Discrete Mathematics with Applications 4th Edition Susanna Solutions Manual

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Chapter1:SpeakingMathematically

Manycollegestudentsappeartohavedifficultyusingandinterpretinglanguageinvolvingif-then statementsandquantification.Section1.1isagentleintroductiontotherelationbetweeninformal andformalwaysofexpressingimportantkindsofmathematicalstatements.Experiencewiththe exercisesinthesectionismeantasawarm-uptopreparestudentstomasterthelinguisticaspects oflearningmathematicstohelpthemcometounderstandthemeaningofmathematicalstatements andevaluatetheirtruthorfalsityinlaterchapters.Sections1.2and1.3formabriefintroductionto thelanguageofsets,relations,andfunctions.Coveringthematthebeginningofthecoursecanhelp studentsrelatediscretemathematicstothepre-calculusorcalculustheyhavestudiedpreviously whilebroadeningtheirperspectivetoincludealargerproportionofdiscreteexamples.

Proofsofsetproperties,suchasthedistributivelaws,andproofsofpropertiesofrelationsand functions,suchastransitivityandsurjectivity,areconsiderablymorecomplexthantheexamples usedinthisbooktointroducestudentstotheideaofmathematicalproof.Thussettheoryas atheoryislefttoChapter6,propertiesoffunctionstoChapter7,andpropertiesofrelationsto Chapter8.InstructorswhowishtodosocouldcoverSection1.2justbeforestartingChapter6and Section1.3justbeforestartingChapter7.

Anaspectofstudents’backgroundsthatmaysurprisecollegeanduniversitymathematicsinstructorsconcernstheirunderstandingofthemeaningof“realnumber.”Whenaskedtoevaluate thetruthorfalsityofastatementaboutrealnumbers,itisnotunusualforstudentstothinkonly ofintegers.Thusaninformaldescriptionoftherelationshipbetweenrealnumbersandpointsona numberlineisgiveninSection1.2onpage8toillustratethattherearemanyrealnumbersbetween anypairofconsecutiveintegers,Examples3.3.5and3.3.6onpage121showthatwhilethereisa smallestpositiveintegerthereisnosmallestpositiverealnumber,andthediscussiononpages433 and434(precedingtheproofoftheuncountabilityoftherealnumbersbetween0and1)describes aprocedureforapproximatingthe(possiblyinfinite)decimalexpansionforanarbitrarilychosen pointonanumberline.

Section1.1

2. a.aremainderof2whenitisdividedby5andaremainderof3whenitisdividedby6

b.aninteger n; n isdividedby6theremainderis3

4. a.arealnumber;greaterthan rb.realnumber r;thereisarealnumber s

6. a s isnegative b.negative;thecuberootof s isnegative(Or : 3 √s isnegative)

c.isnegative; 3 √s isnegative(Or :thecuberootof s isnegative)

7. b.Thereisarealnumberwhosesquareislessthanitself.

True.Forexample,(1/2)2 =1/4 < 1/2.

d.Theabsolutevalueofthesumofanytwonumbersislessthanorequaltothesumoftheir absolutevalues.

True.Thisisknownasthetriangleinequality.ItisdiscussedinSection4.4.

9. a.haveatmosttworealsolutions b.hasatmosttworealsolutions c.hasatmosttworeal solutions d.isaquadraticequation;hasatmosttworealsolutions e E hasatmosttwo realsolutions

11. a.havepositivesquareroots b.apositivesquareroot c r isasquarerootfor e

SolutionsforExercises:SpeakingMathematically

13. a.realnumber;productwitheveryrealnumberequalszero

b.witheveryrealnumberequalszero c ab =0 Section1.2

2. b.Thesetofallrealnumbers x suchthat x islessthanorequaltozeroor x isgreaterthan orequalto1

d.Thesetofallpositiveintegers n suchthat n isafactorof6

4. a.Yes: {2} isthesetwhoseonlyelementis2. b.One:2istheonlyelementinthisset c.

Two:Thetwoelementsare0and {0} d.Yes: {0} isoneoftheelementslistedintheset.

e.No:Theonlyelementslistedinthesetare {0} and {1},and0isnotequaltoeitherof these.

5. Theonlysetsthatareequaltoeachotherare A and D. A containstheintegers0,1,and2andnothingelse.

B containsalltherealnumbersthataregreaterthanorequalto 1andlessthan3.

C containsalltherealnumbersthataregreaterthan 1andlessthan3.Thus 1isin B butnotin C.

D containsalltheintegersgreaterthan 1andlessthan3.Thus D containstheintegers0, 1,and2andnothingelse,andso D = {0, 1, 2} = A

E containsallthepositiveintegersgreaterthan 1andlessthan3.Hence E containsthe integers1and2andnothingelse,thatis, E = {1, 2}

6. T2 and T 3 eachhavetwoelements,and T0 and T1 eachhaveoneelement.

7. b T = {m ∈ Z | m =1+( 1)k forsomeinteger k} = {0, 2} ExercisesinChapter4explorethe factthat( 1)k = 1when k isoddand( 1)k =1when k iseven.So1+( 1)k =1+( 1)=0 when k isodd,and1+( 1)k =1+1=2when k iseven.

e.Therearenoelementsin W becausetherearenointegersthatarebothgreaterthan1and lessthan 3.

f X = Z becauseeveryinteger u satisfiesatleastoneoftheconditions u ≤ 4or u ≥ 1

8. b.Yes,becauseeveryelementin C isin A c.Yes,becauseeveryelementin C isin C

d..Yes,becausealthougheveryelementin C isin A, A containselementsthatarenotin C

9. c.No:Theonlyelementsin {1, 2} are1and2, and {2} isnotequaltoeitherofthese.

d.Yes: {3} isoneoftheelementslistedin {1, {2}, {3}}.

e.Yes: {1} isthesetwhoseonlyelementis1.

g.Yes:Theonlyelementin {1} is1,and1isanelementin {1, 2}.

h.No:Theonlyelementsin {{1}, 2} are {1} and2,and1isnotequaltoeitherofthese.

j.Yes:Theonlyelementin {1} is1,whichisisanelementin {1}.Soeveryelementin {1} is in {1}

10. b.No:Fortwoorderedpairstobeequal,theelementsineachpairmustoccurinthesame order.Inthiscasethefirstelementofthefirstpairis5,whereasthefirstelementofthesecond pairis 5,andthesecondelementofthefirstpairis 5whereasthesecondelementofthe secondpairis5.

d.YesThefirstelementsofbothpairsequal 1 2 ,andthesecondelementsofbothpairsequal 8.

12. Allfoursetshavenineelements.

2 3 ,whichisnotaninteger.

4.

.2 V 6because 2 6 4 = 4 4 = 1,whichisaninteger.

V 6because 2 ( 6) 4 = 4 4 =1,whichisaninteger.

6 4 = 6 4 ,whichisnotaninteger.

4 ,whichisnotaninteger.

6)}

SolutionsforExercises:SpeakingMathematically

U isnotafunctionbecause(4,y)isnotin U forany y in B,andso U doesnotsatisfyproperty (1)ofthedefinitionoffunction.

V isnotafunctionbecause(2,y)isnotin V forany y in B,andso V doesnotsatisfyproperty (1)ofthedefinitionoffunction.

W isnotafunctionbecauseboth(2, 3)and(2, 5)arein W and3 =5,andso W doesnot satisfyproperty(2)ofthedefinitionoffunction.

10. Thefollowingsetsarerelationsfrom {

12. T isnotafunctionbecause,forexample,both(0, 1)and(0, 1)arein T but1 = 1. Many otherexamplescouldbegivenshowingthat T doesnotsatisfyproperty2ofthedefinitionof function.

14. a.Domainof G = {1, 2, 3, 4}, co-domainof G = {a,b,c,d}

b G(1)= G(2)= G(3)= G(4)= c

15. c.Thisdiagramdoesnotdetermineafunctionbecause4isrelatedtoboth1and2,which violatesproperty(2)ofthedefinitionoffunction.

d.Thisdiagramdefinesafunction;bothproperties(1)and(2)aresatisfied.

e.Thisdiagramdoesnotdetermineafunctionbecause2isinthedomainbutitisnotrelated toanyelementintheco-domain.

Solutions forExercises:TheLogicofCompoundStatements

Chapter2:TheLogicofCompoundStatements

Theabilitytoreasonusingtheprinciplesoflogicisessentialforsolvingproblemsinabstractmathematicsandcomputerscienceandforunderstandingthereasoningusedinmathematicalproofand disproof.Becauseasignificantnumberofstudentswhocometocollegehavehadlimitedopportunitytodevelopthisability,aprimaryaimofChapters2and3istohelpstudentsdevelopaninner voicethatspeakswithlogicalprecision.Consequently,thevariousrulesusedinlogicalreasoningare developedbothsymbolicallyandinthecontextoftheirsomewhatlimitedbutveryimportantuse ineverydaylanguage.ExercisesetsforSections2.1–2.3and3.1–3.4containsentencesforstudents tonegate,writethecontrapositivefor,andsoforth.Virtuallyallstudentsbenefitfromdoingthese exercises.AnotheraimofChapters2and3istoteachstudentstherudimentsofsymboliclogicas afoundationforavarietyofupper-divisioncourses.Symboliclogicisusedin,amongothers,the studyofdigitallogiccircuits,relationaldatabases,artificialintelligence,andprogramverification.

Suggestions

1. InSection2.1asurprisingnumberofstudentsapplyDeMorgan’slawtowritethenegationof, forexample,“1 <x ≤ 3”as“1 ≥ x>

3.”Youmayfindthatittakessomeefforttoteachthemto avoidmakingthismistake.

2. InSections2.1and2.4,studentshavemoredifficultythanyoumightexpectsimplifyingstatement formsandcircuits.Onlythroughtrialanderrorcanyoulearntheextenttowhichthisisthecase atyourinstitution.Ifitis,youmighteitherassignonlytheeasierexercisesorbuildinextratimeto teachstudentshowtodothemorecomplicatedones.Discussionofsimplificationtechniquesoccurs againinChapter6inthecontextofsettheory.Atthislaterpointinthecoursemoststudentsare abletodealwithitsuccessfully.

3. InordinaryEnglish,thephrase“onlyif”isoftenusedasasynonymfor“ifandonlyif.”But itispossibletofindinformalsentencesinwhichtheintuitiveinterpretationisthesameasthe logicaldefinition,anditishelpfultogiveexamplesofsuchstatementswhenyouintroducethe logicaldefinition.Forinstance,itisnothardtogetstudentstoagreethat“Theteamwillwinthe championshiponlyifitwinsthesemifinalgame”meansthesameas“Iftheteamdoesnotwinthe semifinalgamethenitwillnotwinthechampionship.”Oncestudentsseethis,youcansuggestthat theyrememberthistranslationwhentheyencountermoreabstractstatementsoftheform“A only if B.”

Throughguideddiscussion,studentswillalsocometoagreethatthestatement“Winningthe semifinalgameisanecessaryconditionforwinningthechampionship”translatesto“Iftheteam doesnotwinthesemifinalgamethenitwillnotwinthechampionship.”Theycanbeencouraged tousethis(orasimilarstatement)asareferencetohelpdevelopintuitionforgeneralstatementsof theform“A isanecessaryconditionfor B.”

Withstudentswhohaveweakerbackgrounds,youmayfindyourselftyingupexcessiveamounts ofclasstimediscussing“onlyif”and“necessaryandsufficientconditions.”Youmightjustassign theeasierexercises,oryoumightassignexercisesonthesetopicstobedoneforextracredit(putting correspondingextracreditproblemsonexams)andusetheresultstohelpdistinguishAfromB students.Itisprobablybestnottoomitthesetopicsaltogether,though,becausethelanguageof “onlyif”and“necessaryandsufficientconditions”isastandardpartofthetechnicalvocabularyof textbooksusedinupper-divisioncourses,aswellasoccurringregularlyinnon-mathematicalwriting.

4. InSection2.3,manystudentsmistakenlyconcludethatanargumentisvalidif,whenthey computethetruthtable,theyfindasinglerowinwhichboththepremisesandtheconclusionare true.Thesourceofstudents’difficultyappearstobetheirtendencytoignorequantificationandto misinterpretif-thenstatementsas“and”statements.Sincethedefinitionofvalidityincludesboth auniversalquantifierandif-then,itishelpfultogobackoverthedefinitionandtheproceduresfor

testingforvalidityandinvalidityafterdiscussingthegeneraltopicofuniversalconditionalstatements inSection3.1.Asapracticalmeasuretohelpstudentsassessvalidityandinvaliditycorrectly,the firstexampleinSection2.3isofaninvalidargumentwhosetruthtablehaseightrows,severalof whichhavetruepremisesandatrueconclusion.Inaddtition,tofurtherfocusstudents’attention onthesituationswhereallthepremisesaretrue,thetruthvaluesfortheconclusionsofarguments aresimplyomittedwhenatleastonepremiseisfalse.

5. InSection2.3,youmightsuggestthatstudentsjustfamiliarizethemselveswith,butnotmemorize, thevariousformsofvalidargumentscoveredinSection2.3.Itiswise,however,tohavethemlearn thetermsmodusponensandmodustollens(becausetheseareusedinsomeupper-divisioncomputer sciencecourses)andconverseandinverseerrors(becausetheseerrorsaresocommon).

Section2.1

2.commonform:If p then q.

∼ q

Therefore, ∼ p.

b. allprimenumbersareodd;2isodd

4.commonform:If p then q. If q then r.

Therefore,if p then r.

b. apolynomial;differentiable;iscontinuous

5. b. Thetruthorfalsityofthissentencedependsonthereferenceforthepronoun“she.”Consideredonitsown,thesentencecannotbesaidtobeeithertrueorfalse,andsoitisnota statement.

c. Thissentenceisfalse;henceitisastatement.

d. Thisisnotastatementbecauseitstruthorfalsitydependsonthevalueof x.

Solutions forExercises:TheLogicofCompoundStatements

15. p q r ∼ q ∼ q ∨ r p ∧ (∼ q ∨ r) T T T F T T

T T F F F F

T F T T T T

T F F T T T

F T T F T F

F T F F F F

F F T T T F

F F F T T F

17.

q p ∧ q ∼ p ∼ q ∼ (p ∧ q) ∼ p ∧ ∼ q T T T F F F F

T F F F T T F ←

F T F T F T F ←

F F F T T T T

differenttruthvaluesinrows2and3

Thetruthtableshowsthat ∼ (p ∧ q)and ∼ p ∧∼ q donotalwayshavethesametruthvalues. Thereforetheyarenotlogicallyequivalent.

20.

same truthvalues

Thetruthtableshowsthat p ∧ t and p alwayshavethesametruthvalues.Thustheyare logicallyequivalent.Thisprovestheidentitylawfor ∧

c p ∧ c p ∨ c T F F T ← F F F F

differenttruthvaluesinrow1

Thetruthtableshowsthat p ∧ c and p ∨ c donotalwayshavethesametruthvalues.Thus theyarenotlogicallyequivalent.

same truthvalues

Thetruthtableshowsthat p ∧ (q ∨ r)and(p ∧ q) ∨ (p ∧ r)alwayshavethesametruthvalues. Thereforetheyarelogicallyequivalent.Thisprovesthedistributivelawfor ∧ over ∨

differenttruthvalues

Thetruthtableshowsthat(p ∨ q) ∨ (p ∧ r)and(p ∨ q) ∧ r havedifferenttruthvaluesinrows 2,3,and6.Hencetheyarenotlogicallyequivalent.

26.SamisnotanorangebeltorKateisnotaredbelt.

28.Theunitsdigitof467 isnot4anditisnot6.

29.Thiscomputerprogramdoesnothavealogicalerrorinthefirsttenlinesanditisnotbeing runwithanincompletedataset.

30.Thedollarisnotatanall-timehighorthestockmarketisnotatarecordlow.

31.Thetrainisnotlateandmywatchisnotfast.

Thusanegationis(num orders ≥ 50or num instock ≤ 300)and((50 > num orders or num orders ≥ 75)or num instock ≤ 500).

Since allthetruthvaluesof((∼ p ∧ q) ∧ (q ∧ r))∧∼ q are F ,((

) ∧ (q

r))∧∼ q isa contradiction.

Solutions forExercises:TheLogicofCompoundStatements

45.Let b be“Bobisadoublemathandcomputersciencemajor,” m be“Annisamathmajor,” and a be“Annisadoublemathandcomputersciencemajor.”Thenthetwostatementscan besymbolizedasfollows:a

Thetruthtableshowsthat(b ∧ m)∧∼ a and ∼ (b ∧ a) ∧ (m ∧ b)alwayshavethesametruth values.Hencetheyarelogicallyequivalent.

Thetruthtableshowsthat(p ⊕ q) ⊕ r and p ⊕ (q ⊕ r)alwayshavethesametruthvalues.So theyarelogicallyequivalent.

≡

≡

≡

≡ p ∧ (∼ q ∨ q) distributivelaw

≡ p ∧ (q∨∼ q) commutativelawfor ∨

≡ p ∧ t negationlawfor ∨

≡ p identitylawfor ∧

Section2.2

2.IfIcatchthe8:05bus,thenIamontimeforwork.

4.Ifyoudon’tfixmyceiling,thenIwon’tpaymyrent.

Solutions forExercises:TheLogicofCompoundStatements

15.

Thetruthtableshowsthatthethreestatementforms p → q ∨ r, p ∧∼ q → r,and p ∧∼ r → q alwayshavethesametruthvalues.Thustheyarealllogicallyequivalent.

b. If n isprimeand n isnotodd,then n is2.

And:If n isprimeand n isnot2,then n isodd.

p q r q → r p → q p → (q → r) (p → q) → r

T T T T T T T

T T F F T F F

T F T T F T T

T F F T F T T

F T T T T T T

F T F F T T F ←

F F T T T T F ←

F F F T T T F ←

differenttruthvalues

Thetruthtableshowsthat p → (q → r)and(p → q) → r donotalwayshavethesametruth values.(Theydifferforthecombinationsoftruthvaluesfor p, q,and r showninrows6,7, and8.)Thereforetheyarenotlogicallyequivalent.

17.Let p represent“2isafactorof n,” q represent“3isafactorof n,”and r represent“6isa factorof n.”Thestatement“If2isafactorof n and3isafactorof n,then6isafactorof n” hastheform p ∧ q → r.Andthestatement“2isnotafactorof n or3isanotafactorof n or6isafactorof n”hastheform ∼ p ∨∼ q ∨ r

Thetruthtableshowsthat p ∧ q → r and ∼ p ∨∼ q ∨ r alwayshavethesametruthvalues. Thereforetheyarelogicallyequivalent.

18. Part1 :Let p represent“Itwalkslikeaduck,” q represent“Ittalkslikeaduck,”and r represent “Itisaduck.”Thestatement“Ifitwalkslikeaduckandittalkslikeaduck,thenitisaduck” hastheform p ∧ q → r.Andthestatement“Eitheritdoesnotwalklikeaduckoritdoesnot talklikeaduckoritisaduck”hastheform ∼ p ∨∼ q ∨ r

Solutions forExercises:TheLogicofCompoundStatements

Thetruthtableshowsthat p

alwayshavethesametruthvalues. Thusthefollowingstatementsarelogicallyequivalent:“Ifitwalkslikeaduckandittalkslike aduck,thenitisaduck”and“Eitheritdoesnotwalklikeaduckoritdoesnottalklikea duckoritisaduck.”

Part2 :Thestatement“Ifitdoesnotwalklikeaduckanditdoesnottalklikeaduckthen itisnotaduck”hastheform

Thetruthtableshowsthat p ∧ q → r and(∼ p ∧∼ q) →∼ r donotalwayshavethesame truthvalues.(Theydifferforthecombinationsoftruthvaluesof p, q,and r showninrows2 and7.)Thustheyarenotlogicallyequivalent,andsothestatement“Ifitwalkslikeaduck andittalkslikeaduck,thenitisaduck”isnotlogicallyequivalenttothestatement“Ifit doesnotwalklikeaduckanditdoesnottalklikeaduckthenitisnotaduck.”Inaddition, becauseofthelogicalequivalenceshowninPart1,wecanalsoconcludethatthefollowing twostatementsarenotlogicallyequivalent:“Eitheritdoesnotwalklikeaduckoritdoesnot talklikeaduckoritisaduck”and“Ifitdoesnotwalklikeaduckanditdoesnottalklikea duckthenitisnotaduck.”

20. b. TodayisNewYear’sEveandtomorrowisnotJanuary.

c. Thedecimalexpansionof r isterminatingand r isnotrational.

e.x isnonnegativeand x isnotpositiveand x isnot0.

Or : x isnonnegativebut x isnotpositiveand x isnot0.

Or : x isnonnegativeand x isneitherpositivenor0.

g.n isdivisibleby6andeither n isnotdivisibleby2or n isnotdivisibleby3.

21.Bythetruthtablefor →,p → q isfalseif,andonlyif, p istrueand q isfalse.Underthese circumstances,(b) p ∨ q istrueand(c) q → p isalsotrue.

22. b. IftomorrowisnotJanuary,thentodayisnotNewYear’sEve.

c. If r isnotrational,thenthedecimalexpansionof r isnotterminating.

e. If x isnotpositiveand x isnot0,then x isnotnonnegative.

Or :If x isneitherpositivenor0,then x isnegative.

g. If n isnotdivisibleby2or n isnotdivisibleby3,then n isnotdivisibleby6.

23. b. Converse:IftomorrowisJanuary,thentodayisNewYear’sEve.

Inverse:IftodayisnotNewYear’sEve,thentomorrowisnotJanuary.

c. Converse:If r isrationalthenthedecimalexpansionof r isterminating.

Inverse:Ifthedecimalexpansionof r isnotterminating,then r isnotrational.

e. Converse:If x ispositiveor x is0,then x isnonnegative.

Inverse:If x isnotnonnegative,thenboth x isnotpositiveand x isnot0.

Or:If x isnegative,then x isneitherpositivenor0.

differenttruthvalues

Thetruthtableshowsthat p → qand ∼ p →∼ q havedifferenttruthvaluesinrows2and3, sotheyarenotlogicallyequivalent.Thusaconditionalstatementisnotlogicallyequivalent toitsinverse.

Thetruthtableshowsthat q → pand ∼ p →∼ q alwayshavethesametruthvalues,sothey arelogicallyequivalent.Thustheconverseandinverseofaconditionalstatementarelogically equivalenttoeachother.

Theif-thenformof“ImeanwhatIsay”is“IfIsaysomething,thenImeanit.”

Thus“ImeanwhatIsay”istheconverseof“IsaywhatImean.”Thetwostatementsarenot logicallyequivalent.

28.Theif-thenformof“IsaywhatImean”is“IfImeansomething,thenIsayit.”

30.Thecorrespondingtautologyis

Solutions forExercises:TheLogicofCompoundStatements

33.Ifthisintegeriseven,thenitequalstwicesomeinteger,andifthisintegerequalstwicesome integer,thenitiseven.

35.IfSamisnotanexpertsailor,thenhewillnotbeallowedonSigne’sracingboat. IfSamisallowedonSigne’sracingboat,thenheisanexpertsailor.

36.ThePersonnelDirectordidnotlie.Byusingthephrase“onlyif,”thePersonnelDirectorset forthconditionsthatwerenecessarybutnotsufficientforbeinghired:ifyoudidnotsatisfy thoseconditionsthenyouwouldnotbehired.ThePersonnelDirector’sstatementsaidnothing aboutwhatwouldhappenifyoudidsatisfythoseconditions.

38.Ifitdoesn’train,thenAnnwillgo.

39. b. Ifasecuritycodeisnotentered,thenthedoorwillnotopen.

41.Ifthistrianglehastwo45◦ angles,thenitisarighttriangle.

43.IfJimdoesnotdohishomeworkregularly,thenJimwillnotpassthecourse.

IfJimpassesthecourse,thenhewillhavedonehishomeworkregularly.

45.Ifthiscomputerprogramproduceserrormessagesduringtranslation,thenitisnotcorrect. Ifthiscomputerprogramiscorrect,thenitdoesnotproduceerrormessagesduringtranslation.

46. c.mustbetrue d.notnecessarilytrue e.mustbetrue f .notnecessarilytrue

Note: Tosolvethisproblem,itmaybehelpfultoimagineacompoundwhoseboilingpointis greaterthan150◦ C.Forconcreteness,supposeitis200◦ C.Thenthegivenstatementwould betrueforthiscompound,butstatements a, d,and f wouldbefalse.

q) ∨ r]

∧∼ [∼ (p ∧ q) ∨ r] ∨ [∼ p ∨ (∼ q ∨ r)]

b. Bypart(a),DeMorgan’slaw,andthedoublenegativelaw, (p → (q → r)) ↔ ((p ∧ q) → r) ≡∼ [∼ p ∨ (∼ q ∨ r)] ∨ [∼ (p ∧ q) ∨ r]

∧∼ [∼ (p ∧ q) ∨ r] ∨ [∼ p ∨ (∼ q ∨ r)]

≡∼ [∼ p ∨ (∼ q ∨ r)]∧∼ [∼ (p ∧ q) ∨ r]

∧∼∼ [(p ∧ q)∧∼ r]∧∼ [∼ p ∨ (∼ q ∨ r)]

≡∼∼ [p ∧∼ (∼ q ∨ r)] ∧ [(p ∧ q)∧∼ r]

∧∼∼ [(p ∧ q)∧∼ r

Thestepsintheanswertopart(b)wouldalsobeacceptableanswersforpart(a).

51.Yes.Asinexercises47-50,thefollowinglogicalequivalencescanbeusedtorewriteany statementforminalogicallyequivalentwayusingonly

2.1 0 99999 islessthaneverypositiverealnumber.

Solutions forExercises:TheLogicofCompoundStatements

Rows2,3,and4ofthetruthtablearethecriticalrowsinwhichallthepremisesaretrue,but row3showsthatitispossibleforanargumentofthisformtohavetruepremisesandafalse conclusion.Hencetheargumentformisinvalid.

Rows1,3,5,7,and8ofthetruthtablerepresentthesituationsinwhichallthepremisesare true,andinalloftheserowstheconclusionisalsotrue.Therefore,theargumentformisvalid.

Rows2,3,6,7,and8ofthetruthtablerepresentthesituationsinwhichallthepremisesare true,butrow3showsthatitispossibleforanargumentofthisformtohavetruepremises andafalseconclusion.Hencetheargumentformisinvalid.

Rows3,and4ofthetruthtablerepresentthesituationsinwhichallthepremisesaretrue, butrow3showsthatitispossibleforanargumentofthisformtohavetruepremisesanda falseconclusion.Hencetheargumentformisinvalid. 13.

Row4ofthetruthtablerepresentstheonlysituationinwhichallthepremisesaretrue,and inthisrowtheconclusionisalsotrue.Therefore,theargumentform(modustollens)isvalid.

15.

16.

The truthtableshowsthatinthetwosituations(representedbyrows1and3)inwhichthe premiseistrue,theconclusionisalsotrue.Therefore,thethesecondversionofgeneralization isvalid.

The truthtableshowsthatintheonlysituation(representedbyrow1)inwhichbothpremises aretrue,theconclusionisalsotrue.Therefore,thethefirstversionofspecializationisvalid.

The truthtableshowsthatintheonlysituation(representedbyrow1)inwhichbothpremises aretrue,theconclusionisalsotrue.Therefore,thesecondversionofspecializationisvalid. 19.

The truthtableshowsthatintheonlysituation(representedbyrow3)inwhichbothpremises aretrue,theconclusionisalsotrue.Therefore,thethesecondversionofeliminationisvalid.

Discrete Mathematics with Applications 4th Edition Susanna Solutions Manual

Solutions forExercises:TheLogicofCompoundStatements

Thetruthtableshowsthatinthefoursituations(representedbyrows1,5,7,and8)inwhich bothpremisesaretrue,theconclusionisalsotrue.Therefore,theargumentform(transitivity) isvalid.

The truthtableshowsthatinthethreesituations(representedbyrows1,3,5)inwhichall threepremisesaretrue,theconclusionisalsotrue.Therefore,proofbydivisionintocasesis valid.

Rows1,3,5,and6representthesituationsinwhichbothpremisesaretrue,butinrow3the conclusionisfalse.Hence,itispossibleforanargumentofthisformtohavetruepremises andafalseconclusion,andsothegivenargumentisinvalid.