o f A x B appears ex actly once in this table. This table is reproduced in Figure 10.2(b), w here the directed lines indicate the order in w hich we w ill encounter the entries is the table. T hat is, we encounter the elem ents of A x B in the order (au b 1), (au b2), (a2, b x), (a \ , b3), (a2, b2) ......... Since every elem ent o f A x B occurs in this list exactly once, this describes a bijective function / : N —*■ A x B , where / ( l ) = (ai, h ) , / ( 2 ) = (alt b2), / ( 3 ) = (a2, b x), / ( 4 ) = (au b3), / ( 5 ) = (a 2, b2), . . . . T herefore, A x B is denum erable.
■
We can use a technique sim ilar to that used in proving Result 10.6 to show that another fam iliar set is denum erable. Result 10.7 P roof
The set Q + o f positive rational numbers is denumerable. C onsider the table show n in Figure 10.3(a). In row i, colum n j , w e place the rational num ber j / i . Certainly, then, every positive rational num ber appears in the table o f Figure 10.3(a); indeed, it appears infinitely often. For exam ple, the num ber 1/ 2 appears in row 2, colum n 1, as w ell as in row 4, colum n 2. The table o f Figure 10.3(a) is reproduced in Figure 10.3(b), where the arrows indicate the order in w hich w e w ill consider the entries in the table. T hat is, w e now consider the positive rational num bers in the order
248
Chapter 10
Cardinalities of Sets
2
3
1
1
1
2 2
3 2
4 2
2 3
3 3
4 3
2
3 4
4 4
4
4
(a)
(b)
Figure 10.3
A table used to show that Q is denumerable
2 13
2 1
1' 2 ’ 1’ 2 ’ 3 ’
W ith the aid o f this list, we can describe a bijective function / : N —> Q + . In particular, w e define / ( l ) = 1/1 = 1, / ( 2 ) = 2 /1 = 2 , / ( 3 ) = 1 /2 and / ( 4 ) = 3 /1 = 3 as expected. However, since 2 /2 = 1 and we have already defined / ( l ) = 1, w e do not define / ( 5 ) = 1 (since / m ust be onetoone). We bypass 2 /2 = 1 and, follow ing the arrows, go directly to the next num ber on the list, nam ely 1/3. In fact, w henever we encounter a num ber on the list that we have previously seen, we m ove to the next num ber on the list. In this m anner, the function / being described will be onetoone. The function / is show n in Figure 10.4. Because every elem ent of Q + is eventually encountered, / is onto as w ell and so f is bijective. Consequently, Q + is denum erable, The function / described in Figure 10.4 is by no m eans unique. There are many w ays to traverse the positive rational num bers in the table described in Figure 10.3(a). The tables show n in Figure 10.5 indicate two additional methods. Som e care m ust be taken w hen proceeding about the entries in the table of Figure 10.3(a).For exam ple, traversing the positive rational num bers by row s (see F ig ure 10.6) ju st w o n 't do. Since the first row never ends, we w ill only encounter the positive integers. The set Q + can also be show n to be denum erable w ith the aid o f the table in Figure 10.7. In the first row, all positive rational num bers j / i w ith i — 1 are shown. In the second row, all positive rational num bers j / i w ith i = 2 and such that j / i has been reduced to low est term s are shown. This results in the rational num ber (2 j — l ) / 2 in row 2, colum n j . We continue in this m anner with all other rows. In this way, every 1 /:
Figure 10.4
2
3
4
5
•••
!
I
I
I
I
I
1
2
5
3

•••
A bijective function / : N > Q
10.2
4
1
1 4 2
2
4 3
3
4 4
4
Figure 10.5
1 2r 2
2
1
1 1
2 1
3
4_
1
1
2
1 2
2 2
3
4
2
2>
1
2 3
3 3
4
3
3 4
4 4
4
Figure 10.6
3 2
Traversing the positive rational numbers
1
3
Denumerable Sets
1
2
4
4
3
4
How not to traverse the positive rational numbers
Figure 10.7
Another bijective function g : N â€”Âť Q +
249
250
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f
1 2 3 1 1 4 0 <21 ~Ql
Figure 10.8
4 i q2
5 1 fis.
A bijective function / : N
Q
positive rational num ber appears exactly once in the table. Thus w hen we proceed through the entries as the arrows indicate, w e obtain the positive rational num bers in the order 12 13 3 14 1’ 1’ 2 ’ \ ' V 3 ’ 1’ ’ ’ ' and the corresponding bijective function g : N • Q + . Therefore, f p ) = 1, g(2)  2, g(3) = 1/2, g (4 ) = 3, g(5) = 3 /2 and so on. N ow that we have show n that Q + is denum erable, it is not difficult to show that the set Q o f all rational num bers is denum erable. Result 10.8 Proof
The set Q o f all rational numbers is denumerable. Since Q is denum erable, we can write Q + = {ci\,q2, i[$, ■■•}• Thus, Q = {0} U \q i, q%, q3, . . . } U {  q u —qi, ~</.s, • •.}. T herefore, Q = {0, q \,  q \ . q i   q i , ■■.}, and the fu n c tion / : N » Q show n in Figure 10.8 is bijective and so Q is denum erable. h
10.3 U ncountable Sets A lthough w e have now given several exam ples of denum erable sets (and consequently countably infinite sets), we have yet to give an exam ple o f an uncountable set. We w ill do this next. First though, le t’s review a few facts about decim al expansions o f real num bers. Every irrational num ber has a unique decim al expansion and this expansion is nonrepeating, w hile every rational num ber has a repeating decim al expansion. For exam ple, = 0.272727 • • •. Som e rational num bers, however, have two (repeating) decim al expansions. For exam ple, \ = 0.5000 ■• ■and  = 0.4999 ■• (The num ber ~ has only one decim al expansion.) In particular, a rational num ber a / b , w here a , b e N, that is reduced to low est term s has two decim al expansions if and only if the only prim es that divide b are 2 or 5. If a rational num ber has two decim al expansions, then one of the expansions repeats the digit 0 from some point on (that is, the decim al expansion term inates), w hile the alternate expansion repeats the digit 9 from som e point on. We are now prepared to give an exam ple o f an uncountable set. R ecall that for real num bers a and b with a < b, the open interval ( a , b ) is defined by (a, b) — {x e R : a < x < b). A lthough, as it w ill turn out, all open intervals (a, b) of real num bers are uncountable, we will prove now only that (0, 1) is uncountable. T h e o re m to P ro v e
The open interval (0, 1) o f real num bers is uncountable.
10.3 PROOF STRATEGY
Theorem 10.9
Proof
Uncountable Sets
251
Since uncountable m eans not countable, it is not surprising that w e should try a pro o f by contradiction here. So the proof w ould begin by assum ing that (0, 1) is countable. Since (0, 1) is an infinite set, this m eans that w e are assum ing that (0, 1) is denum erable, w hich im plies that there m ust exist a bijective function / ? N > (0, 1). T herefore, for each 7? e N, / ( « ) is a num ber in the set (0, 1). It m ight be convenient to introduce some notation for the num ber / ( / ; ) , say f ( n ) = a„, w here then 0 < an < 1. Since f is assum ed to be onetoone, it follow s that a; / aj for distinct positive integers i and j . Each num ber an has a decim al expansion, say a„ — 0 .a„\a„2 a „3 ■■■, w here a„\ is the first digit in the expansion, a,a is the second digit in the expansion and so on. We have to be a bit careful here, however, for as w e have seen, some real num bers have two decim al expansions. To avoid possible confusion, w e can choose the decim al expansion that repeats the digit 0 from some point on. T hat is, no real num ber a„ has a decim al expansion that repeats 9 from som e point on. B ut w here does this lead to a contradiction? From w hat we have said, (0, 1) = {ci\,ci2, 03, . ..}. If we can think o f som e real num ber b e (0 , 1 ) such that b £ {%, a2, <23, . ..}, then this w ould give us a contradiction because this w ould say that / is not onto. So we need to find a num ber b e (0 ,1 ) such that b 7^ an for each 77 e N. Since b e (0, 1), the num ber b has a decim al expansion, say b = 0 .b ib 2bi ■• ■. How can we choose the digits b \ , b2, / >; , . . . so that b / an for every n e N? We could choose b\ ^ (/j ; , />2 ^ fl22, etc. But w ould this m ean that b ^ a \ , b ^ a2, etc? We m ust be careful here. For exam ple, 0.500 ■• ■and 0.499 • • • are tw o equal num bers w hose first digits in their expansions are not equal. O f course, the reason for this is that one is the alternate decim al expansion of the other. Thus, provided we can avoid selecting a decim al expansion for b that is the alternate decim al expansion for som e num ber an, w here n e N, we w ill have found a num ber b e (1, 0) such that b £ {<21, a2, <2 3 , ■• •}• T his w ill give us a contradiction.
The open interval (0, 1) o f real numbers is uncountable.
A ssum e, to the contrary, that (0, 1) is countable. Since (0, 1) is infinite, it is denum erable. T herefore, there exists a bijective function / : N —> (0, 1). For n e N, let / ( « ) = an. Since an e (0, 1), the num ber a„ has a decim al expansion, say 0 .a„ia„2 a lt3 ■■•, where alu e {0, 1. 2, . . . , 9} for all 7 e N. If a„ is irrational, then its decim al expansion is unique. If an e Q . then the expansion m ay be unique. If it is not unique, then, w ithout loss of generality, w e assum e that the digits o f the decim al expansion 0 .an\a„ 2 a n 2 ■■■are 0 from som e position on. For exam ple, since / is bijective, 2 /5 is the im age o f exactly one positive integer and this im age is w ritten as 0.4000 • • ■ (rather than as 0.3999 ■• •)• To sum m arize, w e have / ( l ) = ui = Q.au a u a n  ■ / ( 2) = a 2 = 0.a21«22«23 • ■• / ( 3 ) — «3 = 0 .(731^ 32^33 * ' ‘
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We show that the function / is not onto, however. Define the num ber b — O.bi hzbz • w here b, e {0, 1, 2 , . . . , 9} for all i e N, by bi =
4 5
if an = 5 i f an ^ 5 .
(For exam ple, le t’s suppose that a\ ~ 0.31717 = 0.151515 ■ and a 3 = 0.04000 ■• •. T hen the first three digits in the decim al expansion o f h are 5, 4 and 5, that is, b = 0.545 • ■■.) For each i e N, the digit bt ± an, im plying that b / a„ for all n e N since b is not the alternate expansion o f any rational num ber, as no digit in the expansion o f b is 9. Thus, b is not an im age o f any elem ent o f N. T herefore, / is not onto and, consequently, not bijective, producing a contradiction. b In the p ro o f of T heorem 10.9, each digit in the decim al expansion o f the num ber b constructed is 4 or 5. We could have selected any two distinct digits that did not use 9. It is now easy to give exam ples o f other uncountable sets w ith the aid o f the follow ing result. Theorem 10.10 Proof
Coroliary 10.11 Proof
L e t A and B be sets such that A C B . I f A is uncountable, then B is uncountable. L et A and B be tw o sets such that A c ® and A is uncountable. N ecessarily then A and B are infinite. A ssum e, to the contrary, that B is denum erable. Since A is an infinite subset o f a denum erable set, it follow s by T heorem 10.4 that A is denum erable, producing a contradiction. ■ The set R o f real numbers is uncountable. Since (0, 1) is uncountable by T heorem 10.9 and (0, 1) C R , it follow s by T heorem 10.10 that R is uncountable. h L e t’s pause for a m om ent to review a few facts that w e ’ve discovered about infinite sets (at least about certain infinite sets). First, recall that two nonem pty sets A and B are defined to have the same cardinality ( sam e num ber o f elem ents) if there exists a bijective function from A to 8 . W e’re especially interested in the situation w hen A and B are infinite. O ne fam ily o f infinite sets w e ’ve introduced is the class o f denum erable sets. R ecall too that a set S is denum erable if there exists a bijective function from N to S. Suppose that A and B are tw o denum erable sets. T hen there exist bijective functions f : N —►A a n d g : N —> B . Since / is bijective, / has an inverse function f ~ x : A —>■ N, w here / “ ' is also bijective (Theorem 9.15). Since / _1 : A —» N and g : N —> B are bijective functions, it follow s that the com position function g o f ~ l : A —> B is also bijective (Corollary 9.12). This tells us that A = B ; that is, A and B have the same num ber o f elem ents. We state this as a theorem for em phasis.
Theorem 10.12
Every two denumerable sets are numerically equivalent. N ext, let B be an uncountable set. So B is an infinite set that is not denum erable. A lso, let A be a denum erable set. Therefore, there exists a bijective function f : N > A. We claim that A ^ \B\; that is, A and B do not have the sam e num ber of elem ents.
10.3
I Jncountable Sets
253
L e t’s prove this. A ssum e, to the contrary, that A  — \B\. H ence there exists a bijective function g : A —> B . Since the functions / : N —> A and g : A —> B are bijective, the com position function g o / : N —> B is bijective. B ut this m eans that B is a denum erable set, w hich is a contradiction. We also state this fact as a theorem . Theorem 10.13
I f A is a denumerable set and B is an uncountable set, then A a nd B are not numerically equivalent. T heorem s 10.12and 10.13can also be considered as consequences o fT h eo rem 10.1. In particular, T heorem 10.13 says that Z and R are not num erically equivalent and so IZ 7^ R . So, here are tw o infinite sets that do not have the same num ber o f elem ents. In other w ords, there are different sizes o f infinity. This now brings up a num ber of questions, one o f w hich is: D o there exist three infinite sets so that no tw o o f them have the same num ber of elem ents? A lso, if A is a denum erable set and B is an uncountable set, is one o f these sets "bigger” than the other in som e sense? In other w ords, we w ould like to be able to com pare A  and 5 [ in some precise manner. Since Z  ^ R  and Z C R, it is tem pting to conclude that Z  < Rj but we have yet to give a m eaning to A  < \B\ for sets A and B . This idea w ill be addressed in Section 10.4. We should rem ind ourselves, however, that for infinite sets C and D , it is possible that both C c D and C  = \D\. For exam ple, Z C Q and jZ = Q  since Z and Q are both denum erable. Before leaving our discussion o f Z and R , one other observation is useful. R ecall that, according to T heorem 10.4, if B is an infinite subset o f a denum erable set A. then B is also denum erable. B ut w hat if A is uncountable? T hat is, if B is an infinite subset o f an uncountable set A, can w e conclude that B is uncountable? The sets Z and R answ er this question since Z is infinite, R is uncountable and Z C R . However. Z is not uncountable. We have now seen two exam ples o f uncountable sets, nam ely the open interval (0. 1) o f real num bers and the set R o f all real num bers. N either o f these sets has the same num ber o f elem ents as any denum erable set. But how do they com pare w ith each other? We w ill show, in fact, that these tw o sets have the sam e num ber o f elem ents. P rior to verifying this, w e show that the open interval ( —1, 1) and R have the sam e num ber of elem ents.
Theorem to Prove PROOFSTRATEGY
The sets (—1, 1) and R are num erically equivalent. The obvious approach to proving this theorem is to locate a bijective function / : R Actually, there are several such functions w ith this property. W ith each such function, w e are faced w ith the problem o f determ ining how involved it is to show that the function is bijective. We describe one o f these here. A nother is given in Exercise 10.25. C onsider the function / : (—1, 1) R defined by v fix ) = 1 1 * 1
(“ 1 1)
(See Figure 10.9.) This function is defined for all x e ( —1, 1). O bserve that / ( 0 ) = 0, f i x ) > 0 w hen 0 < .v < I and f ( x ) < 0 w hen —1 < .v < 0 . This function also has the property that lim
x
 — = + o o and lim i i 'i x*—1+ 1 —
254
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y
If you recall enough inform ation about continuous functions from calculus, you m ight see that this function is continuous on the interval (—1, 1). From this inform ation, it follow s that / ( ( —l, 1)) = R and that / is onto. Also, the derivative o f this function on the interval (—1, 1) is l
if x G (0, 1)
1if A' = 0
/ '( * ) =
u T rf i f *
g
(1 ,0 ).
T his says that f ' ( x ) > 0 for all x G (—1, 1) and so / is an increasing function on the interval (—1, 1). This inform ation tells us that / m ust be onetoone and so / is bijective. W hile the argum ent ju st given relies on calculus and you m ay not recall all o f this, an argum ent can be given that is o f the type w e have been discussing. ♦ Theorem 10.14 Proof
The sets (—1, 1) and R are numerically equivalent. C onsider the function / : (—1, 1) —> R defined by A n 
A . 1  x 
(10.2)
We show that / is bijective. First, we verify that / is onetoone. L et f ( a ) = f ( b ) , w here If ^ ^ = 0, then a = b = 0. If ^ = a . b e (  1 , 1). T hen ^ > 0, then a > 0 and b > 0. Thus
— j§ ^ . H ence a ( l — b) = b( 1 — a) and so
a = b. If \r^7 = i— "ttt < 0,’ then a < 0 and b < 0. Thus 1rf= 1+b A t . H ence a (vl + b ) = —\a\ 1—\b\ +a b{ 1 + a) and so a — b. Therefore, / is onetoone. Next, w e show that / i s onto. L e tr e R. S in c e /( 0 ) = 0, we m ay assum e that r / 0. I f r > 0, then g (0, 1) and f ( j ^ ) — r . l f r < 0, then ^ € (  1 , 0 ) and f ( j ^ ) = >'■ Thus / is onto. Since / is a bijective function, the sets (—1, 1) and R are num erically equivalent, h It is straightforw ard to show that the function g : (0, 1) •+ (—1, 1) defined by g (x ) = 2x — 1 is bijective. For this function g and the function / in (10.2) in the pro o f of
10.4
Com paring Cardinalities of Sets
255
T heorem 10.14, it therefore follow s that g o / : (0, 1)  * R is also bijective. T his gives an im m ediate corollary. Corollary 10.15
The sets (0, 1) and R are numerically equivalent. N ot only are (0, 1) and R num erically equivalent, (as w ell as (—1, 1) and R), but every open interval (a, b) o f real num bers w ith a < b and R are num erically equivalent. (See Exercise 10.23.)
10.4 Com paring C ardinalities o f Sets A s w e know, tw o nonem pty sets A and B have the sam e cardinality if there exists a bijective function f : A *■ B . L e t’s illustrate this concept one m ore tim e by showing that two fam iliar sets associated w ith a given set are num erically equivalent. R ecall that the pow er set V ( A ) of a set A is the set o f all subsets o f A and that 2A is the set of all functions from A to {0, 1}. If A = [a, b, c}, then \V (A )\ = 23 = 8. A lso, the set 2 A contains 2 IAI = 23 = 8 functions. So in this case, V ( A ) and 2A have the same num ber o f elem ents. This is not a coincidence. T h e o re m to P ro v e PROOF STRATEGY
For every nonem pty set A, the sets V ( A ) and 2A are num erically equivalent. If we can construct a bijective function </> : V ( A ) —> 2 A, then this will prove that V { A ) and 2 a are num erically equivalent. We use <p for this function since 2A is a set o f functions and it is probably better to use m ore standard notation, such as / , to denote the elem ents o f 2 A. But how can such a function (p be defined? L e t’s take a look at V ( A ) and 2A for A = {a, b}. In this case, 7>(A) = {0, { a } ,[ b ) , [ a ,b } ) ; w hile
2
a = [fa, fa, fa, fa}, w here
fa = fa =
{(a,0), (6,0)}, fa = {(a, 1), (b, 0)}, {(a, 0), (b , 1)}, fa = {(a, 1), (b , 1)}.
Since each o f V ( A ) and 2 A has four elem ents, w e can easily find a bijective function from V ( A ) to 2A. B ut this is not the question. W hat w e are looking for is a bijective function (p : V ( A ) —» 2A for A = {a, b } that suggests a w ay for us to define a bijective function from V ( A ) to 2 A for any set A (finite or infinite). N otice, for A = {a, b], the connection betw een the follow ing pairs o f elem ents, the first elem ent belonging to V ( A) and the second belonging to 2 A: 0,
fa}, {ib}, {a, b],
/ , = { ( a ,0 ) ,( 6 ,0 ) }
fa = {{a, 1), (b. 0)} fa = {(a, 0), (b, 1)} fa = {(a, 1), (b, 1)}.
For exam ple, the subset {a} o f {a. b } contains a but not b , w hile fa m aps a to 1 and b to 0. For an arbitrary set A, this suggests defining (p so that a subset S' o f A is m apped
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into the function in w hich 1 is the im age o f elem ents of A that belong to S and 0 is the im age o f elem ents o f A that do not belong to S. ♦ Theorem 10.16 Proof
For every nonempty set A , the sets 'P(A ) and 2A are numerically equivalent. We show that there exists a bijective function </>from V { A ) to 2 A. Define <fi : V ( A ) —> 2A such that for S e 'P(A ), w e have cj){S) = f s , w here, for x e A, , . . \ 1 fs(x) = } 0
if if
S xiS.
A' €
C e rta in ly ,/j e 2 14.F irst, we show that <p is onetoone. L et Thus, f s = f r , w hich im plies that f s ( x ) — f r ( x ) for every x e A. T herefore, f s ( x ) = 1 if and only if f r ( x ) = 1 for every x e A; that is, x e S if and only if x e T and so S = T . It rem ains to show that 4> is onto. L et / e 2 A. Define S = [x € A : f ( x ) = 1}. H ence f s = f and so <p{S) = / . Thus <p is onto and, consequently, <p is bijective.
■
It is clear that A = {x, y, z ) has few er elem ents than B = {a, b, c, d, e }, that is, A  < \B\. A nd it certainly seems that £  < N  and that, in general, any finite set has few er elem ents than any denum erable set (or than any infinite set). Also, our discussion about countable and uncountable sets appears to suggest that uncountable sets have m ore elem ents than countable sets. B ut these assertions are based on intuition. We now m ake this m ore precise. A set A is said to have sm a lle r c a rd in a lity than a set B , w ritten as A  < £ , if there exists a onetoone function from A to B but no bijective function from A to B . T hat is,  A  <  B\ if it is possible to pair o ff the elem ents o f A w ith som e o f the elem ents o f B but not w ith all of the elem ents of B . If  A  <  B \, then w e also write \B\ >  A . For exam ple, since N is denum erable and R is uncountable, there is no bijective function from N to R. Since the function / : N — R defined by / ( « ) = n for all n e N is injective, it follow s that N  < R . M oreover, A  < \B\ m eans that A  = £  or A  < \B\. Hence to verify that A  < £ , w e need only show the existence o f a onetoone function from A to B. T he cardinality o f the set N o f natural num bers is often denoted by (often read “aleph null”); so N = X0. Actually, K is the first letter o f H ebrew alphabet. Indeed, if A is any denum erable set, then A  — ^ 0 T he set R o f real num bers is also referred to as the c o n tin u u m , and its cardinality is denoted by c. H ence R  = c and from w hat w e have seen, < c. It w as the G erm an m athem atician G eorg C antor w ho helped to put the theory o f sets on a firm foundation. A n interesting conjecture o f his becam e know n as: T h e C o n tin u u m H y p o th esis
T here exists no set S such that K0 < \ S  < c. O f course, if the C ontinuum H ypothesis w ere true, then this w ould im ply that every subset o f R is either countable or num erically equivalent to R. However, in 1931 the
10.4
Comparing Cardinalities of Sets
257
A ustrian m athem atician K urt G odel proved that it was im possible to disprove the C on tinuum H ypothesis from the axiom s on w hich the theory o f sets is based. In 1963 the A m erican m athem atician Paul C ohen took it one step further by show ing that it was also im possible to prove the C ontinuum H ypothesis from these axiom s. Thus the C ontinuum H ypothesis is independent o f the axiom s o f set theory. A nother question that m ight occur to you is the follow ing: Is there a set S such that 15 1 > c? This is a question we can answer, however, and the answ er m ight be surprising. Theorem to Prove PROOFSTRATEGY
If A is a set, then A  < \V{A)\. First, it is not surprising that  A < P (A ) if A is finite, for if A has n elem ents, w here n e N, then V { A ) has 2" elem ents and 2" > n (w hich was proved by induction in Result 6.9). O f course, we m ust still show that A  < V ( A ) \ w hen A is infinite. First w e show that there exists a onetoone function / : A > V { A ) for every set A. L e t’s give ourselves an exam ple, say A = [a, b}. Then V { A ) = {0, {a}, {b}, {a, b }}. A lthough there are m any injective function from A to V ( A ) , there is one natural injective function: / = {(a, {«}), (b, {b})}, in other w ords, define / : A > V ( A ) by / ( x ) = {a}. O nce we have verified that this function is onetoone, then w e know that A  < V (A ) \. To show that the inequality is strict, however, w e m ust prove that there is no bijective function from A to V ( A). The natural technique to use for such a p ro o f is p ro o f by contradiction. ♦
Theorem 10.17 Proof
I f A is a set, then A  < [P{A)\. If A = 0, then A  = 0 and \V (A )\ = 1; so A  < \V (A )\. H ence we m ay assum e that A ^ 0. First, w e show that there is a onetoone function from A to V { A ). Define the function / : A > V ( A ) by f ( x ) — {jc} for each x e A. L et f ( x i) = f ( x z ) . Then {jfi} — {X2}. So *1 = X2 and / is onetoone. To prove that A < V ( A ) \ , it rem ains to show that there is no bijective func tion from A to 'P(A ). A ssum e, to the contrary, that there exists a bijective function g : A —*■ V ( A ) . For each x e A, let g (x) — A x , w here A x c A. We show that there is a subset o f A that is distinct from A x for each x e A. Define the subset B o f A by B = [x e A : x
A.v).
By assum ption, there exists an elem ent y e A such that B — A y. If y € A y , then y £ B by the definition o f B . On the other hand, if y £ A y , then, according to the definition of the set B , it follow s that y e B . In either case, y belongs to exactly one o f A y and B. H ence B ^ A y , producing a contradiction. ■ A ccording to T heorem 10.17, there is no largest set. In particular, there is a set S with  S' > c.
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10.5 The S ch rod erB ern stein Theorem For tw o nonem pty sets A and B J e t / be a function from A to B and let D be a nonem pty subset o f A. By the re s tric tio n / j of / to D , w e m ean the function
f i = {(*, v) fc / : .v c. £>}. H ence a restriction o f / refers to restricting the dom ain o f / . For exam ple, for the sets A = {a, b, c, cl] and B = {1, 2, 3}, let / = [{a, 2), (.b , 1), (c, 3), (d , 2)} be a function from A to S . For Z) = [a, c}, the restriction o f / to D is the function / i : D B given by {(a, 2), (c, 3)}. Som etim es, w e m ight also consider a new codom ain B' for such a restriction f \ o f / . O f course, we m ust have r a n g e (/j) C S '. Next, consider the function g :R [0, oo) defined by g(.v) = x 2 for x e R. A lthough g is onto, g is not onetoone since g ( l) = g( 1) = 1, for exam ple. O n the other hand, the restriction gi o f g to [0, oo) is onetoone, and so the restricted function gi : [0, oo) > [0, oo) defined by .«,<I <v > = g ( x )  x 2 for all x € [0, oo) is bijective. O n the other hand, if / : A  * B is a onetoone function, then any restriction o f / to a subset o f A is also onetoone. L et / : A —$■ B and g : C + D be functions, w here A and C are disjoint sets. We define a function h from A U C to B U D by h ( x ) = l f{x) 1 uix)
lfx€A if x e C .
R ecalling that a function is a set o f ordered pairs, w e see that h is the union o f the two sets / and g. O f course, it is essential for A and C to be disjoint in order to be guaranteed that h is a function. If / and g are onto, then h m ust be onto as well; however, if / and g are onetoone, then h need not be onetoone. The follow ing result does provide a sufficient condition for h to be onetoone, however. Lem ma 10.18
L et f : A h :A UC
B and g : C —*■ D be onetoone functions, where A f l C = 0, and define
> B U I.) hy h u ) = { f ,(x! % *6 [ .gC4 i f x e C .
1,
I f B C \ D = 0 , then h is also a onetoone function. Consequently, i f f a nd g are bijective functions, then h is a bijective function. Proof
A ssum e that h ( x i) = h ( x 2) = y , w here x \ . x ? e A U B. Then y e B U D . So y e B m y r D , say the former. Since B n D — 0, it follow s that y £ D. H ence ,vt , x 2 e A and so h (x i) = f i x i) and h ( x 2) = f ( x 2). Since j \ x  J ~ f ( x 2) and / is onetoone, it follow s that x\ = x 2. a L et A and B be nonem pty sets such that B c A and let f : A > B. Thus for x e A, the elem ent f i x ) e B . Since B C A, it follow s, of course, that f i x ) e A and so f i f i x ) ) € S . It is convenient to introduce som e notation in this case. L et f 1f.v) = f i x ) and le t f 2 ( x ) = f i f i x ) ) . Ingeneral, for an integer/: > 2 , let f k(x) = f ( f k~ l(A')).Hence f l {x), f 2{a ), / 3( x ) , . . . is a recursively defined sequence o f elem ents of B (and o f A as well). Thus f n(x) is defined for every positive integer n.
10.5
The SchroderBem stein Theorem
259
For exam ple, consider the function / : Z —> 2Z defined by f ( r i) = 4 « f o r a ll« e Z. Then / ’(3) = / ( 3 ) = 4 • 3 = 12 and f \ 3) = / ( / ( 3 ) ) = /( 1 2 ) = 4 • 12 = 48. If A and B are nonem pty sets such that B C A, then the function </> : £ > A defined by 0 (x ) = X for all ,v e B is injective. This gives us the expected result that \B  <  A . On the other hand, if there is an injective function from A to B , a m ore interesting consequence results. Theorem 10.19
Let A and B be nonempty sets such that B c A. I f there exists an injective function from A to B , then there exists a bijective function fro m A to B .
Proof
If B = A , then the identity function i A : A ^ B — A is bijective. T hus w e can assum e that B c A and so A  6 ^ 0. L et / : A —» B be an injective function. If / is bijec tive, then the proof is com plete. T herefore, w e can assum e that / is not onto. Hence r a n g e (/) C B and so B — r a n g e ( /) ^ 0. C onsider the subset B ' o f B defined by = { .fix ) : x e A  B . n e N } . Thus S ' c ra n g e (/) . H ence, for each a e A — B , its im age f ( x ) belongs to B ' . M ore over, for x e A — S , the elem ent f 2(x) — f ( f ( x ) ) e B ', f 3(x) = f ( f 2(x)) e B ' and so on. L e tC = (A — B) U B ' and consider the restriction f \ : C —*■ B ' o f / t o C . W e s h o w that / i is onto. L et y e B ' . T hen v = f n(x) for som e x e A — B and som e n e N. This im plies that y = f ( x ) for som e x € A — B or y = f { x ) for som e x e B ' . Therefore, f \ ( x ) = y for som e x e C and so f \ is onto. Furtherm ore, since / is onetoone, the function f \ is also onetoone. H ence f : C —»• B ' is bijective. Let D = B — B ' . Since B — r a n g e ( /) / 0 and B — r a n g e ( /) C B — B \ it follow s that D ^ 0. A lso, D and B ' are disjoint, as are D and C. Certainly, the identity function io '■D > D is bijective. Let h : C U D —*■ B ' U D be defined by
[
i d (x )
if x e D.
By L em m a 10.18, h is bijective. However, C U D = A and B ' U D = B \ so h is a m bijective function from A to B . From w hat w e know o f inequalities (of real num bers), it m ight seem that if A and B are sets w ith A  < B  and B  < A , then A  = B . This is indeed the case. This theorem is often referred to as the S chroderB em stein Theorem . Theorem 10.20
(The S chroderB em stein Theorem) I f A an d B are sets such that A  < B  and \B\ < A, then A = 8.
Proof
Since  A <  B  a n d  B  < A , there are injective functions / : A —> B a n d g : B  y A . Thus gi : B > range(^) defined by g\(x) — g(x) for all x e B is a bijective function. By T heorem 9.15, g ^ 1 exists and gj"1 : range(g) —>■ B is a bijective function. Since f : A  > B and gi : B ^ range(g) are injective functions, it follow s by Theorem 9.11 th a tg ] o f : A —^ range(g) is an injective function. Because range(g) c A,
260
Chapter 10
Cardinalities of Sets
we have by T heorem 10.19 that there exists a bijective function h \ A range(g). Thus h : A —►range(g) and g'j“1 : range(g) —» B are bijective functions. By C orollary 9.12, o h : A > B is a bijective function and  A\ = \B \.
m
T he S chroderB ernstein Theorem is referred to by som e as the C a n to rS ch ro d erB ernstein Theorem . A lthough the history o f this theorem has never been fully d ocu m ented, there are several substantiated facts. A m athem atician w ho w ill forever be associated w ith the theory o f sets is G eorg Cantor (18451918). B orn in R ussia, C antor studied for and obtained his Ph.D in m ath em atics from the U niversity o f B erlin in 1867. In 1869 he becam e a faculty m em ber at the U niversity of H alle in Germany. It was w hile he was there that he becam e interested in set theory. In 1873 Cantor proved that the set o f rational num bers is denum erable. Shortly afterw ard, he proved that the set of real num bers is uncountable. In this paper, he essen tially introduced the idea o f a onetoone correspondence (bijective function). D uring the next several years, he m ade num erous contributions to set theory— studying sets of equal cardinality. There w ere, however, a num ber o f problem s that proved difficult for Cantor. C onsider the follow ing tw o theorem s: T h e o re m A
For any two cardinal numbers a and b, exactly one o f the follow ing occurs: (1) a — b, (2) a < b, (3) a > b.
T h e o re m B
I f A a nd B are two sets f o r which there exist a onetoone function fro m A to B and a onetoone fu nc tion f r o m B to A , then A = \B\. C antor observed that once Theorem A had been proved, T heorem B could be proved. O n the other hand, there has never been any evidence that C antor was able to prove T heorem A. Ernst Zerm elo (18711953) was able to prove T heorem A in 1904. However, Z erm elo ’s proof m ade use o f an axiom form ulated by Zerm elo. This axiom , w hich was controversal in the m athem atical w orld for m any years, is know n as the A xiom o f C hoice. T h e A xiom o f C hoice. For every collection o f pairwise disjoint nonempty sets, there exists a t least one set that contains exactly one element o f each o f these nonempty sets. A s it turned out, not only can the A xiom o f C hoice be used to prove T heorem A, but T heorem A is true if and only if the A xiom o f C hoice is true. E rnst S chroder (18411902), a G erm an m athem atician, was one o f the im portant figures in m athem atical logic. D uring 18971898 Schroder presented a “p ro o f” o f T h e orem B, w hich contained a defect however. A bout the same tim e, Felix B ernstein (1878— 1956) gave his own p ro o f o f T heorem B in his doctoral dissertation, w hich becam e the first com plete proof o f T heorem B. H is p ro o f did not require know ledge of T heorem A. You m ay be surprised to learn that R and the pow er set o f N are num erically equivalent. B ut how could one ever find a bijective function betw een these tw o sets? T heorem 10.20 tells us that discovering such a function is unnecessary.
10.5
Theorem 10.21 Proof
The SchroderBernstein Theorem
261
The sets 'P(N ) and R are num erically equivalent. First w e show that there is a onetoone function / : (0, 1)  » 'P(N ). R ecall that a real num ber a e (0, 1) can be expressed uniquely as a = O.aiaoai ■■• , w here each aj e {0, 1 , ___ 9} and there is no positive integer N such that an — 9 for all n > N . Thus we define f ( a ) = {10"_1a„ : n £ N} = A. For exam ple, / (0.1234) = {1, 20, 300, 4000} and / ( 1 / 3 ) = {3, 30, 3 0 0 ,. ..}. We now show that / is onetoone. A ssum e that f ( a ) — f i b ) , w here a, ft £ (0, 1) and a = 0 .a ia 2 a 3 ■■■ and b = 0.ftift2ft3 • ■• w ith a,, ft; £ {0, 1, . . . . 9} for each i £ N such that the decim al expansion of neither a nor b is 9 from som e p o in t on. Therefore, A = {10"_1fl„ : n £ N} = {lO"1 ^,, : n £ N} = B. C onsider the i th digit, nam ely , in the decim al expansion o f a . T hen 1 0 '" 1at £ A . If at / 0 , then 10,_1a, is the unique num ber in the interval [1 0 '_1, 9 • 10 '1 ] belonging to A. Since A  B , it follow s that 10‘^ 1a, e 5 . H o w e v e r , lO '^ 1/?,is the unique n u m b erin the interval [1 0 '_1, 9 • 10'” 1] belonging to B; so 10' ’a, = \0 '~ 'b ,. Thus a t = bt . If a, = 0, then 0 6 A and there is no num ber in the interval [1 0 '_1, 9 • 1 0 '~ '] belonging to A. Since A = B . it follow s that 0 £ B and there is no num ber in the interval [1 0 '_1, 9  1 0 ' '] belonging to B . Thus bj = 0 and so a, = b,. H ence a, = b, for all i £ N, and so a = ft. T herefore, / is onetoone and (0, 1) < N ext w e define a function g : P (N ) > (0, l) .F o r S c N, d efin eg ( S ) — 0 .51^253 • • *■» where
T hus g(S ) is a real num ber in (0, 1), w hose decim al expansion consists only o f Is and 2s. We show that g is onetoone. A ssum e that g (S ) = g ( T ) , w here S . T C N. Thus g (S ) = s — 0.5js2^3 • • • = 0 t\t2h ■■■ — t = g [ T ), where 5,!
_}1 j 2
ifn e S iiniS
fl n~ \ 2
ifn e T if n £ T .
Since the decim al expansions o f s and t contain no 0s or 9s, both s and t have unique decim al expansions. We show that S = T . First, we verify that S c T . L et k £ S. Then sk — 1. Since s — t, it follow s that tt = 1, w hich im plies that k e T . H ence S ^ T . The pro o f that T C S is sim ilar and is therefore om itted. Thus S = T and g is onetoone. T herefore. I'P(N)! < (0, 1). B y the S chroderB ernstein T h e o re m .\ V (N)J = (0, 1 ). By C orollary 10.15, (0, 1) = R . Thus, P (N ) = R . As a corollary to Theorem s 10.16 and 10.21, w e have the follow ing result. Corollary 10.22
The sets 2N and R are num erically equivalent.
■
262
Chapter 10
Cardinalities of Sets We have already m entioned that A  =
for every denum erable set A and that
R » # . If A is denum erable, then we represent the cardinality o f the set 2A by 2K . By C orollary 10.22, 2K° = c.
E X E R C IS E S FO R C H A PT E R 10 Section 10.1: Num erically Equivalent Sets 10.1. Let S = {Ai, A i , As} be a collection of five subsets of the set A = { 5 ,  4 ........ 5}, where A] = {.y e A : 1 < x 2 < 10} A2 = {.v e A : (x + 2)(.v  4) > 0} A3 = (.v e A : x + 2 + \x — 3 < 5} A4 = {x 6 A : jJ p j x 1} A5 = {x 6 A : sin ^ = 0}. A relation R is defined on S by A,: R A j (1 < i , j < 5) if A, and A , are numerically equivalent. According to Theorem 10.1. R is an equivalence relation. Determine the distinct equivalence classes for this equivalence relation. 10.2. (a) Let S be a collection of n > 2 numerically equivalent sets. Prove that these sets can be shown to be numerically equivalent by means of n  1 bijective functions between pairs of sets in S. (b) W hat other question is suggested by the problem in (a)?
Section 10.2: Denum erable Sets 10.3. Prove that if A and B are disjoint denumerable sets,then A
U
B is
10.4. Let R + denote the set of positive real numbers and let A and B be C = [x e R : —x e 5}. Show that A U C is denumerable.
denumerable. denumerable subsets of R + . Define
10.5. Prove that Z = Z — {2} . 10.6. (a) Prove that the function / : R  {1} >■ R — {2} defined by f ( x ) = is bijective. (b) Explain why R — {1} = R — (2}. 10.7. Let n“ r + 72 S = { x e R : .v = — , n e N
Define / : N ^ (a) (b) (c) (d)
S by f ( n ) =
List three elements that belong to S. Show that / is onetoone. Show that / is onto. Is S denumerable? Explain.
10.8. Prove that the function / : N
Z defined in (10.1) by / ( « ) =
js bijective.
10.9. Show that every denumerable set A can be partitioned into two denumerable subsets of A. 10.10. Let A be a denumerable set and let B = {x, y}. Prove that A x B is denumerable. 10.11. Let 5 be a denumerable set and let A be a nonempty set of unspecified cardinality. If / : A — B is a onetoone function, then what can be said about the cardinality of A? Explain. 10.12. Prove that the set of all 2element subsets of N is denumerable.
Exercises for Chapter 10
263
10.13. A G aussian integer is a complex number of the form a + bi, where a , b e Z and i = V —I . Show that the set g of Gaussian integers is denumerable. 10.14. Prove that S = {(a, b) : a , b & N and b > 2a) is denumerable. 10.15. Let S c N x N b e defined by S = {(*, j ) : i < j }. Show that S is denumerable. 10.16. Let A i , A i , A3, . . . be pairwise disjoint denumerable sets. Prove that U“ j A, is denumerable. 10.17. Let A = {a 1, a2, 03, . •
Define B = A — {a„2 : n e N}. Prove that  A\ =
B .
10.18. A function / : N x N —> N is defined by f ( m , n ) = 2m~ '(2 n — 1). (a) Prove that / is onetoone and onto. (b) Show that N x N is denumerable. 10.19. Prove that every denumerable set A can be partitioned into a denumerable number of denumerable subsets of A.
Section 10.3: Uncountable Sets 10.20. Prove that the set of irrational numbers is uncountable. 10.21. Prove that the set of complex numbers is uncountable. 10.22. Prove that the open interval (—2, 2) and R are numerically equivalent by finding a bijective function h : (—2, 2) —>• R. (Show that your function is, in fact, bijective.) 10.23. (a) Prove that the function / : (0, 1) —> (0, 2), mapping the open interval (0, 1) into the open interval (0 , 2) and defined by f(x') = 2 x, is bijective. (b) Explain why (0, 1) and (0, 2) have the same cardinality. (c) Let a , b 6 R, where a < b. Prove that (0, 1) and (a, b) have the same cardinality. 10.24. Prove that R and R + are numerically equivalent. 10.25. Consider the function g : (—1, 1) —*■ R defined by g(x) = y rjr(a) Prove that g is onto. (b) Prove that g is onetoone. (c) From the information obtained in (a) and (b), what conclusion can be made?
Section 10.4: Comparing Cardinalities of Sets 10.26. Prove or disprove the following: (a) If A is an uncountable set, then (A  = R. (b) There exists a bijective function / : Q —> R. (c) If A, B and C are sets such thatA c f i c C and A and C are denumerable, then B is denumerable. (d) The set S =
: n e n J is denumerable.
(e) There exists a denumerable subset of the set of irrational numbers. (f) Every infinite set is a subset of some denumerable set. (g) If A and B are sets with the property that there exists an injective function / : A —> B ,th e n A  = S . 10.27. Let A and B be nonempty sets. Prove that A < A x B\. 10.28. Prove or disprove: If A and B are two sets such that A is countable and  A < \B , then B is uncountable. 10.29. How do the cardinalities of the sets [0, f ] and [1,3] compare? Justify your answer. f0.30. Let A = {a, b, c}. Then P (A ) consists of the following subsets of A: A a — 0, A;, = A, A c = {a , b}, Ad = {a , c}, A e = {b, c], A f = {a}, A g = {b}, A h = {c}.
264
Chapter 10
Cardinalities of Sets
In one part of the proof of Theorem 10.17. it was established (using a contradiction argument) that  A < \V(A)\ for every nonempty set A. In this argument, the existence of a bijective function g : A > V ( A ) is assumed, where g(x) = A x for each x € A. Then a subset 5 of A is defined by
B = [x e A : x <£ Ax}. (a) For the sets A and 'P(A) described above, what is the set S ? (b) W hat does the set B in (a) illustrate? 10.31. Prove or disprove: There is no set A such that 1A is denumerable.
Section 10.5: The SchroderBemstein Theorem 10.32. Prove that if A, B and C are nonempty sets such that A c B C C and Aj = C , then A = JBj. 10.33. Use the SchroderBem stein Theorem to prove that ](0, 1) = j[0, 1]. 10.34. Prove that jQ — {<7} = No for every rational number q and R — {r} = c for every real number r . 10.35. Let R* be the set obtained by removing the number 0 from R. Prove that R* = Rj, 10.36. Let f : Z —» 2Z be defined by f ( k ) = 4k for all k e Z. (a) Prove that f n(k) = 4"k for each k e Z and each n e N. (b) For this function / , describe the sets B', C and D and functions f \ and h given in Theorem 10.19. 10.37. Express each positive rational number as m / n , where m , n e N and m / n is reduced to lowest terms. Let da denote the number of digits in a e N. Thus dn — 1, d\x = 2, and dioo = 3. Define the function / : Q + —> N so that f ( m / n ) is the positive integer with 2{dm + dn) digits whose first dm digits is the integer m, whose final dn digits is the integer n, and all of whose remaining dm + dn digits are 0. Thus /( 2 / 3 ) = 2003 and /(1 0 /2 7 1 ) = 1000000271. (a) Prove that / is onetoone. (b) Use the SchroderBem stein Theorem to prove that Q+ is denumerable.
ADDITIONAL EXERCISES FOR CHAPTER 10 10.38. Evaluate the proposed proof of the following result. R esult
Let A and B be two sets with  A ] = Z?. If a e A and b e B, then  A — (a}[ = \B — {b}\.
P roof Since A and B have the same number of elements and one element is removed from each of A and B, it follows that \A — {a} = \B — {b}\, s 10.39. Evaluate the proposed proof of the following result. R esult The sets (0, oo) and [0, oo) are numerically equivalent. P roof Define the function / : (0, oo) —^ [0, oo) by f ( x ) = x. First, we show that/ is onetoone. Assume that f ( a ) = f ( b ) . Then a = b and so f is onetoone. Next, we show that / is onto. Let r e [0. oo). Since /(;•) = r , the function / is onto. Since / is bijective, (0, oo) = [0, oo). ■ 10.40. For a real number x , the floor [SfJ of x is the greatest integer less than or equal to x. Therefore, L55J = 5, _3J = 3 and L—5.5J = —6. Let / : N V Z be defined by f ( n ) = (—1)" [_n /2J ■ (a) Prove that / is bijective. (b) What does (a) tell us about Z? (See Result 10.3.)
Additional Exercises for Chapter 10
265
10.41. Show that the following pairs of intervals are numerically equivalent. (a) (0, 1) and (0, oo) (b) (0, 1] and [0, oo) (c) [b, c) and [a , oo), where a, b, c
€R and b
<
c.
10.42. Let S and T be two sets. Prove that if \S — T\ =  r — S, then [S = T . 10.43. Prove each of the following statements: (a) A nonempty set S is countable if and only if there exists a surjective function / : N *■ S. (b) A nonempty set 5 is countable if and only if there exists an injective function g : S > N. 10.44. Prove that  A < N for every finite nonempty set A. 10.45. Let A = (0, 1) be the open interval of real numbers between 0 and 1. For each number ;• e A, let 0.ri/'2/'3 • • ■denote its unique decimal expansion in which no expansion has the digit 9 from some point on. For (a, b) e A x A, let / ( ( a , b)) = f ( ( 0 .aia 2 a 3
0
.b\b 2 b 3 ...) ) =
0
.aibia 2 b 2 . . . ;
while for a e A, let g(a) = g (0 .a ia 2«3 • ■■) = (0 .a ia3a 5 . . . ,
0
.a2 aAa 6 . . .).
With the aid of / and g, which of the following are we able to conclude? Explain your answer. (a) (b) (c) (d) (e) (f)
\A x A\ < A. \A\ < \A x A. \ A x A \ = IA . Nothing because neither / : A x A  > A nor j : A ^  A x A i s a function. Nothing because both / : A x A > A and g : A * A x A are functions but neither is injective. Nothing, for a reason other than those given in (d) and (e).
10.46. Prove, for every integer n > 2 , that if A \ , A2, • . . , A„ are denumerable sets, then Ai x A2 x • • ■ x A„ is denumerable. 10.47. As a consequence of Exercise 10.20 in Section 10.3, the set of all irrational numbers is uncountable. Among the (many) irrational numbers are 4 2 , \/3 , V5 , 4 2 , 4^3, 4^5 and 4 2 . Prove that the set S = { 4 k : k , n e N and k, n > 2} is denumerable. 10.48. Let b , c & Z. A number rbx (real or complex) belongs to a set S if x 2 + bx 4 c. Prove that S is denumerable.
is a root of the polynomial
10.49. We have seen that R is an uncountable set. (a) Show that R can be partitioned into a denumerable number of uncountable sets. (b) Show that R can be partitioned into a uncountable number of countable sets.
I ll Proofs in N um ber Theory
um ber theory is that area o f m athem atics dealing w ith integers and their properties. It is one o f the oldest branches of m athem atics, dating back at least to the Pythagoreans (500 B.C.). N um ber theory is considered to be one o f the m ost beautiful branches of m athem atics. Indeed, it has been said that m athem atics is the queen o f the sciences, w hile num ber theory is the queen o f m athem atics. In large m easure, this subject is characterized by the appeal, clarity and sim plicity o f m any of its problem s and by the elegance and style exhibited in their solutions. The m ain goal o f this chapter is to expand on some of the things w e have learned in order to illustrate the kinds o f proofs that occur in num ber theory.
N
11.1 D ivisib ility P roperties of Integers You m ay already know that every integer n > 2 can be expressed as a product o f prim es and in only one way, except for the order in w hich the prim es are w ritten. We w ill see later how to prove this fact, but first we w ant to return to divisibility of integers (introduced in C hapter 4) and establish several elem entary divisibility properties. R ecall that a p rim e is an integer p > 2 w hose only positive integer divisors are 1 and p . A n integer n > 2 that is not prim e is called a co m p o site n u m b e r (or sim ply com posite). The first ten prim es are 2, 3, 5, 7, 11. 13. 17, 19, 23 and 29. The first ten com posite num bers are 4, 6, 8, 9, 10, 12, 14, 15, 16 and 18. If an integer n > 2 is com posite, then there exist integers a and b such that n â€” ab, w here 1 < a < n and 1 < b < n. Certainly, if there exist integers a and b such that n â€” ab, w here 1 < a < n and \ < b < n, then n is com posite. We sum m arize these observations in the lem m a below. Lem ma 11.1
A n integer n > 2 is composite i f and only i f there exist integers a and b such that n  ab, where 1 < a < n and 1 < b < n. In C hapter 4, a num ber o f basic divisibility properties were presented. We recall some o f these in the follow ing theorem , w here the proofs are also repeated to review the proof techniques that we used.
266
1 1 .2
Theorem 11.2
The Division Algorithm
267
L et a , b and c be integers with a ^ 0. (i) I f a  b, then a  be. (ii) I f a \b a n d b \c, where b ^ 0, then a \ c. (iii) I f a \b and a \c, then a  (b x + c y ) f o r all integers x and y.
Proof
We begin w ith part (i). Since a \ b, there exists an integer q such that b = aq. Therefore, be = a(qc). Since q c is an integer, a \ be. For part (ii), let a \ b and b \ c. So there exist integers <71 and q 2 such that b = aq\ and c = bq2. Consequently, c  bq 2 = (a q \)q 2 = a ( q \q 2). Since q \q 2 is an integer, a \ c. For part (iii), let a \ b and a \ e. T hen there exist integers q\ and q 2 such that b = aq\ and c — a q 2. H ence, for integers x and y, b x + c y = ( a q x)x + (a q 2)y = a ( q \x + q 2 y). Since q\X + q2y is an integer, a \ (bx + ey).
PROOF ANALYSIS
m
In all three parts o f the previous theorem , we were required to show that r \ s for som e integers r and s, w here r ^ 0. To do this, w e show ed that w e could w rite s as r t for som e integer t. O f course, this is sim ply the definition o f w hat it m eans for r to divide s. ♦
Proving the tw o parts o f the next theorem relies on the definition o f r  s again, as w ell as m aking use o f certain observations. For exam ple, in the second part, we use the fact that \xy\ — x y  for every tw o real num bers x and y. Theorem 11.3
L e t a and b be nonzero integers. (i ) I f a  b and b \ a, then a = b or a = —b. (ii) I f a  b, then \a\ < \b\.
Proof
We first prove (i). Since a \ b and b \ a, it follow s that b — aq\ and a = b q 2 for some integers q\ and q2. T herefore, a — bq 2 = (a q \) q 2 = a ( q \q 2). D ividing by a, w e obtain 1 = q 1q2. Hence q\ = q 2 = 1 or qi = q 2 = —1. T herefore, a = b or a — —b. N ext we prove (ii). Since a \ b, it follow s that b = a q for som e integer q. Further m ore, q ^ 0 since b ^ 0. So \q\ > 1. Hence \b\ = \aq\ = \ a \  \ q \ > \ a \  \ = \a\.
■
11.2 The D ivision Algorithm We have discussed the concept o f divisibility a num ber o f tim es now. O f course, w hen we use that term , w e are referring to the statem ent a \ b , w here a , b e Z and a / 0 . Surely, the term division is m ore fam iliar to us. For positive integers a and b, it is an elem entary problem to divide b by a and ask for the quotient q and rem ainder r . For exam ple, for a = 5 and b = 17, we have q = 3 and r = 2; that is, if 17 is divided by 5, a quotient o f
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3 and a rem ainder o f 2 result. This division can be expressed as 17 = 5 ■3 + 2. If a = 6 and b = 42, then q = 7 and r = 0; so 42 = 6 ■7 + 0 or 6 j 42. M ore generally, for positive integers a and b, it is always possible to w rite b = aq + r , w here 0 < r < a. T he num ber q is the q u o tie n t and r is the r e m a in d e r w hen b is divided by a. In fact, not only do the integers q and r exist, they are unique. T his is the essence o f a theorem called the D ivision A lg o rith m . A lthough this theorem m ay seem rather obvious because y o u ’ve probably used it so often, it is im portant and its p ro o f is not obvious. T h e o re m to P ro v e
PROOF STRATEGY
(T h e D ivision A lg o rith m ) For positive integers a and b , there exist unique integers q and r such that b = a q + r and 0 < r < a. We start the proof then w ith tw o positive integers a and b. We have tw o problem s facing us. First, we need to show that there are integers q and r such that b — a q + r and 0 < r < a. Second, w e m ust show that only one integer q and one integer r satisfy the equation b — aq + r and inequality 0 < r < a. H ow can w e com e up w ith integers q and r satisfying these conditions? O ur m ain concern, it turns out, is show ing that there exist integers q and r such that b = a q + r and 0 < r < a. If a  b, then w e know that b = aq for som e integer q. So b = aq + 0 and r — 0 satisfies 0 < r < a. If a / b, then b ^ aq for every integer q and so b — aq ^ 0 for every integer q. However, w hen we perform the operation o f dividing b by a, w e obtain a quotient q and a nonzero rem ainder r. The integer q has the properties that b — a q > 0 and b — aq is as sm all as possible. W hether a \ b or a J(b, this suggests considering the set S — {b — a x : x e Z and b — a x > 0}, w hich is a set of nonnegative integers. O nce w e show that S ^ 0, then w e can apply T heorem 6.7 (w hich states, for every integer m , that the set [i e Z : i > m] is wellordered) to conclude that S has a sm allest elem ent r, w hich m eans that there is an integer q such that b — a q = r . So b = aq + r and w e have the beginning o f a proof. ♦
Theorem 11.4
(T h e D ivision A lg o rith m ) For positive integers a and b, there exist unique integers q and r such that b — aq + r and 0 < r < a.
Proof
First we show that there exist integers q and r such that b = aq + r w ith 0 < r < a. We w ill verify the uniqueness later. C onsider the set S = [b — a x : x e Z and b — a x > 0 } . By letting x = 0, we see that b e S and S is nonem pty. Therefore, by T heorem 6.7, S has a sm allest elem ent r and, necessarily, r > 0. A lso, since r e S, there is som e integer q such that r = b — a q . Thus b = aq + r w ith r > 0. N ext w e show that r < a. A ssum e, to the contrary, that r > a. Let t = r — a. Then t > 0. Since a > 0, it follow s that t < r . M oreover, t = r — a = (b — a q ) — a = b — {aq + a) = b — a{q + 1), w hich im plies that t e S, contradicting the fact that r is the sm allest elem ent o f S. T herefore, r < a, as desired.
11.2
T he Division Algorithm
269
It rem ains to show that q and r are the only integers for w hich b = a q + r and 0 < r < a. Let q' and r' be integers such that b = a q ' + r ’, w here 0 < r' < a. We show that q = q' and r = r ' . A ssum e, w ithout loss of generality, that r' > r ; so r' — r > 0. Since a q + r = a q ' + r', it follow s that a(q  q') = r'  r. Since q — q' is an integer, a \ (r 1 — r). B ecause 0 < r' — r < a, we m ust have;' — r = 0 and so r ' = r . However. a(q — q') = r' — r = 0 and a 0; so q — q' = 0 and q — q . ■ In T heorem 11.4 w e restricted a and b to be positive. W ith m inor m odifications in the proof, w e can rem ove these restrictions, although, of course, we m ust still require that a # 0. Exercise 11.20 asks for a pro o f o f the follow ing result. Corollary 11.5
(T h e D ivision A lg o rith m , G e n e ra l F o rm ) For integers a and b with a ± 0, there exist unique integers q and r such that b — aq + r and 0 < r < \a\. The proof o f T heorem 11.4 is an existence p ro o f since it establishes the existence of the integers q and r but does not provide a m ethod for finding q and r. However, there is an im plicit connection betw een the p ro o f o f the theorem and the m anner in w hich you w ere taught to divide one positive integer by another to find the quotient and rem ainder (as w e m entioned earlier). For instance, w hen dividing 89 by 14 you first determ ine the num ber o f tim es 14 goes into 89, nam ely 6. M ore form ally, you found the largest nonnegative integer w hose product w ith 14 does not exceed 89. This num ber is 6. You then subtracted 14 • 6 = 84 from 89 to find the rem ainder 5. This determ ines the least nonnegative value o f 89  14c/, w here q is an integer, w hich corresponds to the least nonnegative value o f the set S for a = 14 and b = 89 in the p ro o f of T heorem 11.4.
Example 11.6
Consistent with the notation o f Corollary 11.5, fin d the integers q and r f o r the given integers a and b. (i) (ii) (iii) (iv)
Solutions
a = 17, b = 78 a = —17, b = 78
a  17. b — —78 a = “ 17, b =  7 8
(i) By sim ple division, w e see that dividing 78 by 17 results in a quotient of 4 and a rem ainder o f 10, that is, 78 = 17 • 4
+ 10;
(H I)
so q = 4 and r = 10. (ii) By replacing 17 and 4 in (11.1) by  1 7 and  4 , respectively, w e obtain 78 = ( —17)(—4) + 10; so q — —4 and r = 10. (iii) M ultiplying (11.1) through by  1 . w e have
 7 8 = (1 7 ■4) + (1 0 ) = 17(—4) + (1 0 ).
(11.2)
270
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Proofs in Number Theory Since every rem ainder is nonnegative, we subtract 17 from and add 17 to the right side o f (11.2), producing  7 8 = 1 7 ( 5 ) + 7;
(11.3)
so q — —5 and r = 7. (iv) If, in (11.3), we replace 17 and —5 by —17 and 5, respectively, we have  7 8 = (  1 7 )  5 + 7; so q = 5 and r — 7.
#
We now discuss some consequences o f the D ivision A lgorithm . If this algorithm is applied to an arbitrary integer b w hen a = 2, then we see that b m ust have one o f the two form s 2q or 2q + 1 (according to w hether r — 0 or r = 1). O f course, if h — 2c/. then b is even; w hile if b — 2q + 1, then b is odd. We have seen this earlier. This, of course, shows that every integer is even or odd. If we apply the D ivision A lgorithm to an arbitrary integer b w hen a — 3, then b has exactly one o f the three form s 3q, 3q + 1 or 3q + 2 (according to w hether r — 0, 1. 2). We saw this as w ell in C hapter 4. In general, for an arbitrary integer b and arbitrary positive integer a, the rem ainder r is one o f 0, 1 , 2 , . . . , a — 1. H ence b is expressible as exactly one o f a q , a q + 1, . . . , a q + (a — 1). This last rem ark should also sound familiar. In C hapter 8 w e considered, for an integer n > 2, a relation R defined on Z by a R b if a = b (m od n ), that is, if n  (a  b). This relation w as found to be an equivalence relation and the set Z n = {[0], [1].........[» — 1]} o f distinct equivalence classes was referred to as the set o f integers m odulo n. W ith the aid o f the D ivision A lgorithm , we now take a closer look at this relation. F or each integer b , there exist, by the D ivision A lgorithm , unique integers q and r such that b = nq + w here 0 < /■ < n. Thus b — r — nq and n \ (b — r), so b = r (m od «)■ Since b R r, it follow s that b e [/•]. However, since r is the unique integer w ith 0 < r < n  1, w e see that b belongs to exactly one o f the classes [0], [ 1 ] , . . . , [ « — 1]. T hese observations show that (i) the classes [0], [1]............[n — 1] are pairw ise disjoint and (ii)
Z = [0] U [1] U • • • U [« — 1],
neither o f w hich should seem surprising once we recall that the equivalence classes always produce a partition o f the set on w hich the equivalence relation is defined. F ur therm ore, for each r = 0, 1 , . .. , n — 1, [r] = {nq + r : q e Z ) ; that is, [;•] consists of all those integers having a rem ainder o f r w hen divided by n. For this reason, these equivalence classes w ere also referred to as residue classes m odulo n. The special case w here n = 3 was considered in C hapter 8 and the resulting residue classes w ere exhibited. In the present context, these residue classes are [0] = [3q : q e Z} = { . . .  6.  3 , 0, 3 , 6 . . . .} [1] = {3q + 1 : q e Z} = { . . .  5,  2 , 1 . 4 , . . .} [2] = {3q + 2 \ q
e
Z} = { . . . — 4. —1, 2, 5, . . .}.
11.3
Greatest Common Divisors
271
11.3 Greatest Com m on D ivisors We now m ove from the divisors o f an integer to the divisors o f a pair o f integers. An integer c ^ 0 is a com m on divisor o f tw o integers a and b if c  a and c \ b. We are prim arily interested in the largest integer that is a com m on divisor o f a and b. Formally, the greatest com m on divisor o f tw o integers a and b, n o t both 0, is the greatest positive integer that is a com m on divisor o f a and b. The requirem ent that a and b are not both 0 is needed since every positive integer divides 0. We denote the greatest com m on divisor o f two integers a and b by gcd(a, b). although {a, b) is com m on notation as well. If a and b are relatively sm all (in absolute value), it is norm ally easy to deter m ine gcd(a, b). F or exam ple, it should be clear that gcd(8, 12) = 4, gcd(4, 9) = 1 and gcd( 18, 54) = 18. The definition o f g cd (a, b ) does not require a and b to be p o si tive; indeed, it only requires at least one of a and b to be nonzero. F or exam ple, gcd (—10, —15) “ 5, gcd(16. —72) = 8 and gcd(0, —9) = 9. There are tw o useful properties o f the greatest com m on divisor o f tw o integers, particularly from a theoretical point of view, that we w ant to m ention. For integers a and b, an integer o f the form a x + by, w here x , y e Z, is called a linear com bination of a and b. U sing this term inology, we can now restate Theorem 11.2(iii): E very nonzero integer that divides tw o integers b and c divides every linear com bination o f b and c. A lthough there appears to be no apparent connection betw een linear com binations o f a and b and gcd(a, b), w e are about to see that there is, in fact, a very close connection. For exam ple, let a — 10 and b = 16. T hen 6, —4 and 0 are linear com binations o f a and b since, for exam ple, 6 = 10 • (  1 ) + 16 • (1),  4 = 10 • (—2) + 16 • (1) and 0 = 10 ■0 + 16 • 0. The integer 4 is also a linear com bination of a and b since 4 = 10(2) + 16(—1). Further m ore, 2 is a linear com bination of a and b since 2 = 10(—3) + 16(2). O n the other hand, no odd integer can be a linear com bination of a and b since if n is a linear com bination o f a and b, then there exist integers x and y such that n = a x + by = 10x + 16y = 2(5x + 8y). Since 5;v + 8;y is an integer, n is even. Consequently, 2 is the least positive integer that is a linear com bination of 10 and 16. C uriously enough, gcd(10, 16) = 2. We now show that this observation is no coincidence. A gain, the W ellO rdering Principle w ill prove to be useful. Theorem 11.7
Let a and b be integers that are not both 0. Then gcd(a, b) is the least positive integer that is a linear combination o f a and b.
Proof Let S denote the set o f all positive integers that are linear com binations o f a and b, that is, S = {ax + by : x . y e Z and a x + by > 0}. First w e show that S is nonem pty. By assum ption, at least one of a and b is nonzero; so a ■a + b ■b = cr + b: > 0 . Thus a • a + b ■b e S and, as claim ed, S / 0.
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Since S is a nonem pty subset o f N, it follow s by the W ellO rdering Principle that S contains a least elem ent, w hich we denote by d. Thus there exist integers .v0 and y0 such that d = a x q + by®. We now show that d = g cd (a. b). A pplying the D ivision A lgorithm to a and d , we have a — d q + r , w here 0 < r < d. Consequently, r = a  d q = a — q ( a % + byo) = o ( l  q x 0) + b (—q y o); that is, r is a linear com bination o f a and b. If r > 0, then necessarily r e 5, w hich would contradict the fact that d is the least elem ent in S. Therefore, r — 0, w hich im plies that d  a. By a sim ilar argum ent, it follow s that d \ b and so d is a com m on divisor o f a and b. It rem ains to show that d is the greatest com m on divisor o f a and b. L et c be a positive integer that is also a com m on divisor o f a and b. By T heorem 11.2 (iii), c divides every linear com bination of a and b and so c divides d = a x 0 + by0. B ecause c and d are positive and c j d. it follow s that c < d and therefore d = gedU/. b). ■ P R O O F AN ALYSIS
The proof o f T heorem 11.7 illustrates a com m on p ro o f technique involving divisibility of integers. At one point o f the proof, w e w anted to show that d \ a. A com m on m ethod used to show that one integer d divides another integer a is to apply the D ivision A lgorithm and divide a by d , obtaining a — d q + r , w h e r e 0 < r < j</jor. il'</ > 0 ,th e n 0 < r < d. The goal then is to show that r — 0. # T here is another characterization o f the greatest com m on divisor o f two integers, not both 0, that is useful to know. This characterization provides an alternative definition o f the greatest com m on divisor, w hich, in fact, is used as the definition on occasion.
Theorem 11.8
Proof
L et a and b be two integers, not both 0. Then d = g c d (a . b) i f a nd only i f d is that positive integer which satisfies the fo llow ing W o conditions: (1)
d is a common divisor o f a and b;
(2)
i f c is any common divisor o f a and b, then c \ d .
First, assum e that d — gcd(a, b). We show that d satisfies (1) and (2). By definition, d satisfies (1); so it only rem ains to show that d satisfies (2). Let c be an integer such that c \ a and c  b. Since d — gcd(a, b), there exist integers x 0 and yo such that d = a x 0 + byoSince c \ a and c  b, it follow s by T heorem 11.2(iii) that c divides a x 0 + by0 = d. T herefore, d satisfies (2). F or the converse, assum e that d is a positive integer satisfying properties (1) and (2). We show that d = gcd(a, b). Since d is already a com m on divisor o f a and b, it suffices to show that d is the greatest com m on divisor o f a and b. L et c be any positive integer that is a com m on divisor of a and b. Since d satisfies (2). c \ cl. Since c and d are both positive, it follow s by T heorem 11.3(ii) that r < d. w hich im plies that d = g cd (a, b). *
11.4 The E u c lid ea n A lgorithm A lthough w e know the definition of the greatest com m on divisor d o f tw o integers a and b, not both 0, and tw o characterizations, none of these is useful in com puting d. For this reason, we describe an algorithm for determ ining d = gcd(a, b) that is attributed to the
11.4
The Euclidean Algorithm
273
fam ous m athem atician Euclid, best know n for his w ork in geom etry. First, we note that if b = 0, then a / 0 and gcd(a, b) — gcd(a, 0) = \a\. H ence w e m ay assum e that a and b are nonzero. Furtherm ore, since gcd(cr, b) = gcd(a, —b) = g cd(—a , b ) = g cd (—a , —b), we can assum e that a and b are positive. Therefore, gcd(0, —12) = gcd(0, 12) = 12 and gcd(—12,  5 4 ) = gcd(12,  5 4 ) = g c d (  1 2 , 54) = gcd(12, 54) = 6. In general then, we m ay assum e that 0 < a < b. T he procedure for com puting d = gcd(a, b ), w hich we are about to describe and w hich is called the E u c lid e a n A lg o rith m , m akes use o f repeated applications o f the D ivision A lgorithm and the follow ing lemma. Lem ma 11.9Let a and b be positive integers. I f b — a q + r f o r some integers q a n d r , then g cd (a, b) = gcd(;', a). Proof
Let d = gcd(a, b ) and e = gcd(r, a). We show that d = e. First, note that b = a q + r = a q + r • 1; that is, b is a linear com bination o f a and r . Since e = gcd(r, a), it follow s that e  a and e \ r. By T heorem 11.2 (iii), e \ (aq + r • 1) and so e  b. H ence e is a com m on divisor o f a and b. B ecause d = gcd(tf, b ), we have e < d. Since b = aq + r, we can write r = b — a q — b ■1 + a ( —q), and so r is a linear com bination o f a and b. From the fact that d — gcd(o. b ), we obtain d \ (b ■ 1 + a ( —q)% that is, d  r. So d is a com m on divisor o f r and a. Since e = gcd(r, a), it follow s that d < e. Thus e = d. ■ We are now prepared to describe the E uclidean A lgorithm . We begin w ith two integers a and b, w here 0 < a < b. By the D ivision A lgorithm , b = aq\ + r\. w here 0 < /'i < a. By L em m a 11.9, gcd(a, b) = g c d ^ , a). So if r t = 0, then gcd(a, b) = gcd(0, a) = a. H ence w e m ay assum e that r i ^ 0 and apply the D ivision A lgorithm to n and a , obtaining a = r\q% + f‘%, w here 0 < /'2 < >'\■ At this point, w e have gcd(a, b) = g cd (ri, a) — g c d ( ^ , r ^ a n d O < r 2 < r\ < a .I f/'2 = 0 , then gcd(a, b) = g c d ( n , a) = gcd(0, r x) = r\. B y now we should see the usefulness of L em m a 11.9 and also w hat we m ean by repeated applications of the D ivision A lgorithm . We continue this process and obtain the follow ing sequence of equalities and inequalities: b = aq\ + r\
0 < r\ < a
fl =
0 < t'2 < I' 1
+ >'2
>'k1 = rkqk+1 + rk+l
0 < rk+ j < rk
By L em m a 11.9, gcd(a. b) = g cd (ri, a) = gcd (r2, r ,) = • • • = g cd(/> + i, rk)  ■■■ and • • • < rk+\ < / > < • • • < r 2 < r i < a. Since these rem ainders are nonnegative, the strictly decreasing sequence r \ , r2, . . . of rem ainders contains at m ost a term s. L et r„_ i
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be the last nonzero rem ainder. Thus r„ = 0. We then have:
rn 
b = aq\ + fi
w here 0 < n < a
a — /'i<?2 +
w here 0 < r 2 < >‘i
r2
= r,j_ 3^„_2 + r n  2
w here 0 < r „_2 < r„ ?
r „_3 = /> ?</„...■ + r„_!
w here 0 < r B.j < r „_2
4
F n —2
=
(11.4)
“I 0
1n —ltfn
and w e know that gcd(a, b) = gcd(;'i, a) = gcd(r2, r r) = ■■• = g cd (r„_ i, r„ _ 2)  gcd(0 , r „ _ ,) = r„ _ i. T he E uclidean A lgorithm can now be described. We start w ith tw o integers a and b, w here 0 < a < b. If a \ b, then g cd (a, b) = a ; w hile if a J b , then we apply the D ivision A lgorithm repeatedly until a rem ainder o f 0 is obtained. In this latter case, the last nonzero rem ainder is then g c d (a , b). L e t’s see how the E uclidean A lgorithm w orks in practice. Example 11.10 Solution
Use the Euclidean Algorithm to determine d = gcd(374, 946). D ividing 946 by 374, we find that 946 = 374 • 2 + 198. Now, dividing 374 by 198, we have 374 = 198 ■1 + 176. C ontinuing in this m anner, we obtain 198 = 1 7 6  1 + 2 2 176 = 2 2  8 + 0. So gcd(374, 946) = 22.
♦
G iven integers a and b, not both 0, we know that there exist integers s and t such that g cd(a, b) = as + bt. We now describe an algorithm for finding such integers 5 and t, using the notation in the calculations done after the p ro o f o f L em m a 11.9. Since gcd(a, b) = r n \ , our goal is to find integers s and t such that r „ _ i = as + bt. We begin w ith the equation r „_3 = rn 2 <in1 + rni and rew rite it in the form 1 n —\
—
1n—
3
1 n —2C[n— 1 
N ext, using equations (11.4), w e solve for r„_ 2, obtaining t~n—2 =
Fn—4
^ n — 3 ^ n — 2
( 1 1 .5 )
11.5
Relatively Prime Integers
275
N ow substituting this expression for r„ _ 2 into equation (11.5), we obtain t'n—i = j it—3
0_n—\ {]n — 4
t ti—i Qn—2)
= (1 + q „ iq n2)rn3 + {—q n  \) rn4At this point, r„ _ i is represented as a linear com bination of /■„_3 and r„_ 4. N otice that r„_o no longer appears in the expression. A s w e continue w ith this backw ard substitution m ethod, w e elim inate the rem ainders r„ _ 3, r„_ 4, . . . , r2, n one at a tim e and eventually arrive at an equation of the form rn \ — as + bt. Example 11.11 Solution
For a = 374 a nd b — 946, fin d integers s and t such that as + bt = g cd (a, b). U sing the com putations from E xam ple 11.10, we have 22 = 198  176■ 1 = 1 9 8 • 1 + 176 • (  1 ) 176 = 374  198 ■1 — 374 • 1 + 198 • (  1 ) 198 = 946  374 • 2 = 946 • 1 + 374 • (  2 ) . Therefore. 22 = 198 1 + 176 • (—1) = 198 • 1 + [374 • 1 + 198 • (  1 ) ] • (  1 ) = 198 • 1 + 374 (  1 ) + 198 • 1 = 1982 + 37 4  (  1 ) = [946 ■1 + 374 • (  2 ) ] • 2 + 374 • (  1 ) = 946 • 2 + 374 • (  4 ) + 374 • (  1 ) = 946 • 2 + 374 • ( —5). Hence s =  5 and t — 2.
▼
T he integers 5 and t that w e have ju st found are not unique. Indeed, if gcd(<z, b) = d and d = a s + bt, then d = a(s + b ) + b(t  a) as well.
11.5 R elatively Prim e Integers For tw o integers a and b, not both 0, we know that if gcd(a, b) = 1, then there exist integers 5 and t such that as + bt = 1. W hat m ay be surprising is that the converse holds in this special case as well. Theorem 11.12
L e t a and b be integers, not both 0. Then gcd(a, b) = 1 i f a nd only i f there exist integers s and t such that 1 = as + bt.
Proof
If g c d (a , b) — 1, then by T heorem 11.7 there exist integers 5 and t such that a s + bt — 1. We now consider the converse. L et a and b be integers, not both 0, for w hich there exist integers s and t such that a s + bt — 1. By T heorem 11.7, g c d (a , b) is the sm allest positive
276
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integer that is a linear com bination o f a and b. Since 1 is a linear com bination o f a and b , it follow s that gcd(a, b) — 1. ■ Two integers a and b, not both 0, are called relativ ely p rim e if g cd (a, b) = 1. By T heorem 11.12 then, two integers a and b are relatively prim e if and only if 1 is a linear com bination o f a and b. T his fact is extrem ely useful, as w e are about to see. If a , h and c are integers such that a \ be, then there is no reason to believe that a \ b or a  c. For exam ple, let a = 4, b — 6 and c — 2. Then 4  6 • 2 but 4 / 6 and 4 / 2 . However, if a and b are relatively prim e, then we can draw another conclusion. The follow ing result is often called E uclid’s Lemma. T h e o re m to P rove
(E u c lid ’s L em m a)
Let a , b and c be integers, w here a ± 0. If a \ be and g cd(<2, b ) — 1,
then a \ c. PROOFSTRATEGY
Theorem 11.13
If we use a direct proof, then we assum e that a  be and gcd(a, b) — 1. In order to show that a  c, we need to show that c be can be expressed as a r for som e integer r. Because a  be, we know that be = a q for som e integer q. A lso, because gcd(a, b ) = 1, there are integers 5 and t such that as + bt = 1. If we were to m ultiply as + bt = 1 by c, then we w ould have c = a cs 4 bet. However, be = a q and we could factor a from aes + bet. This is the plan. ♦ (E u c lid ’s L em m a) L et a, b and e be integers, where a ^ 0. I f a \ be and g cd (a, b) — 1, then a \ c.
Proof
Since a \ be, there is some integer q such that be = aq. Since a and b are relatively prim e, there exist integers s and t such that 1 — as + bt. Thus c = c ■ 1 = c(as + bt) — a(es) + (bc)t = a(es) + (a q ) t = a(cs + q t). Since cs + q t is an integer, a \ c.
■
E uclid’s L em m a is o f special interest w hen the integer a is Corollary 11.14 Proof
L et b and c be integers and p a prime. I f p \ be, then either p \
a prime. b o r p  e.
If p divides b, then the corollary is proved. Suppose then that p does n ot divide b. Since the only positive integer divisors o f p are 1 and p , it follow s that ged(p, b) — 1. Thus, by E u clid ’s L em m a, p  c and the proo f is com plete. * T he preceding corollary can be extended to the case w hen a prim e p divides any product of integers.
Corollary 11.15
L et a \ , an.........an, where n > 2 , be integers and let p be a prime. I f p  a i a 2 ■■■a n, then p  at f o r some integer i (1 < i < «).
11.6 Proof
The Fundam ental Theorem of Arithmetic
277
We proceed by induction. For n — 2, this is sim ply a restatem ent o f C orollary 11.14. A s sum e then that if a prim e p divides the product o f k integers (k > 2), then p divides at least one of the integers. Now let a i , ajs , a.k+\ b e k + 1 integers, w here p  a\a ^ ■■■ We show that p \ a, for som e i (1 < i < k + 1). L et b — Q\ai ■■■ak . So p \ h<n j ;. By Corollary 11.14, either p \ b or p  ak+1. If p J ak+\, then the p ro o f is com plete. O th erw ise, p  b, that is, p  a \ a 2 ■• ■ak. However, by the induction hypothesis, p \ at for som e i (1 < i < k). In any case, p j a, for some i (1 < i < k + 1). By the Principle of M athem atical Induction, if a prim e p divides the product o f any n > 2 integers, then p divides at least one of the integers. ■ There is another useful fact concerning relatively prim es integers. O nce again, we w ill have occasion to use the result that w henever tw o integers a and b are relatively prim e, then 1 is a linear com bination o f a and b.
Theorem 11.16
Let a , b, c e Z, where a and b are relatively prim e nonzero integers. I f a  c and b \ c, then ab  c.
Proof
Since a  c and b \ c, there exist integers x and y such that c = a x and c — by. Further m ore, since a and b are relatively prim e, there exist integers 5 and t such that 1 = a s + bt. M ultiplying by c and substituting, we obtain c = c ■1 = c(as + bt) = c(as) + c{bt) = (b y ) ( a s ) + (ax)(bt) = a b (sy + x t). Since (s y + x t ) is an integer, a b [ c.
■
By T heorem 11.16 then, if w e w ish to show that 12, say, divides som e integer c, we need only show that 3  c and 4  c since 12 = 3  4 and 3 and 4 are relatively prim e.
11.6 T he F u n d a m e n ta l T h eo rem of A rithm etic It is a basic divisibility fact that every integer can be expressed as a product o f prim es. This fact is m ade precise in a fam ous theorem in num ber theory. Its p ro o f serves as one o f the m ost interesting uses o f the Strong Principle of M athem atical Induction. Theorem 11.17
(F u n d a m e n ta l T h e o re m o f A rith m e tic ) Every integer n > 2 is either prim e or can be expressed as a product o f primes; that is, n = P l P % > *• P m ,
where p \ , p 2 .........p m are prim es. Furthermore, this factorization is unique except p o s sibly f o r the order in which the factors occur. Proof
To show the existence of such a factorization, we em ploy the Strong Principle o f M ath em atical Induction. Since 2 is a prim e, the statem ent is certainly true for n = 2. For an integer k > 2. assum e that every integer i, w ith 2 < i < k, is either prim e or can be expressed as a product o f prim es. We show that k + 1 is either prim e or can be expressed as a product of prim es. O f course, if k + 1 is prim e, then there is nothing further to prove. We m ay assum e, then, that k + 1.1® com posite. By L em m a 11.1. there
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exist integers a and b such that k + 1 = ab, w here 2 < a < k and 2 < b < k. Therefore, by the induction hypothesis, each o f a and b is prim e or can be expressed as a product of prim es. In any case, k + 1 = a b is a product o f prim es. By the Strong Principle o f M athem atical Induction, every integer n > 2 is either prim e or can be expressed as a product o f prim es. To prove that such a factorization is unique, we proceed by contradiction. A ssum e, to the contrary, that there is an integer n > 2 that can be expressed as a product o f prim es in tw o different w ays, say n = P 1 P 2 ■■■ p s = q \ q i • • •qt , w here in each factorization, the prim es are arranged in nondecreasing order; that is, P\ < P i < • ■■ < Ps and <7i < q i < • • • < qt ■Since the factorizations are different, there m ust be a sm allest positive integer r such that p r / qr . In other w ords, if r > 2, then P i = qi for every i w ith 1 < i < r — 1. A fter canceling, w e have P r P r + l ■■■ Ps =
q , q r+ 1
(11.6)
C onsider the integer p r . E ither s = r and the left side o f (11.6) is exactly p r or s > r and p r+i p r + 2 ■■■p s is an integer that is the product o f 5 — r prim es. In either case, p r  qrq,+\ q t■ T herefore, by C orollary 11.15, p r \ q, for som e j w ith r < j < t. B ecause q j is prim e, p, = q j. Since qr < q j, it follow s that q, < p r . By considering the integer qr (instead o f p r), w e can show that p, < qr . T herefore, p r = qr . B ut this contradicts the fact that p r ^ qr . H ence, as claim ed, every integer n > 2 has a unique factorization. ■ A n im m ediate consequence o f T heorem 11.17 is stated next. Corollary 11.18
Every integer exceeding 1 has a prim e factor. In fact, we can say a bit more.
Lem ma 11.19 Proof
I f n is a composite number, then n has a p rim e fa c to r p such that p < *Jn. Since n is com posite, we know that n = ab, w here 1 < a < n and 1 < b < n. A ssum e, w ithout loss of generality, that a < b. Then a 2 < ab = n and hence a < ~Jn. Since a > 1, we know that a has a prim e factor, say p. B ecause a is a factor o f n, it follow s that p is a factor o f n as w ell and p < a < *Jn. m If an integer n > 2 is expressed as a product q \ q 2 ■■■qm of prim es, then the prim es q \ , q 2 , ■■ ■, qm need not be distinct. Consequently, w e can group equal prim e factors and express n in the form n = P ? P z ■■■ Pkk’ w here p \ , p i , . . . , pk are prim es such that p \ < p 2 < ■■■ < Pk and each exponent a, is a positive integer. We call this the ca n o n ic al fa c to riz a tio n o f n. From the Fundam ental T heorem o f A rithm etic, every integer n > 2 has a unique canonical factorization. For exam ple, the canonical factorizations o f 12, 210 and 1000 are 12 = 223, 210 = 2 ■3 • 5 • 7 and 1000 = 2353. O f course, it is relatively easy to determ ine w hether a small
11.6
Tlie Fundam ental Theorem of Arithmetic
279
positive integer is prim e or com posite and. if it is com posite, to express it as a product o f prim es. We m ention som e tests for divisibility by certain integers. T hese m ay already be know n to you. 1. D ivisibility by 2 ,4 a n d o th e r p o w ers o f 2: An integer n is divisible by 2 if and only if n is even (or the last digit o f n is even). In fact, n is divisible by 4 if and only if the tw odigit num ber consisting o f the last tw o digits o f n is divisible by 4, the integer n is divisible by 8 if and only if the threedigit num ber consisting o f the last three digits o f n is divisible by 8 and so on. Therefore, the num ber 14220 is divisible by 4 since 20 is divisible by 4 but it is not divisible by 8 since 220 is not divisible by 8. 2. D ivisibility by 3 a n d 9: A n integer is divisible by 3 if and only if the sum o f its digits is divisible by 3. Indeed, an integer is divisible by 9 if and only if the sum of its digits is divisible by 9. This procedure stops w ith 9, however; that is, it d o esn ’t extend to 27. For exam ple, the sum of the digits o f 27 itself is not divisible by 27 but certainly 27 is divisible by 27. T he sum o f the digits o f the integer 4278 is 21, w hich is divisible by 3 but not by 9. Consequently, 4278 is divisible by 3 but not by 9. Clearly, 4278 is divisible by 2 but it is not divisible by 4 since 78 is not divisible by 4. Thus, 4278 is divisible by 6 by T heorem 11.16. 3. D ivisibility by 5: An integer is divisible by 5 if and only if it ends in 5 or 0; that is, an integer is divisible by 5 if and only if its last digit is divisible by 5. 4. D ivisibility by 11: Start w ith the first digit o f n and sum alternate digits (every other digit). Suppose that the resulting num ber is a. T hen sum the rem aining digits, obtaining b. T hen n is divisible by 11 if and only if a — b is divisible by 11. For exam ple, consider the num ber 71929. O bserve that a = 7 + 9 + 9 = 25, w hile b — 1 + 2 = 3. Since a — b — 25 — 3 = 22 is divisible by 11, the num ber 71929 is divisible by 11. In fact, 71929 = 11 6539. However, since (6 + 3) — (5 + 9) = —5 is not divisible by 11, the integer 6539 is not divisible by 1 1; that is, 71929 is not divisible by l l 2 = 121. A lthough there are tests for divisibility by other prim es such as 7 (see E xercise 11.66) and 13, none o f these are sufficiently practical to m erit inclusion here. If w e apply the tests listed above to the num ber n — 4 71240, then w e find that n is divisible by 5 ,8 (but not 16), 9 and 11. Indeed, n = 5 ■8 • 9 • 11 ■119 = 5 • 8 • 9 • 11 • 7 • 17 = 23 • 32 • 5 • 7 • 11 • 17. We are now in a position to describe an infinite class o f irrational num bers. Theorem 11.20 Proof
Let n be a positive integer. Then
is a rational num ber i f a n d only i f \ / n is an integer.
Certainly, if sfn is an integer, then is rational. H ence w e need only verify the converse. A ssum e, to the contrary, that there exists som e positive integer n such that *fn is a rational num ber but J h is not an integer. H ence J n = a / b for som e positive integers a and b. Furtherm ore, w e m ay assum e that a and b have no com m on factors, that is, gcd(a, b) — 1. Since a / b is not an integer, b > 2 . T herefore, n = a 2 / b 2 and so a 2 — n b 2. By C orollary 11.18, b has a prim e factor p. Thus p \ n b 2 and so p \ a 1 . By C orollary 11.14, p  a. B ut then p \ a and p  b, w hich contradicts our assum ption that gcd(a,b)=l. ■
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Corollary 11.21 Proof
I f p is a prime, then v[ p is irrational. A ssum e, to the contrary, that there exists som e prim e p for w hich y p is rational. By T heorem 11.20, ^ p = n for som e integer n > 2 .T h e n p = n 2. B ecause n 2 is com posite, this is a contradiction.
*
A lthough our rem arks im ply that there are infinitely m any prim es, w e have not yet proved this. We do now. Since our goal is to prove that the num ber o f prim es is not finite, a proof by contradiction is the expected technique. Theorem 11.22
The num ber o f prim es is infinite.
Proof A ssum e, to the contrary, that the num ber o f prim es is finite. Let P = [ p i, p 2,  P the set of all prim es. C onsider the integer m — p \ p 2 ■■• P n + !• Clearly, m > 2. Since m has a prim e factor and every prim e belongs to P , there is a prim e />, (1 < i < n) such that pi  m. H ence m = p , k for som e integer k. L et i = p \ p 2 ■■■P i  i P , + i ■• • Pn T hen I = in  p i p 2  ■■P„ = Pik  p i t = Pi(k ~ I). Since k  I is an integer, p t ] 1, w hich is im possible.
■
Two prim es p and q, w here p < q , are called tw in p rim e s if q = p + 2. N ecessarily, tw in prim es are odd. For exam ple, 5 ,7 and 11,13 are tw in prim es. A lthough we have ju st verified that there are infinitely m any prim es, the num ber o f tw in prim es is not known. Conjecture 11.23
There are infinitely many twin primes.
11.7 C oncepts Involving Sum s of D ivisors F or an integer n > 2, a positive integer a is called a p r o p e r d iv iso r of n if a n and a < n. Thus the proper divisors o f 6 are 1 ,2 and 3, w hile the proper divisors o f 28 are 1, 2, 4, 7 and 14. N ote also that 1 + 2 + 3 = 6 and 1 + 2 + 4 + 7 + 14 = 28. A positive integer n> 2 is called p e rfe c t if the sum o f its proper divisors is n. H ence 6 and 28 are perfect integers— indeed, they are the tw o sm allest perfect integers. T he third sm allest perfect integer is 496. T he largest prim e divisors o f 6, 28 and 496 are 3, 7 and 31, respectively. Sum m ing the integers from 1 to each o f these prim es yields possibly unexpected results: 1+ 2 + 3 = 6 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28 1 + 2 + ■•• + 31 = 496. The integers 6, 28 and 496 can also be expressed as 6 = 2 1(22  1), 28 = 22(23  1) and 496 = 24(25  1).
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281
In fact, Euclid, the fam ous geom eter w ho lived m ore than 2000 years ago, show ed that w henever 2P — 1 is a prim e, then 2p~ l(2p — 1) is a perfect integer. D uring the 18th century, the brilliant Swiss m athem atician L eonhard E uler proved that every even perfect integer is o f the form 2P~ ](2P — 1) w here 2P — 1 is a prim e. Prim e num bers o f the form 2P — 1 are referred to as Mersenne primes. As o f June 2010, 47 M ersenne prim es were know n and therefore 47 even perfect integers w ere known. M uch m ystery surrounds perfect num bers. A re there any odd perfect num bers? No one know s. A re there infinitely m any even perfect num bers? No one know s that either. L e t’s consider the first few prim es— the first seven to be exact: 2, 3, 5, 7. 11, 13, 17. The 1st, 2nd, 4th and 7th prim es are 2, 3, 7 and 17, and the rem aining prim es (the 3rd, 5th and 6th) are 5, 11 and 13. If we add the integers 1, 2 ,4 and 7, w e get the same result as w hen w e add 3, 5 and 6, nam ely, 1 + 2 + 4 + 7 = 3 + 5 + 6 = 14. W hile this fact m ay not appear to be anything special, it is also a fact that sum m ing the prim es that correspond to these tw o sets of integers also gives the sam e result: 2 + 3 + 7 + 17 = 5 + 11 + 13 = 29. W hile the sum s o f these prim es are equal, it is im possible for the products o f these prim es, nam ely 2 ■3 • 7 ■17 and 5 ■11 • 13, to be equal. T his is a consequence, o f course, o f the Fundam ental T heorem o f A rithm etic. On the other hand, these products are surprisingly close since 2  3  7 • 17 = 714 and 5 • 11 • 13 = 715. The serious baseball fan w ill recognize these num bers. For years, the num ber 714 stood as the record for hom e runs in a career. This record was held by Babe Ruth. However, this record was broken in 1974 w hen H ank A aron hit his 7 15th hom e run. Two consecutive integers n , n + 1 are called R u th A a r o n p a irs o f integers if the sums of their prim e divisors are equal. Thus 714 and 715 are a R u th A aro n pair, as are 5 and 6. A lthough such pairs o f integers may appear to be rare, the fam ous H ungarian m athem atician Paul Erdos proved that there are, in fact, infinitely m any R u th A aro n pairs o f integers.
EXERCISES FOR CHAPTER 11 Section 11.1: Divisibility Properties of Integers 11.1. Let a,
b , c , d e Z with a, c ^ 0. Prove that if a \ b
and c 
d, then ac
11.2. Let a ,
b e Z with a ^ 0. Prove that if a  b. then a
 (—b )
and (—a) [ b.
11.3. Let a,
b, c e Z with a , c ^ 0. Prove that if ac \be,
then a
 b.
11.4. Prove
that 3  («3 — n) for every integer n.
11.5. Prove that if n = k 3 + 1 > 3, where k e Z, then n is not prime. [Hint: Recall that k 3 + 1 = (k + 1)(k 2 — k + 1).] 11.6. Find all primes that are 1 less than a perfect cube. 11.7. Prove that 8  (52" + 7) for every positive integer n.

(ad + be).
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11.8. Prove that 5  (33"+1 + 2"+1) for every positive integer n. 11.9. Prove that for every positive integer n, there exist n consecutive positive integers, each of which is composite. [Hint: Consider the numbers 2 + (n + 1)!, 3 + (n + 1) ! , . . . , / ? + (« + 1)1, n + 1 + (n + 1)!.] 11.10. (a) Prove that 6  ( 5 0 + I n ) for every positive integer n. (b) Observe that 5 + 7 = 12 is a multiple of 6. State and prove a generalization of the problem in (a). 11.11. Prove the following: Let d be a nonzero integer, . a ? , ___o,; and . . ., x„ are 2n > 2 integers such that d  a, for all i ( I < i < n), then d \ Xw=i <*/■*;• 11.12. Let p n denote the n\h prime and cn the «th composite number. Thus p\ = 2 and p 2 = 3, while ci = 4 and Ci = 6. Of course, p„ / c„ for all n e N. Determine all positive integers n such that \pn — cn \ = 1. 11.13. (a) Suppose that there are k distinct positive integers that divide an odd positive integer n . How many distinct positive integers divide 2n? How many divide 4/2? (b) Suppose that there are k distinct positive integers that divide a positive integer n that is not divisible by 3. How many distinct positive integers divide 3/z? How many divide 9«? (c) State and answer a question suggested by the questions in (a) and (b). 11.14. For an integer n > 2, let m be the largest positive integer less than n such that m \ n. Then n = m k for some positive integer k. Prove that k is a prime.
Section 11.2: The Division Algorithm 11.15. Illustrate the Division Algorithm for: (a) (c) (e) (g)
a a a a
= = = =
17, b = 125 8. b = 96 22, b =  1 7 15, b = 0
(b) (d) (f) (h)
a= a= u a=
 1 7 , b = 125 —8. b = 96 22. h 17  1 5 , b = 0.
11.16. Give an example of a prime p of each of the forms: (a) 4k + 1 (b) 4/. + 3 (c) 6 k + 1 (d) 6 k + 5. 11.17. Let p be an odd prime. Prove each of the following. (a) p is of the form Ak + 1 or of the form Ak + 3 for some nonnegative integer k. (b) p > 5 is of the form 6 k + 1 or of the form 6 k + 5 for some nonnegative integer k. 11.18.
Show that, except for 2 and 5, every prime can be expressed as 10k + 1, 10/: + 3, 10k + 7or 10/: where k e Z.
+ 9,
11.19. (a) Prove that if an integer n has the form 6^ + 5 for some q e Z, then n has the form 3k + 2 for some k e Z. (b) Is the converse of (a) true? 11.20. Prove the General Form of the Division Algorithm (Corollary 11.5): For integers a and b with a / 0. there exist unique integers q and r such that b = aq + r and 0 < r < a. 11.21.
Prove that the square of every odd integer is of the form 4k + 1, where k e Z (that is, for each odd integer a e Z, there exists k e Z such that a = Ak + 1).
11.22. (a) Prove that the square of every integer that is not a multiple of 3 is of the form 31 + 1, where k e Z. (b) Prove that the square of no integer is of the form 3m — 1, where m e Z. 11.23. Complete the following statement in a best possible manner and give a proof. (See Exercise 11.22(a).) The square of an integer that is not a multiple of 5 is either of the fo rm ______ o r _______ .
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283
11.24. (a) Prove that for every integer m, one of the integers m , m + 4, m + 8, m + 12, m + 16 is a multiple of 5. (b) State and prove a generalization of the result in (a). 11.25. Prove that fflai, . . . , % are n > 2 integers such that a, = 1 (mod 3) for every integer i (1 < i < n), then Vt\a% ■■■a„ = 1 (mod 3). 11.26. Let a, b and c be integers. Prove that if abc = 1 (mod 3), then an odd number of a, b and c are congruent to 1 modulo 3. 11.27. Prove or disprove: If a and b are odd integers, then 4  (a  b) or 4  (a + b). 11.28. Prove for every positive integer n that rr + 1 is not a multiple of 6. 11.29. It is known that there are infinitely many positive integers whose square is the sum of the squares of two positive integers. For example, 5 = 32 + 4~ and 132 = 5~ + 12". (a) Prove that there are infinitely many positive integers whose square is the sum of the squares of three positive integers. For example, 592 = 502 + 302 + 92. (b) Prove that there are infinitely many positive integers whose square is the sum of the squares of four positive integers. 11.30. (a) Let n e N. Show that for every set S of n distinct integers, there is a nonempty subset T of S such that n divides the sum of the elements of T . [Hint: Let S = {fli, « 2 >■• ■. an] and consider the subsets .S':  ! < / ,.« ; ___ _ ak} for each k (1 < Jc < «),] (b) Is the word "distinct" necessary in (a)? 11.31. For an integer n > 2. let S„ be the set of all positive integers m for which n is the smallest positive integer such that when m is divided by n, the remainder 1 results. (a) lb) (c) (d)
W hat does S2 consist of? To which set Sn does 14 belong? To which set S„ does 16 belong? Prove or disprove: For each integer n > 2, the set Sn either contains infinitely many elements or is empty.
Section 11.3: Greatest Common Divisors 11.32. Give an example of a set S of four (distinct) positive integers such that the greatest common divisor of all six pairs of elements of S is 6. 11.33. Give an example of a set S of four (distinct) positive integers such that the greatest common divisors of all six pairs of elements of S are six distinct positive integers. 11.34. Prove for a e Z and n e N that gcd(a, a + n )  n. 11.35. Let a and b be two integers, not both 0, where gcd(a, b) = d. Prove for a positive integer k that gcd(A:a, kb) = kd. 11.36. For positive integers a, b and c, the greatest common divisor gcd(a, b, c) o f a , b and c is the largest positive integer that divides all of a, b and c. Let d — gcd(a, ft, c), e = gcd(«, b) and / = gcd(<?, c). Prove that d = f .
Section 11.4: The Euclidean Algorithm 11.37. Use the Euclidean Algorithm to find the greatest common divisor for each of the following pairs of integers: (a) 51 and 288 (b) 357 and 629 (c) 180 and 252. 11.38. Determine integers x and y such that (see Exercise 11.37): (a) gcd(51, 288) = S ix + 288y (b) gcd(357. 629) = 357x + 629y (c) gcd(180, 252) = 18Ox + 252y.
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11.39. Let a and b be integers, not both 0. Show that there are infinitely many pairs s, t of integers such that gcd(a, b) = as + bt. 11.40. Let a, b e Z, where not both a and b are 0, and let cl = gcd(a, b). Show that an integer n is a linear combination of a and b if and only if d \ n. 11.41. An integer n > 1 has the properties that n  (35 m + 26) and n \ (7m + 3) for some integer m. What is n l 11.42. Let a, b € Z, where not both a and b are 0. Prove that if cl = gcd(a. b), a = a \d and b = bid, then gcd(fli, hi) = 1. 11.43.
Prove the following: Let a, b, c, m , n e Z, where m, n > 2. If a = b (mod m ) and a cl = gcd(m, n), then b = c (mod d).
= c (mod n)
where
11.44.
In Exercise 11.36, it was shown for positive integers a, b and c that gcd(a, b, c)= gcd(gcd(a, b), c). Show that there are integers x, y and z such that gcd(a, b , c) = ax + by + cz.
11.45. Suppose that the Euclidean Algorithm is being applied to determine gcd(a, b) for two positive integers a and b. If, at some stage of the algorithm, we arrive at a remainder r, that is a prime number, then what conclusion can be made about gcd(a, £>)?
Section 11.5: Relatively Prime Integers 11.46. (a) Let a , b , c e Z such that a / 0 and a \ be. Show that if gcd(a. b) / 1, then a need not divide c. (b) Let a, b, c € Z such that a, b ^ 0, a  c and b \ c. Show that if gcd(a, b) ^ 1, then ab need not divide c. 11.47. Use Corollary 11.14 to prove that V3 is irrational. 11.48. Prove that if p and q are distinct primes, then ^ fp q is irrational. 11.49. Let p be a prime and let n e Z, where n > 2. Prove that p x!n is irrational. 11.50. Let n e N. Prove or disprove each of the following: (a) 2n and 4/j + 3 are relatively prime. (b) 2« + 1 and 3n + 2 are relatively prime. 11.51. (a) Prove that every two consecutive odd positive integers are relatively prime. (b) State and prove a generalization of the result in (a). 11.52. Prove that if p > 2 is an integer with the property that for every pair b . c of integers p  be implies that p  b or p  c, then p is prime. (This result is related to Corollary 11.14.) 11.53. Prove that if p and q are primes with p > q > 5, then 24  ( p 2 — q 2). 11.54. A triple (a, b, c) of positive integers such that a 2 + b 2 = c2 is called a Pythagorean triple. A Pythagorean triple (a, b, c) is called primitive if gcd(a, b) = 1. (In this case, it also happens that gcd(a, e) = ged (b , c ) = l.) (a) Prove that if (a, b. c) is a Pythagorean triple, then (an. bn, cn) is a Pythagorean triple for every n e N. (b) In Exercise 13 of Chapter 4, it was shown that if (a, b. c) is a Pythagorean triple, then 3  ab. Use this fact and Theorem 11.16 to show that 12  ab. (c) Prove that if (a, b, c) is a primitive Pythagorean triple, then a and b are of opposite parity. 11.55. Prove the following: Let a, b, m, n e Z, where m, n > 2. If a = b (mod m) and a = b (mod n), where gcd(m, n ) = 1, then a = b (mod mn). 11.56. Prove the following: Let a, b , c, n e Z, where n > 2. If ac = be (mod n ) and gcd(c, n) = 1, then a = b (mod n). 11.57. For two integers a and b, not both 0, suppose that d = gcd(«, b). Then there exist integers x and y such that d = ax + by; that is, d is a linear combination of a and b. This implies that d is also a linear combination of x and y. Find a necessary and sufficient condition that d = gcd(.t, y).
Exercises for Chapter 11
285
11.58. Suppose that the Euclidean Algorithm is being applied to determine gcd(a, b) for two positive integers a and b. If, at some stage of the algorithm, we arrive at remainders r, and ri+\ for which ri+1 = rt  1, then what conclusion can be made about gcd(a, b)l 11.59. (a) Let a and b be integers different from 0 where d = gcd(a, b) and let c £ Z. Prove that if a \ c and b  c, then ab \ cd. (b) Show that Theorem 11.16 follows as a corollary to the result in (a). 11.60. Prove that there are infinitely many positive integers n such that each of n, n + 1 and n + 2 can be expressed as the sum of the squares of two nonnegative integers. (For example, observe that 8 = 22 + 22, 9 = 32 + 02, 10 = 32 + l 2 and 80 = 82 + 42, 81 = 92 + 02 and 82 = 92 + l 2 and that 3 = 2 + 1 and 9 = 8+1.) 11.61. (a) Give an example of integers m , n > 5 such that x f t y ^ 0 for each x e Z m and y e Z „. (b) State a conjecture that provides conditions under which integers m , n > 2 have the property that j f l v ^ 0 for each x G Z„, and y C Z „.
Section 11.6: The Fundam ental Theorem of Arithmetic 11.62. Find the smallest prime factor of each integer below: (a) 539 (b) 1575 (c) 529 (d) 1601 11.63. Find the canonical factorization of each of the following integers: (a) 4725 (b) 9702 (c) 180625 11.64. Prove each of the following: (a) Every prime of the form 3n + 1 is also of the form 6k + 1. (b) If n is a positive integer of the form 3k + 2, then n has a prime factor of this form as well. 11.65. (a) Express each of the integers 4278 and 71929 as a product of primes. (b) W hat is gcd(4278, 71929)? 11.66. Consider the periodic sequence 1 ,3 ,2 , 1 .,  3 ,  2 , 1. 3, 2,  1 ,  3 ,  2 , . . . which we write in reverse order: . . . .  2  3,  1 , 2 , 3, 1,  2  3,  1 , 2 , 3, 1. Next, consider the 8digit positive integer n = wvrfyrtw.itf «/?</•,</n where each a, is a digit. It turns out that 7  n if and only if 3 • a] + 1 ■<36 + (—2) ■
+ (—3) ■04 + (—1) • a} + 2 • a± + 3 • a\ + 1 • aq
is a multiple of 7. Use this to determine which of the following are multiples of 7: (a) 56 (b) 821,317 (c) 31,142,524. 11.67. In the proof of Theorem 11.22, it was proved that there are infinitely many primes by assuming that there were finitely many primes, say p \. p : ........ />«, where p \ < p 2 < ■■■ < Pn The number m = pt p.i ■■■p n + 1 was then considered to obtain a contradiction. Show that an alternative proof of Theorem 11.22 can be obtained by considering p n \ + 1 instead of m. 11.68. Determine a necessary and sufficient condition that p “lpif • • • p* is the canonical factorization of the square of some integer n > 2. 11.69. For two integers m, n > 2, let p \ , p % , . . . , p, be those distinct primes such that each p\ ( I < / < r ) divides at least one of m and n. Then m and n can be expressed as m = p ° l />"' • • • p°: and n — p h{ p f ■■■p p where the integers a, and bj (1 < i < r ) are nonnegative. Let Cj = m in(a,, hi) for 1 < i < r. Prove that gcd(m, n) = p\' p 2 ■■■Py’ ■
286
Chapter 11
Proofs in Number Theory
Section 11.7: Concepts Involving Sums of Divisors 11.70. Let k be a positive integer. (a) Prove that if 2 — 1 is prime, then k is prime. (b) Prove that if 2k — 1 is prime, then n = 2k~ l (2k — 1) is perfect. 11.71. For a real number r , the floor j / J of r is the greatest integer less than or equal to r . The greatest number of distinct positive integers whose sum is 5 is 2 (5 = 1 + 4 = 2 + 3), while the greatest number of distinct positive integers whose sum is 8 is 3 (8 = 1 + 2 + 5 = 1 + 3 + 4 ) . Prove that the maximum number of distinct positive integers whose sum is the positive integer n is [(V l + %n ~ 1 )/2 J.
ADDITIONAL EXERCISES FOR CHAPTER 11 11.72. Evaluate the proposed solution of the following problem. Prove or disprove the following statement: There do not exist three integers n, n + 2 and n + 4, all of which are primes.
Solution
This statement is true.
P roof Assume, to the contrary, that there exist three integers n, n + 2 and n + 4, all of which are primes. We can write n as 3q, 3q + 1 or 3q + 2, where q e Z. We consider these three cases. Case 1. n = 3q. Then 3  n and so n is not prime. This is a contradiction. Case 2. n. = 3q + 1. Then n + 2 = 3q + 3 = 3(<j + 1). Since q + 1 is an integer, 3 ,] (« + 2) and so n + 2 is not prime. Again, we have a contradiction. Case 3. n = 3q + 2. Hence we have n + 4 = 3q + 6 = 3(q + 2). Since q + 2 is an integer, 3  (n + 4) and so n + 4 is not prime. This produces a contradiction. ■ 11.73. An integer a > 2 is defined to be lucky if / ( « ) = n 2 — n + a is prime for every integer n with 1 < n < a — 1. It is known that (1) 41 is lucky and (2) only nine other integers a > 2 are lucky. (a) Prove that if a is a lucky integer, then a is prime. (b) Give an example of three other lucky integers. <c) If a is a lucky integer, what can be said about /( a ) ? 11.74. Prove that log2 3 is irrational. 11.75. State and prove a more general result than that given in Exercise 11.74. 11.76. Given below is an incomplete result with an incomplete proof. This result is intended to determine all twin primes (primes of the form p and q = p + 2) such that p q — 2 is also prime.
Result
Let p and q = p + 2 be two primes. Then pq — 2 is prime if and only if (complete this
sentence). P ro o f Let p and q = p + 2 be two primes such that p q — 2 is also prime. Since p and p + 2 are both primes, it follows that p is odd. By the Division Algorithm, we can write p = 3k + r, where k e Z and 0 < r < 2. Since p is an odd prime, k > 1. We consider three cases for p, depending on the value of r. Case 1. p = 3k. Therefore, p =
,q=
and pq — 2 =
Case 2. p = 3k + 1. Hence q = 3k + 3. Since k > 1, it follows that q — 3(k + 1). Therefore, Case 3. p = 3k + 2. Then q = 3k + 4. Thus
■
Additional Exercises for Chapter 11
287
11.77. Exercise 11.76 should suggest another exercise to you. State a result related to Exercise 11.76 and give a proof of this result. 11.78. Assume that each positive rational number is expressed as m / n , where m, n <=N and m and n are relatively prime. A function / : Q + —» N is defined by f ( m / n ) = 2m3n. (a) Prove that / is onetoone. (b) If you discussed the SchroderBem stein Theorem (Theorem 10.20) in Chapter 10, what information about Q + can you obtain from part (a)? 11.79. We have seen in Section 11.7 that there is a partition of the set of the first seven primes into two subsets such that the sums of the elements in these two subsets are equal. Show that there is no such partition of the set of the first eight primes but there is such a partition of the set of the first nine primes. 11.80. (a) Show that 5039 = 5040 — 1 is prime, while 5041 = 5040 + 1 is not prime. (b) Show that, except for 5039, there is no prime between 5033 = 5040 — 7 and 5047 = 5040 + 7. 11.81. Let <?o>« i . «2  • • • be a sequence of positive integers for which (1) ao = 1, (2) a2n+1 = a„ for n > 0 and (3) a 2 n + 2 = an + a,i+1 for n >
0.
Prove that a„ and an+i are relatively prime for every nonnegative integer n. 11.82. Every positive integer n can be expressed as n = n — a^ ' 10^
cik—i ’ 10*
 • • • a 2 a\ a o, that is, ('•• I a 2 • 102 I ci\ ' 10 I uq.
It was mentioned in Section 11.6 that 9  n if and only if 9  (a^ + a*_i + ■■■+ a 2 + a\ + ao). For example, for the integer n — 32, 751, 9 ( 3 + 2 + 7 + 5 + l) and so 9  32, 751. Verify this by using the fact that 10 = 9 + 1 and for a positive integer r that 10' = (9 + 1)' = 9s + 1 for some integer s. 11.83. Let A be the set of 2element subsets of N and B the set of 3element subsets of N. Let / : A > B and g : B —> A be functions defined by / ( { '. ;! ) = ft, j , i + j } and g({;\ j , k}) = {2'\ 3'5*}, where i < j < k. With the aid of the functions / and g and possibly the SchroderBem stein Theorem (Theorem 10.20 in Chapter 10), which of the following are we able to conclude? (a) A  < 5  (b) 5  < \A\ (c) A = 5  (d) Nothing. 11.84. Let pi, p 2, p i , .. ■be the primes, where 2 = p i < p 2 < p?, < ■■■. Let A be a denumerable set, where A — [a\, a2, a 3 , . . .}. For any integer n > 2, let A" denote the Cartesian product of n copies of A; that is, A" is the set of ordered ^tuples of elements of A. Define a function / : A'1 —>• N by f ( ( a il, a i2 , . . . , a in)) = p[ 1 p ^    p ni \ (a) Prove that / is injective. (b) Use (a) to show that A" and A are numerically equivalent. (c) For every two denumerable sets A and B and every two integers n, m > 2 , show that A” and B m are numerically equivalent. 11.85. Let pi, p 2, . . . , p n+\ denote the first n + 1 primes. Suppose that (t/, V] is a partition of the set S = [pi, p 2, . . . , p „}, where U = { q \ , q 2, . . . , <?j} and V = { n , r2, . . . , Prove that if M — q\ q 2 ■• • qs + r\v 2 ■■■r, < p 2+1, then M is a prime.
12 Proofs in Calculus our introduction to calculus m ost certainly included a study o f lim its— both lim its o f sequences (including infinite series) and limi ts o f functions (including continuity and differentiability). W hile w e learned m ethods for com puting lim its in these areas, the m ethods presented w ere m ost likely based on facts that were not carefully verified. In this chapter, som e o f the proofs o f fundam ental results from calculus w ill be presented. The proofs that occur in calculus are considerably different than any o f those w e have seen thus far. The functions encountered in calculus are realvalued functions defined on sets o f real num bers. That is, each function that we study in calculus is o f the type / : X —> R, w here X C R. In the study o f lim its, w e are often interested in such functions having the property that either (1) X — N and increasing values in the dom ain N result in functional values approaching som e real num ber L or (2) the function is defined for all real num bers near som e specified real num ber a and values approaching a result in functional values approaching some real num ber L. We begin w ith (1), w here X — N.
Y
12.1 L im its of Sequ en ces A se q u en c e (of real num bers) is a realvalued function defined on the set o f natural num bers; that is, a se q u en ce is a function / : N > R. If f ( n )  a„ for each n e N, then f — {(1, «  ). (2, </2j. (3, </.(),.. .}. Since only the num bers a \ , u?, d$,. . . are relevant in / , this sequence is often denoted by a\, <33, . . . or by {an}. The num bers a\ , <22, #3> etc. are called the te rm s o f the sequence {an}, w ith a\ being the first term , cto the second term, etc. Thus an is the nth term o f the sequence. H ence {   is the sequence 1, 1 /2 , 1 /3 , . . . ; is the sequence 1/3, 2 /5 , 3 / 7 ........In these tw o exam ples, the n th term 2n ~f 1 J o f a sequence is given and, from this, w e can easily find the first few term s and, in fact, any particular term. O n the other hand, finding the /7th term o f a sequence w hose first few term s are given can be challenging. For exam ple, the n th term o f the sequence
w hile
1
1
1
2’ 4’ 6’ 288
12.1
Limits of Sequences
289
is 1/ 2 m; the nth term o f the sequence 1 1 1 + 2’ 1+ 4’ 1+ 8’ is 1 + 1/2 " ; the «th term o f the sequence 3 ’ 5’
1 5 2
3
7
’ IT ’ 7 ’ 17’
is (n + l)/(3 n — 1); the «th term o f the sequence 1,
 1 , 1,  1 , 1,  1 , . . .
is (—l) n+1; w hile the nth term o f the sequence 1, 4, 9, 16, • • • is n 1. For the sequence
n
the larger the integer n, the closer \ / n is to 0; and for the
sequence I —  —  , the larger the integer n, the closer n / (2n + 1) is to 1/2 . On the other I 2n + 1 J hand, for the sequence {«2}, as the integer n becom e larger, n 2 becom es increasingly large and does not approach any real number. W hen w e discuss how close tw o num bers are to each other, we are actually consid ering the distance betw een them. We saw in C hapter 8 that the d ista n c e betw een tw o real num bers a and b is defined as a  b\. R ecall that the absolute value o f a real num ber x is I* I
x if x > 0 —x if x < 0.
H ence the distance betw een a = 3and£> = 5 i s 3 — 5 = 5 — 3 = 2; w hile the distance 1 1, betw een 0 and 1/ n , w here n e N, is 0  = 0 n n For a fixed positive real num ber r, the inequality x  < r is equivalent to th equalities —r < x < r. H ence x  < 3 is equivalent to —3 < x < 3, w hile x is equivalent to —4 < x — 2 < 4. A dding 2 throughout these inequalities, w e obtain —4 + 2 < ( x — 2) + 2 < 4 + 2 and so —2 < x < 6. We have seen in E xercise 4.30 and T heorem 4.17 in C hapter 4 that for real num bers x and y, \xy\ = \x\\y\
and
x + y \ < x  + \y\.
Both o f these properties are useful throughout calculus. We m entioned that for som e sequences {a„}, there is a real num ber L (or at least there appears to be a real num ber L ) such that the larger the integer n becom es, the closer an is to L . We have now arrived at an im portant and fundam ental idea in the study of sequences and are prepared to introduce a concept that describes this situation. A sequence {an} o f real num bers is said to converge to a real num ber L if the larger the integer n, the closer an is to L. Since the w ords larger and closer are vague and consequently are open to interpretation, w e need to m ake these w ords considerably m ore precise. W hat w e w ant to say then is that w e can m ake a„ as close to L as w e w ish (that is, w e can m ake a„ — L\ as sm all as we wish) provided that n is large enough. Let e (the G reek letter epsilon) denote how sm all w e w ant \a„ — L\ to be; that is, w e w ant a„ — L\ < € by choosing n large enough. This is equivalent to —e < an — L < e, that
290
Chapter 12
Proofs in Calculus
® (3, a3) L + € L L e
• (2, a2)
•
• (1, a \ )
1 2 Figure 12.1
iii1 1 11h3 N A sequence {a,,} that converges to L
is, L — e < an < L + e. H ence we require that an be a num ber in the open interval (L — e, L + e) w hen n is large enough. Now we need to know w hat w e m ean by “large enough.” W hat we m ean by this is that there is som e positive integer N such that if n is an integer greater than N , then an € {L — e, L + e). If such a positive integer N can be found for every positive num ber e, regardless o f how sm all e m ight be, then w e say that [an] converges to L. This is illustrated in Figure 12.1. Form ally then, a sequence {an} o f real num bers is said to converge to the real num ber L if for every real num ber e > 0, there exists a positive integer N such that if n is an integer w ith n > JV, then \an — L\ < e. As w e indicated, the num ber e is a m easure of how close the term s an are required to b e to the num ber L and N indicates a position in the sequence beyond w hich the required condition is satisfied. If a sequence {an} converges to L, then L is referred to as the lim it o f {a,,} and w e w rite lim an = L . If n~>oo
a sequence does not converge, it is said to diverge. Consequently, if a sequence {a,,} diverges, then there is no real num ber L such that lim a„ = L . n—>oo Before looking at a few exam ples, we introduce some useful notation. For a real n um ber x , recall that \x] denotes the sm allest integer greater than or equal to x. The integer [a'] is often called the ceiling o f x. Consequently, [8 /3 ] = 3, ~\/2
= 2 , \ —\ . 6 \ ~ —\
and "51 = 5. By the definition o f [ x ] , it follow s that if a: is an integer, then [x] = A'; w hile if x is not an integer, then "x" > x. In particular, if n is an integer such that n > \ x ] , then n > x. We now show how the definition o f convergent sequence is used to prove that a sequence converges to som e number. R esult to Prove
PROOFSTRATEGY
The sequence
— } converges to 0. n
H ere w e are required to show, for a given real num ber e > 0, that there is a positive 1 ' 1 1 1 > N , then   0 = — =  < . € . The inequality —< 6 is n n n n
12.1
Limits of Sequences
291
equivalent to n > 1 /e . H ence if w e let N = f l / e l and take n to be an integer greater than N . then n >  . We can now present a form al proof.
Result 12.1
♦
The sequence j  } converges to 0.
Proof
L et e > 0. Choose N — f 1/e~] and let n be any integer such that n > N . T hus n > 1/e 1 1 and so  0 = — < 6. n
PROOF ANALYSIS
A lthough the proof o f R esult 12.1 is quite short, the real w ork in constructing the proof occurred in the pro o f strategy (our “scratch paper” w ork) that preceded the proof, but w hich is not part o f the proof. This explains w hy w e chose N as w e did and why this choice o f iV was successful. In the proof o f Result 12.1, w e chose N = f 1/e~ and show ed that w ith this value o f N , every integer n w ith n > N yields — — 0 < e, w hich, o f course, was our goal. T here is nothing unique about this choice o f N , however. Indeed, w e could have chosen N to be any integer greater than [ 1 /e" or, equivalently, any integer greater than 1/ e and reached the desired conclusion as well. We could not, however, choose N to be an integer sm aller than We cannot in general choose N = 1 /e since there is no guarantee that N is an integer. ♦ We now consider another illustration o f a convergent sequence.
R e su lt to P rove
PROOFSTRATEGY
[ 2 I The sequence { 3 H— r [ converges to 3.
H ere w e are required to show, for a given e > 0, that there exists a positive integer N such that if « > TV, then 2
„
( 3+ ^ )  3
12 r«
\
ii
(,
=  a c e. nz
2 1 T he inequality —z < e is equivalent to — >  and n > J 2 l e . T herefore, if w e let n2 e
N —
s j 2 /e
and choose n to be an integer greater than N , then n > ^ / 2 / e . We can
now give a proof.
♦
Result 12.2
Proof
L et e > 0. C hoose N =
■y/2/e
and let n be any integer such that n > N . Thus
n > ^ 2 / e and n2 > 2 /e . So — <  and — < e. Therefore,
292
Chapter 12
Proofs in Calculus We now consider a som ew hat m ore com plicated example.
R esult to Prove
PROOFSTRATEGY
n 1 1 The sequence \  > converges to  . 2// *11 2 O bserve that n 2
The inequality
n + 1
1
2 n — 2/7 — 1
1
1
2
2(2 n + 1)
An + 2
An + 2
1 < e is equivalent to An + 2 > 1/e , w hich, in turn, is equivalent
An + 2
1 1 " 1 1 to n > — —  . It m ay appear that the proper choice for N is ; b u t i f e > 1/2, 4e _ 2 then N = 0, w hich is not acceptable since N is required to be a positive integer. However, notice that — >  So if n > — , then n > as well. H ence if w e choose Ae Ae 2 Ae 4e 2 N = [1 / 4 e ] , then w e can obtain the desired inequality. ♦
Result 12.3
The sequence 2
Proof
converges to
n + 1
1
L et e > 0 be given. C hoose N = " l/4 e l and let n > N . T hen n > — > — and Ae Ae 2 so An > 2 and An + 2 > 1 /e . H e n c e  < e. Thus e An + 2 n 2«
+
l
1
2/7 — 2/7 — 1
2
2(2 n + 1)
1 4« +
1 2
An +
< e. 2
Again, the choice m ade for N in the proof o f Result 12.3 is not unique. We could choose N to be any positive integer greater than — . We m entioned that a sequence {a„} is said to diverge if it does not converge. To prove that a sequence {a„} diverges, a proof by contradiction would be anticipated. We w ould begin such a pro o f by assum ing, to the contrary, that {a„} converges, say to some real num ber L . We know that for every e > 0, there is a positive integer N such that if n > N , then  a„ — L \ < e. If we could show for even one choice o f e > 0 that no such positive integer N exists, then we w ould have produced a contradiction and proved the desired result. L et’s see how this works in two examples. R esult to Prove PROOFSTRATEGY
The sequence {(—l) 'l+1} is divergent. In a proo f by contradiction, we begin by assum ing that { (  1)"+1} converges, to the lim it L say. O ur goal is to show that there is some value o f e > 0 for w hich there is no positive integer N that satisfies the requirem ent. We choose e = 1. A ccording to the definition o f w hat it m eans for { (  1 ) " +1} to converge to L, there m ust exist a positive integer N such that if n is an integer w ith n > N , then ( —1)',+1 — L J < e = 1. Let k be an odd integer such that k > N . Then (  1 ) * +1  L \  1  L \ = \ L  l \ < 1.
12.1
Limits of Sequences
293
T herefore, —1 < L — 1 < 1 and 0 < L < 2. N ow let I be an even integer such that I > N . Then (  1 ) <+I — L\ = \  l  L \ = \L + 1 < 1. Thus, —1 < L + 1 < 1 and —2 < L < 0. So L < 0 < L, w hich, of course, is im possi ble. We now repeat w hat we have ju st said in a form al proof. ♦ Result 12.4 Proof
The sequence {(—1)'!+1} is divergent. A ssum e, to the contrary, that the sequence {(—1)"+ I } converges. T hen lim (—1)'!+1 = 1
'
n—>oo
L for som e real num ber L . L et e = 1. T hen there exists a positive integer N such that if n > N , then ( —l) '!+l — L  < e = 1. L et k be an odd integer such that k > N. Then
 ( _ 1 / + 1 _ L  =  i — L\ = \ L  1 < 1. Therefore, —1 < L — 1 < 1 and 0 < L < 2. Next, let I be an even integer such that I > N . Then  (  l / +1 — L  =  —1 — L  = \L + 11 = 1 + L \ < 1. So —1 < L + 1 < l a n d —2 < L < 0. Therefore, L < 0 < L , w hich is a contradiction. ■ PROOF ANALYSIS
O ne question that now occurs is how w e knew to choose e = 1. If e denotes an arbitrary positive integer, then both inequalities \L — 1 < e and \L + 1 < e m ust be satisfied, but these result in the inequalities 1 —e < L < l + e
and —1 — e < L < — 1 + e.
In particular, 1 — e < L < — 1 + e and so 1 — e < —1 + e. This is only possible if 2e > 2 or e > 1. H ence if we choose e to be any num ber such that 0 < e < 1, a contra diction w ill be produced. We decided to choose e = 1. ♦ R e su lt to P rove PROOFSTRATEGY
T he sequence {(—l ) ”+1^ j } is divergent. A s expected, w e w ill attem pt a proof by contradiction and assum e that {(—1)"+1 ^ } is a convergent sequence, w ith lim it L say. For e > 0, there is a positive integer N then such that (  1)
n + 1
< e
for each integer n such that n > N . There are som e useful observations. First, if n > N and n is odd, then
n + 1
—L
< € and so —e < —L < e. n + 1
H ence L — e <  < L + e. n + 1
294
Chapter 12
Proofs in Calculus Second, if n > N and n is even, then n n + 1
 L
n < e and so —e <  L < e. n + 1
H ence e <
— < L + e. n + 1
n 1 A lso, since n > 1, w e have / ? + / ; > / / + 1 and so 2« > « + 1. H e n c e  > ii + 1 2 D epending on w hether L = 0, L > 0 or L < 0, w e are faced w ith the decision as to how to choose € in each case to produce a contradiction. + Result 12.5
Proof
The sequence {(—1)"+ I ^ r j} is divergent.
A ssum e, to the contrary, that { (  l ) '!+14 r} converges. Then lim (  1 ) " +1 — — = L n + 1 for som e real num ber L. We consider three cases, depending on w hether L — 0, L > 0 or L < 0 . 1 Case 1. L = 0. L et le =  . T hen there exists a positive integer N such that if n > N , n then (_ 1 )» + 1_ o n + 1 a contradiction.
1 n 1 <  o r   <  . Then 2n < n 4 1 and so n < 1, w hich is 2 n + 1 2
L Case 2. L > 0. L et e = —. T hen there exists a positive integer N such that if n > N , then n 4 1
L < — . L et n be an even integer such that n > N . Then L n L   <  L < — . n 4 1 2 2
L n 3L H ence — <   < — , w hich is a contradiction. 2 n+ 1 2 L Case 3. L < 0 . L et e = ——. T hen there exists a positive integer N such that if n > N , then
L < ——■Let n be an odd integer such that n > N . Then L n L — <    L < 2 « + l 2
3L n L and so —— <  < —. This is a contradiction. 2 n + 1 2
■
A sequence {an} m ay diverge because as n becom es larger, an becom es larger and eventually exceeds any given real number. If a sequence has this property, then {an} is said to diverge to infinity. M ore formally, a sequence {an } d iv erg es to infinity, w ritten lirn^ an = oo, if for every positive num ber M , there exists a positive integer N such that if n is an integer such that n > N , then a„ > M . The sequence { (  1 ) ',+1} encountered
12.2
Infinite Series
295
in Result 12.4, although divergent, does not diverge to infinity. However, the sequence I n 2 + } does diverge to infinity.
R e su lt to P ro v e PKOOF STRATEGY
lim ( n 2 H— n—>oo \
) = oo.
Yl J
For a given positive num ber M , w e are required to show the existence o f a positive integer N such that if n > A M h en n 2 +  > M . N otice that if n 2 > M .th e n n 2 +  > i r > M . n n Since M > 0, it follow s that n 2 > M is equivalent to n > s / M . A form al p ro o f can now be constructed.
Result 12.6
Proof
^
lim ( r 2 +  ) = oo. nroo y nJ Let M be a positive number. Choose N — j~s/~M~j and let n be any integer such that n > N . H ence n > \ / M and so n 2 > M . Thus n 2 1—
> n2 > M.
®
12.2 Infinite Series An im portant concept in calculus involving sequences is infinite series. F or real num bers OO
a 1 5ci2 ,
, w e w rite ^ ^ &k = ci\ I #2 h #3 H” ** * to denote an in fin ite scries (often k= 1
sim ply called a series). F or exam ple,
^
1
1 1
I t P
= 1 + 23 + ?
^ + "
aM
k
1 2
3 ,
g s ? T T = 3 + 9 + 19
are infinite series.
x
T he num bers a\, a%t a3 ___ are called the te rm s o f the series
= «i + 0 2+ k= 1
(7; i • • .T h e notation certainly seem s to suggest that w e are adding the term s a\, < 1 2 ,(13, B ut w hat does it m ean to add infinitely m any num bers? A m eaning m ust be given to this. For this reason, w e construct a sequence {sn}, called the seq u en ce o f p a r tia l su m s of the series. H ere si = ci\, si = Qi + aii s 3 — a i + a 2 + and, in general, for n € N, n S„ = a
1
+ (32 + ■• ■ + a n —
ak■ k= 1
B ecause sn is determ ined by adding a finite num ber o f term s, there is no confusion in understanding the term s o f the sequence {s,,}. If the sequence {sn} converges, say to the OO
°°'
num ber L , then the series V ' ak is said to co n v erg e to L and w e w rite ^ k= 1
ak = L . This
k= \ OO
num ber L is called the s um o f ^ k= 1
OO
ak. If {s„} diverges, then ^ P i
ak is said to diverge.
296
Chapter 12
Proofs in Calculus
The French m athem atician A ugustinL ouis C auchy was one o f the m ost productive m athem aticians o f the 19th century. A m ong his m any accom plishm ents was his definition o f convergence o f infinite series, a definition w hich is still used today. In his w ork C o w s d ’Analyse, C auchy considered the sequence {s,,} o f partial sums o f a series. H e stated the follow ing: If, f o r increasing values o f n, the sum s„ approaches indefinitely a certain limit s, the series will be called convergent and this limit in question will be called the sum o f the series. We consider an exam ple o f a convergent series.
R e su lt to P rove
PROOFSTRATEGY
oo y The infinite series ^ converges to 1. i= i + 1) First, we consider the sequence {sn} o f partial sum s for this series. Since
V
1
_ J
i
i
k ( k + 1) “ T ? + 2 ^ 3 + 3 ^ 4 + " ' ’ 1 1 1 1 1 1 2 it lollow s that si = =  ,# * =  1 = — (— —  and 1 2 2’ 1  2 2  3 2 6 3
 J_ 53 “
_L
1 _ 1
1
1 _ 3
l  2 + 2 ^ 3 + 3 ^ 4 “ 2 + 6 + T2 = 4' H
B ased on these three term s, it appears that s„ =  for every positive integer n. We n + 1 prove that this is indeed the case. $ Lem ma 12.7
For every positive integer n , 1 1 1 sn —  T + T—~ + 1 2 2 3 3 4
P ro o f of L e m m a 12.7
1
n n(n + 1)
n +1
1 l We proceed by induction. For n — 1, w e have ,Tj =  =  and the result holds.
1
1
1
1
'
X
A ssum e that Sjt = —  +  — +  —  + • ■■+ —  =  , w here k, is a positive 1 2 2 3 3 4 F k ( k + 1) k + \ integer. We show that 1 , 1 , 1 1 A + 1 S£M —  — j—  • —  1 2 2 3 3 4 (k + 1 )(A + 2) k + 2' O bserve that
12.2
By the Principle o f M athem atical Induction, sn —
Infinite Series
n + 1
297
for every positive
integer n. There is another w ay that w e m ight have been able to see that sn
n + 1
If we
had observed that an —
n(n + 1)
n
n + 1’
1 1 1 1 1 then a\ = = 1 , a? =  =  , etc. In particular. 1 2 2 2 3 2 3 sn = a\ + ai +
+ • •  + a„
1 1 + h r  r 2 3
1 x = 1
«3
1
n
n + 1
n + 1
1 1 x  T 3 4
+
In any case, since we now know that s„
n + 1
+
L e m m a to P rove PROOFSTRATECY
lim
n>oo
n—K X n +
n + 1
, it rem ains only to prove that = 1.
lim s„ = lim —
n+oo
n
1
= 1. — n + 1
For a given e > 0, we are required to find a positive integer N such that if n > N , then n < e. Now n + 1 
n + 1
n —n — 1
1
n + 1
n + 1
n + 1
The inequality  < e is equivalent to n + 1 > w hich in turn is equivalent to n + 1 € n >  1. I f n >  , then n >  1. We can now present a p ro o f o f this lem m a. e e € Lem ma 12.8
P ro o f of L e m m a 12.8
♦
n = 1. lim «>oo n + 1 Let € > 0 be given. Choose N = "l/e" and let n > N . T hen n >  >  1. So n >  — 1. T hus n + 1 >  and — — < e. H ence e e n + 1 n 
n + 1
1
1
1
n + 1
n + 1
We are now prepared to give a proof o f the result.
< e.
298
Chapter 12
Proofs in Calculus y
00
Result 12.9
The infinite series
— converges to 1. £=1 + 1) OO
Proof
The /?th term o f the sequence {s,,} o f partial sums o f the series
 — — is
^
1 1 1 1 sn — Z— Z + — —— + T r + • • • + 1 2 2 3 3 4 /?(/? + 1) By L em m a 12.7, 1
1
s n — Z +
1
I
^ ^
1 2
^ 7 +
2 3
3 4
••• +
n
■
n(n + 1)
n +1
n and so s„ = . By L em m a 12.8, n + 1
n lim  = 1. HSOO n + 1 OO Since lim s„ — 1, it follow s that )  = 1. n*oo ^ k (k + 1)
a
°°' j i j We now turn to a divergent series. The series ^ — — 1 4 i (  • • • i s fam ous k=i ^ 2 3 and is called the h a rm o n ic series. Indeed, it is probably the best know n divergent series. o°
Result 12.10
j
The harmonic series 7  diverges. * J kIr k= 1 K OO
Proof
A ssum e, to the contrary, that

 converges, say to the num ber L . For each positive 1=1 k
1 j  H ence the sequence {5,,} o f partial sums converges to L.
integer n, let sn =
k= 1 Therefore, for each e > 0, there exists a positive integer N such that if n > N , s„ — L  < e. L et’s consider € = 1 /4 and let n be an integer w ith n > N . Then —
1
4
Since 2 n > TV, it is also the case that I
1 < s„ — L <  .
4
— L\ <  and so — < so,, — L <  . O bserve 4 4 4
that Sin ~ Sn
"I
1
—
n + 1
"I
1
—   I       1 —
n + 2
1
2 >1
>
S„
+
/ 1 \
H
I —
V2 n )
I =
1 Sn +  .
2
H ence 1 r 1  > s2n  L > sn +   L — (i„ — w hich is im possible.
then
1
1
1
1
12.2
PROOF ANALYSIS
Infinite Series
299
In Result 12.10, w e show ed that a certain series diverges; that is, it does not converge. Consequently, it is not surprising that we proved this by contradiction. By assum ing that the sequence {5,,} converges, this m eant that the sequence has a lim it L . This tells us that an inequality o f the type \sn — L\ < e exists for every positive num ber e and for sufficiently large integers n (w hich depend on e). The goal, o f course, w as to obtain a contradiction. We did this by m aking a choice o f e (e = 1 /4 w orked!) that eventually produced a m athem atical im possibility. ♦ OO j The harm onic series 2 ^  not only diverges, it diverges to infinity; that is, if {s„} is
k= 1
^
the sequence o f partial sum s for the harm onic series, then lim s„ = 00. We also establish n—>oo
this fact. First, we verify a lem m a, w hich shows once again that m athem atical induction can be a useful proof technique in calculus.
Lemma 12.11
Proof
vM 1 1 Let sn — /  = 1 H (   1— , where n e N. Then fk=\ f k 2 n integer n .
We proceed by induction. For n = 1, S21 = 1 4 — A ssum e that s2* > 1 4
k
n , > 1 4 — f o r every positive 2
and so the result holds for n = 1,
2 k + 1 w here k e N. We show that i 2*+> > 1 4 — — • N ow observe
2
that
,
1
1
s 2*+> — 1 4“ — 4~ • • • 42 2k+l 1
1
1
 J2‘ + F T T + F T 2 + ' " + 2 w 1 1 1 > S2 t 4  , . 4~ 2 k+i 2 k+x 2 k+x
2k 1 s2‘ + j n T  Slt + 2
>
,
* 1 , k + 1 4 1  = 1 4 •
2
2
2
n By the Principle o f M athem atical Induction, ,y2» > 1 4   for every positive integer n.
Result 12.12
The harmonic series
Proof
For n e N, let sn —
00
1  diverges to infinity. k= 1 k
« 1  . Thus {.?„} is the sequence o f partial sums for the harm onic k= 1 ^ series. We show that lim sn = 00. Let M be a positive integer and choose N = 22M.
300
Chapter 12
Proofs in Calculus
L et n > N . Then, using L em m a 12.11, w e have _ 1, _ 1 2
1 _ _ 1__  ^ . . . _—1 N N + 1 n 1
1
1
N + 1
iV+2 2
n
— S n "b  +  + • ■ • + —
> SN = S22M > 1 +
M > M .
12.3 Lim its o f F unctions We now turn to another com m on type o f lim it problem (perhaps the m ost com m on). H ere w e consider functions f : X > R , w here X c R, and study the behavior o f such a function / near som e real num ber (point) a. For the present, w e are not concerned w hether a e X , but since w e are concerned about the num bers / ( x ) for real num bers x near a , it is necessary that / is defined in som e “deleted neighborhood” o f a. By a d elete d n e ig h b o rh o o d o f a, w e m ean a set o f the type (a — <5, a) U ( a , a + 8 ) — (a  8 , a + 8 )  {a} c X for som e positive real num ber 8 (the G reek letter delta). (See Figure 12.2.) It m ay actually be the case that (a  8 , a + <5) c X for som e 8 > 0. For
x
exam ple, if / : X —> R is defined by f ( x ) — — and we are interested in the behavior x o f / near 0, then 0 g X . In fact, it m ight very w ell be that X = R  {0}, in w hich case, (  8 , 0) U (0, J ) c i for every positive real num ber 8 . On the other hand, if / : X > R x is defined by / (x) = — and, once again, we are interested in the behavior o f / near 0, then 1 ,  1 £ X . A natural choice for X is R — {1, —1}, in w hich case (<5, S ) c l for every real num ber 8 such that 0 < 8 < 1. We are now prepared to present the definition o f the lim it o f a function. L et / be a realvalued function defined on a set X o f real num bers. A lso, let a e R such that / is defined in som e deleted neighborhood o f a. T hen w e say that the real num ber L is the lim it o f f ( x ) as x approaches a, w ritten lim f ( x ) = L , if the closer x is to a, the closer x —>a
f ( x ) is to L. The vagueness o f the w ord closer here too requires a considerably m ore precise definition. L et the positive num ber e indicate how close / ( x ) is required to be to L ; that is, w e require that / ( x ) — L\ < e. T hen the claim is that if x is sufficiently close to a, then / ( x ) — L \ < e. We use the positive num ber 8 to represent how close X m ust be to a in order for the inequality / ( x )  L\ < e to be satisfied, recalling that we are not concerned about how, or even if, / is defined at a. M ore precisely then, L is the lim it o f / ( x ) as x approaches a , w ritten lim f ( x ) = L , X—>Cl if for every real num ber e > 0, there exists a real num ber 5 > 0 such that for every real num ber x w ith 0 < x — a\ < 8 , it follow s that / ( x )  L\ < e. This im plies that if 0 < )x  a\ < S, then certainly f ( x ) is defined. If there exists a num ber L such that
a —6
a
+
< 3
 (K )s* x a x is
a Figure 12.2
A deleted neighborhood of a
12.3
Le
f(x)
L
Limits of Functions
301
L+ e
— ( •  • 
— )(• ) s a —d a x a +d Figure 12.3
x ax is
A geometric interpretation of lim / (x) = L
lim / ( x ) = L , then we say that the lim it lim f ( x ) exists and is equal to L ; otherw ise,
x* a
x+a
this lim it does not exist. Thus to show that lim f i x ) — L , it is necessary to specify x+a
e > 0 first and then show the existence o f a real num ber 8 > 0. Ordinarily, the sm aller the value o f e, the sm aller the value o f <5. However, w e m ust be certain th at the num ber 8 selected satisfies the requirem ent regardless o f how small (or large) e may be. Even though our choice o f S depends on 6, it should not depend on w hich real num ber x with 0 < \x — a\ < 8 is being considered. A ccordingly, if lim / (x) = L, then for a given e > 0, there exists S > 0 such that x+a
if x is any num ber in the open interval (a  S, a + S) that is different from a, then / ( x ) is a num ber in the interval (L — e, L + e). This geom etric interpretation o f the definition o f lim it is illustrated in F igure 12.3. We illustrate these ideas w ith an example. R e su lt to P ro v e PROOFSTRATEGY
Result 12.13 Proof
lim (3x — 7) = 5.
x± 4
B efore giving a form al proof o f this lim it, let's discuss the procedure w e will use. The pro o f begins by letting € > 0 be given. W hat w e are required to do is to find a num ber 5 > 0 such that if 0 < x — 4 < 8 , then (3x — 7) — 5 1 < e or, equivalently, 3(x  4) < e. This is also equivalent to 3 • \x  4  < e and to x — 4 < e /3 . This ♦ suggests our choice o f 8 . We can now give a proof. lim (3x — 7) = 5. X+A L et e > 0 be given. Choose 5 = e /3 . L et x e R such that 0 < \x  4 < 5 = e /3 . Then l(3x — 7) — 5 = \3x  12 = 3(x  4) = 3\x  4 < 3 (e /3 ) = e.
■
L et’s consider another exam ple. R e su lt to P rove PROOF STRATEGY
lim (—2x + 1) = 7.
x —>—3
First we do som e prelim inary algebra. The inequality (—2x + 1) — 7 < e is equivalent to  — 2x — 6] < e and to 2 x + 3 < e. T his suggests a desired value o f 8 . We can now give a proof. ^
302
Chapter 12 Result 12.14
Proof
Proofs in Calculus
lim ( 2.v + 1) = 7. x— >—3 L e te > 0 be given and choose <5 = e/ 2. L e t.x 6 R such that 0 < ]x  ( —3) < S = e/ 2, so 0 < \x + 3 < e /2 . Then (  2 x + 1)  7 =   2(x + 3) =   2 x + 3 = 2 x + 3 < 2 (e /2 ) = e.
■
The tw o exam ples that w e have seen thus far should tell us how to proceed w hen the function is linear (that is, f i x )  a x + b. w here a, b e R). We now present a slight variation o f this.
R e su lt to P rove
PROOFSTRATEGY
lim — ~ 9 = 6. 2x  3
In this exam ple, j/(.v )  l.\ < e becom es
4x2  9 2x  3
(2.v + 3)(2v  3) 
2x  3
 6 < e or, after sim plifying,
6 < e. However, since the num bers x are in a deleted neighbor
hood o f 3 /2 , it follow s that 2.v  3 ^ 0 and so
(2.i + 3)(2.v  3)
6
< e becom es  3 I(2a + 3)  6j < e or \2x  3 < e. T herefore, 2 x  3 /2  < e and \x  3 /2  < e/ 2. We are now prepared to give a proof. ^ 4x2 2x3
L et e > 0 be given and choose <5 = e /2 . Let £ e R such that 0 < x  3 /2  < S = e/ 2. So 2\x  3 /2  < e and 2x  3 < e. H ence (2x + 3)  6 \ < e. Since 2x  3 ^ 0, it (2.v + 3)(2x  3) 4x2  9 follow s that 6 < e and so 6 < e. 2x  3 to
Proof
lim .1
1
Result 12.15
We now turn to a lim it o f a quadratic function. R e su lt to P rove PROOFSTRATEGY
lim x 2 = 9. x—>3 O nce again for a given e > 0, w e are required to find S > 0 such that if 0 < x  3 1 < <5, then x 2 — 9\ < e. To find an appropriate choice o f <5 in term s o f e, w e begin w ith x 2  9 1 < e. We w ish to w ork the expression ]x  3 into this inequality. A ctually, this is quite easy since x 2  9 < e is equivalent to x  3 x + 3 < e. This m ight m ake us think o f w riting x — 3 1 <
and choosing S H owever, <5 is required I * + 3 “ x + 3 1 to be a positive num ber (a constant) that depends on e but is not a function o f x. The expression \x + 3 1 can be elim inated though, as we now show. Since it is our choice how to select <5, we can certainly require 8 < 1, w hich we do. Thus x — 3 < 1 and so — 1 < x — 3 < 1. H ence 2 < x < 4. Thus 5 < x + 3 < 7 and so x + 3 1 < 7 . So. under this restriction for 5, it follow s that .y  3 jx + 3 < l \ x  3. N ow if 7.v  3 < e,
12.3
Limits of Functions
303
that is, if x — 3 < e /7 , then it w ill certainly follow that x — 3 x + 3 < e. A rriving at this inequality required that both \x — 3 < 1 and \x — 3 < e /7 . T his suggests an appropriate choice o f <5. 4 Result 12.16
Proof
lim x 2 = 9. .Y—>3 Let e > O be given and choose <5 = m in (l, e /7 ). L e tx e R such that 0 < \x  3 < S = m in (l, e /7 ). Since x — 3 < 1, it follow s that —1 < x — 3 < 1 and so 5 < x + 3 < 7. In particular, x + 3 < 7 . Because x — 3] < e /7 , it follow s that x 2  9 = x  3 x + 3 < x  3 • 7 < (e /7 ) • 7 = e.
■
We have now seen four proofs of lim its o f type lim / ( x ) — L. In R esult 12.13, we x —>a
chose 8 = e /3 for the given e > 0 and in Result 12.14, w e chose <5 = e /2 . In each case, if we had considered a different value o f a for the sam e function, then the sam e choice of 8 w ould be successful. This is because the function is linear in each case. In R esult 12.15, for a given e > 0, the selection o f <5 = e /2 w ould also be successful if a ^ 3 /2 , provided 3 /2 ^ {a — 8 , a + 8 ). This is because the function / in Result 12.15 defined by / ( x ) = (4 x 2  9 )/(2 x  3) is “nearly linear” ; that is, / ( x ) = 2x + 3 if x ^ 3 /2 and / ( 3 / 2 ) is not defined. However, our choice o f 8 — e /7 in the pro o f o f R esult 12.16 depended on a = 3; that is, if a ^ 3, a different choice of<5 is needed. For exam ple, if we w ere to prove that lim x 2 = 16, then for a given e > 0, an appropriate choice for 8 is m in (l, e /9 ). x —>4
N ext w e consider a lim it involving a polynom ial function o f a higher degree. R e su lt to P ro v e
PROOFSTRATEGY
lim (x 5 — 2 x 3 — 3x — 7) = 3. x ^ 2 For a given e > 0, w e are required to show that (x 5 — 2 x 3 — 3x — 7) — 3 < e if 0 < x — 2 < 8 for a suitable choice o f 8 > 0. We then need to w ork \x — 2 into the expression x 5 — 2 x 3 — 3x — 10. D ividing x 5 — 2 x 3 — 3x — 10 by x — 2, w e obtain x 5 — 2 x 3 — 3x — 10 = (x — 2 )(x4 + 2 x 3 + 2 x 2 + 4x + 5). H ence w e have x 5  2 x 3  3x  10] = x  2 x 4 + 2 x 3 + 2 x 2 + 4x + 5], Thus we seek an upper bound for x 4 + 2 x 3 + 2 x 2 + 4x + 5. To do this, w e im pose the restriction 8 < 1. Thus x — 2 < <5 < 1. So —1 < x  2 < 1 and 1 < x < 3. H ence x 4 + 2 x 3 + 2 x 2 + 4x + 5 < x 4 + 2 x 3 + 2 x 2 + 4 x  + 5 < 170. We are now prepared to prove R esult 12.17.
Result 12.17
Proof
♦
lim (x 5 — 2 x 3 — 3x — 7) = 3.
.v—>2
Let e > 0 be given and choose <5 = m in (l, e /1 7 0 ). L et x 6 R such that 0 < x — 2 < 8 = m in (l, e /1 7 0 ). Since x — 2 < 1, it follow s that 1 < x < 3 and so x 4 + 2 x 3 + 2 x 2 + 4x + 5 < x 4 + 2 x 3 + 2 x 2 + 4 x  + 5 < 170.
304
Chapter 12
Proofs in Calculus
Since \x  2 \ < e /1 7 0 , w e have (x 5  2 .y3  3,v  7)  3 = a 5  2 x 3  3.v  10 = Jx — 2\ • x 4 f 2 x 3 + 2x~ + 4 x j 5 1 < (e/1 7 0 ) • 170 = e. O ur next exam ple involves a rational function (the ratio o f tw o polynom ials).
R esult to Prove
v x2+ 1 2 hm — =  . X+1 X  + 4 5
PROOFSTRATEGY
F irst observe that x2 + 1
2
5(x2 + 1)  2(x2 + 4)
3 x 2  3 1
3 x  1 x + 1
x2 + 4
5
5(x2 + 4)
5 (x 2 + 4)
5(x 2 + 4)
3 1A' + 1 1
H ence it is necessary to find an upper bound f o r  . O nce again we restrict d so 5 (.v + 4) that S < 1. Then \x — 1 < 1 or 0 < x < 2. H ence 1 < x + 1 < 3 and so 3x + 11 < 9 . 1 1 , A lso, since x > 0, it follow s that 5(x + 4) > 20. Thus . , — < — and so 5 (x 2 + 4) 20 3 1 * + 11 5(x2 + 4)
9_
<
20'
We now present a proof o f this result.
Result 12.18 Proof
4
x2+ 1 2 lim — =  . xyl x  + 4 5 L e te > 0 be given and choose 5 = m in (l, 2 0 e /9 ). L et a e R such that 0 < x — 1 < S. Since x — 11 < 1, w e have 0 < x < 2 and 1 < x + 1 < 3. H ence 3 x + l < 3  3 = 9 a n d 5 (x 2 + 4) > 2 0 ,so
5 (x 2 + 4)
< — .T herefore, < 9 / 2 0 . S in c e \x — 1 < 20' 5 (x 2 + 4)
2 0 e /9 , it follow s that x2 + 1
2
x2 + 4
5
3x2 —3
5(x2 + 1)  2 (x 2 + 4)
5 (x 2 + 4)
5 (x 2 + 4) 3 x — l x + 1 5(x 2 + 4)
20e
9
< ~9~ ' 20 “ 6'
We now present one additional exam ple on this topic.
Example 12.19
Solution
Determine lim
x2  1 2x  1
and verify your answer.
Since it appears that lim (x 2 — 1) = 0 and lim (2 x — 1) = 1, w e w ould expect that X—*■1
12.3
is 8 > 0 such that if 0 < x
Limits of Functions
305
1 < 8, then k2
 0 =
 1
2x — 1
1

2
< €.
x  1
O bserve that x2  1
(X  l)(x + 1)
x + i[
2x — 1
2x — 1
\2 x  1.
\x 
1.
\x + l\ . Ordinarily, w e might 2 jc  1 restrict 8 < 1, as before, but in this situation, we have a problem . If 8 < 1, then 0 < \x — 11 < 5 and so \x — 11 < 1. Thus 0 < x < 2 or x e (0, 2). However, this interval o f Proceeding as before, w e find an upper bound for
real num bers includes 1/2 and  X is not defined w hen x = 1 /2 . Thus w e place 2x — 1 a tighter restriction on 8 . T he restriction 8 < 1 /2 is not suitable either, for if \x — 11 < 8
< 1/2, then 1 /2 < x < 3 /2 . E ven though ^
^
is defined for all real num bers x
2x — 1 in this interval, this expression becom es arbitrarily large if x is arbitrarily close to 1/2 , allow ing 2x — 1 to be arbitrarily close to 0. T hat is, w e cannot find an upper bound for
^
12x — 11
if
8
= 1/2 . H ence we require that
8
< 1 /4 , say, and so x — 11 < 5 <
Thus 3 /4 < x < 5 /4 . H ence x + 1 < 9 /4 . Also, 2x — 1 > 2 (  ) — 1 = 1 /2 and so 1 x __ j  9 9 — < 2. Therefore, — <  ■ ! =  . We now give a form al proof. ♦ 12x — 11 2x — 11 4 2 £F
Result 12.20
Proof
1 lim = 0. ► ! 2x  1 L et e > 0 be given and choose 8 = m in ( l/ 4 ,2 e / 9 ) . L et x e R such that 0 < x — 11 < 5 . Since 8 < 1/4, it follow s that x — 11 < 1 /4 and so 3 /4 < x < 5 /4 . H ence x + 1 < 5 / 4 + 1 = 9 /4 . A lso, 2x — 1 > 2 (  ) — 1 = 1 /2 and so x + 11 9 9 Therefore, —— <  ■2 = Since x — 11 < 2x x2  1
xz  1
2x  1
2x — 1
\x + l\ 2x — 11
8
 1 —  < 2. 2x  1
< 2 e /9 , it follow s that 2
e
II <
N ext w e consider a lim it problem w here the lim it does not exist.
R e su lt to P ro v e
lim  does not exist. X —rO X
PROOFSTRATEGY
A s expected, w e w ill give a p roof by contradiction. If lim — does exist, then there exists x>0 x a real num ber L such that lim — — L . H ence for every e > 0, there exists <5 > 0 such v— >o x
306
Chapter 12
Proofs in Calculus
that if 0 < ]jc <
8
, t h e n  L
< 6 . For num bers x “close” to 0, it certainly appears
X
that — is “large” (in absolute value). H ence, regardless o f the value o f e, it seem s that x 1 there should be a real num ber x w ith 0 < Ixl < 8 such that   L > €. It is our plan x .at to show that this is indeed the case. Thus, we choose e = 1, for exam ple, and show that no desired <5 can be found. ♦
Result 12.21
Proof
lim — does not exist. dHHS x A ssum e, to the contrary, that lim  exists. T hen there exists a real num ber L such that x^ O
X
lim — — L . L et e — I. T hen there exists <5 > 0 such that if x is a real num ber for w hich .v 'li X , t h e n  L < e = 1. C hoose an integer n such that n > P/<5~ > 1, Since x n > 1/5, it follow s that 0 < 1j n < 8 . We consider tw o cases. Case 1. L < 0. Let x — \ f n . So 0 Jf [*] < 8 . Since —L > 0, it follow s that
0 < x  <
8
1  L x
= \n — L \ = n — L > n > 1
w hich is a contradiction. Case 2. L > 0. Let x = —l / n . So 0 < x <   L — x
8
. Thus
 — n ~ L\ =  — (« + L ) — n + L > n > 1
producing a contradiction in this case as well. R e su lt to P ro v e
PROOF STRATEGY
Result 12.22
Proof
L et / ( x ) = \ x \ / x , w here x e R and x / 0. Then lim /(.v ) does not exist.
T he graph o f this function is show n in Figure 12.4. If x > 0, then f ( x ) = \ x \ f x = x / x — 1; w hile if x < 0, then f ( x ) — x / x = —x / x = 1. H ence there are num bers x that are “near” 0 such that f i x ) — 1 and num bers x that are “near” 0 such that f i x ) = —1. This suggests a proof. #
L et f ( x ) — \ x \ / x , where x e R and x / 0. Then lim f i x ) does not exist. x*Q
A ssum e, to the contrary, that lim f i x ) exists. T hen there exists a real num ber L such that lim f i x )  L. L et e — 1. Then there exists
8
> 0 such that if x is a real num ber
satisfying 0 < x — 0j = xj < <5, then . f i x ) ■ L < c — I . We consider two cases. Case 1. L > 0. C o n sid erx = —8/2. T hen \x] — 8 / 2 < 8 . However, f ( x ) = f i —8/2) ( 8 / 2 ) / ( —8/2) =  1 . So  f ( x ) — L\ = J 1  L\ = I + L »■ 1, a contradiction.
12.4
Fundam ental Properties of Limits of Functions
307
y 1
1 <
0  c  1
Figure 12.4
Case
2.
(5 /2 )/(5 /2 )
12.4
The graph of the function f ( x ) = \x\/x
L < 0. L et x = 8 / 2 . T hen j.vj—^>/2<<S. A lso, f ( x ) = f ( 8 / 2 ) = = 1. So  f i x ) — L\ = 11  L ( = 1 — L > 1, a contradiction. ■
F undam ental P roperties o f Lim its o f F unctions If w e w ere to continue com puting lim its, then it w ould be essential to have som e theorem s at our disposal that w ould allow us to com pute lim its m ore rapidly. We now present some theorem s that w ill allow us to determ ine lim its m ore easily. We begin w ith a standard theorem on lim its o f sums o f functions.
T h e o re m to P ro v e
If lim f ( x ) = L and lim g( x) = M , then x —>a
lim ( f ( x ) + 0 ) ) = L + M . PROOFSTRATEGY
In this case, w e are required to show, for a given 6 > 0. that j(/(.v ) + g(x)) — (L + M)\ < e if 0 < \x — a\ < 8 for a suitable choice o f 8 > 0. Now \ i f ( x ) + g i x ) ) — ( L + M )  =  ( / ( x )  L ) + (S(x)  M )  <  f i x )  L \ + g ( x )  M . H ence if w e can show that both f i x ) ~ L\ < e /2 and ,!>(.v I — M \ < e /2 , for exam ple, then we w ill have obtained the desired inequality. However, because o f the hypothesis, this can be accom plished. We now m ake all o f this precise. ♦
Theorem 12.23
/ / lim f i x ) = L and lim g (x ) — M , then x>a
x>a
lim ( f ( x ) + g{x)) = L + M .
x+a
Proof
Let e > 0. \ f ( x ) — L\ M  < e /2 . 0 < x — a
Since e /2 > 0, there exists <5] > 0 < e /2 . A lso, there exists 8 2 > 0 such C hoose 8 — min(<5i, 8 2) and let x e \ < 8 , it follow s that both 0 < jx — a\
such that if 0 < that if 0 < \x — a\ R such that 0 < < <5i andO < \x —
\x — a\ < 8 \, then < 8 2 , then \g(x) — x — a\ < 8 . Since a\ < S2  Therefore,
( f ( x ) + g{x))  (L + M ) 1 = ( / ( x )  L ) + (g(x)  M ) < (f {
x )
 L \ + g (x ) — M \ < e /2 + e /2 = e.
308
C h a p te r 12
Proofs in Calculus
T heorem 12.23 states that the lim it o f the sum of two functions is the sum o f their lim its. N ext w e show that this is also true for products. B efore getting to this theorem , le t’s see w hat w ould be involved to prove it. L et lim f i x ) = L and lim g i x ) = M. x —>a
x>a
This m eans that we can m ake the expressions  f i x ) — L \ and g (x ) — M as sm all as w e wish. O ur goal is to show that w e can m ake / ( x ) • g (x ) — L M j as sm all as we wish, say less than e for every given e > 0. The question then becom es how to use w hat we know about f i x ) — L \ and g (x ) — M  as w e consider / (v) • g i x ) — L M \ . A com m on w ay to do this is to add and subtract the sam e quantity to and from f i x ) ■g ( x) — L M . For exam ple, f i x ) ■g i x )  l . M
= f i x ) ■g{x)  f i x ) ■M + f i x ) ■M  L M \ = \f(x)(g(x)M ) + if( x )  L ) M 1 < \f( x ) \\ g ix )  M \ + \f(x)  L \\M \.
If w e can m ake each o f /(.v)li.e(.v) — M\ and  f i x ) — L \ \ M \ less than e /2 , say, then we w ill have accom plished our goal. Since \M\ is a nonnegative constant and jf i x ) — L\ and g (x )  M \ can be m ade arbitrarily sm all, only / ( x )  is in question. In fact, all that is required to show is that f i x ) can be bounded in a deleted neighborhood o f a , that is, / ( x )  < B for som e constant B > 0. Lemma 12.24
Suppose that lim f ( x ) = L . Then there exists S > 0 such that i f 0 < x — a \ < <5, then l/( x ) l < 1 + IL\.
Proof
L et e = 1. Then there exists <5 > 0 such that if 0 < x — a\ < S, then  / ( x ) — L\ < 1. Thus l / ( x )  =  / ( x )  L + L  <  / ( x )  L\ + \L\ < l + \L\.
b
We are now prepared to show that the lim it o f the product o f tw o functions is the product o f their limits. T heorem to Prove PROOFSTRATEGY
If lim f i x ) — L and lim g (x ) = M , then lim f i x ) x± a
x —>a
x>a
■g{x) — L M .
As w e discussed earlier,  f i x ) ■g i x )  L M  = / ( x ) • g (x )  f i x ) ■M + f i x ) ■M  L M \ = /(x )O K x )  M ) + i f i x )  L ) M \ < \f{x)\\g(x)M \ + \fix )L \\M \.
For a given e > 0, w e show that each o f f ( x ) , , g ( x ) — M \ and  f i x ) — L \ \ M \ can be m ade less than e /2 , w hich w ill give us a p ro o f o f the result. O f course, this follow s im m ediately for f i x ) — L \ \ M \ if M = 0. O therw ise, we can m ake f i x ) — L\ less than e /2 M . By L em m a 12.24, we can m ake / ( x )  less than 1 + L . Thus w „!»(x) — M \ < e / 2 ( l + \L ). Now, le t’s p ut all o f the pieces together. ♦ Theorem 12.25
I f lim f i x ) = L and lim g ( x ) = M, then lim f ( x ) ■g i x ) — L M .
12.4
Proof
Fundam ental Properties of Limits of Functions
309
L et e > 0 be given. By L em m a 12.24, there exists 8 : > 0 such that if 0 < \x — a\ < Si, then \ f { x ) \ < l + \L\. Since lim «(.r) = M , there exists 8 2 > 0 such that if 0 < x —>a
\x  a\ < 8 2, then g(x)  M \ < 6 /2 (1 + \L\). We consider tw o cases. Case 1. M = 0. Choose 8 = min(Si, Sq). L e tx e R such that 0 < \x —a\ <
8
. Then
 f ( x ) ■g(x)  LM\ =  f i x ) ■g(x)  f i x ) ■M + f ( x) ■M —LM\
=  f(x)(g(x)  M ) + i f i x )  L)M\ 5 \fix)\\gix)M \ + \fix )L \\M \
< (1 + / . )f/2 ( 1 + \L\) + 0 = e /2 < e.
Case 2. :VI / 0. Since lim f i x ) = L , there exists <53 > 0 such that if 0 < x —>a
\X
—a\ <
83,
then  f i x ) —L\ < e/2\M\. In this case, we choose <5 = min(c'i;. 8 2 , 8 3 ). N ow let x e R such that 0 < \x — a\ < 8 . Then  / ( 4  • g (x)  LM\ =  f i x ) ■g(x)  f i x ) ■M + f i x ) ■M —LM\ = \f(x)igix)M ) + (f(x )L )M \ < \fix)\[g ( x )  M \ + \ m
L\\M \
< ( l + L ) e /2 ( l +  L  ) + (e /2 M )] M 
= e/2 + e/2 = e.
m
N ext we consider the lim it o f the quotient o f tw o functions. As before, let lim J'(x) = x+a L and lim e ( x) — M . O ur goal is to show that lim ^ = — . O f course, this is not & x^c, g i x ) M . f ix ) L true if M = 0; so w e w ill need to assum e that M / 0. To prove that lim x —>a g ( x ) ~ M ' fix) L can be m ade arbitrarily small. O bserve that we are required to show that g( x) M fix)
L
gix)
M
f i x ) ■M  L ■g i x )
f i x ) ■M  L M + L M  L ■g ix ) g{x) ■M
g {x) • M ( f { x )  L ) M + L ( M  g i x )) g i x ) ■M fix)L \
\g(x)\\M\
, \ L \ \ M — g(x)\ \g(x)\\M\
L?00l Thus to show that
<
\f(x )L \\M \ + \L\\M gix)\
fix)
L
gix)
M
can be m ade less than e for any given positive num ber
\ f ( x ) — L\ \ L \ \ M  g{x)\ e, it is sufficient to show that each o f — — and — — —— can be m ade less IgOOl \gix)\\M\ than e /2 . O nly 1/  t;(.v)' requires study. In particular, w e need to show that there is an upper bound for l/] g ( x )  in som e deleted neighborhood o f a. Lemma 12.26
I f lim "(.v) = M 4 0. then l/f ( jc ) [ < 2 / \ 1 \ f o r all x in some deleted nei ghborhood x>a
o f a.
310
Chapter 12 Proof
Proofs in Calculus
L e te = Af /2 . T hen there exists 5 > 0 such that if 0 < \x — a\ < S, then g (x ) — M \ <  A /1/2. Therefore, M  = \M  g ( x ) + g O ) <  M  g(xi{ + g (x ) H ence g (x ) > \M\  \ M  g (x ) > \M\  \ M \ / 2 = M /2 .T h u s l / f i  x )  < 2 / M . i
Theorem to Prove PHOOFSTRATEGY
/(* ) a g(_x)
L M
If lim f i x ) = L and lim g ( x ) = M ^ 0, then lim
Returning to our earlier discussion, w e now have f i x ) _ L_ <
\L\\M g(x)\
M
g ( x)
\g(x)\\M\
1^(41
This suggests how sm all w e m ust m ake \ f ( x )  L\ and g (x ) — M \ — \ M — g i x ) ' to accom plish our goal.
Theorem 12.27
Proof
♦
I f lim f i x ) = L and lim g( x) — M ^ 0, then lim
fix)
L
g ( x)
M '
L et e > 0 be given. By L em m a 12.26, there exists <5i > 0 such that if 0 < x — a\ < Si, then 1/ , ij(.v) < 2 / \ M \ . Since lim f i x ) = L , there exists S2 > 0 such that if 0 < X —>Cl
\x  a\ < S2. then  f i x )  L\ < \ M \ e /4 . We consider tw o cases. Case 1. L = 0. D efine S = m in(Sj, S2). L et x. G R such that 0 < \x  a\ < S. Then fix)
L
gix)
M
t# W  2 , 0A— —— 6 <6. • 4" 4  Af I 2
C ase 2. L ^ 0. Since lim g (x ) = Af, there exists S3 > 0 such that if 0 < \x  a\ < S3, A'—^(2 then g (x ) — M  < \ M \ 2 e / 4 \ L \ . In this case, define S = m in(Si, S2, S3). Let x e R such that 0 < Ix — a\ < S. T hen fix)
L
\ f j x ) —L j
l Ll  M g( x) l
g ( x)
M
lgC*)l
\gix)\\M\
M e
2
~ 7 ~ ‘ lM 
J L\_
I Mp e
2
M  ' 4 L 
Now, w ith the aid o f T heorem s 12.23, 12.25, 12.27 and a few other general results, it is possible to give sim pler argum ents for som e o f the lim its we have discussed. First, we present some additional results, beginning w ith an observation concerning constant functions and follow ed by lim its o f polynom ial functions defined by f i x ) — x n for some n e N. Theorem 12.28
Let a, c 6 R. I f f i x ) — c f o r all x e R , then lim f i x ) = c. x —*a
12.4
Proof
Theorem 12.29 Proof
Fundam ental Properties of Limits of Functions
311
< L et e > 0 be given and choose <5 to be any positive number. Let x e R such th \x  a\ < 8 . Since f i x ) = c for all x € R, it follow s that \ f ( x )  cj = \c  c\ = 0 < e . ■ L e t f i x ) = x for all x e R. For each a g R. lim f i x ) = a. x>a
L et e > 0 be given and choose  f ( x ) — a\ = \x — a\ <
8
8
= e. Let , t e R
such that 0 < \x — a\ <
8
. T hen
= e.
■
We now extend the result in T heorem 12.29. Theorem 12.30 Proof
Let n g N and let f ( x ) = x n f o r all x e R. Then f o r each a g R, lim / (x) = a". We proceed by induction. The statem ent is true for n — 1 since if f i x ) — X, then lim f ( x ) = a by T heorem 12.29. A ssum e that lim x k — a k, w here k g N. We show
x^a
x+a
that lim x k+x — a k+1. O bserve that lim x k+l = lim ( x k ■x ) . B y T heorem s 12.25 x m
x * a
x^a
and
12.29 and the induction hypothesis, lim
x> a
= lim ( x k * x) = ( lim x k^\ ( lim x ) = (a k) (a) = ak+y. x —*a v
'
\x * a
/
\x * a
/
B y the Principle of M athem atical Induction, lim x n = a ” for every n e N.
■
x —>a
It is possible to prove the follow ing theorem by induction as w ell. We leave its proof as an exercise (Exercise 12.32). Theorem 12.31
Let f i , f z , ■• • , f n be functions in > 2) such that lim f ( x ) = L t f o r 1 < i < n. Then lim
x ^ a
1 M x ) Hb f„(x)) — L i + L 2 Hh L„,
W ith the results w e have now presented, it is possible to prove that if p ( x ) = c„xn + cn Xx n~ x + ■• • + c \ x + Co is a polynom ial, then lim p ( x ) = c„a" + cn_ \ a " ~ ] + • • • + c\ a +
cq
x ta
= pi a) .
( 12. 1)
For exam ple, applying this to Result 12.13, w e have lim(3.\' — 7) = 3 • 4 — 7. x>4
Similarly, Result 12.14 can be established. Result 12.15 cannot be established directly since lim (2 x —3) = 0. A pplying w hat w e now know to Result 12.16, we have 2
lim x 2 — 32 = 9 and in Result 12.17, x> 3
lim (.r5  2 x 3  3x  7) = 25  2 • 23  3 • 2  7 = 3. x —>2 A lso, if r is a rational function, that is, if r ( x ) is the ratio p ( x ) / q ( x ) o f tw o polyno m ials p i x ) and q ( x ) such that q( a) ^ 0 for a e R , then by T heorem 12.27, p(x) lim r ( x) — l i m  = x^a x^a q(x)
lim.,..,.,., p i x ) p ( a)  — = —— = >ia). lim j^ o q i x ) qi a)
(1 2..Z)
312
Chapter 12
Proofs in Calculus So, in R esult 12.18, w e have *2+ l l2 + 1 2 lim — = — x *■i x 2 + 4 l2 + 4 5
A lthough it is sim pler and certainly less tim econsum ing to verify certain lim its w ith the aid o f these theorem s, we should also know how to verify lim its by the e — 8 definition.
12.5 C ontinuity O nce again, let / : X —>• R be a function, w here I C R , and let a be a real num ber such that / is defined in som e deleted neighborhood o f a. R ecall that lim f i x ) — L for x±a
som e real num ber L if for every e > 0, there exists 8 > 0 such that if x € (a — S, a + 8 ) and x / a, then  f ( x ) — L\ < 6 . If / is defined at a and f ( a ) — L , then f is said to co n tin u o u s at a. T hat is, / is continuous at a if lim f i x ) = f { a) . T herefore, a function x —>a
f is continuous at a if for every e > 0, there exists 8 > 0 such that if \x — a \ < 8 , then If i x ) — f { d )  < e. (N otice that in this instance, 0 < \x — a\ < 8 is being replaced by \x — a\ <
8
.) Thus for / to be continuous at a, three conditions m ust be satisfied:
(1) / is defined at a ;
(2) lim / ( x ) exists; x>a
(3) lim f { x ) = f { a ) . x >a
We now illustrate this. Problem 12.32
A function f is defined by f i x ) — {x 2 — 3x + 2 ) / ( x 2  1) f o r all x e R  {—1, 1}. Is f continuous at 1 under any o f the foll owing circumstances: ia) f is not defined at 1; (b) / ( l ) = 0; (c) / ( I ) = —1 /2 ?
Solution
For / to be continuous at 1. the function / m ust be defined at 1. So w e can answ er question (a) imm ediately. The answ er is no. In order to answ er questions (b) and (c), we m ust first determ ine w hether lim f i x ) exists. O bserve that X * l
3x + 2
(x  !)(*  2)
1
( x  l X x + 1)
/(* ) =
since x ^ 1. B ecause f ( x ) =
x + \
x —2 lim am x + 1
x + 1
is a rational function, w e can apply (12.2) to obtain
lim.v^ .i(x
2)
1
lim , ^ i ( x + 1)
2
H ence if / ( l ) = —1 /2 , then / is continuous at 1. Therefore, the answ er to question (b) is no and the answ er to (c) is yes. ♦
For additional practice, w e present an e — 8 pro o f that lim
3x I 2 1
12.5 PROOFSTRATEGY
Continuity
313
O bserve that x 2 — 3x + 2
/
x2  l
1\
V V
1
(x  l)(x  2)
1
(x  2)
(x  l)(x + 1)
2
(x + 1) 1 2
2(x — 2) + (x + 1)1
3x — 3
3 * ~ 1
(x + 1)
2(x + 1)
2  jc + 1 '
2
If [x — 11 < 1, then 0 < x < 2 and \x + 1 [ > 1, so \ / \ x + 11 < 1. We are now prepared to prove that lim f i x ) = —1/2. ♦ 1
Result 12.33
lim x~*l
Proof
x 2 — 3x + 2 x2 —1
L et € > 0 and choose S — m in (l, 2 e /3 ). L et x e R such that \x — 1 < 5 . Since x — 11 < 1, it follow s that 0 < x < 2. So \x + 11 > 1 and l /  x + 11 < 1. H ence x 2 — 3x + 2
/
1\
(x  l)(x — 2)
1
x2  l
i
2j
(x  l)(x + 1)
2
3x — 3 2(x + 1)
x 2 —
3 jc — 11
1
x + 1 ' 2 3
2f
2 x + 11 < 2 ’ T
e.
Indeed, (12.2) states that if a rational function 7 is defined by r ( x ) = p ( x ) / q ( x ) , w here p ( x ) and q ( x) are polynom ials such that q(a) 0, then r is continuous at a . A lso, (12.1) im plies that if p is a polynom ial function defined by p ( x ) = cnx n + c„_ \ x n~ x + ■• ■+ Cix + Co, then p is continuous at every real num ber a. We now present som e exam ples concerning continuity for functions that are neither polynom ials nor rational functions. R e su lt to P rove PROOFSTRATEGY
The function / defined by / ( x ) = y/x for x > 0 is continuous at 4. B ecause / ( 4 ) = 2, it suffices to show that lim V x = 2. Thus  / ( x ) — L \ = \*Jx — 2. To w ork x — 4 into the expression Jx. — 2, w e m ultiply V x — 2 by (Jx + 2 ) /( ^ x + 2), obtaining IV x — 2 =
(Jx  2 )(Vx + 2)
x — 4 a/ x
+ 2
First we require that S < 1, that is, \x — 4 < 1, so 3 < x < 5 . Since V * + 2 > 3, it follow s that 1/( V x + 2) < 1/3. H ence , r ol ^ ~ 21 =
x — 4 [x — 4 rx fi 2o < 3— •
This suggests an appropriate choice for S. Result 12.34
The fun ction f defined by f ( x ) = s / x f o r x > 0 is continuous at 4.
314
Chapter 12
Proofs in Calculus
y A
3 2
l <i>  x
+ 3
2
1
o
1
2

3
Figure 12.5
Proof
The graph of the ceiling function / (.v ) = pr"[
L et € > 0 be given and choose <5 = m in (l, 3e). L et , « e R such that \x — 4 < S. Since \x — 4 < 1, it follow s that 3 < x < 5 and so *Jx + 2 > 3. T herefore, l / ( y f x + 2) < 1/3 . Flence (V x  2 )(v /x + 2) sfx + 2
I* ~ 4  1 —— <  l 3 e ) = e. «Jx + 2 3
F igure 12.5 gives the graph o f the ceiling function / : R — Z defined by f(x') = f x ] . This function is not continuous at any integer but is continuous at all other real num bers. We verify the first o f these rem arks and leave the pro o f o f the second rem ark as an exercise (Exercise 12.37). Result 12.35
The ceiling function f : R —»• Z defined by f ( x ) = [x] is not continuous at any integer.
Proof
A ssum e, to the contrary, that there is som e integer k such that / is continuous at k. T herefore, lim f ( x ) = f ( k ) = [fe] = k. H ence for 6 = 1, there exists S > 0 such that if x — k \ < 5, then / ( x ) — f ( k ) \ = / ( x ) — k \ € € = 1. L et <5j = min(5, 1) and let X\ e (k, k + <5i).Thus/: < X\ < k + S a n d k < .v; < k + I.H e n c e /(.V i) = "xil — k + 1 and / ( x i ) — k  = \(k — 1J — A. — 1 < 1. a contradiction. b
12.6 D ifferen tiab ility We have discussed the existence and nonexistence o f lim its lim / ( x ) for functions x^a f : X —> R w ith X c R , w here / is defined in a deleted neighborhood o f the real num ber a and, in the case o f continuity at a, investigated w hether lim f (x) = f ( a ) if
12.6
Differentiability
315
equation of tangent line: y —f ( a) = m[x — a),
m
where m = f ' ( a )
JC
Figure 12.6
Derivatives and slopes of tangent lines
/ is defined in a neighborhood o f a. If / is defined in a neighborhood o f a , then there is an im portant lim it that concerns the ratio o f the differences / (x) — / (a) and x  a. A function / : X  * R, w here X C R , that is defined in a neighborhood o f a real num ber a is said to be d iffe re n tia b le at a if lim —  — — exists. This lim it is called xro x —a the d eriv a tiv e of / at a and is denoted by f ' ( a ) . Therefore, f ' i a ) = lim
fix )  f(a) x —a
You probably already know that f ( a ) is the slope of the tangent line to the graph of y = f (x ) at the point (a , / ( « ) ) . Indeed, if f ( a )  m. then the equation o f this line is y  f { a ) — m i x  a). See F igure 12.6. We illustrate derivatives w ith an exam ple. Example 12.36
Show that the function f defined by f i x ) = 1/ x 2 f o r x # 0 is differentiable at 1 and determine / ' ( l ) .
Solution
Thus we need to show that lim — —— j — = lim 1— exists. In a deleted neighx^i x —1 »• >i x  ■ l borhood of 1, 4 1 X —1 Since
1— x —1
l  x
(1 — V)(1 + x )
2
x 2(x — 1)
1 !V
(12.3)
x 2(x — 1)
1 4“ x . is a rational function, w e can once again use (12.2) to see that
lim
1+ x
l i m ^ i (1 + x )
x 2
lim * _ * i( s 3 ^ j)
2
=
2
1
and so / ' ( l ) = —2. We present an e — <5 proof o f this lim it as w ell. For a given e > 0, w e are required to 4  1 (  2 ) < e. O bserve find S > 0 such that if x e R w ith 0 < x — 11 < <5, then x —1
316
Chapter 12
Proofs in Calculus
that 2 
( 
x 2 —x — \
2 + 2
2) —
\x  l2 x + 1
X 2
If w e restrict <5so that 8 < 1 /2 , then x  1 < l / 2 a n d s o l / 2 < x < 3 /2 . S in ce* > 1 /2 , it follow s that x 2 > 1/ 4 and 1/ x 2 < 4. A lso, since * < 3 /2 , it follow s that 2x + 11 < 4. H ence \x  l 2 x + 1 \ / x 2 < 16x  11. This shows us how to select S. We now prove that f \ l ) = —2. Result 12.37 Proof
♦
Let f be the function defined by f ( x ) = \ / x 2 f o r x / 0. Then / ' ( l ) =  2 . L e te > 0 be given and choose <5 = m in ( l/2 , e /1 6 ).L e t* e R su c h th a tO < \x — 1 < <5. Since jc — 11 < 1 /2 , it follow s that 1 /2 < * < 3 /2 . Thus x 2 > 1 / 4 and so I / * 2 < 4. A lso, \2x + 11 < 4. Since \x — 1 < e /1 6 , it follow s that i
/(* )/(!) (2) x —1
1
x — 2
1+ x
(  2)
xz
1
x 2 —x — 1
+2
12 x + 11 72
16
F rom Result 12.37, it now follow s that the slope o f the tangent line to the graph of y = l / x 2 at the point (1. 1) is  2 and, consequently, that the equation o f this tangent line is y — 1 = —2 ( x — 1). D ifferentiability o f a function at som e num ber a im plies continuity at a , as w e now show. Theorem 12.38
Proof
I f a function f is differentiable at a, then f is continuous at a. f{x)  fk i)
exists and equals the real x —a num ber f \ a ) . To show that / is continuous at a, w e need to show that lim f ( x ) = f (a). Since / is differentiable at a , it follow s that lim
We w rite f { x ) as fix) =
f i x )  f { a )' (x  a ) + f ( a ) . x —a
Now, using properties o f lim its, w e have lim f i x )
lim
x* a
/(* )  f(a)' x —a
lim (x — a) + lim / ( a )
= /'( « ) ■ 0 + / ( a ) = f { a ) . T he converse o f T heorem 12.38 is not true. For exam ple, the functions / and g defined by / (x) = .r ] and g (x ) = Zfx are continuous at 0 but neither is differentiable at 0. T hat / is not differentiable at 0 is actually established in E xam ple 12.22.
Exercises for Chapter 12
317
EXERCISES FOR CHAPTER 12 Section 12.1: Limits of Sequences 12.1. Give an example of a sequence that is not expressed in terms of trigonometry but whose terms are exactly those of the sequence of (cos(«7r)}. 12.2. Give an example of two sequences different from the sequence {n 2 — n\ + \n — 2} whose first three terms are the same as those of {n2 — n\ + \n — 2). 12.3. Prove that the sequence {^ } converges to 0. 12.4. Prove that the sequence { }
converges to 0.
12.5. Prove that the sequence {1 +
converges to 1.
12.6. Prove that the sequence { 5 ^5 } converges to 12.1. By definition, lim„^oo an = L if for every e > 0, there exists a positive integer N such that if n is an integer with n > N , then \a„ — L  < e. By taking the negation of this definition,write outthe meaning of lim„^oo <2/i L using quantifiers. Then write out the meaning of {a,,} diverges using quantifiers. 12.8. Show that the sequence {n4} diverges to infinity. 12.9. Show that the sequence
j j
diverges to infinity.
12.10. (a) Prove that I + 5 + 5 +  1 ^ < 2y/n for every positive integer n. (b) Let s„ = yt + ^ ^ + ■■■+ ^2 for each n e N. Prove that the sequence {s„} converges to 0. 12.11. Prove that if a sequence {5,,} converges to L, then the sequence {s„2} also converges to L.
Section 12.2: Infinite Series 12.12. Prove that the series YiT=\ (3^2)W+i) converges and determine its sum by (a) computing the first few terms of the sequence {s„} of partial sums and conjecturing a formula for sn\ (b) using mathematical induction to verify that your conjecture in (a) is correct; (c) completing the proof. 12.13. Prove that the series Y1T= 1 F converges and determine its sum by (a) computing the first few terms of the sequence {s„} of partial sums and conjecturing a formula for sn; (b) using mathematical induction to verify that your conjecture in (a) is correct; (c) completing the proof. 12.14. The terms a \ , aj. a 3 , ■■■of the series
1
ak 316 defined recursively by ci\ — g and 2
Cl11 — Cl17—1
n(n + l)(n + 2) for n > 2. Prove that 12.15. Prove that the series 12.16. 12.17.
ak converges and determine its value. (t+W diverges t0 infinity.
(a) Prove that if YlkLi ak is a convergent series, then lim„^oo an = (b) Show that the converse of the result in (a) is false. Let
0.
Y a L\ ak be an infinite series whose sequence of partialsumsis{s„} where s„ = 4 ^ 
(a) W hat is the series Y/aL 1 ak } (b) Determine the sum s of
ak and prove that
ak — s 
318
Chapter 12
Proofs in Calculus
Section 12.3: Limits of Functions 12.18. Give an e —<5proof that limv_>2 (§x + l) = 4. 12.19. Give an e —S proof that lim v^ _ i (3x — 5) = —8. 12.20. Give an e —S proof that lim v_>2 (2x2 — x — 5) = 1. 12.21. Give an e —S proof that lim v^ 2 12.22. Determine limx^ i
* 3
= 8.
and verify that your answer is correct with an e — S proof.
12.23. Give an e — <5 proof that lim v_3 12.24. Determine limA^ 3
'iZ&Zis and verify that your answer is correct with an e — S proof.
12.25. Show that lim t ^ o p  does not exist. 12.26. The function / : R —»■ R is defined by
I
I 1.5
x < 3 x = 3
2
x > 3.
(a) Determine whether limA_*.3 f i x ) exists and verify your answer. (b) Determine whether lim.v^ ^ f ( x ) exists and verify your answer. 12.27. A function g : R —» R is bounded if there exists a positive real number B such that g(x) < B for each X € R. (a) Let g : R —> R be a bounded function and suppose that / : R —►R and a e R such that lim.v^a f i x ) = 0. Prove that limx_>a f ( x ) g ( x ) = 0. (b) Use the result in (a) to determine limA_>ox2 sin (J j). 12.28. Suppose that lim v_„0 f i x ) = L, where L > 0. Prove that lim v>a s/ f i x ) — v T . 12.29. Suppose that / : R » R is a function such that limA_*o f ( x ) = L. (a) Let c 6 R. Prove that lim v_,.c f i x — c) = L. (b) Suppose that / also has the property that / ( a + b) — f i a ) + f i b ) for all a, b e R. Use the result in (a) to prove that limA^ r f i x ) exists for all c e R. 12.30. Let / : R —» R be a function. (a) Prove that if limA^ 0 f i x ) — L, then limA^ a / ( x )  = \L\. (b) Prove or disprove: If limA_»a / ( x )  = L , then limA.^ a f i x ) exists.
Section 12.4: Fundam ental Properties of Limits of Functions 12.31. Use limit theorems to determine the following: (a) lim M  (x3 — 2x2 — 5x + 8) (b) limA_* i (4x + 7)(3x2 — 2) (c) l i m ^ g f } 12.32. Use induction to prove that for every integer n > 2 and every n functions f , f 2, ■■■, f„ such that lim f (x) = Li for 1 < i < n, x*a
lim ( /i( x ) + f 2{x) + • • • + fn ix )) = L \ + L 2 + • • • + L n.
319
Additional Exercises for Chapter 12 12.33. Use Exercise 12.32 to prove that ]irnv_*fl p( x) = p(a) for every polynomial p( x) = cnx n + cn \ x n~ x + ■• ■+ c\ x + Co12.34.
any n > 2 functions such that limA^ a f , ( x ) = L for 1< i <
Prove that if fa, f a , . . . , /„ are
n, then
lim (/i(x ) • fa(x) ■• • f n(x j) = L i ■L 2  ■■L n. X * d
Section 12.5: Continuity 12.35.
The function / : R — (0, 2} > R is defined by f ( x ) = can be defined at 2 such that / is continuous at 2.
■
Use limit theorems to determine
12.36. The function / defined by f ( x ) = is not defined at 3. Is it possible to define / at 3 such that / is continuous there? Verify your answer with an e — 8 proof. 12.37.
Let / : R >• Z be the ceiling function defined by f ( x ) = f x l. Give ane — S proof that if a is a real number that is not an integer, then / is continuous at a.
12.38.
Show that Exercise 12.33 implies that every polynomialis continuous at every real number.
12.39.
Prove that the function / : [1, oo) > [0. oo) defined by f ( x ) = V * — 1 is continuous at jc = 10.
12.40. (a) Let / : R —> R be defined by ft \ _ "
 0 if x is rational { 1 if * is irrational.
In particular, / ( 0 ) = 0. Prove or disprove: / is continuous at x = 0. (b) The problem in (a) should suggest another problem to you. State and solve such a problem.
Section 12.6: Differentiability 12.41. The function / : R —> R is defined by f ( x ) = x 2. Determine / '( 3 ) and verify that your answer is correct with an e — S proof. 12.42. The function / : R — {—2} —>■R is defined by f i x ) = is correct with an e — S proof. 12.43. The function / : R » R is defined by f ( x ) answer is correct with an e — S proof. 12.44. The function / : R
= x 3.
Determine / ' ( l ) and verify that your answer Determine f ' ( a ) for a e R + and verify that your
—»•R is defined by
/ M = P S‘n' I
0
^
if x = 0.
Determine /'( 0 ) and verify that your answer is correct with an e — 8 proof.
ADDITIONAL EXERCISES FOR CHAPTER 12 12.45. Prove that the sequence { }
converges to i .
12.46. Prove that lim„^oo 12.47. Prove that the sequence {1 + (—2)"} diverges. 12.48. Prove that lim ^ o o (V « 2 + 1 —n) = 0.
320
Chapter 12
Proofs in Calculus
12.49. Prove that the sequence {(—1)'!+1^
t j } diverges.
1 12.50. Prove that lim wso 3 n + 1 3 12.51. Let a, Co, Ci R such that ci / 0. Give an e — <5 proof that lim vs.a (cix + Cp) = c\a + Co12.52. Evaluate the proposed solution of the following problem. P roblem The function ./ : R > R is defined by
f i x)
if x = 2.
Determine whether lim f i x ) exists. x —>2
Solution
Consider x2 —4 (x — 2)(x + 2) lim f ( x ) = l i m  = l i m  = lim (x + 2) = 4. x —>2 X — 2 T—>2 x — 2 x + 2 '
However, since lim / ( x ) = 4 / 2 = /( 2 ) , the limit does not exist. x^  2 12.53. Evaluate the proposed proof of the following result. . 2n 1 2 converges to  . Result The sequence 3/7 + 5  3 P ro of
10
Let e > 0 be given. Choose N
10
10 5 and let n > N . Then n > — . 9e 3
97
10
Hence 9n >  15 and 9n + 15 > — . Therefore, e 6 9 n + 15 lO
1 > e
10
and
9 n + 15
Now 2n
2
6n — 2(3n + 5)
1 10
10
3n + 5
3
3(3re + 5)
9 n + 15
9n + 15
< 6.
12.54. Evaluate the proposed proof of the following result. R esult lim — — = —1. Jr> i 2x — 3 Proof L e tf > 0 and follows that 0 < x < 2
choose <5= m in(l, ^ f) .L e t0 < and so 2x 
\x—Ij< S= m in ( l, ?f). Since \x — 1 < 1, it
31< 2x +   3 = 2 x  + 3< 4 + 3 = 7.
Since \x — 1 < 7 e/2 , we have 1 2x  3
+ 1
2x  2
2i.v  11
2x — 3
js  s i < 7' y
2
7e ~ t'
12.55. Let {a„}, {bn} and {c„} be sequences of real numbers such that a„ < bn < c„ for every positive integer n and lim a„ = lim c„ = L. (a) Prove that lim (cn — a,,) = 0. n— >00 (b) Prove that lim bn = L.
Additional Exercises for Chapter 12
321
12.56. In Chapter 10 it was shown that the set Q of rational numbers is denumerable and consequently can be expressed as Q = {<71, q2, q},...} . A function / : R >â– R is defined by fix) =
 i f * = qâ€ž in = 1, 2, 3 , . . . ) 0 if x is irrational.
(a) Prove that / is continuous at each irrational number. (b) Prove that / is not continuous at any rational number. (c) If the function / were defined as above except that / ( 0 ) = 0, then prove that / would be continuous at 0.
13! Proofs in Group Theory
any o f the proofs that w e have seen involve fam iliar sets o f num bers (especially integers, rational num bers and real num bers). A lso, m ost o f the theorem s and exam ples that w e have encountered concern additive properties o f these num bers or m ultiplicative properties (or both). M any im portant properties o f integers (or o f rational or real num bers) com e not from the integers them selves, but from the addition and m ultiplication o f integers. This suggests a basic question in m athem atics: For a given nonem pty set S, can w e describe other, less fam iliar m ethods o f associating an elem ent o f S w ith each pair o f elem ents o f S in such a w ay that som e interesting properties occur? The m athem atical subject that deals w ith questions such as these is a b s tra c t a lg e b ra (also called m o d e rn a lg e b ra or sim ply a lg eb ra). In this chapter, w e w ill look at one o f the m ost fam iliar concepts in abstract algebra. First, however, w e m ust have a clear understanding o f w hat w e m ean by associating an elem ent o f a given set w ith each pair o f elem ents o f that set and w hat properties m ay be considered interesting.
M
13.1 Binary O perations
322
W hen we add tw o integers a and b, we perform an operation (nam ely addition) to produce an integer that we denote by a + b. Similarly, w hen w e m ultiply these tw o integers, we perform another operation (nam ely m ultiplication) to produce an integer that w e denote by a ■b (or ab). Both o f these operations do som ething very similar. E ach takes a pair a. b o f integers, actually an ordered pair (a, b) o f integers, and associates w ith this pair a unique integer. Therefore, these operations are actually functions, nam ely functions from Z x Z to Z. These functions are exam ples o f a concept called a binary operation. By a b in a ry o p e ra tio n * on a nonem pty set S, w e m ean a function from S x S to S; that is, * is a function that m aps every ordered pair of elem ents o f S to an elem ent of S. Thus * : S x S —» S. In particular, if the ordered pair (a. b) in S x S is m apped into the elem ent c in S by a binary operation * (that is, c is the im age o f (a, b) under *), then w e w rite c — a * b rather than the m ore aw kw ard notation * ((a, b)) = c. Consequently, addition + and m ultiplication • are binary operations on Z. For ex am ple, under addition, the ordered pair (3, 5) is m apped into 3 + 5 = 8; w hile under m ultiplication, (3, 5) is m apped into 3  5 = 15. Subtraction is also a binary operation on Z but it is not a binary operation on N since, for exam ple, subtraction m aps the ordered
13.1
Binary Operations
323
pair (3, 5) into 3 — 5 = —2, w hich does not belong to N. Therefore, subtraction is not a function from N x N to N since it is not defined at (3, 5), as w ell as at m any other ordered pairs o f positive integers. Similarly, division is not a binary operation on Z or N because it is not defined for m any ordered pairs, including (1, 0) € Z x Z and (2, 3) e N x N because 1 /0 ^ Z and 2 /3 ^ N. However, division is a binary operation on the set Q + o f positive rational num bers since the quotient o f tw o positive rational num bers is once again a positive rational number. N ot only are addition and m ultiplication binary operations on Z, they are binary operations on Q and R, as w ell as on R + (the positive real num bers) and Q + . For the set R* o f nonzero real num bers, m ultiplication is a binary operation but addition is not (since 1 + ( —1) = 0 ^ R*, for exam ple). If * is a binary operation on a set 5, then, by definition, a * b e S for all a, b 6 S. If T is a nonem pty subset o f S and a , b e T , then certainly a * b e S; however, a * b need not belong to T. A nonem pty subset T o f S is said to be closed under * if whenever, a , b € 7 \th e n a * £ > e T as well. If * is a binary operation on S, then certainly S’is closed under *. A lthough subtraction is a binary operation on Z. the subset N o f Z is not closed under subtraction. A m ong fam iliar sets w ith fam iliar binary operations are: (a) the set Z„ = {[0], [1], • • • , [ « — 1]}, n > 2, o f residue classes m odulo n, under residue class addition [a] + [b ] = [a + b] and under residue class m ultiplication [a] ■[fo] — [ab] (as defined in C hapter 8); (b ) the set M 2(R ) o f all 2 x 2 m atrices over R (that is, w hose entries are real num bers) under m atrix addition a c
b] + d
re g
f h
a + e c + g
b + f d + h
and under m atrix m ultiplication a c
b d
e J
f h .
ae + bg ce + d g
a f + bh c f + dh
(c) the set j r R = R r o f functions from R to R under function addition ( / + g)(x) = f ( x ) + g(x), under function m ultiplication ( / • g) (x) = f ( x ) ■g ( x) and under function com position ( / o g) (x) = f { g ( x ) ) \ (d) the pow er set V ( A') o f a set A under set union, under set intersection and under set difference. For a m ore abstract exam ple o f a binary operation, let S = [a, b, c}. A binary oper ation * on 5 is illustrated in the table in Figure 13.1, w here, then a * a = b, a * b = c, a * c = a, etc. Since every elem ent in the table belongs to S, it follow s that * is indeed a binary operation on S. A lthough it m ay seem relatively clear that a binary operation is defined in each exam ple given above, not all binary operations are so im m ediate. Result 13.1
For a , b £ R — {—2}, define a * b = a b + 2a + 2b + 2,
324
Chapter 13
Proofs in Group Theory
Figure 13.1
*
a
b
c
a
b
c
a
b
a
c
a
c
c
a
b
A binary operation * on S = {a,b,c}
where the operations indicated in a b + 2 a + 2 b + 2 are ordinary addition and multi plication in R. Then * is a binary operation on R — {—2}. Proof
We need to show that if a, b e R — {—2}, then a * b e R — {—2}. A ssum e, to the con trary, that there exists som e pair a , y e R — {—2} such that x * y £ R — {—2}. Thus x * y = x y + 2x + 2 y + 2 =  2 . This equation is equivalent to (x + 2){y + 2) = 0, so either x — —2 or y = —2, w hich is im possible. H ence * is a binary operation on
R ~ {—2}.
■
A nonem pty set S w ith a binary operation * is often denoted b y (S, *). We refer to (S, *) as an a lg e b ra ic s tru c tu re . There are certain properties that (S, *■) may possess that w ill be of special interest to us. In particular, G I (S, *) is associative if a * (b * c) — (a * b) * c for all a , b , c e S; G2 (S , *) has an elem ent e, called an id e n tity elem e n t (or sim ply an id e n tity ), if a * e = e * a = a for each a e S; G3 (S, *) has an identity e and, for each elem ent a € S, there is an elem ent s € S, cafled an in v erse for a, such that a * s = s * a = a; G4 (5, *) is co m m u ta tiv e if a * b = b * a for all a , b e S. Two elem ents a, b e S are said to co m m u te i f a * b = b * a . l i every tw o elem ents o f S com m ute, then (5, *) satisfies property G4. B y property G2, an identity com m utes w ith every elem ent o f S', and by property G3, each elem ent o f S com m utes w ith an inverse o f this elem ent (assum ing, o f course, that it has an inverse). A n algebraic structure (S, *) m ay satisfy all, som e or none o f the properties G I  G4; however, (S, *) cannot satisfy G3 w ithout first satisfying G2. For elem ents a , b , c e S, the expression a * b * c is, strictly speaking, not defined. Since * is a binary operation, it is only defined for pairs o f elem ents o f S. There are two standard interpretations of a * b * c . N amely, does a * b * c m ean a * (b * c) or does a * b * c m ean (a * b) * c? O n the other hand, if (S , *) satisfies property G I (the associative property), then a * ( b * c ) = (a * b ) * c and so either interpretation is acceptable in this case. For this reason, w e often w rite a * b * c (w ithout any parentheses). Ordinarily, however, w e w ill continue to w rite a * (b * c) or (a * b) * c to em phasize the im portance of parentheses, even w hen (S, *) satisfies the associative property. C ertainly (Z. + ) satisfies properties G I  G4, w here 0 is an identity elem ent and —n is an inverse for the integer n. M oreover, (R, •) satisfies properties G I, G2 and G4, w here the integer 1 is an identity elem ent. Turning to property G3, w e see that every
13.1
Binary Operations
325
real num ber r has 1/ r as an inverse, except 0, w hich has no inverse since there is no real num ber j such that 0 • s — s • 0 = 1. Thus (R , •) does not satisfy G3. In the case o f (R*, •), w here, recall, R* is th e s e t o f all nonzero real num bers, all four properties G 1 G 4 are satisfied. The algebraic structure (Z„, + ), n > 2 , also satisfies all o f the properties G1  G4, w here [0] is an identity and [—a] is an inverse o f [a]. On the other hand, (Z „ , •) satisfies only G l, G2 and G4, w here [1] is an identity; how ever (Z„, •) does not satisfy G3 since, for exam ple, there is no elem ent [s] e Z„ such that [0][s] = [1] in Z„ and so [0] does not have an inverse. The algebraic structure (M 2(R), + ) satisfies all o f the properties G l  G4, w here a is an identity and The algebraic is an inverse o f c structure (M 2(R), •) only satisfies G l and G2, w here I =
is an identity. For
0
a b to have an inverse, the num ber a d — be (the determ inant o f A) m ust be c d nonzero. T hus ( Mz i R) , •) does not satisfy G3. A lso, A
'1 0
o' 0
'0 0
1■ '0 — 0 0
1' 0
'0 0
'0 o' — 0 0
1' '1 0 _0
o' 0
shows that (M 2(R), •) does not satisfy property G4. T he algebraic structure (S , *) show n in F igure 13.1 is not associative since, for exam ple, b * (b * c) — b * a = a, w hile ( b * b ) * c = c * c = b and so b * (b * c) (b * b ) * c. Since S contains no elem ent e such that e * x = x * e = x for &\l x e S, it follow s that (S, *) does not have an identity. A lso, a * c ^ c * a since a * c = a and c * a — c. Consequently, (S, *) has none o f the properties G 1G 4. L et’s look at another binary operation defined on S = {a, b, c}. Example 13.2
A binary operation * is defined on the set S = {a , b, c} by x * y  x f o r all x , y e S. Determine which o f the properties G l  G 4 are satisfied by (5 , *).
Solution
L e l x , y and z be any three elem ents o f S (distinct or not). T h e n x * (y * z ) = x * j = x , w hile ( x * y ) * z = x * z = x. T hus (S , *) is associative. Now (S , *) has no identity since for every elem ent e 6 S, it follow s that e * a = e * b = e and so it is im possible for e * a — a and e * b = b. Since (5, *) has no identity, the question o f inverses does not apply here. Certainly, (5, *) is not com m utative since a * b — a w hile b * a — b. ♦ T he verification o f the associative law in Exam ple 13.2 probably w ould have looked better had we w ritten x * (y*z) = x * y = x = x * z = (x*y)*z.
Example 13.3
Let No be the set o f nonnegative integers a n d consider (No, *), where * is the binary operation defined by a * b = \a — b\ f o r all a , b e No Determine which o f the f o u r properties G l  G4 are satisfied by (No, *).
Soluti on
Since 1 * ( 2 * 3 ) = 1 *  2 — 3] = 1 * 1 = 1 — 11 = 0 and (1 * 2) * 3 = 11 — 2  * 3 = 1 * 3 = 11 — 3 1 = 2, it follow s that (1 * 2) * 3 ^ 1 * (2 * 3) and so (No, *) is not
326
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associative. L et a e No Then a * 0 = 0 * a = \a\ = a and so 0 is an identity for (No, *). Since a * a = \a — s [ . s = 0 for all a e N o , it follow s that a is an inverse o f itself. Because \a — b\ = \b — a , we have a * b = b * a and (N o, *) is com m utative. H ence (N o, *) satisfies properties G2  G4. ♦ Example 13.4
Solution
Let * be the binary operation defined on Z by a * b = a + b — 1 f o r a, b e Z, where the operations indicated in a + b — 1 are ordinary addition and subtraction. Determine which o f the f o u r properties G1  G 4 are satisfied by (Z, *). For integers a. b and c,
a * (b * c) = a * (b + c — '1) = a + (b + c — 1) — 1 = a + b + d — 2, w hile (a * b) * c = (a + b — I) * c — (a + b — Y) + c — ] = a + b + c — 2 . Thus a * (b * c) — (a * b) * c and (Z, *) is associative. Since a + b — 1 = b + a — 1, it follow s that a * b — b * a for all a, b e Z and so (Z, *) is com m utative. L et a be an integer. O bserve that a * \ ~ a + \ ^ \ = a. Thus 1 is an identity for (Z, *). For b = —a + 2 e Z, w e have a * b — a * (—a + 2) = a + (—a + 2) — 1 = 1. H ence b is an inverse o f a and every integer has an inverse. Therefore, (Z, *) satisfies all four properties G1  G4. ♦ Analysis
L e t’s discuss this exam ple a bit more. It w as show n that 1 is an identity for (Z, *). How did w e know to choose 1? Actually, this was a natural choice since w e w ere looking for an integer e such that e * a — a for every integer a. Since e * a = a + e — I — a, it follow s that e = I. The choice o f b = —a + 2 for an inverse o f a com es from solving a * b = a + b — 1 = 1 for i>. ♦
13.2 Groups O ne o f the m ost elem entary, yet fundam ental, characteristics o f the algebraic structure (Z, + ) is the ability to solve linear equations, that is, equations o f the type a + x — b. By this, we m ean that given integers a and b, w e seek an integer x for w hich a + x = b. H ow does one solve this equation? First, w e are w ell aw are that (Z, + ) has an identity elem ent 0. A lso, a has —a as an inverse. If we add —a to a + x , w hich is the same as adding —a to b since a + x and b are the sam e integer, then we obtain —a + (a + x ) = —a + b. A pplying the associative law to (13.1), w e now obtain (—a + a) + x = —a + b and so 0 + x = x = —a + b.
(13.1)
13.2
Groups
327
This tells us that i f a + x = b has a solution, then the only possible solution is —a + b. It doesn’t tell us that —a + b is actually a solution, but we can easily take care o f this. L etting x = —a + b, we have a + x = a + {—a + b) = (a + (—a)) + b — 0 + b = b. O f course, since (Z, + ) also satisfies the com m utative property, the solution —a + b can also be w ritten as b + ( —a ) = b — a. L e t’s look at the related linear equation w hen the operation is m ultiplication, say in (R*, ■)• Here, for a , b e R*, w e seek i e R * such that a ■x = b. Recall that 1 is an identity elem ent in (R*, •). M ultiplying both sides o f a ■x = b by ^ (w hich belongs to R*), we obtain — ■(a ■x ) = — • b = a a a
(13.2)
A pplying the associative law to (13.2), w e obtain
T herefore, x — b / a . To show that ^ is. in fact, a solution o f a ■x = b, w e let x —  and obtain
The solutions o f the tw o equations that w e have ju s t discussed should look very fam iliar to you. However, these have been given to illustrate a m ore general situation. Suppose that we have an algebraic structure (S , *) in w hich w e w ould like to solve all linear equations, that is, for a , b e S, we w ish to show that there exists an elem ent x e S such that a * x = b. If (S, *) is associative, has an identity e and a has an inverse s e S, then w e have s * (a * x ) = s * b and so s * (a * x ) = (s * a) * x = e * x = x = s * b. To show that s * b is, in fact, a solution of the linear equation a * x = b, w e let x = s * b and obtain a * x = a * ( s * b ) = (a * s ) * b = e * b = b , and so s * b is a solution. You m ight now have observed that in order to solve all linear equations in an algebraic structure (S , *), it is necessary that (S, *) satisfy the three properties G I  G3 (property G4 is not required). A lgebraic structures that satisfy properties G I  G3 are so im portant in abstract algebra that they are given a special nam e and w ill be the m ajor em phasis of this chapter. A g ro u p is a nonem pty set G together w ith a binary operation * that satisfies the follow ing three properties: G I A ssociative L aw : a * (b * c) = (a * b) * c for all a, b, c e G; G2 E x iste n ce o f Id e n tity : T here exists an elem ent e e G such that a * e = e t a = a for every a e G; G3 E x iste n ce o f In v erses: For each elem ent a e G, there exists an elem ent s e G such that a * s = s * a = e.
328
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+
[0]
[1]
[2]
[0]
[1]
[2]
[1]
[2]
[0]
2]
[0]
Figure 13.2
fl]
+
[0]
[1]
[2]
[3]
[0]
[1]
[2]
[3]
[1]
[2]
[3]
[0]
[2]
[3]
[0]
[1]
[3]
[0]
[1]
[2]
The group tables for (Z3> + ) and (Z4, +)
H ence, a group is a special kind o f algebraic structure (G , *), nam ely, one that satisfies properties G l  G3. If the operation * is clear, then we often denote this group sim ply by G rather than by (G , *). A n elem ent e e G satisfying property G2 is called an id e n tity for the group G , w hile an elem ent s satisfying property G3 is called an in v erse o f a. If a group G also satisfies the com m utative property G4. then G is called an a b e lia n g ro u p , a term nam ed after the N orw egian m athem atician N iels H enrik A bel. If a group G does not satisfy G4, then G is called a n o n a b e lia n g ro u p . We have now seen several abelian groups, namely, (Z, + ), (Q , + ) , (R . + ), (Z„, + ), (Q + , •), (R + •) and (R*, •) as w ell as the algebraic structure (Z. *) described in E xam ple 13.4. The o r d e r o f a g ro u p G , denoted by G , is the cardinality o f G. If the order of G is finite, then G is a finite g ro u p ; w hile if G has an infinite num ber o f elem ents, then G is an in fin ite g ro u p . A ll o f the groups given above are infinite groups except for ( l n, + ) w hich has order n. If a finite group G has relatively few elem ents, then we often describe the operation * by m eans o f a table, called a g ro u p ta b le (or o p e ra tio n tab le). For exam ple, the group tables for (Z 3, + ) and (Z 4, + ) are show n in Figure 13.2. A lthough (Z„, + ) is a group for every integer n > 2, (Z„, •) is not a group for any n > 2, since, as we have m entioned, the elem ent [0] has no inverse. T his suggests con sidering the set Z* = Z„ — {[0]} = {[1], 2 . • • • , [ « — 1]}, n > 2, under m ultiplication. F or som e integers n > 2, m ultiplication is not a binary operation on Z*. For exam ple, [2] e Z^ but [2] • [2] = [0] £ Z \ . On the other hand, m ultiplication is a binary operation on Z i. In fact, ( Z , •) satisfies properties G l, G 2 and G4, w here [1] is an identity. Since [1] • [1] = [1], [2] • [3] = [3] • [2] = [1] and [4] • [4] = [1], every elem ent o f (Z „ •) has an inverse. T herefore, ( Z , •) satisfies property G3 as well and so is an abelian group. This, o f course, brings up the question o f w hich algebraic structures (Z*, •) are groups. Perhaps the exam ples above suggest the answer. T h e o re m to P ro v e PROOFSTRATEGY
The set Z*, n > 2, is a group under m ultiplication if and only if n is a prim e. If n is a com posite num ber, then there exist integers a and b such that 2 < a. b < n 1 and n = ab. So [a], [b ] e Z* and [«][£>] = [11] — [0] ^ Z* and m ultiplication is not a binary operation. Therefore, the only possibilities for (Z*, ■) to be a group is w hen n is a prim e. Suppose then that p is a prim e. First, w e need to verify that m ultiplication is in fact a binary operation on Z * ; that is, if [a], [b] e Z *, then [a][b] e Z * . If [ab] £ Z;,, then
13.2
Groups
329
[ab] —[0];s o a & s O (m o d p), w hich im plies that p  ab. By C orollary 11.14, p \ a or p b and so [a] = [0] or [b] = [0], T hat is. either [a] £ Z* or [b] <£ Z * , a contradiction. To show that (Z * , •) is a group, it rem ains only to verify that property G3 is satisfied. L et r be an integer w ith 1 < r < p — 1. We need to show that [r] has an inverse; that is, there exists [5] e Z* such that [r][s] = [1], Since p is a prim e and 1 < r < p — 1, the integers r and p are relatively prim e. By T heorem 11.12, the integer 1 is a linear com bination o f r and p. So, there exist integers x and y such that 1 = r x + p y . Using the definition o f addition and m ultiplication in Z p and observing that [p] = [0] in Z p , w e have [1] = [rx + py ] = [rx ] + [py] = [r] ■[x] + [p] ■[>:] = [r] ■I x ] + [0] • [y] = [r] • \x] + [0] = [r] • [x]. H ence [x] is an inverse for [r],
♦
We now give a concise proof o f the theorem . Theorem 13.5 Proof
The set Z*, n > 2, is a group under multiplication i f and only i f n is a prime. A ssum e that n is a com posite number. Then there exist integers a and b such that 2 < a, b < n — 1 and n = ab. H ence [a], [b] e Z* and [a][b] = [n] = [0] ^ Z*, w hich im plies that m ultiplication is not a binary operation on Z*. Certainly then, Z* is not a group under m ultiplication. For the converse, assum e that p is a prim e. First, w e show that m ultiplication is a binary operation on Z * . A ssum e, to the contrary, that it is not. Then there exist [a], [b] e Z* such that [a][b] £ Z*p . Since [a][b] i Z*p , it follow s that [«][£] = [ab] = [0], Thus ab = 0 (m o d p ) and so p \ ab. By C orollary 11.14, p  a or p \ b. T herefore, [a] = [0] or [fo] = [0], w hich contradicts the fact that [a], [b] e Z * . Here [1] is the identity. H ence (Z * , •) is an algebraic structure that satisfies properties G1 and G2. It rem ains to show that (Z*, •) satisfies property G3. Let [r] e Z*, w here w e can assum e that 1 < r < p — L Since r and p are relatively prim e, 1 is a linear com bination o f r and p by T heorem 11.12. Thus 1 = r x + p y for som e integers x and y. So [1] = [rx + py] = [rx] + [py]  [r] •[x]+ [p] ■
[y]
= M • [x] + [0] • [>’] = [r] ■[x]. Thus, [x] is an inverse for [r] and (Z * , •) is a group.
■
By T heorem 13.5, for every prim e p , (Z*, ■) is an abelian group o f order p — 1. A nother exam ple o f an abelian group is (G , *), w here G = {a, b, c} and * is defined in F igure 13.3. It is not difficult to see that a is an identity for (G , *) and that a , b and c are inverses for a, c and b, respectively. Since a * b — b * a, a * c = c * a and b * c — c * b , it follow s that G is abelian. We have one additional property to verify to show that G is a group, however, nam ely, the associative property. W hat w e are required to show is that x * ( j * z) = (x * y ) * z for all x , y , z e G. Since there are three choices for each o f x, y and z, w e have 27 equalities to verify. B ecause b * ( c * b ) = b * a = b and (b * c) * b = a * b = b, it follow s that b * (c * b) = (b * c) * b. Since the rem aining 26 equalities can also be verified, G is, in fact, an abelian group.
330
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*
CL,
a
a
b
c
b
c
a
c
a
b
Figure 13.3
An abelian group with three elements
We m entioned earlier that the algebraic structure (Af2(R), •) does not satisfy property G3 even though it does satisfy property G2. The m atrix I —
^ 0
^ 1
is an identity for
a b e M 2(R) has an inverse if and only if c d its determ inant det A = a d — be ^ 0. In this case, (M 2(R ), •)• Furtherm ore, a m atrix A —
B =
1 a d — be
d —c
b a
is an inverse for A and so A B = B A = I . Let M 2*(R) = {A e M 2 {R ) : det A / 0}. Since det(A B ) = det(A ) ■d e t(5 ) for all A, B e M2(R), it follow s that if A, B c M (R ), then A B e (R) and so M 2*(R) is closed under m atrix m ultiplication. This im plies that (M J(R ), •) is a group. On the other hand, since the m atrices A =
and B
1
belong to M ( R ) and AB =
and B A =
it follow s that ( M ( R ) S ■) is a nonabelian group.
13.3 Perm utation Groups O ne o f the m ost im portant classes o f groups concerns a concept introduced in C hapter 9. R ecall that a p e r m u ta tio n o f a nonem pty set A is a bijective function / : A —> A; that is, / is onetoone and onto. In C hapter 9, it was show n that: (1)
the com position o f every two perm utations o f A is a perm utation o f A;
(2)
com position o f perm utations o f A is an associative operation;
(3)
the identity function perm utation o f A ;
(4)
every perm utation o f A has an inverse, w hich is also a perm utation o f A.
:A
A defined by iA(a) = a for all a e A is a
13.3
Perm utation Groups
331
By a p e r m u ta tio n g ro u p , w e m ean a group (G , o ), w here G is a set o f perm utations o f som e set A and o denotes com position. Let 5,4 denote the set o f all perm utations o f A. Then by ( l)  ( 4 ) above, w e have the follow ing result. Theorem 13.6
For every nonempty set A, the algebraic structure (S a , °) is a permutation group. The group (Sa, °) is called the sy m m e tric g ro u p on A. Therefore, every sym m etric group is a perm utation group. We have already noted that for the set JFr of all functions from R to R , (.F r, o ) is an algebraic structure, w here o denotes function com position. By T heorem 13.6, for the set S R o f bijective functions from R to R, the algebraic structure ( S r , o ) is a group, namely, the sym m etric group on R. The identity function z'r in S r , defined by /r ( x ) = x for all x e R , is an identity in the group (S r, o ) . The functions / and g defined by
f ( x ) = x + 1 and g (x ) = 2 x for all x e R belong to S r; however, ( / o g)(x) = f(g(x)) = / ( 2x) = 2x + 1 and
(g ° f)(x) = g(f(x)) = g(x + 1) = 2x + 2. Since / o g ^ g o / , it follow s that ( S r , o ) is a nonabelian group. If A — {1, 2, ■• •, n], w here n e N, the group Sa is com m only denoted by S„. The group (Sn, o) has n! elem ents (see Theorem 9.7) and is called the sy m m e tric g ro u p (of degree n ) . T he sym m etric group (S„, o) is therefore a finite group o f order n l . By the notation introduced in C hapter 9, S 3 — { a j, 0 1 2 , 0 1 3 , <24, a$, a 6}, where 1 a i _ , l
2 2
3\ 3 /
(1 “2 _ l l
2 3
3\ 2
013
2 2
3 1
1
2 1
3\ 3J
fl “s _ ^2
2 3
3\ 1J
( 1 2 a6~ \ 3 1
3 2
014
~ \2
(\ ~ \ 3
Recall that w ith this notation for a perm utation, an elem ent o f {1, 2, 3} listed in the first row m aps into the elem ent in the second row directly below it. Thus a \ is the identity o f S3. L et’s consider the com position a 3 o a 6. For exam ple, ( a 3 o a 6) ( l) = a 3( a 6( l) ) = a 3(3) = 1. A lso, (o?3 o a 6 )(2) = 3 and (a 3 o ag)(3) = 2. Therefore, “3 0 « 6 = [ ,
2 2
3\
/I
3
2 j
3\
(I = ^
2
3
3 2  = « 2
Sim ilarly «6 0 «3 = «4 H ence 03 o a 6 ^ a 6 o 03. Therefore, (S3, o) is a nonabelian group. T his shows that there is a nonabelian group o f order 6. T here is no nonabelian group having order less than 6, however. W hen taking the com position o f two elem ents a, /3 e Sn, w here n e N, we often w rite a o ft as aft and say that w e are multiplying a and yS. W ith this notation, we have «3a6 = a 2, «6«3 = «4, = 0(3013 = a i and a \ = — 0 1 5 . The group table for (S3, o) is show n in Figure 13.4 (containing all 36 products!).
Chapter 13
Proofs in Gi’onp Theory OL\
OL2
Cx3
O'4
tti
Oi2
a3
CI4
Cx.g
CK5
CK5
«6 Q?4
OL2
Q!5
«3
a3
CKl
Q?4 L CK2
CL2
CX1
Ct4
04
05
«5
Q!4
OL2
«6
OL6
03
(I4
Figure 13.4
^3
C(!2
a\ Ctl
Q!2
OL5
The group table for (53, o)
A perm utation group need not consist o f all perm utations o f som e set A . For exam ple, if w e consider the subsets G i = {ai, 0 1 2 } and G t = {ai, 0(5, a^} o f S3, then ( G 1, o) and (G2, o) are both perm utation groups. Their group tables are show n in Figure 13.5. A lso, let
a
:)
fto
= 0
1
3
1
3
AC
2
5 5) aG 5
4
;)
1)
be perm utations o f the set {1. 2, 3, 4} and let G3 = { ft. f t , f t , f t} . T hen (G3, o) is an abelian perm utation group, w hose group table is show n in Figure 13.6. D uring the latter part o f the 18th century, a m ajor problem in m athem atics co n cerned w hether the roots o f every fifth degree polynom ial w ith real coefficients could be o.\
a2
Oil
Ck!2
0 ;^
«5
<^6
Q ?6
:W S .
C*2
Ot§
Figure 13.5
cc6
O '!
& 5
The group tables for (G lf o) and (G2, o) ft
ft2
ft
0A
ft
ft
ft
ft
ft
ft
ft
ft
ft
ft
/?3
ft
ft
ft
ft
/?4
ft
ft
ft
ft
Figure 13.6
The group table for (G3, o)
13.4
Fundam ental Properties of Groups
333
expressed in term s o f radicals and the usual operations o f arithm etic. It w as w ell know n that the roots o f a quadratic polynom ial a x 2 + b x + c, w here a , b , c e R and a ^ 0, are (—b + \ f b 2 — 4 a c) / 2 a and {—b — *Jb 2 — 4 a c ) / 2 a , w hich is a consequence o f the quadratic form ula. Furtherm ore, it had been know n since the 16th century th at the roots o f all third degree (cubic) and fourth degree (quartic) polynom ials w ith real coefficients could be described in term s o f radicals and the standard operations o f arithm etic. But fifth degree polynom ials proved to be another story. However, in 1824 N iels H enrik Abel proved that the roots o f fifth degree polynom ials w ith real coefficients could not. in gen eral, be expressed in such a way. F rom this, it follow s that for every integer n > 5, there exist polynom ials of degree n w ith real coefficients w hose roots cannot be expressed in term s o f radicals and the standard operations o f arithm etic. H is w ork w ent unnoticed, however, until after his death at age 26. Som e tim e later the F rench m athem atician E variste G alois characterized those p o ly nom ials o f degree 5 and greater w hose roots can b e expressed in term s of radicals and ordinary arithm etic. Like A bel, G alois died very early (at age 20), but in his case from a unlikely cause: a duel. G alois’ w ork also was n ot recognized until 11 years after his death when Joseph Liouville addressed the A cadem y o f Sciences in Paris: “I hope to interest the A cadem y in announcing that am ong the papers o f Evariste G alois I have found a solution, as precise as it is profound, o f this beautiful problem : w hether or not it is solvable by radicals.” In developing his theory, G alois associated, w ith a given polynom ial, a set G o f perm utations o f the roots o f the polynom ial. This set G had the property that w henever s , t 6 G, then the com position s o t e G; that is, G was closed under com position. H e referred to G as a group, a term that enjoys a perm anent and prom inent place in abstract algebra.
13.4 F undam en tal Properties of Groups We now consider som e properties possessed by all groups. O f course, any property satisfied by all groups m ust be a consequence o f properties G I  G3. U nless stated otherw ise, the sym bol e represents an identity in the group under consideration. One sim ple, but im portant, property satisfied by every group (G , *) allow s us to cancel a in a * b = a * c and conclude that b — c. A ctually, since a group need not be abelian, there are tw o such cancellation properties. Theorem 13.7
Proof
Every group ( G , * ) satisfies: (a)
T h e L e ft C a n c e lla tio n L a w Let a, b, c e G. I f a * b = a * c, then b — c.
(b)
T h e R ig h t C a n c e lla tio n L aw Let a , b , c e G. I f b * a = c * a , then b — c.
We prove (a) only. (The p ro o f o f (b) is similar. See E xercise 13.21.) A ssum e that a * b = a * c. L et 5 be an inverse for a. T hen s * (a * b) = s * {a * c). So s * (a * b) = (s * a) * b = e * b = b,
334
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w hile s * (a * c) = (s * a) * c = e * c = c. Therefore, b = c.
■
The last tw o sentences o f this proof could have been replaced by: Then b = e * b = (s * a) * b = s * (a * b) ~ s * (a * c) = (s * a) * c = e * c = c. The next result w ill com e as no surprise. Theorem 13.8
Let (G , *) be a group and let a, b e G. The linear equations a * x = b and x * a = b have unique solutions in G.
Proof
We prove only that a * x — b has a unique solution. (The rem aining p ro o f is left as E xercise 13.22.) L et e be an identity for G, let 5 be an inverse o f a and let x = s * b. Then a * x = a * ($ * b) = (a * s) * b = e * b = b. So x = s * b is a solution o f the equation a * x — b. It rem ains to show that s * b is the only solution o f a * x = b. Suppose that Xi and x 2 are solutions of a * x = b. Then a * x x = b and a * x 2 = b. Thus a * x \ = a * x 2. A pplying the Left C ancellation Law (Theorem 13.7(a)), we have jq = x3. ■ The preceding theorem provides som e interesting inform ation for us. Suppose that we have a group table for a group G and w e are looking at the row corresponding to the elem ent a. T hen this row contains the elem ents a * g for all g e G. Let b e G. By Theorem 13.8, there exists x e G such that a * x — b. T hat is, the elem ent b m ust appear in the row corresponding to a. This is illustrated in Figure 13.7. On the other hand, the elem ent b cannot appear tw ice in this row since the equation a * x — b has a unique solution. H ence we can conclude that every elem ent o f G appears exactly once in every row in the group table o f G. B y considering the equation x * a — b , we can likew ise conclude that every elem ent o f G appears exactly once in every colum n in the group table o f G. A s w ith com position in a sym m etric group, it is custom ary in a group G to refer to the binary operation * as multiplication and to indicate the product o f elem ents a and b in G by ab rather than a * b to sim plify the notation. We thus write a * a = aa — a 2. The lone exception to this practice is w hen w e have a group w hose operation is addition, in w hich case we continue to use + as the operation. It is also com m on practice never to use + as an operation w hen the group is nonabelian. L et’s use this new ly adopted notation to present a theorem that shows that every group G has a unique identity and x
Figure 13.7
The equation a * x = b
13.4
Fundam ental Properties of Groups
335
every elem ent o f G has a unique inverse, two facts that you m ay have already suspected w ere true. Theorem 13.9
Let G be a group. Then (a) G has a unique identity and (b) each element in G has a unique inverse.
Proof
A ssum e that e and / are tw o identities in G. Since e is an identity, e f = / ; and since / is an identity, e f = e. Thus e = e f = f . This verifies (a). N ext let g e G and suppose that s and t are both inverses of g. So gs = sg = e and g t = tg = e. Thus s = se = s( gt) = (s g ) t — et = t. w hich verifies (b).
"
It is custom ary to denote the (unique) inverse o f an elem ent a in a group by a If the operation in a group under consideration is addition, then w e follow the standard practice of denoting the identity by 0 and the inverse o f a by  a . We now present two theorem s involving inverses in a group. Theorem 13.10
Let G be a group. I f a e G, then ( a ” 1) 1 = a.
Proof
Since a a ~ ' = a ~ l a = e, the elem ent a is the inverse o f a  '1, that is. (a *)
= a.
m
For elem ents a and b in a group, the next theorem establishes a connection am ong the inverses a  1 , b ~ x and ( a b ) ~ l . T heorem to Prove
L et G be a group. For a, i e G , ( a b )  1 = b ~ la ~ x.
PRO O FSTRATEG Y
Suppose that the group under consideration is G. F or a . b e G , their product ab € G. Since ab £ G , the elem ent ab has an inverse— indeed a unique inverse. The inverse of ab is denoted by (ah) The theorem claim s that the inverse of a b is the elem ent b a in G. To show that an elem ent i e G is an inverse o f x e G , w e are required to show that s x = .w = e. T herefore, to show that b ~ l a ~ l is the inverse o f ab, w e need to show that (ab)
Theorem 13.11
= ( b ~ 1 a ~ ' ) (ab) — e.
^
Let G be a group. I f a, b e G, then ( ab ) ~l = b ~ xa ~ l .
Proof
To show that b ~ la ~ l is the inverse o f a b ,it suffices to show that (ab) (b 'a 1) = e and (ab) = e. We verify the first of these as the p ro o f o f the second equality is
336
Chapter 1 3
Proof's in Group Theory
similar. O bserve that (ab)( b~l a ~ }') = ((a b ) b ~ l )ci~l = ( a ( b b ~ ' ) ) a ~ l = ( a e ) a ~l = a a ~ l = e. In w ords, w hat T heorem 13.11 states is that the inverse o f the product o f two elem ents in a group is the product o f their inverses in reverse order. O f course, if G is an abelian group and a , b e G, then (a b r 1 = b ~ la ~ l = a ~ ' b ~ l . (See Exercise 13.25.)
13.5 Subgroups There have been occasions w hen we w ere considering a group (G, *) and a subset H o f G such that H is a group under the same operation *; that is, ( H, *) is also a group. If (G , *) is a group and H is a subset o f G such that (H, *) is a group, then ( H, * ) is called a su b g ro u p o f G. For exam ple, (Z, + ) is a subgroup o f (Q , + ), which, in turn, is a subgroup o f (R, + ). A lso, the groups G j and Go in Figure 13.5 are subgroups o f S3, w hile the group G 3 in Figure 13.6 is a subgroup o f £g. If (G , * ) is a group with identity e , then ({e}, *) and (G , *) are always subgroups o f (G , *). H ence if G has at least two elem ents, then G has at least tw o subgroups. T he group (2Z, + ) o f even integers under addition is a subgroup o f (Z, + ). To see this, first observe that 2Z C Z. Since the sum o f two even integers is an even integer, 2Z is closed under addition. Since the associative law o f addition holds in Z, it holds in 2Z as well. The identity in (Z. + ) is 0. Since 0 is an even integer, 0 e 2Z. Finally, the additive inverse (the negative) o f an even integer is an even integer. Thus (2Z, + ) is a group. T herefore, show ing that (2Z. + ) is a subgroup of (Z, + ) is sim pler than showing that it is group. This observation is true for every subgroup. Theorem 13.12
Proof
(T h e S u b g ro u p Test) A nonempty subset H o f a group G is a subgroup o f G i f and only i f (I) a b e H f o r all a , b e H and (2) a ~ [ e H f o r all a c //. We first show that if H is a subgroup o f G. then properties (1) and (2) are satisfied. Since H is closed under m ultiplication, property (1) is certainly satisfied. We now show that the identity e o f G is also the identity o f H . L et / be the identity o f H . Thus / ■ / = / . Since e is the identity o f G, it follow s that / • e = f . Therefore, / • / = / • e. B y the L eft C ancellation Law (Theorem 13.7(a)), f = e. H ence, as claim ed, the identity e of G is also the identity o f H . N ext, let a e H . Since a 6 G , it follow s that the inverse a o f a belongs to G and so a a ~ ] = e. It rem ains to show that a ~ l € H . Since H is a subgroup o f G, a has an inverse a' in H . Thus aa' = e in H and aa' — e in G as well. T herefore, a a ' — aa 1. B y the L eft C ancellation Law again, a' = a 1 and so property (2) is satisfied. N ext, w e verify the converse, nam ely if H is a nonem pty subset o f G satisfying properties (1) and (2), then II is a subgroup o f G. Let a , b , c £ H . Since / / C G , it follow s that a , b, c e G. Since G is a group, a(bc) = (ab) c by property G l. Thus m ultiplication is associative in H as well. From property (1), it follow s that H is closed under m ultiplication and from property (2), every elem ent o f H has an inverse. It rem ains only to show that H contains an identity. Since H ^ 0, there exists an elem ent a e H . B y (2), a 1 e H \ and by (1), a a ~ l = e £ H . Since H contains the identity e o f G, it follow s that x e = e x ~ x for all x e H and so e is the identity o f H as well. ■
13.5
Subgroups
337
We now illustrate the Subgroup Test.
R e su lt to P ro v e
PROOFSTRATEGY
L et H =
a c
b 0
: a, b, c e R }. Then (H , + ) is a group.
The elem ents o f H are m atrices, in fact, m atrices in M 2 (R). Indeed, a m atrix in M 2 (R) belongs to H if and only if the entry in row 2, colum n 2 is 0. We have already seen that (M 2 (R ), + ) is a group. Since H uses the sam e operation as in M 2 (R ), nam ely addition, it is appropriate to prove that (H , + ) is a group by the Subgroup Test (Theorem 13.12). To use the Subgroup Test, w e first need to know that the set H is nonem pty. Since the zero m atrix
q
q
satisfies the requirem ent for it to belong to H, w e need only to
show that conditions (1) and (2) o f Theorem 13.12 are satisfied, nam ely, that H is closed under addition and that if A is a m atrix in H , then its inverse (its negative in this case) —A belongs to H as well. This w ill be relatively routine to show. ♦
Result 13.13
Proof
Let H =
: a, b, c € R [. Then ( H, + ) is a group.
We show in fact that (H , + ) is a subgroup o f (M 2 (R ), + ). Certainly, H is a nonem pty “0 O' subset o f M 2(R) since the zero m atrix belongs to H . L et A, B e H. Then 0 0 A =
ai 0 2,
Cl2 o
and B =
w here a,, bt e R (1 < i < 3). Then A + B =
b\
Cl\ b\ ^3 + Z?3
b2 0 Cl2 “t ^2 0
e H and the inverse
—a\ —ci2 e H . Consequently, by the Subgroup Test, H is a subgroup —«3 0 o f M 2(R) and so (H , + ) is a group. o f A is —A =
If G is an abelian group, then we know that every tw o elem ents o f G com m ute. B ut even if G is nonabelian, w e know that its identity com m utes w ith every elem ent o f G. H owever, there m ay very w ell be other elem ents o f G that com m ute w ith all elem ents o f G. The set o f all elem ents in a group G that com m ute with every elem ent in G is called the c e n te r of G and, in fact, is always a subgroup o f G. This subgroup is often denoted by Z (G ). Since Z (G ) = G if G is abelian, the center is m ost interesting w hen G is nonabelian. R e su lt to P ro v e
F or a group G, the center Z (G ) = {a e G : ga = ag for all g e G] is a subgroup o f G.
PROOFSTRATEGY
To prove this result, it seem s natural to use the Subgroup Test. Since e e Z (G ), it follows that Z (G ) ^ 0. We are now required to show that Z (G ) satisfies the tw o properties required o f the Subgroup Test.
338
Chapter 13
Proofs in Gl'Oup Theory
First, w e show that Z (G ) is closed under m ultiplication; that is, if a, b e Z ( G ) , then ab e Z (G ). We em ploy a direct proof. Let a , b e Z (G ). To show that ab e Z (G ), we need to show that ab com m utes w ith every elem ent o f G. So let g e G. We m ust show that (ab)g = g(ab). This suggests starting w ith (ab)g. B y the associative law, (ab)g = a(bg). However, since b e Z (G ), it follow s that a(bg) = a(gb). We can continue in this m anner to com plete the proof o f this property. Second, w e need to show that if a e Z(G), then a 1 e Z(G). A gain, we use a direct proof. Let a e Z ( G) . T hen a com m utes w ith every elem ent o f G. To show that a~x e Z ( G ) , we need to verify that a ~ 1 com m utes w ith every elem ent o f G. L et g e G. We m ust show then that a ~ l g = g a 1 to com plete the proof. But how do we do this? T heorem 13.11, w hich deals w ith inverses o f elem ents, m ay be helpful. We know that (.r.yj 1 —• y ~ lx ~ l for all a , y e G. So (ag) 1 — g ~ 1 a ~ 1. This, however, involves g ~ l . B ut if we start w ith a g ~ l = £ _1a , then we have ( a g " 1) ‘ = (.g” 1^ ) 1• Result 13.14
♦
For a group G, the center Z ( G ) = {a £ G : ga — ag f o r all g e G} is a subgroup o f G.
Proof
Since eg = ge for all g e G , it follow s that e e Z (G ) and so Z (G ) is nonem pty. First, w e show that Z (G ) is closed under m ultiplication. L et a, b e Z ( G) . T hus ag = ga and bg = gb for all g e G. We show that ab e Z (G ). Since I a b ) g = a (bg) = a (gb) = (ag ) b = (ga ) b = g ( ab) , ab e Z (G ). H ence Z (G ) is closed under m ultiplication. N ext we show that every elem ent o f Z (G ) has an inverse in Z ( G) . L et a € Z (G ) and g e G. We show that a 1 € Z (G ); that is, a " 1 and g com m ute. Since a com m utes w ith all elem ents o f G, it follow s that a and g ~ l com m ute and so a g = g ~ la. Since every elem ent o f G has a unique in verse, ( a g 1 ) " 1 = ( g ^ a ) " 1. By T heorem 13.11, ( a g  1 ) * = (g  1 ) 1 a " 1 = g a 1 and ( g _1a ) 1 = a 1 (g _1) 1 = a ~ l g. Therefore, a ~ l g = g a ~ l .
■
For A = {1, 2, • • •, n }, n > 2andA' e A, let G/. consist o fth o se perm utations a in the sym m etric group (Sn, o ) such that a ( k) — k (that is, G k consists o f all those perm utations o f A that “stabilize” or fix k). The set G/; is called the sta b iliz e r o f k in Sn. Result 13.15
Proof
For integers k and n with 1 < k < n and n > 2, the stabilizer G* o f k in S„ is a subgroup o f S„. We use the Subgroup Test. Surely, the identity a j in S„ belongs to G k, so Gu / 0. L et a, P e Gk  Thus a( k) = /3(k) — k. H ence (a o j3)(k)  a(fi(k)) = a ( k) = k and so a o 0 e Gk C onsider the inverse a " 1 o f a . Thus a 1 o a = a \ . Therefore, ( a  1 q a) (k) = a i (k) — k. H ence ( a " 1 o a) (k) = a _1(a(fe)) = u \ k ) = a j (k) — k. Thus a ” 1 e Gk B y the Subgroup Test, G* is a subgroup o f Sn. ■ The group ( G i , o) show n in Figure 13.5 is the stabilizer o f 1 in (S3, o). However, (Go, o) show n in Figure 13.5 is not the stabilizer o f 2 in (S3, o).
1 3 .5
Subgroups
339
We have already m entioned that the set 2Z o f even integers is a subgroup o f (Z, + ). In fact, for every integer n > 2, the set n Z = [nk : k e Z} of m ultiples o f « is a subgroup o f (Z, + ) (see E xercise 13.31). In C hapter 8, we saw that the relation R defined on Z by a R b i f a = b (m od n) is an equivalence relation. This relation can also be described in another manner, nam ely a R b if a — b e n Z and so a — b = h for som e elem ent h e n Z or a — b + h. It turns out that this equivalence relation is a special case o f a m ore general situation. Suppose that H is a subgroup o f a group (G , •) and a relation R is defined on G by a R b if a = bh for som e h e H . (Note that b + h in Z is replaced by bh here since the operation in G is multiplication.) T hen this relation is also an equivalence relation.
Theorem 13.16
Proof
Let H be a subgroup o f a group (G , ■)• The relation R defined on G by a R b i f a — bh f o r some h 6 H is an equivalence relation.
First, we show that R is reflexive. L et a e G. Since a = ae (w here e is the identity of G and therefore o f H) , a R a and so R is reflexive. N ext, we show that R is sym m etric. A ssum e that a R b , w here a , b e G. T hen a = bh for som e h e H . Since H is a group, h ~ l € H and so cih~l = (b h ) h ~ [ = b ( h h ~ l ) = be = b, or b = a h ~ x. T herefore, b R a and R is sym m etric. Finally, w e show that R is transitive. A ssum e that a R b and b R c, w here a , b , c e G. Then a — bh\ and b = c h 2 for elem ents h\ and h j in H. Therefore, a = bh i = ( ch 2 )hi — c{h, 2 h\). Since h%, h\ e H , it follow s that e i f as well. Thus a R c and so R is transitive. m
For a subgroup H o f a group (G , ■), the equivalence relation defined in T heo rem 13.16 gives rise to equivalence classes. For each elem ent g e G, the equivalence class [g] is defined by [<?] = {* ^ G : i S ^ ) = { i e G :
x = g h for som e h e H]
= { g h : h e H} . The set {gh : h e H ] is often denoted by g H and is called a left coset o f H in G, that is, [g] = g H . We saw in C hapter 8 that for an equivalence relation defined on a set S, the distinct equivalence classes form a partition o f S. Consequently, the equivalence relation defined in Theorem 13.16 results in a partition o f G into the distinct left cosets o f H in G. An im portant characteristic o f a left coset g H o f H in G is that g H and H have the same num ber o f elem ents, that is, \gH\ = \H\. In order to see this, w e show for an elem ent g e G that there is a bijection from H to g H . L et <p : H ^ g H be defined by (j){h) = gh. First, we show that 0 is onetoone. A ssum e that <j>(h\) = <p(h2 ). Then gh \ = g h 2  By the Left Cancellation Law, h\ — ho T herefore, </> is onetoone. Next, we show that <p is onto. Let gh e g H . Since <p(h) = g h , it follow s that (f>is onto and so <p is a bijection and \gH\ = \H\. T herefore, every tw o left cosets o f H in G have the sam e num ber o f elem ents. W hat w e have ju st observed provides all the inform ation that is needed to prove a fundam ental theorem o f group theory, one due to JosephL ouis L agrange, probably the greatest F rench m athem atician of the 18th century.
340
Chapter 13
Theorem 13.17
Proofs in Group Theory
(L a g ra n g e ’s T h eo rem ) I f H is a subgroup o f order m in a (finite) group G o f order n, then m  n.
Proof
We have already seen that the distinct left cosets of H in G form a partition o f G and that every tw o left cosets have the sam e num ber o f elem ents. Suppose that there are k left cosets of G. Then n — ink and so m \ n. a L agrange’s theorem first appeared in 17701771 in connection w ith the problem o f solving the general polynom ial o f degree 5 or higher. W hile this theorem was not presented in this general form by Lagrange and in fact group theory had yet to be invented, it is universally referred to as L ag ran g e’s theorem . We have seen that the group consisting o f the nonzero elem ents o f Z7 form s a group under m ultiplication, that is, G = Z7 = {[1], [2], . . . , [6]}. Since H = {[1], [6]} is a subgroup o f order 2 in G, the distinct left cosets o f H in G are [1 ] H = H , [2]H = {[2], [5]} and [3] H = {[3], [4]}.
1 3.6 Iso m o r p h ic G roups
~]
Suppose that w e w ere asked to give exam ples o f tw o groups o f order 3. One possible exam ple is (Z 3, + ). O n the other hand, w e m ight try to construct tw o groups o f order 3, say G = {a , b, c} and H = { x , y , gj. O f course, we m ust also describe binary operations for both G and H . Let us denote the binary operation for G by * and the binary operation for H by o. So w e have tw o groups (G , *) and ( H , o), both o f order 3. O ne o f the elem ents o f G is the identity for G and one o f the elem ents o f H is the identity for H . Suppose that w e decide on a as the identity for G and x as the identity for H . H ence the operations * and o in G and H, respectively, satisfy the partial tables show n in F igure 13.8. B ecause every elem ent in each of the groups G and H m ust occur exactly once in every row and colum n in the tables show n in F igure 13.8, the com plete tables for * and o m ust be those show n in Figure 13.9. We can now see readily that in G w e have a ~ l = a , b ~ { — c and c ~ l — b\ w hile in H, x ~ ] = x, y ~ [ = z and = y. Verifying the associative laws requires m ore effort, but it can be show n that the associative law holds in each case. T hus (G , *) and ( / / , o) are both groups, and we have ju st given exam ples o f two groups o f order 3. O r have w e? There is som ething very sim ilar about these tw o exam ples. They are not really two different groups at all. Indeed, the group (H , o) is m erely a disguised form of the group
a
b
c
0
a
a
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b
b
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*
Figure 13.8
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Partial tables for groups (G, *) and (H , o)
13.6
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Figure 13.9
Isomorphic Groups
X
341
Complete tables for groups (G, *) and (H , a) z
y
X
z
y
z
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y
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*
a
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Figure 13.10
X
Groups (G, *) and (H , o)
(G , *). L et’s describe w hat w e m ean by this. If the elem ents a , b, c in G are replaced by x, y, z, respectively, then w e have the identical table. W h at’s im portant here is not only that x, y and z in H are playing the roles o f a, b and c in G b ut that the operations in the tw o groups are doing the sam e thing. For exam ple, if we m ultiply b and c in G (obtaining the elem ent a), then m ultiplying the corresponding elem ents y and z in H gives us the elem ent corresponding to a, nam ely x. A lthough it m ay appear that this is the natural correspondence betw een the elem ents o f G and the elem ents o f H , w e should n ot be m isled by the order in w hich the elem ents o f these tw o groups are listed. For exam ple, suppose that w e consider the tw o groups (G , *) and ( H, o) once again (in Figure 13.10), w here the elem ents o f H are listed in the order x, z , }’■ We can see that the elem ents a , b, c in G also correspond to x, z, y, respectively. We consider the two groups (G , *) and ( H, o) to be actually a single group, as these two groups have the sam e order (though the sets are different) and their operations perform the sam e functions (though different sym bols are used for the operations). T he technical term for this is that they are isomorphic groups (groups having the sam e structure). In general, tw o groups (G , *) and ( H , o) are iso m o rp h ic if there exists a bijective function (p '■G *■ H satisfying the property cp(a * b ) = (p(a)
o 4 >(b)
(13.3)
for all a , b e G. Any such function cp that satisfies (13.3) is said to be o p e ra tio n p re se rv in g . Thus for (G , *) and ( H. o) to be isom orphic, there m ust be a bijective, operationpreserving function <p : G —►H.l£(j> has these properties, then cp is called an iso m o rp h ism . I f 0 : G —>• H is an isom orphism , then <p is also a bijective function. Thus <p has an inverse function : H —> G , w hich is also an isom orphism (Exercise 13.51). For isom orphic groups (G , * ) and ( H , o ), there are certain properties that every isom orphism from G to H m ust have. We look at tw o exam ples o f these.
342
Chapter 13
Theorem 13.18
Proofs in Group Theory
Let (G , *) and (H , o) be isomorphic groups, where the identity o f G is e and the identity o f H is f . lf(p '■G —>■ H is an isomorphism, then (a) 0 (e) = f and (b ) <p ( g ' 1) = (0 (g ))1 M
P roof
all g € G.
First w e prove (a). Let h e H . Since 0 is onto, there exists g e G such that (pig) = h. Since e * g = g * e = g and 0 is operationpreserving, it follow s that 4>(e) o <j){g) = 0 (e * g) = (p(g)  0 ( g * e ) = (pig) o <p(e) and so 0 (e ) o h — h o 0 (e ) = h. T his im plies that (p{e) is the identity o f H and so 0 (e) = /. N ext w e prove (b). Let g 6 G. Since g * g 1 = g ~ ] * g — e, it follow s that 0 (g = 0 (g ” 1 * g ) = 0 (e ) and so 0 (g ) o 0 ( g  1) = 0 (g 1 ) o 0 (g ) = (pie) = f . This says that 0 (g  1 ) is the inverse o f 0 (g ), that is, 0 (g * 1) = ( 0 ( g ) ) " 1.
■
By definition, if tw o groups (G , *) and (H , o) are isom orphic, then there exists an isom orphism 0 : G > H . Since 0 is a bijective function, it follow s that G   \H\. Actually, it’s not surprising that isom orphic groups have the sam e num ber o f elem ents since, w hen w e say that G and H are isom orphic, we are technically saying that these groups are the sam e, except for w hat the elem ents and binary operations are called. On the other hand, consider the group tables o f tw o groups show n in Figure 13.11. You m ight notice that the first group is (Z 4, + ). The second group G is abelian, has identity e and x 1 = x for all x 6 G. In addition, o f course, G has order 4. So Z 4 and G both have order 4. Yet Z 4 and G are not isom orphic; for assum e, to the contrary, that they are isom orphic. T hen there exists an isom orphism 0 : Z 4 > G. By T heorem 13.18, we know that 0([O]) = e. L et 0 ( [ 1]) = x e G. So 0 ( [ 2]) = 0([1 + 1]) = 0([1] + [1]) = 0 ([1 ]) • 0 ([1 ]) = x • x = x 2 = e. Thus 0 ( [ 2]) = 0([O]) = e but this contradicts the fact that 0 is onetoone. Consequently, these tw o groups o f order 4 are not isom orphic. T herefore, if tw o groups have the same num ber o f elem ents, they need not be isom orphic. However, it turns out that any group o f order 4 is isom orphic to one o f the tw o groups o f order 4 described in Figure 13.11.
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b
a
e
+
[0]
[1]
[2]
[3]
[0]
[01
[1]
[21.
[3]
[1]
[1]
[2]
[3]
[2]
[2]
[3]
[0]
[1]
b
[3]
[3]
[0]
[1]
[2]
c
Figure 13.11
Two groups of order 4
13.6
Isomorphic Groups
343
We saw in C hapter lO that Z  = Q  even though Z is a proper subset o f Q. However, (Z. + ) and (Q . + ) are not isom orphic. Result 13.19 Proof
The groups (Z, + ) and (Q, + ) are not isomorphic. A ssum e, to the contrary, that (Z , + ) and (Q , + ) are isom orphic. T hen there exists an isom orphism 0 ■Z >• Q. L et 0 (1 ) = a e Q . Since 0 (0 ) = 0, it follow s that a / O.Thus a / 2 e Q a n d a / 2 / 0. Since 0 is onto, there exists an integer n ^ 0 such that 0 (« )  a /2 . Then 0 (2 « ) = 0 ( n + n ) = 0(/O + 0 (« ) =  +  = « • Since 0 is onetoone, 2n = 1. H ow ever then, n = 1 /2 ^ Z, w hich is a contradiction. ■ On the other hand, the set 2Z of even integers is a proper subset of Z; yet (2Z, + ) and (Z, + ) are isom orphic.
Result 13.20 Proof
The groups (2Z, + ) and (Z, + ) are isomorphic. Define the function 0 : Z > 2Z by (pin) = 2n for each n e Z. First, w e show that / is onetoone. A ssum e that 0 ( a ) = <p(b). T hen 2a — 2b. D ividing by 2, w e obtain a = b and so 0 is onetoone. N ow we show that 0 is onto. L et n e 2Z. Since n is even, n = 2k for some integer k. Then 0(fe) = 2k = n. This shows that 0 is onto. Finally, w e show that 0 is operationpreserving. Let a , b e Z. Then 0 ( a + b) — 2 (a + Z?) = 2a + 2fo = 0 ( a ) + <p{b).
■
It should be obvious that every group G is isom orphic to itself. In fact, the identity function /(, : (< —> G defined by i c i g ) = g f ° r all g e G is an isom orphism . However, other perm utations o f G can be isom orphism s. Result 13.21
Let G a group and let g e G. The function 0 : G  * G defined by 0 ( a ) = g a g ~ [ f o r all a e G is an isomorphism.
Proof
First, we show that / is onetoone. A ssum e that 0 ( a ) = (p{b). T hen g a g ~ 1 = g b g ~ ]. C anceling g on the left and g ” 1 on the right, w e obtain a = b. N ext, w e show that 0 is onto. L et c e G. Then 0 { g ~ 'c g ) = g C r 'c g ) g ~ l = (g g _1) c ( g ^ g ) = ece = c. Finally, w e show that 0 is operationpreserving. Let a, b e G. Then 4>{ab) = g (ab)g 1 = (gag ') (gbg !) = (p(a)<p(b).
344
Chapter 13
Proofs in Group Theory
E X E R C IS E S FO R C H A PT E R 13 Section 13.1: Binary Operations 13.1. Consider the algebraic structure (S, *), where S = [x, y, z} and * is described in the table in Figure 13.12. Compute (a) (b) (c) (d)
x * (v * z) and (x * y ) * z. x * (x * x ) and i.v * x ) * x . . y * (y * v) and (y * y ) * y . W hat conclusion can you draw from (a)(c)?
13.2. For every pair a, b of elements in the indicated sets, the element a * b is defined. W hich of these are binary operations? For those that are binary operations, determine which of the properties G1  G4 are satisfied, (a) a * b = 1 on the set Z (b ) a * b = a / b on the set N (c) a * b = a b on the set N (d) a * b = max{a, b} on the set N (e) a * b= a + b + ab on the set Z (f) a * b = a + b  1 on the set Z (g) a * b = ab + 2a on theset Z (h) a * b = ab  a  b + 2 on the set R  {1} (i) a * b = s / a b on Q (j) a * b = a + b on the set S of odd integers. a —b b a (a) matrix addition?
: a, b € R j. Is T is closed under
13.3. L e tT
(b)
matrix multiplication?
13.4. Suppose that * is an associative binary operation on a set S. Let T = {a e S : a * x = x * a for all x € 5}. Prove that T is closed under *. 13.5. Suppose that * is an associative and commutative binary operation on a set S. Let T = [a e S : a * a = a). Prove that T is closed under *. 13.6. For matrices A =
a c
b d
e g
and B =
A *B =
a c
b d
f h *
in M2(R), the binary operation * is defined on M jiR ) by a + e —1 c+ g
e h
_8
b+ f d + h 1
Which of the properties G1  G4 are satisfied? 13.7. For n > 2 and [a], [b] € Z„, the binary operation * is defined on Z„ by [a] * [£>] = [a + b + 1]. W hich of the properties G1  G4 are satisfied?
Figure 13.12
¥
X
y
z
X
y
z
y
y
y
X
X
z
z
z
y
A binary operation on the set S = .v. v, z}
Exercises for Chapter 13
a
a
c
b
a
c
d
d Figure 13.13
345
a b
A binary operation on the set S = {a , b, c, d } in Exercise 8
Section 13.2: Groups 13.8. Let S = {a, b, c, d). Figure 13.13 shows a partially completed table for an associative binary operation * defined on S. (a) Complete the table. (b) Is the algebraic structure (S , *) a group? 13.9. Let (G, *) be a group with G = { a ,b , c , d), where a partially completed table for (G, *) is given in Figure 13.14. Complete the table. 13.10. None of the following binary operations * on the given set result in a group. Which is the first property among G l, G2, G3 that fails? (a) Let * be defined on R + by a * b = \fab. (b) Let * be defined on R* by a * b — a/b. (c) Let * be defined on R + by a * b= a + b+ ab. 13.11. (a) Determine whether for all fa], [£>] € Z£ = {[1], [2], [3], [4], [5]}, there exists [*] e Z \ such that M M = [b](b) Why is the answer to the question posed in (a) not surprising? 13.12. Let G = (a) (b) (c) (d)
Prove Prove Prove Prove
b : a, b e R and a ^ 0 0 that G is closed under matrix multiplication. that there exists E e G such that E ■A = A for allA e G. for each A e G that there is A' e G such that A • A'= E. or disprove: (G, •) is a group.
13.13. Let * be an associative binary operation on the set G such that the following hold: (i) There exists e e G such that g * e = g for all g e G. (ii) For each g e G, there exists g' e G such thatg * g' = e. Prove that (G, *) is a group. *
a
b
a
d
c
b
c
d
a
c d Figure 13.14
A partially completed table for (G, *) in Exercise 9
Chapter 13
346
Proofs in Group Theory
Section 13.3: P erm utation Groups 13.14. For I < i < 6. each function /} is a permutation on the set Q — {0, 1}: f \ U) = x f 2(x) = 1  x / 3(x) =  /! < » = ^ /> U ) = /e (x ) = jry Show that the set /•' {/ i , / ; . . . . . /(,} is a group under composition. 13.15. Prove that if A is a set with at least three elements, then the symmetric group (S a , p ) is nonabelian. 13.16. Give examples of the following (if they exist): (a) (b) (c) (d)
a finite abelian group a finite nonabelian group an infinite abelian group an infinite nonabelian group.
13.17. Determine all elements x in the group S3 such that x 2 = a\ and all those elements y in S3 such that y3 = a i13.18. For the permutations: 1 2 Yl 11 2 3 4
34 \ T2
\ 2
f\ 3
23 4 4 1
1 2 3 1\ /I 2 3 4 73 ~ \ 3 4 1 2) T4 ^41 2 3 of the set {1, 2, 3, 4}, show that the set G = [yi, yn, K3, Ya] under composition is an abelian group. 13.19. Consider the following permutations on the set A = {1, 2, 3, 4, 5}.
1 1
2
2
3
3 4
5 ^\
1 1
2 2
3 4 3 5
5\
4
5/
ft ~ ( X 2 3
4
y2
3
1
4
/ 1
2
3
4
5
5\ /
5\
ft / 1
2
3 4
ft ~\ 3
1
2
/I
2
3 4 2 5
4) m ~ \ 2 3 1 5 4J \3 1 For G = [fi 1, f t , • • •, Pb), show that (G, o) is a group of permutations on A.
13.20. For a permutation group G on a set A, a relation R is defined on A by a R b if there g(a) = b.
4
5
5 5 4
exists g e
(a) Prove that R is an equivalence relation on A. (The equivalence classes resulting from this equivalence relation R are called the orbits of A under G. ) (b) For the group G in Exercise 13.19, determine the orbits of A under G .
Section 13.4: Fundam ental Properties of Groups 13.21. Prove Theorem 13.7(b) (The Right Cancellation Law): Let (G, *) be a group. If b * a = c * a , where a, b, c e G, then b = c. 13.22. Prove the following (see Theorem 13.8): Let (G , *) be a group and let a . b e G. The linear equation x * a = b has a unique solution x in G. 13.23. Let (G, *) be a group and let a, b, c e G. Prove that each of the following equations has a unique solution for x in G and determine the solution. (a ) a * x * b = c (b) a * b * x = C. 13.24. Let a and b be two elements in a group G. Prove that if a and b commute, then a 1 and b~l commute. 13.25. Let G be a group. Prove that G is abelian if and only if (ab)~ l = a ~ l b~ i for all a. b € G.
Gs
Exercises for Chapter 13
347
+ _____________________
Figure 13.15
Constructing a table for the group (Z9, + ) in Exercise 13.26
13.26. Construct a table for the group ( Zg, + ) by listing the elements of Zg across the top of the table in some order and along the left side of the table in some order in such a way that every element of Zg appears in each of the nine 3 x 3 regions indicated in the table in Figure 13.15. 13.27. By Theorem 13.9, every group G has a unique identity. That is, G contains only one element e such that ae = ea = a for all a e G. Suppose that e' is an element of G such that e'b = b for some element b € G. Prove or disprove: e is the identity of G. 13.28. Let (G, *) be a group. Prove that if g * g = e for all g e G, then G is abelian. 13.29. Let (G, *) be a finite group of even order. Prove that there exists j e G
such that g / e and g * g = e.
13.30. Suppose that G is a finite abelian group of order n, say G = {gj, g2, . . ., g„}. Let g = g ig 2    g ng ig 2   g n W hat is g?
Section 13.5: Subgroups 13.31. For an integer n > 2, prove that the set n Z = {nk : k e Z j of multiples o f n is a subgroup of (Z, + ). 13.32. Which of the following are subgroups of the given group? (a) (b) (c) (d)
The The The The
subset N in (Z, + ) subset {[0], [2], [4]} in (Z 7, + ) subset {[1], [2], [4]} in {Z*, •) subset {2" : n e Z} in (Q*. ■)•
13.33. Let H and K be two subgroups of a group G. Prove or disprove: (a) H D K is a subgroup of G. (b) H U K is a subgroup of G. 13.34. For each of the following subsets H of M (R ), prove or disprove that (H , •) is a subgroup of (iV/(R), •)■ (a) H =
a c
b 0
: a, b, c e R, be ^ 0
(b) H =
a 0
b c
: a, b, c e R, ac ^ 0
13.35. Let H — {a + b s / 3 : a, b e Q. a / 0 or b / 0}. Prove that H is a subgroup of (R*, •). 13.36. For n e N, let T be a nonempty subset of {1, 2 ,■ ••,« } and define G T = [a e S„ : a{t) — t for all t e T}. Prove that G t is a subgroup of (S„, o).
Chapter 13
348
Proofs in Group Theory
13.37. Recall that M (R ) = { A e M2(R) : det(A) / 0}. Let H = {A € .V/:(R) : det(A) = 1 or det(A) =  l j . Prove that H is a subgroup of (M*_(R), •)• 13.38. Let G be an abelian group and let H = {a2 : a e G}. Prove that H is a subgroup of G. 13.39. Let G be an abelian group and let H = {a e G : a 2 = e}. Prove that H is a subgroup of G. 13.40. W hat are all the subgroups of a group of order p, where p is a prime? 13.41. Prove or disprove: There exists a group of order 372 containing a subgroup of order 22. 13.42. Prove that a nonempty subset H of a group G is a subgroup of G if and only if a b ~ l e H
for all a, be
13.43. (a) Let (G, *) be a finite group. Prove that if H is a nonempty subset of G that is closed under *,then is a subgroup of G. (b) Show that the result in (a) is false if (G, *) is an infinite group. 13.44. Let B be nonempty subset of a set A, S = { / e S a '■ f ( B ) = B } and T = { / e SA : f ( b ) = b for each b e B }. (a) Prove that (b) Prove that
(S, o) is a subgroup of ( S a , o). (T, 6) is a subgroup of (S , o).
13.45. A group G has order 48. If there are six distinct left cosets of a subgroup H in G, then what is the order of HI ( 1 2 3\ /1 2 3\ 13.46. For the subgroup H = {«•. « 2) of (S 3 , o), where a\ = I ^ 0 ^ I and a 2 = I j ^ 2 ) ’ ^eter™ ne the distinct left cosets of H in (S3, o). 13.47. For a subgroup H of a group (G, •)» let g H be a left coset distinct from H . By the element g2 e G, we mean g ■g. Prove or disprove: g 2 e g H .
Section 13.6: Isom orphic Groups 13.48. Let H =
1 0
n 1
n e Z
(a) Prove that H is a subgroup of (A /(R ), ■)• is an isomorphism. ^ 0 (c) Parts (a) and (b) should suggest another question to you. Ask and answer a related question.
(b) Prove that the function / : (Z. + ) —> (H, •) given by f ( n ) =
13.49. In each of the following, determine whether the function </> is an isomorphism from the first group to the second group. (a) (b) (c) (d)
<p : (Z, + ) —> (Z, + ) defined by <p(n) = 2n. <fi : (Z, + ) —> (Z, + ) defined by <p(n) = n + 1. 4> '■(R + ) *• (R + , ■) defined by <p(r) = 2''. 4> : (Af(R), •) * (R*. ■) defined by <p(A) = det(A).
13.50. Obviously, (R + , •) and (R + , •) are isomorphic groups. Consider the function 0 : R + —> R + defined by <p(r) = r 2 for all r e R + . Is <p an isomorphism? 13.51. Let (G, *) and (H. o) be two groups. Prove that if <fi : G — H is an isomorphism, then the inverse function 0 1 of <p is an isomorphism from H to G. 13.52. Let G, H and K be three groups. Prove that if (f>i ■G —>■ H and (p2 : H composition 4>2 0 4>\ ■G K is an isomorphism.
* K are isomorphisms, then the
H
349
Additional Exercises for Chapter 13 13.53. Let (G, *) be a group. Define a binary operation o on G by a o b = b * a. (a) Prove that (G, o) is a group. (b) Prove that (G. *) and (G, o) are isomorphic. [Hint: Consider the function 0 (g ) = g  1 .] 13.54. Explain why the groups (Q, + ) and (R, + ) are not isomorphic.
13.55. We saw in Example 13.4 that with the binary operation * defined o n Z b y a * b = a + b — 1, (Z, *) is an abelian group. Prove that (Z, *) is isomorphic to (Z. + ). 13.56. (a) Let G and H be isomorphic groups. Prove that if G is abelian, then H is abelian. (b) Show that the groups (Zg, + ) and (S3, o) are not isomorphic. 13.57. Let B = {i : n e Z  {0}} and let A = R  B. (a) Prove for each n e Z that the function /„ : A —> A defined by f n(x) = (b) Let P = { f n : w e Z}. Prove that (P, o) is a group of permutations on A. (c) Prove that the groups (Z, + ) and (P , o) are isomorphic.
is bijective.
13.58. Let (G, o) and ( //, *) be groups with identities e and e', respectively.Suppose that f : G ^ H is function with the property that f ( a o b) = f ( a ) * f ( b ) for all a, b e G.
a
(a) Let M = ran g e (/). Prove that (M, *) is a subgroup of (H, *). (b) Let K = {a € G : f ( a ) = e'}. Prove that (K , o) is a subgroup of (G, o). 13.59. Let A = {— : m and n are odd integers, gcd(m, n) — 1 and n > 1}. Let R* = R — {0} and for a e A, let f a : R* > R* be defined by f a(x) = x a. (a) Prove that (A, ■) is a subgroup of (Q*, ■), where the product of every two elements lowest terms. (b) Show for each a e A that f a is a permutation on R*. (c) Let F = { fa : a e A}. Prove that (F, o) is a subgroup of (S r., o). (d) Prove that (A, •) and (F , o) are isomorphic groups.
of A is reduced
ADDITIONAL EXERCISES FOR CHAPTER 13 13.60. Let (G, *) be a group. An element g of G is an idem potent for * if g * g = g. Prove that there exists exactly one idempotent in G. 13.61. Let G be a group and let a e G. The set Z(a) = {g e G : ga = a g } is called the cen tralizer of a. Prove that the centralizer of a is a subgroup of G. 13.62. Let a , i e Z , where a, b 7 ^ 0 and let H = {am + bn : m, n e Z} be the set of all linear combinations of a and b. (a) Prove that H is a subgroup of (Z, + ). (b) Let d = gcd(a, b). Prove that H = dZ. 13.63. Define * o n R — {\} by a * b = a + b — ab. (a) Prove that (R — {1}, *) is an abelian group. (b) Prove that (R — {1}, *) is isomorphic to (R*. •)• 13.64. Let G be a group of order p q , where p and q are distinct primes. W hat are the possible orders of a subgroup of G?
350
Chapter 13
Proofs in Group Theory
13.65. Let H be a subgroup of (Z, + ) with at least two elements and let m be the smallest positive integer in H . Prove that H = m Z. [Hint: Use the Division Algorithm.] 13.66. Evaluate the proposed proof of the following statement. R esult
There exists no group containing exactly two distinct elements that do not commute.
P roof Assume, to the contrary, that there exists a group G containing exactly two distinct elements, say x and y, that do not commute. Thus x y # yx. Since x and y are the only two elements of G that do not commute, x _1 and y do commute. Thus x _ Iy = y x ~ ]. Multiplying by x on both the left and right, we obtain x (x _1 y) x = x (.yx1 ) x. Simplifying, we have y x = x y . This is a contradiction. ■ 13.67.
A group G of order n > 2 contains a subgroup H . In a left coset decomposition of H in G,the number of distinct left cosets is the same as the order of H . If some left coset contains p elements, where p is a prime, then what is n?
13.68. Evaluate the proposed proof of the following statement. R esult
There exists no abelian group containing exactly three distinct elements x such that x = e.
P ro o f Assume, to the contrary, that there exists an abelian group G such that x 2 = e for exactly three distinct elements x of G. Certainly, e2 = e, so there are two nonidentity elements a and b such that a 2 = b2 — e. Observe that (a b)2 = a 2b2 = ee = e. Hence either ab = a, ab = b or ab = e, which implies, respectively, that b = e, a = e or a = b, producing a contradiction.
■
13.69. Prove or disprove the following: For each odd integer k > 3, there exists no abelian group containing exactly k elements x such that x = e. 13.70. For a function / : N * N, we let f l = f and f 2 = / defined recursively by f k = f o f k~ l . Thus / " : N elements / and g in SN such that f 2 = g2 = is but
o
/ . More generally, for k > 2, the function f k is —»■ N for each n e N. Give an example of two ( / o g)m 7= ;'N for all m e N.
Answers to OddNumbered Section Exercises
EXERCISES FOR CHAPTER 1 Section 1.1: Describing a Set 1.1 1.3 1.5
Only (d) and (e). (a) A  = 5, (b) B = 11, (c) C = 51, (d) \D\ = 2, (e) \E\ = 1, ( f )  F  = 2 (a) A = {—1, —2,  3 , ...) = {* e Z : x < —1) (b) B = { 3 ,  2 , . . . , 3} = [x € Z :  3 < x < 3} = {x e Z : x < 3) (c) C = { 2 ,  1 , 1, 2) = {x e Z :  2 < x < 2, x ^ 0} = (x e Z : 0 < x < 2} 1.7 (a) A = {• • •, —4, —1 ,2 ,5 , 8, • • •} = {3x + 2 : x e Z} (b) B = {•••, 1 0 .  5 ,0 ,5 ,1 0 , •■•} = {5x : x e Z) (c) C = {1, 8, 27, 64, 125, • • ■} = {x3 : x e N ) 1.9 A = {2, 3 ,5 ,7 , 8, 10, 13} B = {x e A : x = y + z, where y , z e A] = {5, 7, 8, 10, 13} C = {;• e B : r + s e B for some s e B} = {5, 8}
Section 1.2: Subsets 1.11 Let r = min(c — a , b — c) and let / = (c —r, c + r). Then / is centered at c and I C (a , £>). 1.13 See Figure 1.
Figure 1
Answer for Exercise 1.13
351
352 1.15 1.17 1.19 1.21
Answers to OddNmnbered Section Exercises V (A ) = {0, {0}, {{0}}, A } V (A ) = {0. {0}, {0}, {{0}}, {0, 0}, {0, {0}}, {0, {0}}, A}; \V{A)\ = 8 (a) S. = {0, {1}}. (b) S = {1}. (c) 5 = {0, {1}, {2}, {3}, {4, 5}}. (d) S = {1,2, 3,4, 5). B = {1,4,5}.
Section 1.3: Set Operations 1.23
1.25
Let A = {1. 2 ....... 6} and B = {4, 5 , . . . , 9}. Then A  B = {1, 2, 3), B  A = {7, 8, 9} and A n B = {4,5,6}. Thus \A — B \ = ( A n S  = \B — A = 3. See Figure 2.
(a)
A = {1}, B = {{1}}, C =
{1, 2}. = {1,2}. (c) A = {1}, B — {{1}}, C = {{1}, 2}. Let C/ = {1.2, . . . , 8 } be a universal set, A = {1,2, 3,4} and B = {3, 4. 5. 6}. Then A ~ B = {1, 2}, B  A = {5, 6}, A n B = {3, 4} and AAJB = {7, 8}. See Figure 3.
(b) A = {{1}. 1}, B = {1}, C 1.27
Figure 3
Answer for Exercise 1.27
1.29 (a) The sets 0 and {0} are elements of A. (b) A = 3. (c) All of 0, {0} and {0, {0}} are subsets of A. (d) 0 n A = 0. (e) {0} n A = {0}. (f) {0, {0}} n A = {0. {0}}. (g) 0 U A = A. (h) {0} U A = A. (i) {0, {0}} U A = A. 1.31 A = {1, 2}. B = {2}, C = {1, 2, 3}, D = {2, 3}. 1.33 A = {1}, B = {2}. Then {A U B, A fl 6 , A —B, B — A} is the power set of {1, 2}. 1.35 Let U = { 1 ,2 ,..., 8}, A = { 1 ,2 ,3.5}, B = {1,2,4, 6} and C = {1, 3, 4, 7}. See Figure 4.
Figure 4
Answer for Exercise 1.35
Exercises for Chapter 1
353
Section 1.4: Indexed Collections of Sets 1.37 U x e s * = A U B U C = ( 0 ,1 ,2 ........5) a n d f Xe.sx = A n B n C = {2}. 1.39 Since A = 2 6 and  Aa  = 3 for each a e A, we need to have at least nine sets of cardinality 3 for their union to be A; that is, in order for Uaes Aa = A, we must have S > 9. However, if we let S = [a,d, g , j , m, p, s ,v, then U ass Aa = A. Hence the smallest cardinality of a set S with Aa = A is 9. 1.41 (a) {A„}„eN, where A„ = {x e R : 0 < x < 1/n] = [0, 1/n], (b) {A„},!ely, where A„ = [a e Z : \a\ < n] = {—n,  ( n — 1 ) , . . . , ( « — 1), n). 1.43 1.45
Ur€R+ A '' = UreR+(—r ’ ' ) = n / sR+ A r = n reR+(—r ’ = U „£n = U«6n(  ; . 2  f ) = (  1 , 2); n „ sN A„ = n ^  J . 2  i ) = [0, 1].
Section 1.5: Partitions of Sets 1.47
1.49 1.51 1.53
1.55
(a) Si is not a partition of A since 4 belongs to no element of S i. (b) S2 is a partition of A. (c) S3 is not a partition of A because 2, for example, belongs to two elements of S3. (d) S4 is not a partition of A since S4 is not a set of subsets of A. A = {1, 2, 3, 4}. S, = {{1}, {2}, {3, 4}} and S2 = {{1,2}, {3}, {4}}. Let S = {Ai, A2, A3}, where Ai = {x e Q : x > 1), A2 = {x e Q : x < 1} and A3 = {1}. Let S = {Ai, A2, A3, A4}, where A\ = [x e Z : x is odd and x is positive}, A2 = {x € Z : x is odd and x is negative}, A3 = {x e Z : x is even and x is nonnegative}, <44 = {x e Z : x is even and x is negative}. \Vi\ = 2, 17^21 = 3, \V3\ = 5, P 4 = 8, P 5 = 13, \V6\ = 21.
Section 1.6: Cartesian Products of Sets 1.57 1.59 1.61
A x B  {(x, x), (x, y), (y, x), (y, y), (z, x), (z, y)}. V (A ) = {0, {a}, {b}, A}, A x V(A) = {(a, 0), (a , {a}), (a, {&}), (a.A), (ft, 0), (b , {a}),(b, {b}), (ft, A)}. P(A ) = {0, {1}. {2}, A}, V (B ) = {0, B}, A x B = {(1, 0), (2, 0)}, P(A ) x V{B) = {(0, 0), (0, S), ({1}, 0), ({1}, B), ({2}, 0), ({2}, fi), (A, 0), (A. B)}. 1.63 S = {(3, 0), (2, 1), (1,2), (0, 3), (  3 , Oj. ( 2 , 1), (  1 , 2), (2.  1 ) , (1.  2 ) , (0,  3 ), (  2 ,  1 ) , (  1 , 2)}.
354 1.65
Answers to OddNumbered Section Exercises A x B = [—1, 3] x [2, 6], which is the set of all x = 3, y = 2 and y — 6.
points on and within the square bounded by .v = —1,
EXERCISES FOR CHAPTER 2 Section 2.1: Statements 2.1 2.3 2.5 2.7 2.9
(a) A false statement (b) A true statement (c) Not a statement (d) Not a statement (an open sentence) (e) Not a statement (f) Not a statement (an open sentence) (g) Not a statement (a) False. 0 has no elements, (b) True (c) True (d) False. {0} has 0 as its only element, (e) True (f) False. 1 isnot a set. (a) {x e Z : x > 2} (b) {x e Z : x < 2) 3 .5 .1 1 .1 7 ,4 1 ,5 9 P (n) : is even. P(n) is true only for n — 5 and n = 9.
Section 2.2: The Negation of a Statement (a) (b) (c) 2.13 (a) (b) (c) (d) (e) 2 .1 1
\/2 is not a rational number. 0 is a negative integer. 111 is not a prime number. The real number r is greater than ~Jl. The absolute value of the real number a is at least 3. At most one angle of the triangle is 45a. The area of the circle is less than 9jt. The sides of the triangle have different lengths. (f) The point P lies on or within the circle C.
Section 2.3: The Disjunction and Conjunction of Statements 2.15
See Figure 6.
P
Q
~Q
P a (~Q )
T
T
F
F
T
F
T
T
F
T
F
F
F
F
T
F
Figure 6 2.17 (a) P v Q: 15 is odd or 21 is prime. (True) (b) P A Q\ 15 is odd and 21 is prime. (False) (c) ( ~ P ) v Q'. 15 is not odd or 21 is prime. (False) (d) P A Q): 15 is odd and 21 is not prime. (True)
Section 2.4: The Im plication 2.19 (a) ~ P: 17 is not even (or 17 is odd). (True) (b) P v Q 17 is even or 19 is prime. (True) (c) P A Q: 17 is even and 19 is prime. (False) (d) P => Q: If 17 is even, then 19 is prime. (True)
Answer for Exercise 2.15
Exercises for Chapter 2
355
2.21 (a) P => Q: If V 2 is rational, then 22/7 is rational. (True) (b) Q => P: If 22/7 is rational, then V2 is rational. (False) (c) (~ P ) => (~ 2 ): If \/2 is not rational, then 22/7 is not rational. (False) (d) (~ Q) =4> (~ P): If 22/7 is not rational, then V2 is not rational. (True) 2.23 (a), (c), (d) are true. 2.25 (a) true, (b) false, (c) true, (d) true, (e) true. 2.27 Cindy and Don attended the talk. 2.29 Only (c) implies that P V Q is false.
Section 2.5: More on Im plications 2.31 (a)
(b) (c) 2.33 (a) (b) (c)
P(x) => Q(x): If jc = 4, then x = 4. P (—4) => Q (—4) is false. P ( —3) => Q (—3) is true. P (l) =>■ g ( l) is true. P (4) => 2(4) is true. P(5) =>■ 2(5) is true. P (x) => Q(x): If x 2 = 16, then .r = 4. True for all x e S. P (x) => Q(x)\ I f v > 3, then 4x — 1 > 12. True for all x € S. True for (x, y) = (3, 4) and (x, y) = (5, 5), false for (x, y) = (1, —1). True for (x, y) = (1, 2) and (x, j ) = (6, 6), false for (x, _y) = (2, —2). True for ( x ,y ) e {(1, —1), (—3, 4), (1, 0)} and false for (x, y) = (0, 1 ).
Section 2.6: The Biconditional 2.35 2.37
P ■£> Q : The integer 18 is odd if and only if 25 is even. (True) The real number x —3 < 1 if and only if x e (2, 4). The condition \x —3 < 1 is necessary and sufficient for x e (2, 4). 2.39 (a) True for all x e S — {—4}. (b) True forx e S — {3}.(c) True fo rx e S — {—4, 0}. 2.41 True if n = 3. 2.43 P (l) =>■ 2(1) is false (since P( 1) is true and 2(1) is false). 2 (3) => P(3) is false (since 2(3) is true and P (3) is false). P(2) 2(2) is true (since P (2) and 2(2) are both true). 2.45 True for all n G S — {11}.
Section 2.7: Tautologies and Contradictions 2.47
2.49
The compound statement (P A (~ 2 )) A (P a Q) is a contradiction since it is false for all combinations of truth values for the component statements P and Q. See the truth table below. p
Q
~Q
Pa Q
T T F F
T F T F
F T F T
T F F F
P
A
(~ Q) F T F F
(P
A
(~ Q) ) a (P
a
Q)
F F F F
The compound statement ((P Q) A (Q =» P)) =>(/ >=>/ ?) is a tautology since it is true for all combinations of truth values for the component statements P, Q, and R. See the truth table below. P
Q
R
T T F F T T F F
T F T F T F T F
T T T T F F F F
P
T F T T T F T T
Q => R
( P = > Q ) a ( Q= > R)
P => R
T T T T F T F T
T F T T F F F T
T T T T F F T T
((/> => Q )
A
(Q
=►R)) =» (P => R) T T T T T T T T
((P => 2 ) A ( 2 => R)) =$■ (P => R)' If P implies Q and Q implies R, then P implies R.
356
Answers to OddNumbered Section Exercises
Section 2.8: Logical Equivalence 2.51 (a) See the truth table below.
(b)
p
Q
T T F F
T F T
F
P^Q
~ P F F T T
P)
T F T T
F T F T
=>
Q)
T T F T
Since P =>■ Q and (~ P) => (~ Q) do not have the same truth values for all combinations of truth values for the component statements P and Q, the compound statements P => Q and (~ P) =>• (~ Q) are not logically equivalent. Note that the last two columns in the truth table are not the same. The implication Q => P is logically equivalent to (~ P) => (~ Q ).
2.53 (a) The statements P =>■ Q and (P A Q) <£■ P are logically equivalent since they have the same truth values for all combinations of truth values for the component statements P and Q. See the truth table. P
Q
P=>Q
Pa Q
T T F F
T
T F T T
T
F T F
(P A Q)
P
T F T T
F F F
(b) The statements P =>• (Q
V R) and ( ~ Q) ■> ((~ P I V R) are logically equivalent since they have the same truth values for all combinations of truth values for the component statements P, O and R. See the truth table.
2.55
2.57
P
Q
R
1' T F F T T F F
1 F T F T F T F
1' T T T F F F F
~ P F F T T F F T T
QvR
P = > ( QV R )
(~P)VR
T
T T T T T F T T
(~ Q) =►((~ P ) v R)
T T T T F F T T
T T T T T
F T F T F T F T
T
T T T F T F
F T T
The statements ( P V Q) => R and (P =#• R ) A (Q R) are logically equivalent since they have the same truth values for all combinations of truth values for the component statements P, Q and R. See the truth table. P
Q
R
PvQ
(P v Q) =►R
P =» R
Q => R
(? = » /? ) a (Q =* R)
1
1
T F F T T F F
F T F T F
T T T T F F F F
T T T F T
T T T T F F F T
T T T T F F
T T T T F T F T
T T T T F F F T
T
F
T T
F
T T
Since there are only four different combinations of truth values of P and Q for the second and third rows of the statements S., S. ,V?. S4) and Ss, at least two of these must have identical truth tables and so are logically equivalent.
Section 2.9: Some Fundam ental Properties of Logical Equivalence 2.59 (a) Both x ^ 0 and y ^ O . (b) Either the integer a is odd or the integer b is odd. 2.61 Either x 2 = 2 and x ^ y/2 or x = V2 and x 2 ^ 2. 2.63 If 3n + 4 is odd. then 5n —6 is odd.
Exercises for Chapter 3
357
S ection ‘2 .10: Q uantified Statem ents 2.65 Vx € S, P (x ): For every odd integer x, the integer x 2 + 1 is even. 3x £ S, Q(x): There exists an odd integer a such that x 2 is even. 2.67 (a) There exists a set A such that A n A ^ H . (b) For every set A, we have A £ A. 2.69 (a) False, since P( l ) is false, (b) True, for example, P (3) is true. 2.71 (a) 3a, b e Z, ab < 0 and a + b > 0. (b) Vx, y £ R, x y implies that x 2 + y 2 > 0. (c) For all integers a and b either ab > 0 or a + b < 0. There exist real numbers x and y such that x ^ y and x 2 + y 2 < 0. (d) Va, b £ Z, ab > 0 or a + b < 0. 3x, y e R, x ^ y and x 2 + y 2 < 0. 2.73 (b) and (c) imply that P(x) =>■ Q(x) is true for all x e T. 2.75 Let S = {3, 5, 11} and P(s, t) : st — 2 is prime. (a) Vs, t e S, P{s, t). (b) False since P (11, 11) is false. (c) 3s, t 6 5, ~ P(s, t). (d) There exist s , t e S such that st — 2 is not prime. (e) True since the statement in (a) is false. 2.77 (a) There exists a triangle T\ such that for every triangle T2,r(T2) > r(T\). (b) v r , £ A, 3T2 e ~ P(Tl , T 2). (c) Forevery triangle T\, there exists a triangle T2 such that r(T2) < r(Ti). 2.79 (a) There exists b e B such that for every a e A, a — b < 0. (b) Let b = 10. Then 3 — 10 = —7 < 0, 5 — 10 = —5 < 0 and 8 —10 = —2 < 0. S e c t i o n 2 .1 1 : C h a r a c t e r i z a t i o n s o f S t a t e m e n t s 2.81 2.83
An integer n is odd if and only if n2 is odd. (a) a characterization, (b) a characterization, (c) a characterization. (d) a characterization. (Pythagorean theorem) (e) not a characterization. (Every positive number is the area of some rectangle.)
EXERCISES FOR CHAPTER 3 S e c t i o n 3 .1 : T r i v i a l a n d V a c u o u s P r o o f s 3.1
P roof Since x 2 —2x + 2 = (x — l)2 + 1 > 1, it follows that x 2 —2x + 2 ^ 0 for all x £ R. Hence the statement is true trivially. ■ 3.3 P roof Note that r ^  = r + j . If r > 1, then r + j: > 1; while if 0 < r < 1, then j > 1 and so r + ; > 1. Thus r ~  < 1 is false for all r e Q and so the statement is true vacuously. ■ P roof Since rr —2n + 1 = (n — l)2 > 0, it follows that n2 + 1 > 2n and so n + 1 > 2 . Thus the statement is true vacuously. ■ 3.7 Proof Since (x —y)2 + (x —z)2 + (y — z)2 > 0, it follows that 2x2 + 2y2 + 2z2 —2xy —2x z —2yz > 0 and so x 2 + y 2 + z 2 > x y + x z + yz. Thus, the statement is true vacuously. ■
3.5
S e c t i o n 3 .2 : D i r e c t P r o o f s 3.9
Proof Let x be an even integer. Then x = 2a for some
integer a . Thus
5x — 3 = 5(2a)  3 = 10a  4 + 1 = 2(5a  2) + 1.
Since 5a —2 is an integer, 5x —3 is odd.
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358
3.11 3.13 3.15
Answers to OddNumbered Section Exercises Proof Let 1 —n2 > 0. Then n = 0. Thus 3n —2 = 3 ■0 —2 = —2 is an even integer. Proof Assume that (n + 1)2{n + 2)2/4 is even, where n S S. Then n = 2 and (n + 2)2(n+ 3)2/4 = 100, which is even. Proof Let n e A 0 B = {3, 5, 7, 9). Then 32  2 = 7. 52  2 = 23, 72  2 = 47 and 92 2 = 79are all primes.
n h
■
Section 3.3: Proof by Contrapositive 3.17
3.19
3.21 3.23 3.25
First, we prove a lemma. Lemma Let n e Z. If 15« is even, then n is even. (Use a proof by contrapositive to verify this lemma.) Then use this lemma to prove the result. Proof of Result Assume that 15» is even. By the lemma, n is even and so n = 2a for some integer a. Hence 9/7 = 9(2a) = 2(9a). Since 9a is an integer, 9n is even. ■ [Note: This result could also be proved by assuming that 15n is even (and so 15/7 = 2a for some integer a) and observing that 9n = 15ft —6n = 2 a — 6n.] Lemma Let x e Z. If 7x + 4 is even, then x is even. (Use a proof by contrapositive to verify this lemma.) Proof of Result Assume that 7x + 4 is even. Then by the lemma, x is even and so x = 2a for some integer a. Hence 3a  1 1 = 3(2a) II = 6 ,; 1 2  I 2(3a 6) + 1. Since 3a —6 is an integer, 3x — 11 is odd. ■ To verify the implication “If n is even, then (/? + l)2 — 1 is even”, use a direct proof. For the converse, “If (n + l)2 — 1 is even, then n is even.", use a proof by contrapositive. Proof Assume that n £ A U B. Then n = 3 and n(n  1)(« —2)/6 = 1 is odd. a Proof Assume that n £ A. Then n e B = [2, 3, 6, 7). If n = 2, then (n2 + 3/f  4)/2 = 3 is odd. If n = 3, then (n2 + 3n — 4)/2 = 7 is odd. If n = 6, then (n2 + 3n  4)/2 = 25 is odd. If n = 7, then 0n2 + 3n —4)/2 = 33 is odd. ■
Section 3.4: Proof by Cases 3.27
Let n e Z. We consider two cases. Case 1. n is even. Then n = 2a for some integer a. Thus n3 —n = 8a3 —2a = 2(4 a3 — a). Since 4a3 —a is an integer, n3 —n is even. Case 2. n is odd. Then n = 2b + 1 for some integer b. The remainder of the proof is similar to that of Case 1. 3.29 Assume that a .b e Z such that ab is odd. By Exercise 3.28, a and b are both odd and so a2and b2 are both odd by Theorem 3.12. Thus a2 + b2 is even. 3.31 Proof Assume that a or b is odd, say a is odd. Then a = 2x + 1 for some integer x. We consider two cases. Case 1.b is even. Then b = 2y for some integer y. Thus ab = a(2v) = 2 (ay). Since ay is an integer, ab is even. Also, a f b = (2x t 1) ~ 2y = 2(x 4 y) ( 1. Since x + y is an integer, a + b is odd. Hence ab and a + b are of oppositeparity. Case 2 .b is odd. Then b = 2y + 1 for some integer y. Thus a + b — (2a + 1) ~t~ (2y f 1) = 2x *12y + 2 = 2(x f y j 1). Since x + y + 1 is an integer, a + b is even. Furthermore, ab = (2x + 1)(2y 4 1) = 4x y
3.33
2x j 2y t 1 = 2(2xy —.v •t• _v) i I.
Since 2xy + x + y is an integer, ab is odd. Hence ab and a + b are of opposite parity. Proof Assume that n g A n B. Then n = 1 or n = 4. If n = 1, then 2n2  5n =  3 is negative and odd; while if n = 4, then 2n2 — 5n = 12 is positive and even.
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Exercises for Chapter 4
359
For the converse, assume that n e A n B . Then n = 2 or n = 3. If n = 2, then 2n2 — 5 n= —2 is negative and even; while if n = 3, then 2«2 —5« = 3 is positive and odd. Thus if n e A n B, then neither (a) nor (b) occurs. ■ 3.35 Proof Let n be a nonnegative integer. We consider two cases. Case 1. n = 0. Then 2" + 6" = 2° + 6° = 2, which is even. Case 2. « is a positive integer. Then n  1 is a nonnegative integer. Therefore, 2" + 6« = 2" + (2 . 3)" = 2" + 2" • 3" = 2(2"“ ' + 2"~l ■3"). Since 2"~’ + 2"_1 ■3" is an integer, 2" + 6" is even.
■
Section 3.5: Proof Evaluations 3.37 3.39 3.41
(3) is proved. The converse of the result has been proved. No proof has been given of the result itself. From the first sentence of the proposed proof and the final sentence, it appears that the result in question is the following: Let x, y G Z. If x or y is even, then x y 2 is even. If this, in fact, is the result, then the proof is not correct. A proof by cases should be given, namely Case 1. x is even, and Case 2. y is even. 3.43 Result Let x £ Z. If 7x  3 is even, then 3x + 8 is odd. A direct proof of the result is given with the aid of the lemma: Let x e Z. If 7x —3 is even, then x is odd.
EXERCISES FOR CHAPTER 4 Section 4.1: Proofs Involving D ivisibility of Integers 4.1
Proof Assume that a \ b. Then b = ac for some integer c. Then b2 = (ac)2 = a2c2. Since c2 is an integer,
a 2  b2. " 4.3 (a) Proof Assume that 3  m. Then m = 3q for some integer q. Hence m2 = (3q)2 = 9q2  3{3q2). Since 3q 2 is an integer, 3  m 2. ■ (b) Let m € Z. If 3 j(m2. then 3 / m . (c) Start with the following: Assume that 3 /m . Then m= 3q + 1 or m = 3q + 2, whereq £ Z. Consider these two cases. (d) Let m £ Z. If 3  m 2, then 3  m. (e) Let m £ Z. Then 3  m if and only if 3 [ m 2. 4.5 Proof Assume that a \ b or a \ c, say the latter. Then c = ak for some integer k. Thus be = b(ak) = a{bk). Since bk is an integer, a \ be. 4.7 For the implication “If 3 J(n, then 3  (2n2 + 1).”, use a direct proof. Assume that 3 / « . Then n = n = 3q + 2 for some integer q. Then consider these two cases. For the converse “If 3  (2n2 + 1), then 3 / n.” use a proof by contrapositive. 4.9 (a) Proof Assumethat2 ] (x2  5).T henx2  5 = 2y for some integer y a n d so x 2 = 2y + 5 = 2(y + 2) + 1. Since y + 2 e Z, it follows that x 2 is odd. By Theorem 3.12, x is also odd and so x = 2a + 1 for some integer a. Hence x 2  5 = (2a + l)2  5 = 4a2 + 4a  4 = 4(a2 + a  1). Since a2 + a  1 is an integer, 4  (x2  5). (b) x = 3. 4.11 Proof Let n £ Z with n > 8. Then n = 3q where q> 3, n = 3q + 1where q > 3 or n = 3q + 2 where q > 2 . We consider these three cases. Case 1. n = 3q where q > 3 . Then n = 3a + 5b, where a = q > 3 and b = 0. Case 2. n = 3q + I where q > 3 . Then n = 3(q  3) + 10, where q  3 > 0. Thus n = 3a+ 5b, a = q  3 > 0 and b = 2. Case 3. n = 3q + 2 where q > 2. Then n = 3(q — 1) + 5, where q — 1 > 1. Thus n = 3a + 5b, where a = q — 1 > 1 and b = 1.
■
wher
■
360
Answers to OddNumbered Section Exercises
4.13 First, we prove the following lemma. Lem ma If c £ Z, then c2 = 0 (mod 3) or c2 = 1 (mod 3). Proof If 3  (c2  1), then we have the desired conclusion. Otherwise, 3 / (c2 — 1) and so 3  c by Result 4.6. Then 3  c2 by Exercise 3 of this chapter. Thus, either c2 e= 0 (mod 3) or c2 = 1 (mod 3). ■ We are now prepared to prove the main result. Proof Assume that 3 /ab. By Result 4.5, 3 / a and 3 J[b. Thus, 3  (a1  1) and 3  (b2 1) by Result 4.6. Therefore, 3  (a2 + b2  2). Since c2 ^ 2 (mod 3) for any positive integer c by the lemma, it follows that a2 + b2 cannot equal c2. a
Section 4.2: Proofs Involving Congruence of Integers 4.15 Proof Assume that a = b (mod n) and a = c (mod n). Then n \ (a — b) and n \ (a —c). Hence a — b = nx and a  c = ny, where x, y g Z. Thus b = a  nx and c = a  ny. Therefore, b  c = {a  nx)  (a  ny) = ny — nx = n(y —x). Since y  x is an integer, n \ (b — c) and so b = c (mod n). ■ 4.17 (a) Proof Assume that a = 1 (mod 5). Then 5  (a  1). So a  1 = 5k for some integer k. Thus a = 5k + 1 and so a2 = (5k + l) 2 = 25 k2 + 10A + 1 = 5(5k? + 2k) + 1. Thus a2  1 = 5(5k2 + 2k). Since 5k2 + 2k is an integer, 5  (a2  1) and so a2 = 1 (mod 5). [Note: We could alsoobserve that a2 — 1 = (a — l)(fl + 1). This is also a consequence of Exercise 14 in this chapter.] h (b) We can conclude that b2 = 1 (mod 5). 4.19 Proof Assume that a = 5 (mod 6) and b = 3 (mod 4). Then 6  (a  5) and 4  (b — 3). Thus a —5 = 6x and b — 3 = 4y, where x, y e Z. So a = 6x + 5 and b = 4y + 3. Observe that 4a + 6b = 4( 6,\ + 5) + 6(4y + 3) = 24x + 20 + 24y + 18 = 24x + 24_v + 38 = 8(3.v + 3y + 4) + 6. Since 3x + 3y + 4 is an integer, 8](4a + 6b — 6) and so 4a + 6 b = 6 (mod 8). ■ 4.21 Either a = 3q, a = 3q + 1 or a = 3q + 2 for some integer q. We consider these three cases. Case I. a = 3q. Then a3 —a = (3q f — (3q) = 21q'i  3 q = 3(9q 3q). Since 9q 3 — q is an integer, 3  (a3 —a) and so a 3 = a (mod 3). The other cases are handled in a similar manner. 4.23 Proof Since 6  a, it follows that a = 0 (mod 6) and so a = 6q for some q e Z. Therefore, a + i = i (mod 6) for i = 1, 2 , . . ., 5. First, assume that x , y 6 S = {a, a + 1 , . . . , a + 5), where one of x and y is congruent to 1 modulo 6 and the other is congruent to 5 modulo 6. We may assume that x = a + 5 and y = a + 1. Then x = 6q + 5 and y = 6q + 1. Thus x 2  y2 = (6q + 5)2  (6q + l) 2 = (36<?2+ 60q + 25)  (36q2 + 12q
+ 1)
= 48 q + 24 = 24(2q + 1). Since 2q + 1 is an integer, 24  (x2 —y2). For the converse, assume that x and y are distinct odd integers in S such thatone of x and y is not congruent to 1 or 5 modulo 6. Since a is even, either x or v is a + 3. There aretwo cases. Case 1, x = a + 5 and y — a + 3. Thus x = 6^ + 5 and y = 6q + 3.Now •T2 ~ y 2 = (6q + 5)2  (6q + 3)2 = (36q 2 + 60q + 25)  (36q 2 + 36q + 9) = 24g + 16. Since q is an integer, 24 / (x2 — y 2). Case 2 .x = a + 3 and y = a + \ . (The proof here is similar to the proof of Case 1.)
»
Exercises for Chapter 4
361
Section 4.3: Proofs Involving Real Numbers Proof Assume that x 2 —4x = y 2 — 4 y and x ^ y. Thus x 2 — y 2 —4(x — y) = 0 and so (x —;y)[(x 4 y) —4] = 0. Since x ^ y, it follows that (x + y) — 4 = 0 and so x 4 y = 4. m 4.27 A proof by contrapositive can be used: Assume that x < 0. Then 3x4 4 1 > 1 and x 7 + x 3< 0.Thus 3x4 41 > 1 > 0 > x 1 4 x 3. 4.29 Proof Let r e R such that r — 11 < 1. Since r — 1 < 1, it follows that 0 < r < 2. Because(r— 2)2 > 0, we have 4.25
r 2  4r + 4 > 0. Thus 4 > 4r — r 2 = r(4 —r). Since 0 < r < 2, it follows that r(4 —r) > 0. Dividing both sides by r{4 — r), we obtain r~ j > 1. 4.31 Proof Since x = (x + y) + (—v) < x 4 y \ 4  1 y = x + y\ + \y\, it follows that x 4 y\ > x — y. 4.33 Observe that r 3 4 s 3 4 ?3 —3rst = j( r 4 .s 4 0t('" —s)2 4 (s —t)2 4 (t — r)2]. 4.35 Proof Assume that x(x 4  1) > 2. Then x 2 4 x > 2 and so x 2 + x —2 > 0. Thus (x 4 2)(x — 1) > 0. Therefore, either (a) x 4 2 and x — 1 are both positive or (b) x 4 2 and x — 1 are both negative. If (a) occurs, then x > 1; while if (b) occurs, then x < —2. 4.37 Proof Since (x —y)2 4 (x —z)2 4 (y —z)2 > 0, it follows that 2x2 + 2y2 4 2z2 > 2xy 4 2xz 4 2yz. Dividing by 2 produces the desired inequality. 4.39 P roof Observe that 0a2 + c2)(b2 + d 2) = a2b2 4 crd2 + b2c2 + c2d 2 = (ab + cd)2 + (ad —be)2 > (ab + cd)2. m
Section 4.4: Proofs Involving Sets First, we show that if A U B = A, then B c A. Assume that A U B = A. Let x e B. Then x £ A l l B . Since A U B = A, it follows that x e A. Thus B C A. Next we show that if B C A, then A U B = A. A proof by contrapositive begins by assuming that A U B ^ A. 4.43 (a) Consider A = {1, 2}, B = {2, 3) and C = {2, 4}. (b) Consider A = {1, 2}, B = {1} and C = {2}. (c) Suppose that B ^ C. We show that either A fi B ^ A H C or A U 5 ^ A U C . Since B ^ C, it follows that B g C or C g B , say the former. Thus there exists b e B such that b £ C. We consider two cases, according to whether b e A or b £ A. Case I. b 6 A. Since b e B and b e A, it follows that b e A n B. On the other hand, b <£ C and so b £ A n C. Thus A n g ^ A f l C . Case 2. b £ A. Then show that A U B ^ A U C . 4.45 Proof Let n e B. Then n e Z and n = 3 (mod 4). So n= 4q + 3 for some integer q. Therefore, ■ n = 2(2q + 1) + 1. Since 2q + 1 e Z, it follows that 2  (n — 1) and so n = 1 (mod 2). Thus n e A. 4.47 (a) Each element n e A — B can be written as n = 3a + 2 for some integer a, where n is even. Thisimplies that a is even, say a = 2b for some integer b. Thus n = 3a + 2 = 3(2b) 4 2 = 6b 4 2. (b) Proof Let n € A n B . Then n = 3a + 2 for some integer a and n is odd. Thus a is odd, say a = 2b + 1 for some integer b. Thus n = 3a 4 2 = 3(2b 4 1) + 2 = 6b + 5. Therefore, 4.41
n2  1 = (6b 4 5)2  1 = 36b2 + 60b 4 24 = 12(3fo2 45b + 2). Since 3b2 4 5b + 2 is an integer, 12  (n2 — 1) and so n2 = 1 (mod 12). ■ First, we show that A c (A — B ) U (A n B) . Let x € A. Then x ^ B or x e B . If x ^ B , then x e A — 5 and x e (A — £ ) U (A fi £). If x e 5 , then x e A D B and so x e (A — B ) U (A D 5 ). Therefore, A c (A — B ) U (A n 5). Next, show that (A — B ) U (A n S ) c A. 4.51 (e) is a necessary condition for A and B to be disjoint. 4.49
362
Answers to OddNumbered Section Exercises
Section 4.5: Fundamental Properties of Set Operations 4.53 First, we show that A n f S U C) C ( A n f i ) U ( A n C). Let x e A n ( B U C). Then x e A and x e B U C . Since x e f i U C . i t follows that x e B or x g C, say x e B. Because x e A and x e B , it follows that x e A n B . Hencex e ( A n B ) U (A D C). Next, show that (A n B ) U (A n C ) C A n ( B U C). 4.55 We first show that ( A  B ) n ( A — C) C A  ( B U C). Let x e (A  B ) n (A  C ) . Then x € A  B and x e A  C . Since x e A ~ B , it follows that x e A and x £ B . Because x e A  C, we have x e A and m £ C . Since x <£ B and x <£ C , we have x B U C . Thus x 6 A  ( B U C ). Next, show that A  ( f i U C ) C f A  B ) f l ( A  C). 4.57 P roof By Theorem 4.22. a u (B n C ) = a n (b n c ) = a n ( § u c ) = A n ( B u C ) = (A n b ) u (A n C ) = (A n B ) U (A  C),
as desired. B 4.59 First, we show that A  (B  C) c (A n C) U (A  B). Let x e A  (B  C). Then x e A and x £ B  C. Since x <£ B  C . it is not the case that x € B and x <£C. Thus either x <£ B or x e C. Since x E A, either x e A  B o v x e A H C . Thus x e ( A f l C ) U ( A  f i ). Therefore, A  (B  C) C (A n C) U (A —5 ). Next, show that ( A n C ) U ( A  S ) C A —(B  C).
Section 4.6: Proofs Involving Cartesian Products of Sets 4.61 For A = {1} and B = {2}, 'P(A) = {0, A} a n d W ) = {0, B}. Thus V ( A ) x P { B ) = {(0, 0), (0, B) , (A, 0), (A, 5)}. Since A x B = ( 1. 2)}, it follows that P ( A x B ) = {0, A x B}. 4.63 Let A and B be sets. Then A x B = 5 x A if and only if A = B or one of A and B is empty. P roof First, we show that if A = B or one of A and B is empty, then A x B = B x A. If A = B , then certainly A x B = B x A: while if one of A and B is empty, say A = 0, then A x B = 0 x B = 0 = B x 0 = BxA. For the converse, assume that A and B are nonempty sets with A ^ B . Since A ^ 5 , at least one of A and B is not a subset of the other, say A % B. Then there is an element a 6 A such that a/>’. Since 5 ^ 0 , there exists an element b € B. Then (a, b) e A x B but (a , b) g B x A. Hence A x B =£ B x A. m 4.65 First, assume that A x C C B x C. Show that A c j . For the converse, assume that A C S . Show that A x C c B xC. 4.67 We first show that A x ( B n C) c (A x B) n (A x C). Let (x, y) e A x ( B n C ) . Then x e A and y e B n C . Thus y e B and y 6 C. Thus (x> y) £ A x B and (x, y) e A x C. Therefore, (x,y) e (A X B ) n ( A x C). It remains to show that (A x B ) fl (A x C ) C A x ( S ft C). 4.69 Proof Let (x, y) e (A x 5 ) U (C x D ) . Then (x, y) e A x B or (x, j )  C x Z). Assume, without loss of generality, that (x, y) e A x 5 . Thus x e A and y e B . This implies thatx e A U C and y e B U D. Therefore, (x, y) e (A U C ) x ( B U D). E
E X E R C IS E S FOR C H A PT E R 5 Section 5.1: Counterexamples 5.1
Let a = b =  1. Then log(<;/>! = log 1 = 0 but logiV/) and logc/j) are not defined. Thus a = b = 1 is a counterexample. 5.3 If n = 3, then I n 2 + 1 = 19. Since 3 / 19, it follows that n = 3 is a counterexample. 5.5 If a = 1 and b = 2, then (a + b f = 33 = 27, but a3 + 2a2b + la b + l a b 2 + f t3 = l + 4 + 4 + 8 + 8 = 25. Thus a = 1 and b = 1 form a counterexample.
Exercises for Chapter 5 5.7 5.9
363
There is no counterexample. It can be shown that if a, b e R + and (a + b) (£ +  ) = 4, then a — b. Let x = 3 and n = 2. Then x" + (x + l)'7 = 32 + 42 = 25 = 52 = (jc + 2.)*. Then .v = 3, n = 2 is a counterexample.
Section 5.2: Proof by Contradiction 5.11 Assume, to the contrary, that there exists a smallest positive irrational number r. Then r /2 is a positive irrational number and r /2 < r. 5.13 Let a and b be odd integers and assume, to the contrary, that 4  (a2 + b2). Then a2 + b2 = Ax for some integer x. Starting with this, a contradiction can be obtained. 5.15 Assume, to the contrary, that 1000 can be expressed as the sum of three integers a, b and c, an even number of which are even. There are two cases. Case 1. None o f a, b and c is even. Then a = 2x + 1, b = 2y + 1 and c = 2z + 1, where x, y , z e Z. Thus 1000 = (2x + 1) + (2y + 1) + (2z + 1) = 2(x + y + z + 1) + 1.
5.17
5.19
5.21
5.23
5.25 5.27
5.29
Since x + y + z + l i s an integer, 1000 is odd, which is a contradiction. Case 2. Exactly two o f a, b and c are even, say a and b are even and c is odd. (The argument is similar to that in Case 1.) Proof Assume, to the contrary, that there exist an irrational number a and a nonzero rational number b such that a /b is a rational number. Then a /b = p /q , where p , q e Z and p , q ^ 0. Since b is a nonzero rational number, b = r/s, where r,s 6 Z and r, s j= 0. Thus a = (bp)/q = (rp )/(sq ). Since rp, sq € Z and sq ^ 0, it follows that a is a rational number, which is a contradiction. 9 Lemma Let a be an integer. Then 3  a2 if and only if 3  a. (See Exercise 3 in Chapter 4.) Proof of Result Assume to the contrary, that V3 is rational. Then ~/3 = p/q , where p , q € Z and q ^ 0. We may assume that p / q has been reduced to lowest terms. Thus 3 = p 2/ q 2 or p 2 = 3q 2. Since 3  p 2, it follows by the lemma that 3  p. Thus p = 3x for some integer x. Thus p 2 = (3a)2 = 9x2 = 3q 2. So 3x2 = q2. Since x 2 is an integer, 3  cy2. By the lemma, 3  q and so q = 3y, where y e Z. Hence p = 3x and q = 3 y, which contradicts our assumption that p / q has been reduced to lowest terms. ■ (a) One possible way to prove this is to use the fact that for integers a and b, the product ab is even if and only if a is even or b is even. Proof Assume, to the contrary, that s/6 is rational. Then ~/6 = a /b for nonzero integers a and b. We can further assume that a /b has been reduced to lowest terms. Thus 6 = a2/ b 2\ so a2 = 6b2 = 2(3b2). Because 3b2 is an integer, a2 is even. By Theorem 3.12, a is even. So a = 2c, where c e Z . Thus (2c)2 = 6b2 and so 4c2 = 6b2. Therefore, 3b2 = 2c~. Because c2 is an integer, 3b2 is even. By Theorem 3.17, either 3 is even or b2 is even. Since 3 is not even, b2 is even and so b is even by Theorem 3.12. However, since a and b are both even, each has 2 as a divisor, contradicting the fact that a/b has been reduced to lowest terms. ■ (b) We can use an argument similar to that employed in (a) to prove that is irrational for every odd positive integer k. Proof Assume, to the contrary, that there is some integer a such that a = 5 (mod 14) and a = 3 (mod 21). Then 14  (a —5) and 21  (a — 3); so a = 5 + 14* and a = 3 + 21_y for some integers x and y. Therefore, 5 + 14,v = 3 + 21 y, which implies that 2 = 21y — 14x = 7(3y — 2x). Since 3y  2x is an integer, 7  2, which is a contradiction. ■ Proof Suppose that there exist three distinct positive integers a , b and c such that each divides the difference of the other two. We may assume that a < b < c. Thus c  (b — a). Since 0 < b — a < c, this is a contradiction. ■ Proof Assume, to the contrary, that there exist positive real numbers .v and y such that J x + y = *Jx + s/y . Squaring both sides, we obtain x + y = x + 2^/x^/y + >'andso2v/F v/y = 2 s/ x y = 0. This implies that x y = 0. Thus x = 0 or y = 0, which is a contradiction. ■ Assume, to the contrary, that there exist positive integers .t and y such that x 2 — y 2 = m = 2s. Then (x + y)(x —y) = 2s, where s is an odd integer. We consider two cases, according to whether * and j are of the same parity or of opposite parity. Note that if x and y are of the same parity, then both x + y and x  y are even, while if x and y are of opposite parity, then both x + y and x  y are odd. Produce a contradiction in each case.
364 5.31 5.33
Answers to OddNumbered Section Exercises Assume, to the contrary, that there exists an integer m suchthat3 /( m 2  l ) and3 Jm.Thus/w = 3q + 1 or m = 3q + 2 for some integer q. Produce a contradiction in each case. (a) Proof Assume, to the contrary, t h a t #  3x + 1 = 0 has a rational number solution p /q , where p , q € Z and q # 0. We may assume that p /q is expressed in lowest terms. Thus K — + 1 = 0 and so p 2 ~~3pq + q2 = 0. We consider two cases. Case 1. Exactly one of p and q is even, say p is even and q is odd. Then p = 2r and q = 2s + 1, where r.s e Z. Hence p 2  3pq + q 2 = (2r) 2  3(2r)(2.s + 1} + (2s + l)2 = 4 r2 — 12;s —6r + 4s2 + 4s + 1 = 2(2 r 2 —6rs —3; + 2s2 + 2s) + 1 = 0 . Since 2r2 —6rs —3r + 2s2 + 2s € Z, it follows that p 2  3/x/ * q2 is odd and equals 0, which is a contradiction. Case 2. Both p and q are odd. (The proof in this case is similar to the proof of Case 1.) ■ (b) For positive integers k and n with k < n and odd integers a, b and c, the equation ax" + bxk + c = 0 has no rational number solution.
Section 5.3: A Review of Three Proof Techniques 5.35
(a) Proof Assume that x  2 > 1. Since x > 0, it follows, by multiplying by x, that x2  2 > x and so x2 —x —2 > 0. Hence x  2)(x + 1) > 0. Dividing by the positive number x + 1, we have x —2 > 0 and so x > 2 . ■ (b) Proof Assume that 0 < x < 2. Thus x 2  x  2 = (x  2)(x + 1) < 0 and so x 2 2 < x. D positive number x, we have X —§ < 1. ■ (c) Proof Assume, to the contrary, that there exists a positive number x such that x —  > 1 an Thus x 2 —x —2 = (x  2)(x + 1) < 0 and so x 2 —2 < x. Dividing by the positive number x, we have x — 2 < 1, producing a contradiction. ■ 5.37 (a) Proof Let i , j e R + such that x < y, Multiplying both sides by x and y, respectively, we obtain x2 < xy and xy < y 2. Therefore, x2 < xy < y2 and so x 2 < y2. ■ (b) Proof Assume that x 2 > y 2. Thus x2  y 2 > 0 and so (x + y)(x  y) > 0. Dividing by the positive number x + y, we obtain x —y > 0 and X > y. ■ (c) Proof Assume, to the contrary, that there exist positive numbers x and y such that x < y and x2 > y 2. Since x < y, it follows that x 2 < x y and xy < y 2. Thus x 2 < y2. producing a contradiction. b 5.39 Proof (Direct Proof) Let a. h. r e Z. We show that exactly two of ab. ac and be cannot be odd. If all of a, b and c are odd, then ab, ac and be are all odd; otherwise, at least one of a , b and c is even, say a is even. Then ab and ac are even. ■ P roof (Proof by Contradiction) Let a, b, c e Z and assume, to the contrary, that exactly two of ab, ac and be are odd, say ab and ac are odd. Then a, b and c are odd, which implies that be is odd. a contradiction, h
Section 5.4: Existence Proofs 5.41
P roof Consider the rational number 2 and the irrational number ^ is irrational, then a = 2 and b =
. If 2
1 have the desired properties. If, on the other hand, 2 ^ is rational, then / 1\^ V2 1 /— \2 ^ \ = 2^5 = 23 = V2
5.43
______ ______■ is irrational and so a = 2^5 and b = V 2 have the desired properties. P roof Assume, to the contrary, that there exist nonzero real numbers a and b such that s/a2 + b2 = s/a3 + b3.
Exercises for Chapter 6
365
Raising both sides to the 6th power, we obtain a6 + 3a4b2 + 3a2b4 + b6 = a 6 + 2a3/)3 + b6. Thus 3a2  2ab + 3b2 = (a  b f + 2a2 + 2b2 = 0. Since this can only occur when a — b = 0, we have a contradiction. ■ Let W = S — T . Since T is a proper subset of S, it follows that 0 ^ W c S. Then R(x) istrue for every x e W, that is, Vx e W , R(x) is true. 5.47 Proof Suppose that S = [a, b, c}. The nonempty subsets of 5 are {a}, {ft}, {c}, {a, b }, {a, c}, {b, c) and {a, b, c}. For each such subset A of S, aA is congruent to 0, 1, 2, 3, 4 or 5 modulo 6. Since there are seven nonempty subsets of S, there must be two of these seven subsets, say B and C, such that aB = ac (mod 6). ■ 5.45
Section 5.5: Disproving Existence Statements 5.49 We show that if a and b are odd integers, then 4 / (3a2 + l b 2). Let a and b be odd integers. Then a — 2x 4 1 and b = 2y + 1 for integers x and y. Then 3a2 + l b 2 = 4(3.r2 + 3x + l y 2 + l y + 2) + 2. Since 2 is the remainder when 3a2 + l b 2 is divided by 4, it follows that 4 / (3a2 + l b 2). 5.51 Show that if n is an integer, then n4 + r? + n2 + n is even. Let n e Z. Then n is even or n is odd.We consider these two cases. Case 1. n is even. Then n = 2a for some integer a. Then n4 + n3 + n2 + n = n{n + 1)(n2 + 1) = 2 a(n + 1)(n2 + 1) = 2[a(n + 1)(«2 + 1)]. Since a(n + 1)(n2 + 1) is an integer, n4 + n3 + n2 + n is even. Case 2. n is odd. (The argument is similar here.)
EXERCISES FOR CHAPTER 6 Section 6.1: The Principle of Mathematical Induction The sets in (b) and (d) are wellordered. Proof Let 5 be a nonempty set of negative integers. Let T = {n : —n e S } . Hence T is a nonempty setof positive integers. By the WellOrdering Principle, T has a least element m. Hence m < n for all n 6 T . Therefore, —m € S and —m > —n for all —n 6 S. Thus —m is the largest element of S. ■ 6.5 Proof We use induction. Since 1 = 2 • 12 — 1, the formula holds for n = 1. Assume that the formula holds for some integer k > 1, that is,
6.1 6.3
1 + 5 + 9 + . . . + (4k  3) = 2k2  k. We show that 1 + 5 + 9 + . . . + [4(k + 1)  3] = 20k + 1)2 ~ ( k + 1). Observe that 1 + 5 + 9 + . . . + [4(k + 1)  3] = [1 + 5 + 9 + . . . + (4k  3)] + 4(k + 1)  3 = (2k1  k ) + (4k + l) = 2k2 + 3 k + l = 2(k + l) 2 —(k + 1).
6.7
The result then follows by the Principle of Mathematical Induction. One possibility: 1 + 7 + 13 + . . . + (6n — 5) = 3n2 — 2n for every positive integer n.
■
366 6.9
Answers to OddNumbered Section Exercises Proof We proceed by induction. For n = 1, we have 1 3 = 3 = lll±Me±7), which is true. Assume that 1 • 3 + 2 • 4 + 3 11 + • •  + k(k + 2) == t(*+1)6(—— where k E N. We then show that (,k + 1)(A + 2)[2(k + 1) + 7] 1 • 3 + 2 ■4 + 3 • 5 + ■■■+ (it + l){k + 3) =  £  D (k + 1)(k t 2)(2 k + 9) 6 Observe that l  3 + 2 4 + 3 5 +   + (& + 1)(& + 3) = [1  3 + 2 4 + 3 5 +   + k{k + 2)] + (k + l){k + 3) = k(k + w k + J 1 + ( k + m + 3 ) 6 k(k + 1)(2k + 7) + 6 (k + \)(k + 3)
6
_ (k + l)(2fc2 + 7k + 6k + 18) _ (k + 1 ) M 2 + 13A + 18) 6 6 (k + l)(k + 2K2A + 9) 6 By the Principle of Mathematical Induction. trin + 1)(2« + 7) 1 3 + 2  4 + 3 5 H+ n(n + 2) =  ^— — 6 for every positive integer n. 6.11 We proceed by induction. Since
^
, the formula holds for /? = 1. Assume that
1 1 + 3 4 4 5
1 a + 2)(A + 3)
k 3k + 9
where k is a positive integer. Then show that 1 1 1 + — + ■■• + 3 4 4 5 (k + 3)(k + 4)
k+ 1 3(^ + 1 ) + 9
k +1 3(k + 4)
6.13 Proof We proceed by induction. Since 1 ■1! = 2!  1. the statement is true for n = 1. Assume that 1 ■11 + 2 2 ! +  + k k\ = ( £ + 4 ) !  1, where k e N. We show that 1 ■1! + 2  2! + • • ■+ (k + 1) ■(k + ) )! = (k + 2)!  1. Now 1 . U + 2 2! + . . . + (£ + 1)  (k + 1)! = (1 • 1! + 2 2! + ■■■+ k ■k\) + (k + I) ■(k + l)\ = f l + 1)!  1 + (£ + 1) • (k + 1)! = ( k + l ) \ ( k + 2 )  l = ( k + 2 ) \  1. By the Principle of Mathematical Induction, 1 ■1! + 2 ■2! + ■  + n  nl = (n + 1)! — 1 for all n £ N. 6.15 Proof We proceed by induction. Since ^ < 2VT  1. the inequality holds for n = 1. Assume that
Exercises for Chapter 6
367
for a positive integer k. We show that
1
1
1
1
,
 = +  p + 7= + • • • + . < 2 V k + 1  1. VI V2 V3 v^+T Observe that 1
1
1
1
/
1
1
1
7 l + V2 + V l+ " + VF+I _ \7T + V5 + v5 + " '+ VI7 ' v'F+T s /F T T
V FTT
Since 4(&2 + k) < (2/r + l)2, it follows that 2V k2 + A: < 2k + 1 and so 2V&2 + k + 1 < 2(k + 1) = 2(s/k + l)2. Therefore, 2V FTX + 1 ,7 = = = — 5 2 V F T T . V FTI Thus 1 1 1 1 , p + ^ + ^ + + . < 2 V * + 1  1. VI V2 V3 V F+T By the Principle of Mathematical Induction, ^7f + ^5 + ^  + " ’ + ^ integer n.
—
— 1f°r every positive m
Section 6.2: A More General Principle of Mathematical Induction 6.17
Proof We need only show that every nonempty subset of Shas a least element. So let T be a nonempty subset of S. If T is a subset of N, then, by the WellOrdering Principle, T has a least element. Hence we may assume that T is not a subset of N. Thus T — N is a finite nonempty set and so contains a least element t. Since t < 0, it follows that t < x for all x € T ; so t is a least element of T . m 6.19 Proof We use induction. We know that if a and b are two real numbers such that a b = 0, then a = 0 or b = 0. Thus the statement is true for n = 2. Assume that: If 1
a\, a 2,
. . .,
ak
are any k > 2 real numbers whose product is 0, then a, = 0 for some integer
i
with
< i <k.
We wish to show the statement is true in the case of k + 1 numbers, that is: If
bi , b i , . • . , bk+\
i (1
<
i
are
k + 1
real numbers such that
Let b\ , b2 ..........bic+1 be k + 1 real numbers such that / (1 < i < k + 1). Let b = b\b2 ■■■b k. Then b i b 2 ■■■b k+]
6.21
b\b2 ■■■£>a+i =
0, then
bt =
0 for some integer
< k + 1).
=
b i b 2 ■■■b k+\ =
( b ,b 2 ■• ■b k)b k+]
=
0. We show that b = 0 for some integer b b k+ ,
= 0.
Therefore, either b = 0 or bk+l = 0. If bk+l = 0, then we have the desired conclusion. On the other hand, if b = b\b2 ■■■bk = 0 , then, since b is the product of k real numbers, it follows by the inductive hypothesis that b, = 0 for some integer / (1 < / < k ). In any case, bt = 0 for some integer i (1 < i < k + 1). The result then follows by the Principle of Mathematical Induction. ■ Proof We proceed by induction. Since 4  (5° — 1), the statement is true for n = 0. Assume that 4  (5* — 1), where k is a nonnegative integer. We show that 4  (5*'+1 — 1). Since 4  (5* — 1), it follows that 5* = 4a + 1 some integer a. Observe that 5/'+1 — 1 = 5  5* — 1 = 5(4a + 1)  1 = 20a + 4 = 4(5a + 1).
368
Answers to OddNumbered Section Exercises
Since (5a + 1) e Z, it follows that 4  (5i + 1  1). By the Principle of Mathematical Induction, 4  (5"  1) for every nonnegative integer n . m 6.23 We employ mathematical induction. For n = 0, we have 7  0, which is true. Assume that 7  (32k  2 k) for some integer k > 0. We show that 7  (32(i+1)  2('t+1)) . Since 7  (32*  2k), it follows that 32*  2k = la for some integer a. Thus 32k = 2k + la . Then show that 32(i+1) —2l*+1* = 7(2* + 9a). 6.25 Proof We use induction. Since 4! = 24 > 16 = 24, the inequality holds for n = 4. Suppose that k\ > 2k f an arbitrary integer k > 4. We show that (k + 1)! > 2k+l. Observe that (k + 1)! = (k + 1)k\ > (k + 1) ■2* > (4 + 1)2*' = 5 • 2k > 2 ■2k = 2*+'.
6.27
Therefore, (k + 1)! > 2k+l. By the Principle of Mathematical Induction, n! >2" for every integer n > 4 . m P roof We proceed by induction. Since 1 < 2 the inequality holds for n = 1. Assume that I + I 4 . I ^ 2 —  for some positive integer k. We show that 1 + j +  4 4 p j g < 2  ^py. Observe that 1
1
4
/ I I 1 \ 7 + K  + 77 I + ' V 4 9 k1)
1
1 4  4  4 1 ,,
9
, ,^
(k + l) 2
1
= I 1+
— 
11
0
(k + l)2
—(k 4~ 1)” 4 k
~ + T + ( T + T)2 = ~ +
k(k 4 l)2
= 2  k2+ k+ } < 2  ^ ± ^ = 2 m + l f k{k + 1 ) 2 k
6.29
+
1
By the Principle of Mathematical Induction, 1 4  ^ 4  1 4  f ^ < 2  § for every positive integer n . P roof We proceed by induction. By De Morgan’s law, if A and B are any two sets, then A n B = A U B . Hence the statement is true for n — 2. Assume, for any k sets A\, A2, . . . , Aa, where k > 2 , that A, n a 2 n • • ■ n A t = Aj u a 2 u • ■ ■ u A*.
Now consider any k 41 sets, say />’ . B j , . . ., B*+i. We show that B1n B 2 n    n B i+1 =
b
1u
b
2u 
Let B = Bi n B2 n • ■• n B*. Observe that b , n b 2 n ■ • • n b. . : = (Bj n b 2 n •  • n B k) n B k+i = b n b m
= B U B m = ( B , U B 2 U • ■ • U B k) U B t+i
=
6.31
1 U 5 2 U • • • U B/<_i.
The result then follows by the Principle of Mathematical Induction. i P roof We proceed by induction. Since a (^) = l 2 for every positive real number a, the inequality is true for n = 1. Assume for each k positive real numbers ax, a2  ak that
Let b\, b i........bk+\ be k + 1 positive real numbers. We show that
Observe that
(x>) (f$  (x>) (t £) ♦  ( ±
l ) +
U
P )
—
Exercises for Chapter 6
Since
369
> 2 (see Exercise 4.78), it follows that
By the Principle of Mathematical Induction, (X]"=1 ai)
^ j ^ f‘2 f°r every n positive real numbers
a l , a 2, . . . , a n.
m
Section 6.3: Proof by Minimum Counterexample 6.33
Assume, to the contrary, that there is a positive integer n such that 6 / 7 « (n2 — l). Then there is a smallest positive integer n such that 6 /7 « («2  l). Let m be this integer. Since 6 [ 0 and 6  42, itfollows that 6  In (n2 — l) when n = 1 and n = 2. So m > 3 and we can write m = k + 2, where 1 < k < m. Consequently, 6  Ik (k2 — l) and so I k (k2 — I) = 6x for some integer x. Then show that 1m (m 2 — 1) = 6(x + I k 2 + 14k + 7). 6.35 Proof Assume, to the contrary, that there is a positive integer n such that 1 + 3 + 5 + ■■• + (2n — 1) ^ n2. Let m be the smallest such integer. Since 2 • 1 — 1 = l 2, it follows that m > 2. Thus m can be expressed as m = k + 1, where 1 < k < m. Therefore, 1 + 3 + 5 + • ■■+ (2k — 1) = k2. Then 1 "t 3
5 + ■• • + (2i7i — 1) = 1 + 3 + 5 H 1 (2k { 1) = [1 + 3 + 5 + • • • + (2k  1)] + (2k + 1) —
k2 + (2k + 1) = (&+ l)2 = m2,
which is a contradiction. ■ Assume, to the contrary, that there is some nonnegative integer n such that 3 / (2" + 2'!+1). Then there is a smallest nonnegative integer n such that 3 / (2" + 2'1+1). Let m be this integer. Since 2° + 21 = 3, we have m > 1. So we can write m = k + 1, where 0 < k < m. Thus 3  (2k + 2k+l) and so 2k + 2k+l = 3x for some integer x. Then show that 2m + 2m+l = 3(2x). 6.39 Assume, to the contrary, that there is some positive integer n such that 12 / (n4 —n2). Then there is a smallest positive integer n such that 12 / («4 —n2). Let m be this integer. It can be shown that if 1 < n < 6, then 12  (n4 —n2). Therefore, m > 7. So we can write m = k + 6, where 1 < k < m. Consider (k + 6)4 —(k + 6)2.
6.37
Section 6.4: The Strong Principle of Mathematical Induction 6.41
Conjecture A sequence {«„} is defined recursively by a\ = 1 and a„ = 2a„_i for n > 2 . Then a„ = 2n~' for all n > 1. Proof We proceed bymathematical induction. Since <2! = 21_1 = 2° = 1, it follows that a„= 2"~* when n = 1. Assume that ak= 2k~l for some positive integer k. We show that ak+\ = 2k. Since k > 1,it follows that k + 1 > 2 . Therefore, ak+1 = 2ak = 2 ■2^
6.43
= 2k.
The result follows by the Principle of Mathematical Induction. Conjecture A sequence {a,,} is defined recursively by a\ = 1, a2 = 4, a3 = 9 and Cln — &n—i
Q/i—2 “f @n—3
2(2n
3)
for n > 4. Then a„ = n2 for all n > 1. Proof We proceed by the Strong Principle of Mathematical Induction. Since a\ = l 2 = 1, it follows that an = n2 when n = 1. Assume that a, = i2, where 1 < i < k for some positive integer k. We show that
■
370
Answers to OddNumbered Section Exercises
ak+1 = (k + I)2.Since a2 = fli+i = (1 + l)2 = 4 and a3 = a2+i = (2 + l) 2 = 9, it follows that ak+] = (k + i f for k = 1,2. Hencewe may assume that k > 3. Since it + 1 > 4, ak+l = fift —ak\ + Ok~2 + 2[2(/r + 1) —3] = k2  ( k  1)2 + (A  2)2 + (4k  2) = k2  (k2  2k + 1) + (£2  4k + 4)+ (4k 
2)
= k2 + 2k + 1 = (k + l)2. The result then follows by the Strong Principle of Mathematical Induction. ■ 6.45 Proof We use the Strong Principle of Mathematical Induction. Since 12 = 3 • 4 + 7 ■0. the statement is true when n = 12. Assume for an integer k > 12 that for every integer i with 12 < i < k, there exist nonnegative integers a and b such that i = 3a + lb. We show that there exist nonnegative integers x and y such that k + 1 = 3jt + l y . Since 13 = 3 ■2 + 7 • 1 and 14 = 3 • 0 + 7 • 2, we may assume that k > 14. Since k —2 > 12, there exist nonnegative integers c and d such that k — 2 = 3c + Id. Hence k + 1 = 3(c + 1) + Id. By the Strong Principle of Mathematical Induction, for each integer n > 12, there are nonnegative integers a and b such that n = 3a + lb. n 6.47 We show that every odd integer n > 15 can be expressed as 3a 4  11 b or as 5c + I d for nonnegative integers a .b , c and d. Proof We use the Strong Principle of Mathematical Induction. First, observe that 15 = 3  5 + 11 0 = 5 ■3 + 7 0, 17 = 3  2 + 11 ■1, 19 = 5 • 1 + 7 2 , 21 = 3 ■7 + 11 ■0 and 23 = 3 ■4 + 11 • 1. Thus the statement is true for 15, 17, 19, 21 and 23. Assume that the statement is true for every odd integer i with 15 < i < k. where k > 23 is an odd integer. We show that the statement is true for the integer k + 2. Suppose that k = 3a + 11b for some nonnegative integers a and b. Sincek > 23, eithera > 3 or b > 2. If a > 3, then k + 2 = 3(a  3) + 1 l(b + 1); while if b > 2, then k + 2 = 3(a + 8) + 11(6  2). Hence we may assume that k = 5c + I d for some nonnegative integers c and d. If c > 1, then k + 2 = 5(c — 1) + l(d + 1). The remaining situation is where c = 0 and so k > 23 is an odd integer multiple of 7. So k = Id, where d > 5. In this case, k + 2 = 5 • 6 + l( d — 4). The result then follows by the Strong Principle of Mathematical Induction. ■
E X E R C IS E S FOR C H A PT E R 7
Section 7.1: Conjectures in M athematics 7.1
(a) 1 7 + 1 8 + • ■ • + 2 5 = 6 4 + 1 2 5 . (b) Conjecture For every nonnegative integer n, ( n 2 + 1) + 0?2 + 2) H K » +
(c) Proof
l ) 2 = « 3 + {n + l ) 3.
Observe that (,n2 + 1) + (n2 + 2) + ■• • + (n + l)2 = (n2 + 1) + (n2 + 2) + ■• • + (n2 + 2n +
1)
= (2ft + 1)n2 + [1 + 2 + • ■■+ (2n + 1)] = (2n + 1)n2 + (n + = 2n} + n2 + 2n2 + 3n + 1 = i f + (n3 +
1)(2m + 1)
3n2 + 3n + 1)= n3
+ (n+ I)3. ■
7.3 (a) a2 = 3, a3 = 8, a4 = 54. (bl For each n e N, a„ is an integer. 7.5 (a) The ordered partitions of 4 are 4, 3 + 1, 1 + 3, 2 + 2, 2 + 1 + 1. 1 + 2 + 1, 1 + 1 + 2 and 1 + 1 + 1 + 1. So there are 8 ordered partitions of 4. (b) Conjecture For each positive integer n. there are 2"_1 ordered partitions of n. [This conjecture is true.]
371
Exercises for Chapter 7.7
(a) 4  2  1 1 = 4 + 2 + 1 + 1 . (b) 3 • 3 • 1 ■1 ■1 = 3 + 3 + 1 + 1 + 1. (c) Conjecture For every integer n > 3, there exist n positive integers whose sum equals their product. [This conjecture is true.] 7.9 Conjecture There exists such a checkerboard if and only if mn is even. [The conjecture is true.]
Section 7.2: Revisiting Quantified Statements 7.11
7.13
7.15
7.17
7.19
7.21
7.23
7.25
7.27
(a) Let S be the set of all positive even integers and let P { n ) : 3n + 2"~2 is odd. 3n e S, P(n). (b) Proof For n = 2 e S , 3n + 2 '^ 2 = 7 is odd. ■ (a) Let P ( n ) : 3n2 —5n + 1 is an even integer. 3n e Z, P(n). (b) We show the following: For all n e Z, 3n2 — 5n + 1 is odd. This can be proved by a direct proof with two cases,namely n even and n odd. (a) Let P (m , n): m(n —3) < 1. 3n e Z, Vm € Z, P (m , n). (b) Proof Let n = 3. Then m(n —3) = m  0 = 0 < l . ■ (a) Let P(m, n): —nm < 0. 3n e N, Vm e Z, P(m, n). (b) 'in e N, 3/77 £ Z, ~ P(m, n). (c) Let n be a positive integer. For m = 0, we have —nm = —n ■0 = 0. (a) Let P(a, b, .r): a < x < b and b — a = 1. Vx e R , 3a,b e Z , P (a ,b ,x ). (b) Proof Let xe R. If x is an integer, then let a = x and b = x + 1. Then a < x< b and b — a = 1. Thus we may assume that x is not an integer. Then there exists an integer a such that a < x < a + 1.Let b = a + 1. ■ (a) Let S be the set of even integers, let T be the set of odd integersand let P(a, b, c): a < c <b or b < c < a. Va e S,Vb e T, 3c e Q, P(a, b, c). (b) Proof For a e S and b e T, let c = (a + b)/2. If a < b, then a < c < b;while if b< a, then b < c < a. m (a) Let S be the set of odd integers and P(a, b, c): a + b + c = 1. 3a, b ,c 6 S, P (a, b, c). (b) Proof Let a = 3 and b = c = —1. Then a + b + c = 1. ■ (a) 3L e R, V<? e R+, 3d e R +, Wx e R, P{x, d) => Q(x, L, e). (b) Proof Let L = 0 and let e be any positive real number. Let d = e/3. Let x € R such that ]x\ < e/3. Then \3x  L \ = \3x\ = 3x < 3(e/3) = e. m Proof Let a e Z. Then b = a — I and c = 0 are integers such that a —b\ = 1 > cd = 0 • d = 0 for every integer d. *
Section 7.3: Testing Statements 7.29
(a) The statement is true. Proof. Assume that k2 + 3£ + 1 is even where k e N. Then k2 + 3k + 1 = 2x for some integer x. Observe that {k + I)2 + 3(k + 1) + 1 = k2 + 2k + 1 + 3k + 3 + 1 = (k2 + 3k + 1) + 2k + 2x + 2k + 4 = 2(x +
4 k + 2).
Since x + k + 2 is an integer, (k+ l)2+ 3(k + 1) + 1 is even. ■ (b) The statement is false sinceP( l ) is false. 7.31 This statement is false. Let n = 0 and let k be any nonnegative integer. Since k > 0 = n, the integer n = 0 is a counterexample.
372 7.33 7.35 7.37 7.39 7.41 7.43 7.45 7.47 7.49 7.51 7.53 7.55 7.57
7.59 7.61 7.63 7.65
Answers to OddNumbered Section Exercises
This statement is false. Let x = 99 and y = z = 1. Then x + y + z = 101, while no two of x, y and z are of opposite parity. Thus, x = 99, y = 1, z = 1 is a counterexample. The statement is true. Proof Assume that A ^ 0. Since A ^ 0, there is an element a e A. Let B = {a}. Then Afl B ^ 0.■ The statement is false. Let A = {1}, which is nonempty, and let B be an arbitrary set. Since1 s AU follows that A UB 0. The statement is true. Proof Let A be a proper subset of S and let B = S — A. Then 5 ^ 0 , A U B = S and A n B = 0. ■ The statement is true. Observe that 0 • c = 0 for every integer c. The statement is false. Let x = 1 and y — —2. Then x 1 < y 1 but x > y. The statement is true. Proof Let a be an odd integer. Then a = a + 1 + (—l ) i s a sumof three oddintegers. ■ The statement is true. Let ft = c —a. The statement is true. Consider r = (a + b)/2. The statement is false. Let A ^ 0 and B = 0. Then A U fi ^ 0. The statement is true. Proof Let a be an odd integer. Then a + 0 = a, where ft = 0 is even and c — a is odd. ■ The statement is true. Let f ( x ) = x 3 + x 2 — 1. Observe that / ( 0) = —1 and / ( l ) = 1. Now apply the Intermediate Value Theorem of Calculus. The statement is true. Proof Assume that A — B 0. Then there exists x e A — B. Thus x € Aand x £B . Since x B , it follows that x £ B — A. Therefore, A — B =£ B  A. u The statement is true. Proof Let ft e Q+. Then a = ft/V 2 is irrational and 0 < a < b. m The statement is false. For A = 0, B = {1} and C = {1, 2}, we have A fl B = A H C — 0, but B ^ C. Thus A, S and C form a counterexample. The statement is true. Consider B = 0. Since A U fi ^ 0 , this requires that A ^ 0 . The statement is false. Note that x 2 + x + 1 = (x + j ) 2 +  >  > 0 for every x £ R.
7.67 7.69 7.71 7.73 7.75
The statement is true. For a nonzero rational number r, observe that r = (;• V 2) • ^=. The statement is false. The sets 5 = {1, 2, 3} and T = {{1, 2}, {1, 3}, {2, 3}} form a counterexample. The statement is false. The numbers a = ft = 0 and c = 1 form a counterexample. The statement is true. Let a = 2, ft = 16 and c = 4. The. statement is true. Proof Let n e Z. If /? ^ 0, then n = n + 0 has the desired properties. If /? = 0. then n = 0 = 1 + (—1). ■ 7.77 The statement is false. For n = 11, n2 — n + 11 = l l 2. 7.79 The statement is true. Proof Let a and ft be two consecutive integers such that 3 /a ft. Since 3 / aft, it follows by Result 4.5 that 3 / a and 3 /ft. Therefore, a ^ 0 (mod 3) and ft ^ 0 (mod 3). Thus a = 3g + 1 and ft = 3<? + 2 for some integer q and so a + ft = (3q + 1) + (3q + 2) = 6q + 3 = 3(2(7 + 1). Since 2q + 1 is an integer, 3  (a + ft). 7.81 The statement is true. Proof Let a = 6/5, ft = 10/3 and c — 15/2. Then aft = 4, ac = 9, ftc = 25 and abc — 30. ■
EXERCISES FOR CHAPTER 8 Section 8.1: Relations dom(ft) = {a, ft} and range(/?) = {s, t). Since A x A = {(0, 0), (0, 1), (1, 0), (1, 1)} and jA x A = 4, thenumberof subsets of A x A andhence the number of relations on A is 24 = 16. Four of these 16relations are 0, A x A, {(0, 0)}, and{(0, 0),(0, 1), (1, 0)). 8.5 R  ] = {(1, 1), (2, 1), (2, 2), (3, 1), (3, 2), (3, 3)}.
8.1
8.3
Exercises for Chapter 8
373
8.7 F ora, b e N, a R~ l b if and only if b R a, while b R a if b + 4a is odd. That is, R~' = {(x, y) : y + 4x is odd}. 8.9 (a) The statement is false. Let A = {1, 2, 3, 4), B = {1, 2, 3, 5}, and C = {1, 2, 3}. Then A = 5  = 4. Then R = C x C is a relation from A to B with i? = 9 and R = R~' but A ^ B. Thus A, B and R constitute a counterexample. (b) Suppose that \R\ = 9 is replaced by /? = 10. Then the statement would be true. P roof Let A and B be sets with A = \B\ = 4 . Assume, to the contrary, that there exists a relation from A to B with \R\ = 10 and R = R ~ ] but A ^ B. Since A and B have the same number 4 of elements, there is an element x e A — B and an element y e B — A. Since R = /?1, it follows that x is not related to any element of B by R and no element of A is related to y. This implies that R C (A — {x}) x (B — {y}). Since A — {x} ■jB —{>’} = 3  3 = 9, this is a contradiction. ■
Section 8.2: Properties of Relations 8.11 The relation R is reflexive and transitive. Since (a, d) e R and (d, a) R , it follows that R is not symmetric. 8.13 The relation R is transitive but neither reflexive nor symmetric. 8.15 The relation R is reflexive and symmetric. Observe that 3 R 1 and 1 R 0 but 3/? 0. Thus R is not transitive. 8.17 The relation R is symmetric and transitive but not reflexive. 8.19 The relation R is reflexive and symmetric. Observe that —1 ^ 0 and 0 R 2 but —1 /? 2. Thus R is not transitive. 8.21 The statement is true. Proof Let A = [cii, a2] and suppose that R is a relation on A that has none of the properties reflexive, symmetric, and transitive. Since R is not symmetric, we may assume that a\ R a2 but a2 ft a\. Since at most one of (a i, a \) and (a2, a2) belongs to R, it follows that R is transitive, which is a contradiction. Since the hypothesis of the implication is false, the statement is true vacuously. ■ 8.23 Since a  a for every a e N, the relation R is reflexive. The relation R is symmetric for suppose that a R b. Then a \ b or b  a. This, however, says that b R a. The relation R is not transitive since for a = 2, b = 1, and c = 3, a R b and b R c but a fl c.
Section 8.3: Equivalence Relations 8.25 There are three distinct equivalence classes, namely [1] = {1, 5}, [2] = {2, 3, 6}, and [4] = {4}. 8.27 P roof Since a 3 = a 3 for each a e Z, it follows that a R a and R is reflexive. Let a, b e Z such that a R b. Then a 3 = b3 and so b3 = a 3. Thus b R a and R is symmetric. Let a, b, c e Z such that a R b and b R c. Thus a 3 = ft3 and b3 = c3. Hence a 3 = c3 and so a R c and R is transitive. ■ Let a, b e Z. Note that a3 = b3 if and only if a = b. Thus [a] = {a} for every a e Z. 8.29 P roof Assume that a R b, c R d and a R d. Since a R b and R is symmetric, b R a. Similarly, d R c. Because b R a, a R d and R is transitive, b R d. Finally, since b R d and d R c, it follows that b R c, as desired. ■ 8.31 P roof First assume that R is an equivalence relation on A. Thus R is reflexive. It remains only to show that R is circular. Assume that x R y and y R z. Since R is transitive, x R z. Since R is symmetric, z R x . Thus R is circular. For the converse, assume that R is a reflexive, circular relation on A. Since R is reflexive, it remains only to show that R is symmetric and transitive. Let x, y € A such that x R y. Since R is reflexive, y R y. Because (1) x R y and y R y and (2) R is circular, it follows that y R x and so R is symmetric. Let x, y, z e A such that x R y and y R z. Since R is circular, z R x. Now because R is symmetric, we have x R z. Thus R is transitive. Therefore, R is an equivalence relation on A. a 8.33 (a) P roof Let a e Z. Since a —a = A ■0 e H ,\t follows that a R a and R is reflexive. Next, assume that a R b, where a , b e Z. Then a —b e H and so a — b = Ak, where k e Z. Then b — a = 4(—k). Since —k e Z, it follows that b — a e H and b R a. Therefore, R is symmetric. Finally, assume that a R b and b R c where a, b, c e Z. Then a — b e H and b —c e H . So a — b = Ak and b — c = Ai for k, I € Z. Therefore, a — c = (a —b) + (b —c) = Ak + 4t = A(k + I). Since A' + I e Z, it follows that a — c e H and so a R c. Thus R is transitive and R is an equivalence relation, ss
374
Answers to OddNumbered Section Exercises
(b) Let a e Z. Then [a] = {x e Z : x P a} = {x £ Z : x — a e H} = { . t e Z : i  a = 4 t for some integer £} = [a + Ak : (' 6 Z). Since every integer can be expressed as Ak + r where r is an integer with 0 r < 3. it follows that the distinct equivalence classes are [0]. [1], [2], and [3], where [r] = {Ak + r : k e Z} fo rr = 0, X, 2, 3. 8.35 The statement is false. Suppose that there are equivalence relations Pi a n d P ^ n S = {a, b, f} such that Pi g P 2, P 2 g R\ and Pi U P 2 = S x S. Since Pi and P2 are both reflexive, it follows that (a, a), (.b, b), (c, c) € Pi p P 2. Because P i g P 2, there exists some element of P  that is not in P 2, say (a, b) £ Pi —P 2. Necessarily then, (b , a) £ Pi —P? as well. Because P 2 g P i. there exists some element of P 2 that is not in P i . We may assume that (b, c) £ P 2  P i. Thus (c, b) e R2  Pi Since P j U P 2 = S x S, it follows that (a, c) £ P, U P 2. We may assume that (a, c) £ P i . Since (/?. a) 0 /\’:. it follows by the transitive property that (b.c) e R\, which is not true.
Section 8.4: Properties of Equivalence Classes 8.37 Let a e N. Then a2 + a2 = 2(a2) is an even integer and so a R a . Thus P is reflexive. Assume that a R b, where a, be. N. Then a2 + b2 is even. Since b2 + a2 = a2 + b2, it follows that b2 + a2 is even. Therefore, b R a and P is symmetric. Finally, show that R is transitive. There are two distinct equivalence classes: [1] = {x e N : x 2 + 1 is even} = {x e N : x2 is odd} = {x € N : x is odd} [2]= {x £ N : x2 + 4 is even} = {x e N : x2 is even} = {x 6 N : x is even}. 8.39 (a) Proof First, we show that P is reflexive. Let x e S. Then x + 2x = 3x. Since 3  (x + 2x), it follows that x P x and P is reflexive. Next, we show that P is symmetric. Let x P y, where x, y £ S. Then 3 (x + 2y) and so x + 2_y = 3a, where a e Z and so x = 3a — 2y. Thus y, + 2x = y + 2(3a  2y) = 6a  3y = 3(2a  y). Since 2a  y is an integer, 3  fj> + 2x). Thus J P x and P is symmetric. Finally, we show that P is transitive. Let x R y and y R z, where x, y, z e S. Then 3  (x + 2y) and 3  (v + 2z). So x + 2y = 3a and y + 2z = 3b, where a , b e Z. Thus (x + 2y) + (y + 2z) = 3a + 3b and so x + 2z = 3a + 3b  3y = 3(a + b  y). Since a + ft —y is an integer, 3  (x + 2z) and so x P z. Therefore, P is transitive. ■ (b) There are three distinct equivalence classes: [0] = (0, 6 }, [1] = {1.  2 , 4, 7}, and [7 ] = {7 , 5}. 8.41 (a) Suppose that P , and P 2 are two equivalence relations defined on a set S. Let P = Pi n P 2. First, we show that P is reflexive. Let a £ S. Since P i and P 2 are equivalence relations on 5, it follows that (a, a) € Pi and (a, a ) e P 2. Thus (a, a) e R and so P is reflexive. Assume that a R b , where a . b e S. Then (a. b) e R = Pi D P 2. Thus (a, b) g Pi and (a, b) 6 P 2. Since Pi and P 2 are symmetric, (b, a) e Pi and (b. a) e P 2. Thus (b, a) € R and sob R a. Hence P is symmetric. Finally, show that P is transitive. (b) Let a e Z. For x £ Z, it follows that x P i a if and only if x P2 a and x P 3 a. That is, x Pi a if and only if x = a (mod 2) and x = a (mod 3). First, suppose that x = a (mod 2) and x = a (mod 3). Hence x = a + 2k and x = a + 31 for some integers k and I. Therefore, 2k = 31 and so I is even. Thus I = 2m for some integer m, implying that x = a + 31 = a + 3(2m) = a + 6m and so x —a = 6m. Hence x = a (mod 6). If x = a (mod 6), then* = a (mod 2) andx = a (mod 3). Thus [a] = {x £ Z : x = a (mod 6)}.
[0] = {. . .. 1 2 ,  6 , 0, 6, 12, ...}, [1] = {.. .,  1 1,  5 . 1,7, 13,...}, [2] = {. . ., 1 0 .  4 , 2, 8, 14, ...}, [3] = {. . . ,  9 ,  3 , 3, 9, 15, ...}, [4] = {. . ..  8 ,  2 , 4, 10, 1 6 , . ..}, [5] = {.. ..  7 ,  1 , 5 , 1 1 , 17, ...}. 8.43 Proof Fora, £ A, a.j = {x £ A : x P a,} = {x £ A : (x, a,) G P}. Therefore, I counts all those ordered pairs in P for which a, is the second coordinate and so ]T"=1 f e l l counts all ordered pairs in P, that is, J2"=i [«<] = IP [ Since P is reflexive, (a, , a, ) e R for / = 1, 2 , . . . , n and for each pair i, j of distinct integers, either (a, , cij) and (a j , a,) both belong to P or neither belongs to P. Suppose that there are k ordered pairs (a;, a , ) in R with 1 < / < ; < « . Then  P  == n + 2k and so I^ ] I is even if and only if n is even B
Exercises for Chapter 8
375
Section 8.5: Congruence Modulo n 8.45 Let a e Z. Since 3a + 5a = 8a, it follows that 8  (3a + 5a) and so 3a + 5a = 0 (mod 8). Hence a R a and R is reflexive. Next, we show that R is symmetric. Assume that a R b, where a , b € Z. Then 3a + 5b = 0 (mod 8), that is, 3a + 5b = 8A' for some integer k. Observe that (3a + 5b) + (3b + 5a) = 8a + 8b. Thus 3ft + 5a = 8a + 8ft  (3a + 5ft) = 8a + 8ft  8k = 8(a + ft  k). Since a + ft —k is an integer, 8  (3 ft + 5a) and so 3ft + 5 a s 0 (mod 8). Hence b R a and R is symmetric. Finally, show that R is transitive. 8.47 There are two distinct equivalence classes, namely, [0] = {0, ±2, ±4, . . .} and [1] = {± 1 , ±3, ± 5 , . . 8.49 Proof Let a e Z. Since 5a — 2a = 3a, it follows that 3  (5a —2a) and so 5a = 2a (mod 3). Hence a R a and R is reflexive. Next, we show that R is symmetric. Assume that a R ft, where a, ft 6 Z. Then 5a s 2ft (mod 3), that is, 5a —2ft = 3k for some integer k. Observe that (5a —2ft) + (5ft  2a) = 3a + 3ft. Thus 5ft  2a = 3a + 3ft  (5a  2ft) = 3a + 3ft  3k = 3(a + b  k ) . Since a + ft  A is an integer, 3  (5ft —2a) and so 5ft = 2a (mod 3). Hence ft R a and R is symmetric. Finally, we show that R is transitive. Assume that a R ft and ft R c, where a, ft, c e Z. Thus 5a = 2ft (mod 3) and 5ft = 2c (mod 3). So 5a —2ft = 3x and 5ft —2c = 3v, where „r, y e Z. Observe that (5a  2ft) + (5ft  2c) = (5a  2c) + 3ft = 3x + 3y. Thus 5a —2c = 3x + 3y —3ft = 3(x + j —ft). Since x + y  ft is an integer, 3  (5a  2c) and 5a = 2c (mod 3). Therefore, a R c and R is transitive. ■ There are three distinct equivalence classes, namely, [0] = {0, ±3, ±6, ...}, [1] = { . . .,  5 ,  2 , 1, 4, . . . }, and [2] = { . ..,  4 ,  1 , 2, 5, . . 8.51 First, we show that R is reflexive. Let a e Z. Since 2a + 3a = 5a, it follows that 5  (2a + 3a) and so a R a. Hence R is reflexive. Next, we show that R is symmetric. Assume that a R ft, where a, ft e Z. Then 2a + 3ft s 0 (mod 5). Hence 2a + 3ft = 5k for some integer k. Observe that (2a + 3ft) + (2ft + 3a) = 5a + 5ft. Thus 2ft + 3a = 5a + 5ft —(2a + 3ft) = 5a + 5ft —5k = 5(a + b — k). Since a + ft  k is an integer, 5  (2ft + 3a) and so 2ft + 3a = 0 (mod 5). Hence ft R a and R is symmetric. Finally, show that R is transitive. The distinct equivalence classes are [0], [1], [2], [3] and [4], 8.53 The relation R is an equivalence relation. Proof Let a e R. Since a —a = 0 = 0 ■7r, it follows that a R a and R is reflexive. Next, suppose that a R ft, where a, ft e R. Then a  ft = kn for some k e Z. Since ft  a = (  k ) n and —k e Z. it follows that ft R a and so R is symmetric. Finally, suppose that a R b and ft R c, where a, ft, c 6 R. Then a —ft = k n and ft —c = i n for k, I € Z. Thus a — c = (a — ft) + (ft—c) = (k + l ) n . Because k + 1 e L, a R c and R is transitive. Therefore, R is an equivalence relation. ■ [0] = [x e R : x R 0} = e R : x = kn, where k e Zj = {kn : k e Z}. [n] = {x e R : x R jt} — {x 6 R : x — Jt = k n , where k e Z} = {(k + l)?r : k e Z} = {kn :k s Z ) = [0]. [V2] = {V2 + k n : k e Zj.
Section 8.6: The Integers Modulo n 8.55 (a) [2] + [6] = [8] = [0], (c) [1 3 ] + [138] = [125] = [5],
(b) [2] • [6] = [12] = [4], (d) [1 3 ] • [138] = [3][2] =
[6],
8.57 (a) Proof Let a, ft e T. Then a = 4k and ft = 4£ for k, I e Z.Thus a + ft = 4(fc + 1) and aft = 4(4kt). Since k + £, 4k l e Z, it follows that T is closed under addition and multiplication. ■ (b) Yes. Let a e S — T and ft 6 T. Then ft = 4£for£ e Z. Thus aft = 4(a£). Since a i e Z, it follows that ab e T.
376
Answers to OddNumbered Section Exercises
(c) (d) (e) 8.59 (a) (b) (c) 8.61 (a)
No. For example, a = 1e S — T and ft = 4 e T but a + ft = 5 g T. Yes. For example, a = 2 and b = 6 belong to S — T and ab = 12 = 4 ■3 e T. Yes. For example, a = 2 and b = 6 belong to S — T and a = 8 = 4 2 € I . No. Consider [A] = [2] and [ft] = [4], Then [a] ^ [0] and [b] ^ [0], but [a] ■[ft] = [8] = [0], If Z8 is replaced by Z9 or Z i0, then the answer is no; while if Z8 is replaced by Z n , thenthe answer is yes. Let a, ft e Z„, where n > 2 is prime. If [a] ■[ft] = [0], then [a] = [0] or [ft] = [0], Suppose that an element [a] s Z„, also belongs to Z„. That is, [a ] in Z„, is the same set as [ft] in Z„. Since a, a + m e [a] e Z,„, it follows that a, a + m e [ft] e Z„. Therefore, n \ [(a + m) —a] or n \ m. Similarly, m \ n and so n = m, that is, Zm = Z„. (b) If m, n > 2 and m n , then Zm n L„ = 0. For example, 7, n Z3 = 0.
EXERCISES FOR CHAPTER 9 Section 9.1: The D efinition of Function 9.1 9.3 9.5 9.7 9.9
9.11
do m (/) = {a , ft, c, rf] and range(/) = {y, z}. Since R is an equivalence relation, R is reflexive. So (a, a) e R for every a e A. Since R is also a function from A to A, we must have R = {(a, a) : a e A) and so R is the identity function on A. Let A' = [a € A : (a, b) e R for some ft e B). Furthermore, for each element a ' e A', select exactly one element ft' e [ft € B : (a', ft) e }. Then f = {(a', ft') : a' e A'} is a function from A' to B. B = {(3. 4), (17, 6), (29, 60), (45, 22)} and so R is a function from A to B. (a) Since 0 1 and 0 Ri (~ I), A’; is not a function. (b) Since 0 ( ^ ) and 0 (—^ ) , ^2 is not a function. (c) For each a s N, ft = (1 —3a)/5 G Q is the unique element such that 3a + 5ft = 1. So R 3 defines a function. (d) For each x e R. y = 4  \x  2 is a unique element of R. So R 4 defines a function. (e) Since0 A', 1 and 0 R$ (—1), R$ is not a function. (a) f ( C ) = C, f  \ C ) = C U [x e R :  . t 6 C), f ~ \ D ) = R  {0}, /  ’({I}) = {1. 1}. (b) /( C ) = [0, 00), /  ‘(C) = [e, 00), f ~ l(D) = (1, 00), /H {11) = {e}. (c) /( C ) = [e, 0 0 ), f  ' ( C ) = [0, 00), />(£») = R, /' ‘([ I}) = [0}. (d) /( C ) = [—1. 1], f ~ \ C ) = fl + 2nn : n e Zj, f ~ l{D) = U„eZ(2n7r, (2« + 1)tt), / “ , ({l}) = { l + 2 « n  : « eZ}. (e) /( C ) = (  00", 1], f  \ C ) = {!}, t l (D) = (0,2), /  ‘({I}) = {1}.
Section 9.2: The Set of All Functions from A to B 9.13
B a = { / , f 2, . . . , ./«}. where f x = [(1, x), (2, *), (3, ,v), f 2 = {(1, x), (2, x), (3, j)}, fy = [(1, x), (2, y), (3, x)}, / 4 = {(1, x), (2, y), (3, y)}. By interchanging x an dy in / , f 2, f 3, f 4, we obtain fs< /6, f i , /s9.15 For A = {a, b, c} and S = [0, 1}, there are eight different functions from A to B, namely / , = {(a. 0),(ft, 0), (c, 0)}, fy = {(a, 0), (ft, 0), (c, 1)}, fy = {(fl, 0),(ft, 1), (c, 0)}, U = {(a , 0), (ft, 1), (c, 1)}, / 5 = {(a, 1),(ft, 0), (c, 0)}, / 6 = {(a, 1), (ft, 0), (c, 1)}, h
9.17
= { (« .
1 ), ( b, 1 1. (c , 0 )} ,
/ g = { (a , 1 ), (ft, 1 ), (c ,
1)},
(a) A reasonable interpretation of CB‘4 is {/’ : / : B a —>• C}. (b) For A = {0. 1} and B = {a, ft}, = { / , f 2, / 3, / 4}, where /1 = {(0, a), (1, «)}, / 2 = {®, a), (1, ft)}, / a = {(0, ft), (l,a)} and / 4 = {(0, ft), (1, ft)}. Then for C = {x, y}, C fiA = { ju gw}, where ,?1 = {<./'=  v >. ( f y , x ) , ( f y , x), ( f y , x)}, f t = { ( f y , x), ( f y , x), ( / 3, x),
ga = {(fy, y), (fy, y), (fy, y), (fy, >')}•
( f y , y % . . .,
Exercises for Chapter 9
377
Section 9.3: OnetoOne and Onto Functions 9.19
Let A = {1,2} and B = (3, 4, 5}, Then / = {(1, 3), (2, 4)} and g = ((3, 1), (4, 2), (5, 2)} have the desired properties, 9.21 (a) The function / is injective. Proof Assume that f ( a) = f ( b ), where a, b € Z. Then a —3 = b —3. Adding 3 to both sides,we obtain a = b. m (b) The function f is surjective. Proof Let n e Z. Then n + 3 e Z and f ( n + 3) = (n + 3) — 3 = n. m 9.23 The statement is true. The function / : A —>■V(A) defined by f( a ) = [a] has thedesired property. 9.25 Consider the function / : R —> R defined by f ( x ) = x 3 —x = (x + l)x(x — 1). Since /( 0 ) = / ( l ) , it follows that / is not onetoone. One way to show that / is onto is to use the Intermediate Value Theorem. Method #1. Let r e R. Since lim.v^ooCx3 —x) = oo and limv_i._00(x3 —x) = —oo, there exist real numbers a and b such that / ( a ) < r < f(b ). Since / is continuous on theclosed interval[a, b], there exists c such that a < c < b and f( c ) = r. Method #2. Let r e R. If r = 0, then /(0 ) = 0 = r. Suppose that r > 0. Then r + 1> 1 and r + 2 > 1; so f ( r + 1) = r(r + l)(r + 2) > r. Since / ( 0) < r < f ( r + 1), it follows by the Intermediate Value Theorem that there exists c e (0, r + 1) such that /( c ) = r. If r < 0, then s = —r > 0 and, as we just saw, there exists c e (0, s + 1) such that /( c ) = s. Then / ( —c) = —s = r. 9.27 (a) R = {(2, 8), (3, 6), (4, 8), (5, 10)}. The relation R is a function from A to 6. (b) Since range(^) = B, the function R is onto. However, since 2 R 8 and 4 R 8, R is not onetoone. 9.29 Proof By Exercise 9.1 2 (b ),/(C D D) c / ( C ) fl /(£>). So it remains to show that f ( C ) fl f ( D ) C f ( C fl D) under the added hypothesis that f is onetoone. Let y e /( C ) fl f( D ) . Then y e /( C ) and y 6 /(£>). Since y 6 /( C ), there exists x e C such that y = f( x ) . Furthermore, since y e /(£>), there exists z e D such that y = f (z). Since / is onetoone, z = x. Thus x e C n D and so y € f { C n D). Therefore, /( C ) fl f ( D ) C / ( C n D) and so f ( C n D) = /( C ) n /(£>). n
Section 9.4: Bijective Functions 9.31
(a) Proof Let [a], [b] e Z5 such that [a] = [b]. We show that /([a ]) = f([b]), that is, [2a + 3] = [2b + 3]. Since [a] = [b], it follows that a = b (mod 5) and so a — b = 5x for some integer x. Observe that (2a + 3)  (2b + 3) = 2(a  b) = 2(5*) = 5(2x).
Since 2x is an integer, 5  [(2a + 3) — (2b + 3)]. Therefore, 2a + 3 = 2b + 3 (mod 5) and so [2a + 3] = [2b + 3]. ■ (b) Since /([0 ]) = [3], /( [ I ] ) = [0], /([2 ]) = [2], /([3 ]) = [4], and /([4 ]) = [1], it follows that / is onetoone and onto and so / is bijective. 9.33 Define f l(x) = x 2 forx e A and fi( x ) = ^/x for x e A. ( f 3(x) = 1 —* is another example.) 9.35 (a) Consider S = {2, 5, 6}. Observe that for each y e B, there exists x e S such that x is related to y. This says that y(R ) < 3. On the other hand, let S' C A such that for every element y of B, there is an element x € S' such that x is related to y. Observe that S' must contain 6, at least one of 2 and 3 and at least one of 4, 5, and 7. Thus S' > 3. Therefore, y(R ) = 3. (b) If R is an equivalence relation defined on a finite nonempty setA, then y(R ) is thenumber of distinct equivalence classes of R. (c) If / is a bijective function from A to B, then y ( f ) =  A.
Section 9.5: Composition of Functions 9.37 9.39
g o / = {(1, y), (2, x), (3, x), (4, x)}. (a) (g o /)([«]) = * (/([* ])) = *([3a]) = [21a] = [a], ( / o g)([a]) = /(g ([a ])) = /([7 a ]) (b) Each of g o / and / o g is the identity function on Z 10.
= [21a] = [a].
378 9.41
Answers to OddNumbered Section Exercises Proof We first show that / is onetoone. Let a, ft e A such that f ( a ) = f(b). Now a = iA(a) = ( / o f)(a ) = / ( / ( a ) ) = f { f ( b ) ) = ( / o /)(ft) = iA(b) = ft.
Thus / is onetoone. Next, we show that / is onto. Let c e A. Suppose that /( c ) = d e A. Observe that f i d ) = / ( / ( c ) ) = ( / o /)(c ) = i'A(c) = c. Thus / is onto. 9.43 (a) (i) Direct Proof. Assume that g o f is onetoone. We show that / is onetoone. Let f ( x ) = f i y ) , where x, y e A. Since g i f i x ) ) = g ifi y ) ) , it follows that (g o f) ( x ) = (g o f)(y ). Since g o f is onetoone, x = y. (ii) Proof by Contrapositive. Assume that / is not onetoone. Hence there exist distinct elements a , b e A such that f ( a ) = /(ft). Since
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(g o f) (a ) = g i f (a)) = g i f (ft)) = (g o /)(ft), it follows that g o f is not onetoone. (iii) Proof by Contradiction. Assume, to the contrary, that there exist functions / : A —>■B and g : B —»■ C such that g o / is onetoone and / is not onetoone. Since / is not onetoone, there exist distinct elements a, ft € A such that /( a ) = /(ft). However then,
■
(g ° f) ( a ) = g i f (a)) = g( / ( f t » = (g o /)(ft), contradicting our assumption that g o / isonetoone. (b) Let A = {1, 2, 3}, B = {w, x, y, z), and C = {a, ft, c}. Define f : A B by / = {(!■ w), (2, x), (3, _y)} and g : £ > C by g = l(w, a), {x, ft), (y, c), (z, c)}. Then g o / = {(1, a), (2, ft), (3, c)) is onetoone, but g is not onetoone. 9.45 (a) (g o /)(1 8 , 11) = g (/(1 8 , 11)) = *(29, 18) = (47, 29). (b) The function g o / : A x S  > f i x B i s onetoone. Proof Assume that (g o f)(a , ft) = (g o f)(c , d ), where (a, ft), (c,d) e A x B. Then g i f {a, ft)) = g (/(c , d)) and so g(a + ft, a) = g(c + d, c). Therefore, (2a + ft, a + ft) = (2c + d, c + d), which implies that 2a + ft = 2c + and a + b = c + d. Solving these equations, we find that a= c and ft = d; so (a, ft) = (c, rf). Thus g o / is onetoone. (c) The function g o f : A x B —> B x B is onto. Proof Let (m,n) 6 6 x B. Then m and n are odd integers. Therefore, a = m —n e A and ft = 2« —m € B. Hence (g o /) ( a , ft) = g ( /( a , ft)) = g i f i m — n,2n — m)) = g(n, m — n) = (m, n). Thus g o / is onto. 9.47 (a) The statement is true. Proof Let a € R. Suppose that f ( a ) = ft, g(ft) = c, and h(b) = d. Then ((g + h) o f) ( a ) = (g + h)(fia)) = (g + h)(b) = gib) + h(b) = c + d; while [(g o / ) + (h o f)](a) = (g o /)(« ) + (h o f) (a ) = g i f (a)) + h (f(a)) = gib) + hib) = c + d. Therefore, (g + A) o / = (g o / ) + (h o / ) . (b) The statement is false. For example, suppose that f i x ) = x 2, g(x) = x, and h{x) = x for x 6 R. So (g + h)ix) = 2x. Then [ / o (g + A)](l) = /( ( g + /z)(l)) = /( 2 ) = 4; while [ ( / o g) + ( / o A)](l) = ( / o g)(l) + i f o h)(\) = /( g ( l) ) + /( /id ) ) = / ( l ) + / ( l ) = 1 + 1 = 2 . Thus / o (g + h) # ( / o g) + ( / o /i) in general.
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Section 9.6: Inverse Functions 9.49
Let / = {(a, a), (ft, a), (c, ft)}. Then / is a function from A to A. But the inverse relation / 1 = {(a, a), (a, ft), (ft, c)} is not a function.
Exercises for Chapter 9 9.51
379
Proof First, we show that / is onetoone. Assume that f ( a ) — /(ft), where a, ft e R  {3}. Then 5a 5b  —  =  —  • Multiplying both sides by (a —3){b  3), we obtain 5a(b  3) = 5b(a  3). Simplifying, we have 5ab — 15a = 5ab — 15b. Adding —5ab to both sides and dividing by —15, we obtain a = b. Thus / is onetoone. To show that / is onto, let r e R  {5}. We show that there exists i 6 R  {3} such that f( x ) = r. 3/• 3r Consider* = . (S ince ^ 3, it follows that j e R  (31.) Then r —5 r —5 3r \
5(^j)
15,
15,
implying that / is onto. Therefore, / is bijective. Since ( / o /  l ) (*) = * for all l e R  {5}, it follows that
( / ° / “')(*) = /(/'(*))=
5f l(X)
Thus 5 / “ *(*) = * ( / ” '(*) — 3) and 5 f ~ l (x) = x f ~ l{x) — 3x. Collecting the terms involving f ~ l(x) on the same side of the equation and then factoring f ~ ' ( x ) from this expression, we have x f ~ \ x )  5 f ~ \ x ) = 3x; so f ~ '( x ) ( x —5) = 3x. Solving for / 1(x), we obtain
/'(*)=
3A'
* —5
9.53 9.55
Since there are 6 = 3! bijective functions from A to 5 , there are six functions A to B that have inverses. (a) The proof is similar to that in Exercise 9.51. (b) / = /* • (c) f o f o f = f . 9.57 (a) Proof Observe that f i x ) > 0 if and only if x > 1 and that f ( x ) < 0 if and only if * < 1. First, we show that / is onetoone. Assume that /( a ) = f(b). We consider two cases. Case 1. f (a) — /(ft) > 0. Then ~Ja — 1 = Vft — 1. Squaring both sides, we get a — 1 = ft — 1 and so a = ft. Case 2. /( a ) = /(ft) < 0. Then = jzy Therefore, a — 1 = ft  1 and so a = ft. Hence / is onetoone. Next, we show that / is onto. Let r e R. We consider two cases. Case l . r > 0 . Then f ( r 2 + 1) = y/(r2 + 1) — 1 = r. Case 2. r < 0. Then / ( ^ ) = = r. Therefore, / is onto and thus a bijection. ■ *±i
if * < 0
(b) f ~ \ x ) = 9.59
x2+ 1 if * > 0 Proof First, obsen'e that g o f : A C and h o f : A >• C. Let ft e B. Since / is bijective, there is a unique element a e A such that /( a ) = ft. Since g o f — h o / , it follows that (g o f)(a ) = (h o f) (a ) and so g i f (a)) = h ifia )). Therefore, gib) = Mb) and so g = h. ■
Section 9.7: Permutations