SolutionManualfor AFirstCourseinAbstractAlgebra,withApplications ThirdEdition
byJosephJ.RotmanExercisesforChapter1
1.1 Trueorfalsewithreasons.
(i) Thereisalargestintegerineverynonemptysetofnegativeintegers.
Solution. True.If C isanonemptysetofnegativeintegers,then C ={−n : n ∈ C } isanonemptysetofpositiveintegers.If a isthesmallestelement of C ,whichexistsbytheLeastIntegerAxiom,then a ≤−c forall c ∈ C ,sothat a ≥ c forall c ∈ C .
(ii) Thereisasequenceof13consecutivenaturalnumberscontaining exactly2primes.
Solution. True.Theintegers48through60formsuchasequence; only53and59areprimes.
(iii) Thereareatleasttwoprimesinanysequenceof7consecutive naturalnumbers.
Solution. False.Theintegers48through54are7consecutive naturalnumbers,andonly53isprime.
(iv) Ofallthesequencesofconsecutivenaturalnumbersnotcontaining 2primes,thereisasequenceofshortestlength.
Solution. True.Theset C consistingofthelengthsofsuch(finite) sequencesisanonemptysubsetofthenaturalnumbers.
(v) 79isaprime.
Solution. True. √79 < √81 = 9,and79isnotdivisibleby2,3, 5,or7.
(vi) Thereexistsasequenceofstatements S (1), S (2),... with S (2n ) trueforall n ≥ 1andwith S (2n 1) falseforevery n ≥ 1.
Solution. True.Define S (2n 1) tobethestatement n = n ,and define S (2n ) tobethestatement n = n .
(vii) Forall n ≥ 0,wehave n ≤ Fn ,where Fn isthe n thFibonacci number.
Solution. True.Wehave0 = F0 ,1 = F1 ,1 = F2 ,and2 = F3 .Usethesecondformofinductionwithbasesteps n = 2and n = 3(verifyingtheinductivestepwillshowwhywechoose thesenumbers).Bytheinductivehypothesis, n 2 ≤ Fn 2 and n 1 ≤ Fn 1 .Hence,2n 3 ≤ Fn .But n ≤ 2n 3forall n ≥ 3, asdesired.
(viii) If m and n arenaturalnumbers,then (mn )!= m !n ! Solution. False.If m = 2 =
1.2(i) Forany n ≥ 0andany r = 1,provethat
4.
Solution. Weuseinductionon n ≥ 1.When n = 1,bothsides equal1 + r .Fortheinductivestep,notethat
(ii) Provethat
Thisisthespecialcaseofthegeometricserieswhen r
2;hence,thesumis
alsoprovethisdirectly,byinductionon
1.3 Show,forall n ≥ 1,that10n leavesremainder1afterdividingby9. Solution. Thismayberephrasedtosaythatthereisaninteger qn with 10n = 9qn + 1.Ifwedefine q1 = 1,then10 = q1 + 1,andsothebasestep istrue.
Fortheinductivestep,thereisaninteger qn with 10n +1 = 10 × 10n = 10(9qn + 1) = 90qn + 10 = 9(
Define qn +1 = 10qn + 1,whichisaninteger.
1.4 Provethatif0 ≤ a ≤ b ,then a n ≤ b n forall n ≥ 0. Solution. Basestep a 0 = 1 = b 0 ,andso a 0 ≤ b 0 Inductivestep.Theinductivehypothesisis
Since a ispositive,Theorem1.4(i)gives
;since b is positive,Theorem1.4(i)nowgives
1.9 Findaformulafor1 + n j =1 j ! j ,anduseinductiontoprovethatyour formulaiscorrect.
Solution. Alistofthesumsfor
Thesearefactorials;better,theyare2
theguess
Wenowuseinductiontoprovethattheguessis always true.Thebasestep S (1) hasalreadybeenchecked;itisonthelist.Fortheinductivestep,we mustprove
Bytheinductivehypothesis,thebracketedtermis (n + 1)!,andsotheleft sideequals
Byinduction, S (n ) istrueforall n ≥ 1.
1.10 (M.Barr)ThereisafamousanecdotedescribingahospitalvisitofG.H. HardytoRamanujan.Hardymentionedthatthenumber1729ofthetaxi hehadtakentothehospitalwasnotaninterestingnumber.Ramanujan disagreed,sayingthatitisthesmallestpositiveintegerthatcanbewritten asthesumoftwocubesintwodifferentways.
(i) ProvethatRamanujan’sstatementistrue.
Solution. First,1729isthesumoftwocubesintwodifferent ways:
1729 = 13 + 123 ; 1927 = 93 + 103 .
Second,nosmallernumber n hasthisproperty.If n = a 3 + b 3 , then a , b ≤ 12.Itisnowamatterofcheckingallpairs a 3 + b 3 for such a and b .
(ii) ProvethatRamanujan’sstatementisfalse.
Solution. Onemustpayattentiontohypotheses.Consider a 3 + b 3 if b isnegative:
1.11 Derivetheformulafor n i =1 i bycomputingthearea (n + 1)2 ofasquare withsidesoflength n + 1usingFigure1.1.
Solution. Computethearea A ofthesquareintwoways.Ontheonehand, A = (n + 1)2 .Ontheotherhand, A
D isthediagonal and S isthe“staircase.”Therefore,
1.12(i) Derivetheformulafor n i =1 i bycomputingthearea n (n + 1) of arectanglewithheight n + 1andbase n ,aspicturedinFigure1.2.
Solution. Computethearea R oftherectangleintwoways.On theonehand, R = n (n + 1).Ontheotherhand, R = 2| S |,where S istheshadedregion(whoseareaiswhatweseek).
(ii) (Alhazen)For fixed k ≥ 1,useFigure1.3toprove
AsindicatedinFigure1.3,arectanglewithheight
+ 1
canbesubdividedsothattheshadedstaircase
1.13(i) Provethat2n > n 3 forall n ≥ 10.
Solution. Basestep.210 = 1024 > 103 = 1000.(Notethat 29 = 512 < 93 = 729.)
Inductivestep Notethat n ≥ 10implies n ≥ 4.Theinductive hypothesisis2n > n 3 ;multiplyingbothsidesby2gives
(ii) Provethat2n > n 4 forall n ≥ 17.
Solution. Basestep.217 = 131, 072 > 174 = 83, 521.(Notethat 164 = (24 )4 = 216 .)
Inductivestep.Notethat n ≥ 17implies n ≥ 7.Theinductive hypothesisis2n > n 4 ;multiplyingbothsidesby2gives
1.14 Around1350,N.Oresmewasabletosumtheseries
/2
be
bydissectingtheregioninFigure1.4intwoways.Let A n betheverticalrectanglewithbase 1 2n andheight
1.15 Let g1 ( x ),..., gn ( x ) bedifferentiablefunctions,andlet f ( x ) betheirproduct: f ( x ) = g1 ( x ) gn ( x ).Prove,forallintegers n ≥ 2,thatthederiva-
Basestep.If n = 2,thisistheusualproductruleforderivatives.
1.16 Prove,forevery n ∈ N,that (1 + x )n ≥ 1 + nx whenever x ∈ R and 1 + x > 0.
Weprovetheinequalitybyinductionon n ≥ 1.Thebasestep n = 1says1 + x ≥ 1 + x ,whichisobviouslytrue.Fortheinductivestep,
werecordtheinductivehypothesis:
Multiplyingbothsidesofthisinequalitybythepositivenumber1 + x preservestheinequality: (
because nx 2 ≥ 0.
1.17 Provethateverypositiveinteger a hasauniquefactorization a = 3k m , where k ≥ 0and m isnotamultipleof3.
Solution. ModelyoursolutionontheproofofProposition1.14.Replace “even” by “multipleof3” and “odd” by “notamultipleof3.” Weprove thisbythesecondformofinductionon a ≥ 1.Thebasestep n = 1holds, for1 = 30 1isafactorizationofthedesiredkind.
Fortheinductivestep,let a ≥ 1.If a isnotamultipleof3,then a = 30 a isagoodfactorization.If a = 3b ,then b < a ,andsotheinductive hypothesisgives k ≥ 0andaninteger c notdivisibleby3suchthat b = 3k c .Itfollowsthat a = 3b = 3k +1 c ,whichisafactorizationofthedesired kind.Wehaveprovedtheexistenceofafactorization.
Toproveuniqueness,supposethat n = 3k m = 3t m ,whereboth k and t arenonnegativeandboth m and m arenotmultiplesof3;itmustbe shownthat k = t and m = m .Wemayassumethat k ≥ t .If k > t ,then canceling3t frombothsidesgives3k t m = m .Since k t > 0,theleft sideisamultipleof3whiletherightsideisnot;thiscontradictionshows that k = t .Wemaythuscancel3k frombothsides,leaving m = m .
Solution. Theproofisbythesecondformofinduction.
Basestep: F0 = 0 < 1 = 20 and F1 = 1 < 2 = 21 .(Therearetwobasesteps becausewewillhavetousetwopredecessorsfortheinductivestep.)
Inductivestep:
BaseStep.If n = 1,then
4n +1 + 52n 1 = 16 + 5 = 21, whichisobviouslydivisibleby21.Sinceourinductivestepwillinvolve twopredecessors,weareobligedtocheckthecase n = 2.But43 + 53 = 64 + 125 = 189 = 21 × 9.
Nowthelasttermisdivisibleby21;the firstterm,bytheinductivehypothesis,andthesecondbecause52 4 = 21.
1.21 Foranyinteger n ≥ 2,provethatthereare n consecutivecompositenumbers.Concludethatthegapbetweenconsecutiveprimescanbearbitrarily large.
Solution. Theproofhasnothingtodowithinduction.If2 ≤ a ≤ n + 1, then a isadivisorof (n + 1)!;say, (n + 1)!= da forsomeinteger d .It followsthat (n + 1)!+ a = (d + 1)a ,andso (n + 1)!+ a iscompositefor all a between2and n + 1.
1.22 Provethatthe firstandsecondformsofmathematicalinductionareequivalent;thatis,provethatTheorem1.4istrueifandonlyifTheorem1.12is true.
Solution. Absent.
1.23 (DoubleInduction)Let S (m , n ) beadoublyindexedfamilyofstatements, oneforeach m ≥ 0and n ≥ 0.Supposethat
(i) S (0, 0) istrue;
(ii)if S (m , 0) istrue,then S (m + 1, 0) istrue;
(iii)if S (m , n ) istrueforall m ≥ 0,then S (m , n + 1) istrueforall m ≥ 0.
Provethat S (m , n ) istrueforall m ≥ 0and n ≥ 0.
Solution. Conditions(i)and(ii)arethehypothesesneededtoprove,by (ordinary)inductionthatthestatements S (m , 0) aretrueforall m ≥ 0.
Nowconsiderthestatements
T (n ) : S (m , n ) istrueforall m ≥ 0
Weprovethatallthestatements T (n ) aretruebyinductionon n ≥ 0.The basestephasbeenprovedabove,andcondition(iii)ispreciselywhatis neededfortheinductivestep.
1.24 Usedoubleinductiontoprovethat
AccordingtoExercise1.23,therearethreethingstoverify.
Solution. That θ isanacuteangleimpliesthatthenumberssin θ ,cot
, andsec θ areallpositive.Theinequalityofthemeansgives
1.26 IsoperimetricInequality.
(i) Let p beapositivenumber.If isanequilateraltrianglewith perimeter p = 2s ,provethatarea( ) = s 2 /√27. Solution. Thisisanelementaryfactofhighschoolgeometry.
1.27
(ii) Ofallthetrianglesintheplanehavingperimeter p ,provethatthe equilateraltrianglehasthelargestarea.
Solution. Use Heron’sformula:ifatriangle T hasarea A and sidesoflengths a , b ,
1.28 Trueorfalsewithreasons.
(i) Forallintegers r with0 < r < 7,thebinomialcoefficient 7 r isa multipleof7.
Solution. True.
(ii) Foranypositiveinteger n andany r with0 < r < n ,thebinomial coefficient n r isamultipleof n
Solution. False.
(iii) Let D beacollectionof10differentdogs,andlet C beacollection of10differentcats.Therearethesamenumberofquartetsofdogs astherearesextetsofcats.
Solution. True.
(iv) If q isarationalnumber,then e 2π iq isarootofunity. Solution. True.
(v) Let f ( x ) = ax 2 + bx + c ,where a , b , c arerealnumbers.If z is arootof f ( x ),then z isalsoarootof f ( x ). Solution. True.
(vi) Let f ( x ) = ax 2 + bx + c ,where a , b , c arecomplexnumbers.If z isarootof f ( x ),then z isalsoarootof f ( x ) Solution. False.
(vii) Theprimitive4throotsofunityare i and i Solution. True.
1.29 Provethatthebinomialtheoremholdsforcomplexnumbers:if u and v are complexnumbers,then
Solution. Theproofofthebinomialtheoremforrealnumbersusedonly propertiesofthethreeoperations:addition,multiplication,anddivision. Theseoperationsoncomplexnumbershaveexactlythesameproperties. (Divisionentersinonlybecausewechosetoexpand (a + b )n byusingthe formulafor (1 + x )n ;hadwenotchosenthisexpositorypath,thendivision wouldnothavebeenused.Thus,thebinomialtheoremreallyholdsfor commutativerings.)
1.30 Showthatthebinomialcoefficientsare “symmetric”:
1.32(i) Show,forevery n ≥ 1,thatthe “alternatingsum” ofthebinomial coefficientsiszero:
Solution. If f ( x ) = (1 + x )n ,then f ( 1) = (1 1)n = 0;but theexpansionisthealternatingsumofthebinomialcoefficients.
(ii) Usepart(i)toprove,foragiven n ,thatthesumofallthebinomial coefficients n r with r evenisequaltothesumofallthose n r with r odd.
Solution. Bypart(i),
Sincethesignof n r is ( 1)r ,theterms n r with r evenarepositivewhilethosewith r oddarenegative.Justputthosecoefficients withnegativecoefficientontheothersideoftheequation.
1.33 Provethatif n ≥ 2,then
Solution. Again,consider f ( x ) = (1 + x )n .Therearetwowaystodescribe itsderivative f ( x ).Ontheonehand, f ( x ) = n (1 + x )n 1 .Ontheother hand,wecandoterm-by-termdifferentiation:
1.34 If1 ≤ r ≤ n ,provethat
Solution. Absent.
1.35 Let ε1 ,...,εn becomplexnumberswith |ε j |= 1forall j ,where n ≥ 2.
(i) Provethat
Solution. Thetriangleinequalitygives |u + v |≤|u |+|v | forall complexnumbers u and v ,withnorestrictionontheirnorms.The inductiveproofisroutine.
(ii) Provethatthereisequality,
ifandonlyifallthe ε j areequal. Solution. Theproofisbyinductionon n ≥ 2. Forthebasestep,supposethat
Theargumentconcludesasthatofthebasestep.
+1 = 1.Bythebasestep,
+1 ,andtheproofis complete.
1.36 (StarofDavid )Prove,forall
Solution. UsingPascal’sformula,oneseesthatbothsidesareequalto
1.37 Forallodd n ≥ 1,provethatthereisapolynomial gn ( x ),allofwhose coefficientsareintegers,suchthat
Solution. FromDeMoivre’stheorem,
1.38(i)
Whatisthecoefficientof x 16 in (1 + x )20 ?
Solution. Pascal’sformulagives 20 16 = 4845.
(ii) Howmanywaysaretheretochoose4colorsfromapalettecontainingpaintsof20differentcolors?
Solution. Pascal’sformulagives 20 4 = 4845.Onecouldalso haveusedpart(i)andExercise1.30.
1.39 Giveatleasttwodifferentproofsthataset X with n elementshasexactly 2n subsets.
Solution. Therearemanyproofsofthis.Weofferonlythree.
Algebraic
Let X ={a1 , a2 ,..., an }.Wemaydescribeeachsubset S of X bya bitstring;thatis,byan n -tuple
( 1 , 2 ,..., n ),
where
i =
0if ai isnotin S 1if ai isin S .
(afterall,asetisdeterminedbytheelementscomprisingit).Butthereare exactly2n such n -tuples,fortherearetwochoicesforeachcoordinate.
Combinatorial
Inductionon n ≥ 1(takingbasestep n = 0isalso fine;theonlysetwith 0elementsis X = ∅,whichhasexactlyonesubset,itself).If X hasjust oneelement,thentherearetwosubsets: ∅ and X .Fortheinductivestep, assumethat X has n + 1elements,ofwhichoneiscoloredred,theother n beingblue.Therearetwotypesofsubsets S :thosethataresolidblue; thosethatcontainthered.Byinduction,thereare2n solidbluesubsets; denotethemby B .But,thereareasmanysubsets R containingthered astherearesolidbluesubsets:each R arisesbyadjoiningtheredelement toasolidbluesubset,namely, B = R −{red} (eventhesingletonsubset consistingoftheredelementalonearisesinthisway,byadjoiningthered elementto ∅).Hence,thereare2n + 2n = 2n +1 subsets.
BinomialCoefficients.
If X has n elements,thenthenumberofitssubsetsisthesumofthe numberof0-subsets(thereisonly1,theemptyset),thenumberof1subsets,thenumberof2-subsets,etc.But n r isthenumberof r -subsets, aswehaveseeninthetext,andsothetotalnumberofsubsetsisthesum ofallthebinomialcoefficients,whichis2n ,byExercise1.31.
1.40 Aweeklylotteryasksyoutoselect5differentnumbersbetween1and45. Attheweek’send,5suchnumbersaredrawnatrandom,andyouwinthe jackpotifallyournumbersmatchthedrawnnumbers.Whatisyourchance ofwinning?
Solution. Theansweris ”45choose5”,whichis 45 5 = 1, 221, 759.The oddsagainstyourwinningaremorethanamilliontoone.
1.41 Assumethat “term-by-term” differentiationholdsforpowerseries:if f ( x ) = c0 + c1 x + c2 x 2 +···+ cn x n +··· ,thenthepowerseriesforthederivative f ( x ) is f
(i) Provethat f (0) = c0 .
Solution. f (0) = c0 ,foralltheothertermsare0.(Ifone wantstobefussy–thisisthewrongcourseforanalyticfussiness–thenthepartialsumsoftheseriesformtheconstantsequence c0 , c0 , c0 ,....)
(ii) Prove,forall n ≥ 0,that f (n ) (
1 x + x 2 gn ( x ), where gn ( x ) issomepowerseries.
Solution. Thisisastraightforwardproofbyinductionon n ≥ 0. Thebasestepisobvious;fortheinductivestep,justobservethat f n +1 ( x ) = ( f (n ) ( x )) .As f (n ) ( x ) isapowerseries,byassumption,itsderivativeiscomputedtermbyterm.
(iii) Provethat cn = f (n ) ( x )(0)/ n ! forall n ≥ 0.(Ofcourse,thisis Taylor’sformula.)
Solution. If n = 0,thenourconventionsthat f (0) ( x ) = f ( x ) and 0!= 1givetheresult.Fortheinductivestep,useparts(i)and(ii).
1.42 (Leibniz)Provethatif f and g are C ∞ -functions,then ( fg )(n ) ( x ) = n k =0
n k f (k ) ( x ) g (n k ) ( x ).
Solution. Theproofisbyinductionon n ≥ 1.