Discrete Mathematics And Its Applications 2025 Release 1St Rosen Solutions Manual by mmsharp

TheFoundations:LogicandProofs

SECTION1.1PropositionalLogic

2. Propositionsmusthaveclearlydeﬁnedtruthvalues,soapropositionmustbeadeclarativesentencewithnofreevariables.

a) Thisisnotaproposition;it’sacommand.

b) Thisisnotaproposition;it’saquestion.

c) Thisisapropositionthatisfalse,asanyonewhohasbeentoMaineknows.

d) Thisisnotaproposition;itstruthvaluedependsonthevalueof x .

e) Thisisapropositionthatisfalse.

f) Thisisnotaproposition;itstruthvaluedependsonthevalueof n

4.a) JanicedoesnothavemoreFacebookfriendsthanJuan.

b) QuincyisnotsmarterthanVenkat.

c) ZeldadoesnotdrivemoremilestoschoolthanPaola.

d) BrianadoesnotsleeplongerthanGloria.

6.a) JenniferandTejaarenotfriends.

b) Therearenot13itemsinabaker’sdozen.(Alternatively:Thenumberofitemsinabaker’sdozenisnotequalto13.)

c) Abbysentfewerthan101textmessagesyesterday.Alternatively,Abbysentatmost100textmessagesyesterday.

d) 121isnotaperfectsquare.

8.a) True,because12 > 8and12 > 4.

b) True,becauseChas48MPresolutioncomparedtoB’s24MPresolution.Notethatonlyoneoftheseconditionsneeds tobemetbecauseoftheword or

c) False,becauseitsresolutionisnothigher(allofthestatementswouldhavetobetruefortheconjunctiontobetrue).

d) False,becausethehypothesisofthisconditionalstatementistrueandtheconclusionisfalse.

e) False,becausetheﬁrstpartofthisbiconditionalstatementisfalseandthesecondpartistrue.

10.a) Ididnotbuyalotteryticketthisweek.

b) EitherIboughtalotteryticketthisweek,or[intheinclusivesense]IwonthemilliondollarjackpotonFriday.

c) IfIboughtalotteryticketthisweek,thenIwonthemilliondollarjackpotonFriday.

d) Iboughtalotteryticketthisweek,andIwonthemilliondollarjackpotonFriday.

e) IboughtalotteryticketthisweekifandonlyifIwonthemilliondollarjackpotonFriday.

f) IfIdidnotbuyalotteryticketthisweek,thenIdidnotwinthemilliondollarjackpotonFriday.

g) Ididnotbuyalotteryticketthisweek,andIdidnotwinthemilliondollarjackpotonFriday.

h) EitherIdidnotbuyalotteryticketthisweek,orelseIdidbuyoneandwonthemilliondollarjackpotonFriday.

12.a) Theelectionisnotdecided.

b) Theelectionisdecided,orthevoteshavebeencounted.

c) Theelectionisnotdecided,andthevoteshavebeencounted.

d) Ifthevoteshavebeencounted,thentheelectionisdecided.

e) Ifthevoteshavenotbeencounted,thentheelectionisnotdecided.

f) Iftheelectionisnotdecided,thenthevoteshavenotbeencounted.

g) Theelectionisdecidedifandonlyifthevoteshavebeencounted.

h) Eitherthevoteshavenotbeencounted,orelsetheelectionisnotdecidedandthevoteshavebeencounted.Notethat wewereabletoincorporatetheparenthesesbyusingthewords either and else

14.a) Ifyouhavetheﬂu,thenyoumisstheﬁnalexam.

b) Youdonotmisstheﬁnalexamifandonlyifyoupassthecourse.

c) Ifyoumisstheﬁnalexam,thenyoudonotpassthecourse.

d) Youhavetheﬂu,ormisstheﬁnalexam,orpassthecourse.

e) Itiseitherthecasethatifyouhavetheﬂuthenyoudonotpassthecourse,orthecasethatifyoumisstheﬁnalexam thenyoudonotpassthecourse(orboth,itisunderstood).

f) Eitheryouhavethefluandmissthefinalexam,oryoudonotmissthefinalexamanddopassthecourse.

16.a) r ∧¬q b) p

18.a) Thisis �� ↔ �� ,whichistrue.

b) Thisis �� ↔ �� ,whichisfalse.

c) Thisis �� ↔ �� ,whichistrue.

d) Thisis �� ↔ �� ,whichisfalse.

20.a) Thisis �� → �� ,whichistrue.

b) Thisis �� → �� ,whichistrue.

c) Thisis �� → �� ,whichisfalse.

d) Thisis �� → �� ,whichistrue.

22.a) Theemployermakingthisrequestwouldbehappyiftheapplicantknewbothoftheselanguages,sothisisclearlyan inclusive or.

b) Therestaurantwouldprobablychargeextraifthedinerwantedbothoftheseitems,sothisisanexclusive or

c) Ifapersonhappenedtohavebothformsofidentiﬁcation,somuchthebetter,sothisisaninclusive or

d) Thiscouldbearguedeitherway,buttheinclusiveinterpretationseemsmoreappropriate.Thisphrasemeansthat facultymemberswhodonotpublishpapersinresearchjournalsarelikelytobeﬁredfromtheirjobsduringtheprobationaryperiod.Ontheotherhand,itmayhappenthattheywillbeﬁredeveniftheydopublish(forexample,iftheir teachingispoor).

24.a) Thenecessaryconditionistheconclusion:Ifyougetpromoted,thenyouwashtheboss’scar.

b) Ifthewindsarefromthesouth,thentherewillbeaspringthaw.

c) Thesuﬃcientconditionisthehypothesis:Ifyouboughtthecomputerlessthanayearago,thenthewarranty isgood.

d) IfWillycheats,thenhegetscaught.

e) The“onlyif”conditionistheconclusion:Ifyouaccessthewebsite,thenyoumustpayasubscriptionfee.

f) Ifyouknowtherightpeople,thenyouwillbeelected.

g) IfCarolisonaboat,thenshegetsseasick.

26.a) IfIamtoremembertosendyoutheaddress,thenyouwillhavetosendmeanemailmessage.(Thishasbeenslightly rewordedsothatthetensesmakemoresense.)

b) IfyouwerebornintheUnitedStates,thenyouareacitizenofthiscountry.

c) Ifyoukeepyourtextbook,thenitwillbeausefulreferenceinyourfuturecourses.(Theword“then”isunderstoodin English,evenifomitted.)

d) Iftheirgoaltenderplayswell,thentheRedWingswillwintheStanleyCup.

e) Ifyougetthejob,thenyouhadthebestcredentials.

f) Ifthereisastorm,thenthebeacherodes.

g) Ifyoulogontotheserver,thenyouhaveavalidpassword.

h) Ifyoudonotbeginyourclimbtoolate,thenyouwillreachthesummit.

i) Ifyouareamongtheﬁrst100customerstomorrow,thenyouwillgetafreeicecreamcone.

28.a) YouwillgetanAinthiscourseifandonlyifyoulearnhowtosolvediscretemathematicsproblems.

b) Youwillbeinformedifandonlyifyoureadthenewspapereveryday.(Itsoundsbetterinthisorder;itwouldbe logicallyequivalenttostatethisas“Youreadthenewspapereverydayifandonlyifyouwillbeinformed.”)

c) Itrainsifandonlyifitisaweekendday.

d) Youcanseethewizardifandonlyifheisnotin.

e) MyairplaneﬂightislateifandonlyifIhavetocatchaconnectingﬂight.

30.a) Converse:IfIstayhome,thenitwillsnowtonight.Contrapositive:IfIdonotstayathome,thenitwillnotsnow tonight.Inverse:Ifitdoesnotsnowtonight,thenIwillnotstayhome.

b) Converse:WheneverIgotothebeach,itisasunnysummerday.Contrapositive:WheneverIdonotgotothebeach, itisnotasunnysummerday.Inverse:Wheneveritisnotasunnyday,Idonotgotothebeach.

c) Converse:IfIsleepuntilnoon,thenIstayeduplate.Contrapositive:IfIdonotsleepuntilnoon,thenIdidnotstay uplate.Inverse:IfIdon’tstayuplate,thenIdon’tsleepuntilnoon.

32. Atruthtablewillneed2n rowsifthereare n variables.

a) 22 = 4 b) 23 = 8 c) 26 = 64 d) 25 = 32

34. Toconstructthetruthtableforacompoundproposition,weworkfromtheinsideout.Ineachcase,wewillshowthe intermediatesteps.Inpart(d),forexample,weﬁrstconstructthetruthtablesfor p ∧ q andfor p ∨ q andcombinethem togetthetruthtablefor(p ∧ q) → (p ∨ q).Forparts(a)and(b)wehavethefollowingtable(columnthreeforpart(a), columnfourforpart(b)).

Forparts(c)and(d)wehavethefollowingtable.

Forpart(e)wehavethefollowingtable.

Forpart(f)wehavethefollowingtable.

36. Forparts(a)and(b)wehavethefollowingtable(columntwoforpart(a),columnfourforpart(b)).

Forparts(c)and(d)wehavethefollowingtable(columnsﬁveandsix).

Forparts(e)and(f)wehavethefollowingtable(columnsfiveandsix).Thistimewehaveomittedthecolumnexplicitlyshowingthenegationof q .Notethatthefirstisatautologyandthesecondisacontradiction(seedefinitionsin Section1.3).

38. Forparts(a)and(b)wehave

Forparts(c)and(d),wehave

Finally,forparts(e)and(f)wehave

40. Thistimethetruthtableneeds24 = 16rows.

42. Thisstatementistrueifandonlyifallthreeclauses, p ∨¬q , q ∨¬r ,and r ∨¬p aretrue.Suppose p , q ,and r are alltrue.Becauseeachclausehasanunnegatedvariable,eachclauseistrue.Similarly,if p , q ,and r areallfalse,then becauseeachclausehasanegatedvariable,eachclauseistrue.Ontheotherhand,ifoneofthevariablesistrueandthe othertwofalse,thentheclausecontainingthenegationofthatvariablewillbefalse,makingtheentireconjunctionfalse; andsimilarly,ifoneofthevariablesisfalseandtheothertwotrue,thentheclausecontainingthatvariableunnegated willbefalse,againmakingtheentireconjunctionfalse.

44. Theindexeddisjunctions, ¬pi ∨¬pj ,saythatforeachpairofdistinctpropositionsinthelist,atleastoneisfalse.By letting i rangefrom1to n 1and j rangefrom i + 1to n ,wearecheckingallpairsofdistinctpropositionsinthelist. Bytakingtheconjunctionoverallpairsofindexes,wecheckthatnotwodistinctpropositionsarebothtrue.Inother words,theexpressionsaysthatforeverypairofpropositions,atleastoneofthepairisfalse.Thatisequivalenttosaying

thatforeverypairofpropositions,atmostoneofthetwoistrue.Whichisequivalenttothestatementthatatmostone ofthepropositionsinthelististrue.

46.a) Sincetheconditionistrue,thestatementisexecuted,so x isincrementedandnowhasthevalue2.

b) Sincetheconditionisfalse,thestatementisnotexecuted,so x isnotincrementedandnowstillhasthevalue1.

c) Sincetheconditionistrue,thestatementisexecuted,so x isincrementedandnowhasthevalue2.

d) Sincetheconditionisfalse,thestatementisnotexecuted,so x isnotincrementedandnowstillhasthevalue1.

e) Sincetheconditionistruewhenitisencountered(since x = 1),thestatementisexecuted,so x isincrementedand nowhasthevalue2.(Itisirrelevantthattheconditionisnowfalse.)

48.a) 11000 ∧ (01011 ∨ 11011) = 11000 ∧ 11011 = 11000

b) (01111 ∧ 10101) ∨ 01000 = 00101 ∨ 01000 = 01101

c) (01010 ⊕ 11011) ⊕ 01000 = 10001 ⊕ 01000 = 11001

d) (11011 ∨ 01010) ∧ (10001 ∨ 11011) = 11011 ∧ 11011 = 11011

50. Thetruthvalueof“FredandJohnarehappy”is min(0 8,0 4) = 0 4.Thetruthvalueof“NeitherFrednorJohnishappy” is min(0.2,0.6) = 0.2,sincethisstatementmeans“Fredisnothappy,andJohnisnothappy,”andwecomputedthetruth valuesofthetwopropositionsinthisconjunctioninExercise49.

52. Thiscannotbeaproposition,becauseitcannothaveatruthvalue.Indeed,ifitweretrue,thenitwouldbetrulyasserting thatitisfalse,acontradiction;ontheotherhandifitwerefalse,thenitsassertionthatitisfalsemustbefalse,so thatitwouldbetrue—againacontradiction.Thusthisstringofletters,whileappearingtobeaproposition,isinfact meaningless.

54. No.Thisisaclassicalparadox.(Wewillusethemalepronouninwhatfollows,assumingthatwearetalkingaboutmales shavingtheirbeardshere,andassumingthatallmenhavefacialhair.Ifwerestrictourselvestobeardsandallowfemale barbers,thenthebarbercouldbefemalewithnocontradiction.)Ifsuchabarberexisted,whowouldshavethebarber? Ifthebarbershavedhimself,thenhewouldbeviolatingtherulethatheshavesonlythosepeoplewhodonotshave themselves.Ontheotherhand,ifhedoesnotshavehimself,thentherulesaysthathemustshavehimself.Neitheris possible,sotherecanbenosuchbarber.

SECTION1.2ApplicationsofPropositionalLogic

2. Recallthat p onlyif q means p → q .Inthiscase,ifyoucanseethemoviethenyoumusthavefulﬁlledoneofthetwo requirements.Thereforethestatementis m → (e ∨ p).Noticethatineverydaylifeonemightactuallysay“Youcansee themovie if youmeetoneoftheseconditions,”butlogicallythatisnotwhattherulesreallysay.

4. Theconditionstatedhereisthatifyouusethenetwork,theneitheryoupaythefeeoryouareasubscriber.Thereforethe propositioninsymbolsis w → (d ∨ s).

6. ThisissimilartoExercise2:

8.a) “But”means“and”: r ∧¬p

b) “Whenever”means“if”:(r ∧ p) → q

c) Accessbeingdeniedisthenegationof q ,sowehave ¬r → ¬q .

d) Thehypothesisisaconjunction:(¬p ∧ r) → q

10. Wewritethesesymbolically: u → ¬a , a → s , ¬s → ¬u .Notethatwecanmakealltheconclusiontruebymaking a false, s true,and u false.Thereforeiftheuserscannotaccesstheﬁlesystem,theycansavenewﬁles,andthesystemis notbeingupgraded,thenalltheconditionalstatementsaretrue.Thusthesystemisconsistent.

12. Thissystemisconsistent.Weuse L , Q , N ,and B tostandforthebasicpropositionshere,“Thefilesystemislocked,” “Newmessageswillbequeued,”“Thesystemisfunctioningnormally,”and“Newmessageswillbesenttothemessage buffer,”respectively.Thenthegivenspecificationsare ¬L → Q , ¬L ↔ N , ¬Q → B , ¬L → B ,and ¬B .Ifwewant consistency,thenwehadbetterhave B falseinorderthat ¬B betrue.Thisrequiresthatboth L and Q betrue,bythetwo conditionalstatementsthathave B astheirconsequence.ThefirstconditionalstatementthereforeisoftheformF → T, whichistrue.Finally,thebiconditional ¬L ↔ N canbesatisfiedbytaking N tobefalse.Thusthissetofspecifications isconsistent.Notethatthereisjustthisonesatisfyingtruthassignment.

14. ThisissimilartoExample6,aboutuniversitiesinNewMexico.TosearchforhikinginWestVirginia,wecouldenter WEST AND VIRGINIA AND HIKING.Ifweenter(VIRGINIA AND HIKING) NOT WEST,thenwe’llgetwebsites abouthikinginVirginiabutnotinWestVirginia,exceptforsitesthathappentousetheword“west”inadiﬀerentcontext (e.g.,“Followthestreamwestuntilyoucometoaclearing”).

16. ThisissimilartoExercise14,exceptthatproblemaskedaboutconductingageneralBooleansearch,whilethisspecificallyaskedaboutaGooglesearch.Thusweuseahyphenratherthantheword NOT,andwecanomittheword AND sinceGoogledefaultstorequiringthepresenceofallsearchterms.EnteringMEN’S(SHOES OR BOOTS)-WORKis aneﬀectiveandeﬃcientsearch.

18.a) Queencannotsaythis.Inparticular,theinscriptionsonTrunks1and2cannotbothbefalse,becauseifTrunk1’s inscriptionisfalse,thenitisnotempty,butifTrunk2’sinscriptionisfalse,thenthereisnotreasureinTrunk1.These arecontradictory.

b) Queencansaythis;treasuresareinTrunks1and3.Ifexactlyoneoftheinscriptionsistrue,itcannotbethatof Trunk3,bytheargumentinpart(a).TheinscriptiononTrunk1beingtrueisconsistentwithTrunk2bearingafalse statement.Sincetherearetreasuresintwoofthetrunks,thismeansTrunks2and3containthetreasures,whichmeans thattheinscriptiononTrunk3istruemakingtwotrueinscriptions,ratherthanone.Theﬁnalpossibilityisthatthe inscriptiononTrunk2istrue,inwhichcaseTrunk1containsatreasuremakingitsclaimofbeingemptyfalse.Knowing thattheinscriptiononTrunk3isfalsethenmeansthatTrunk2doesnotcontainatreasure,andsotheothertreasureis inTrunk3.

c) Queencansaythis,butonecannotdeterminethelocationofthetreasures.IftheinscriptiononTrunk1isfalse,then thetreasuresareinTrunks1and2.IftheinscriptiononTrunk2isfalse,thenthetreasuresareinTrunks2and3.The inscriptiononTrunk3cannotbefalse,sincetheinscriptionsonTrunks1and2arecontradictory.

d) Queencannotsaythis.TheinscriptionsonTrunks1and2cannotbothbetrue.

20.a) Iftheexplorer(awoman,sothatourpronounswillnotgetconfusedhere—thecannibalswillbemale)encounters atruth-teller,thenhewillhonestlyanswer“no”toherquestion.Ifsheencountersaliar,thenthehonestanswertoher questionis“yes,”sohewilllieandanswer“no.”Thuseverybodywillanswer“no”tothequestion,andtheexplorerwill havenowaytodeterminewhichtypeofcannibalsheisspeakingto.

b) Thereareseveralpossiblecorrectanswers.Oneisthefollowingquestion:“IfIweretoaskyouifyoualwaystoldthe truth,wouldyousaythatyoudid?”Thenifthecannibalisatruthteller,hewillansweryes(truthfully),whileifheisa liar,then,sinceinfacthewouldhavesaidthathedidtellthetruthifquestioned,hewillnowlieandanswerno.

22. Wewilltranslatetheseconditionsintostatementsinsymboliclogic,using j , s ,and k forthepropositionsthatJasmine, Samir,andKantiattend,respectively.Theﬁrststatementis j → ¬s .Thesecondstatementis s → k .Thelaststatement

Chapter1TheFoundations:LogicandProofs is ¬k ∨ j ,because“unless”means“or.”(Wecouldalsotranslatethisas k → j .FromthecommentsfollowingDefinition5 inSection1.1,weknowthat p → q isequivalentto“ q unless ¬p .”Inthiscase p is ¬j and q is ¬k .)First,supposethat s istrue.Thenthesecondstatementtellsusthat k isalsotrue,andthenthelaststatementforces j tobetrue.Butnow thefirststatementforces s tobefalse.Soweconcludethat s mustbefalse;Samircannotattend.Ontheotherhand, if s isfalse,thenthefirsttwostatementsareautomaticallytrue,nomatterwhatthetruthvaluesof k and j are.Ifwe lookatthelaststatement,weseethatitwillbetrueaslongasitisnotthecasethat k istrueand j isfalse.Sotheonly combinationsoffriendsthatmakeeverybodyhappyareJasmineandKanti,orJasminealone(ornoone!).

24. If A isaknight,thenhisstatementthatbothofthemareknightsistrue,andbothwillbetellingthetruth.Butthatis impossible,because B isassertingotherwise(that A isaknave).If A isaknave,then B ’sassertionistrue,sohemust beaknight,and A ’sassertionisfalse,asitshouldbe.Thusweconcludethat A isaknaveand B isaknight.

26. Wecandrawnoconclusions.Aknightwilldeclarehimselftobeaknight,tellingthetruth.Aknavewilllieandassert thatheisaknight.Sinceeveryonewillsay“Iamaknight,”wecandeterminenothing.

28. Supposethat A istheknight.Thenbecausehetoldthetruth, C istheknaveandtherefore B isthespy.Inthiscaseboth B and C arelying,whichisconsistentwiththeiridentities.Toseethatthisistheonlysolution,ﬁrstnotethat B cannot betheknight,becauseofhisclaimthat A istheknight(whichwouldthenhavetobealie).Similarly, C cannotbethe knight,becausehewouldbelyingwhenstatingthatheisthespy.

30. Thereisnosolution,becauseneitheraknightnoraknavewouldeverclaimtobetheknave.

32. Supposethat A istheknight.Then B ’sstatementistrue,sohemustbethespy,whichmeansthat C ’sstatementisalso true,butthatisimpossiblebecause C wouldhavetobetheknave.Therefore A isnottheknight.Nextsupposethat B istheknight.Histruestatementforces A tobethespy,whichinturnforces C tobetheknave;oncemorethatis impossiblebecause C saidsomethingtrue.Theonlyotherpossibilityisthat C istheknight,whichthenforces B tobe thespyand A theknave.Thisworksoutﬁne,because A islyingand B istellingthetruth.

34. Neither A nor B canbetheknave,becausetheknavecannotmakethetruthfulstatementthatheisnotthespy.Therefore C istheknave,andconsequently A isnotthespy.Itfollowsthat A istheknightand B isthespy.Thisworksoutﬁne, because A and B arethenbothtellingthetruthand C islying.

36.a) Welookatthethreepossibilitiesofwhotheinnocentmenmightbe.IfSmithandJonesareinnocent(andtherefore tellingthetruth),thenwegetanimmediatecontradiction,sinceSmithsaidthatJoneswasafriendofCooper,but JonessaidthathedidnotevenknowCooper.IfJonesandWilliamsaretheinnocenttruth-tellers,thenweagaingeta contradiction,sinceJonessaysthathedidnotknowCooperandwasoutoftown,butWilliamssayshesawJoneswith Cooper(presumablyintown,andpresumablyifhewaswithhim,thenheknewhim).Thereforeitmustbethecasethat SmithandWilliamsaretellingthetruth.Theirstatementsdonotcontradicteachother.BasedonWilliams’statement, weknowthatJonesislying,sincehesaidthathedidnotknowCooperwheninfacthewaswithhim.ThereforeJones isthemurderer.

b) Thisisjustlikepart(a),exceptthatwearenottoldaheadoftimethatoneofthemenisguilty.Cannoneofthembe guilty?Ifso,thentheyarealltellingthetruth,butthisisimpossible,becauseaswejustsaw,someofthestatementsare contradictory.Canmorethanoneofthembeguilty?If,forexample,theyareallguilty,thentheirstatementsgiveusno information.Sothatiscertainlypossible.

38. Thisinformationisenoughtodeterminetheentiresystem.Leteachletterstandforthestatementthatthepersonwhose namebeginswiththatletterischatting.Thenthegiveninformationcanbeexpressedsymbolicallyasfollows: ¬K → H ,

.Notethatwewereabletoconvertallofthese statementsintoconditionalstatements.Inwhatfollowswewillsometimesmakeuseofthecontrapositivesofthese conditionalstatementsaswell.Firstsupposethat H istrue.Thenitfollowsthat A and K aretrue,whenceitfollows that R and V aretrue.But R impliesthat V isfalse,sowegetacontradiction.Therefore H mustbefalse.Fromthisit followsthat K istrue;whence V istrue,andtherefore R isfalse,asis A .Wecannowcheckthatthisassignmentleads toatruevalueforeachconditionalstatement.SoweconcludethatKevinandVijayarechattingbutHeather,Randy,and Abbyarenot.

40. NotethatDiana’sstatementismerelythatshedidn’tdoit.

a) Johndidit.Therearefourcasestoconsider.IfAliceisthesoletruth-teller,thenCarlosdidit;butthismeansthat Johnistellingthetruth,acontradiction.IfJohnisthesoletruth-teller,thenDianamustbelying,soshedidit,butthen Carlosistellingthetruth,acontradiction.IfCarlosisthesoletruth-teller,thenDianadidit,butthatmakesJohntruthful, againacontradiction.SotheonlypossibilityisthatDianaisthesoletruth-teller.ThismeansthatJohnislyingwhenhe deniedit,sohedidit.NotethatinthiscasebothAliceandCarlosareindeedlying.

b) Againtherearefourcasestoconsider.SinceCarlosandDianaaremakingcontradictorystatements,theliarmustbe oneofthem(wecouldhaveusedthisapproachinpart(a)aswell).ThereforeAliceistellingthetruth,soCarlosdidit. NotethatJohnandDianaaretellingthetruthaswellhere,anditisCarloswhoislying.

42. Thisisoftengivenasanexerciseinconstraintprogramming,anditisdiﬃculttosolvebyhand.Thefollowingtable showsasolutionconsistentwithalltheclues,withthehouseslistedfromlefttoright.Reportedlythesolutionisunique.

NATIONALITYNorwegianItalianEnglishmanSpaniardJapanese COLORYellowBlueRedWhiteGreen PETFoxHorseSnailDogZebra JOBDiplomatPhysicianPhotographerViolinistPainter DRINKWaterTeaMilkJuiceCoﬀee

InthissolutiontheJapanesemanownsthezebra,andtheNorwegiandrinkswater.Thelogicalreasoningneededtosolve theproblemisratherextensive,andthereaderisreferredtothefollowingwebsitecontainingthesolutiontoasimilar problem: http://web.archive.org/web/20180328154401/mathforum.org/library/drmath/view/ 55627.html

44.a) Eachof p and q isnegatedandfedtotheORgate.Thereforetheoutputis(¬p) ∨ (¬q). b) ¬(p ∨ ((¬p) ∧ q)))

46. Wehavetheinputscomeinfromtheleft,insomecasespassingthroughaninvertertoformtheirnegations.Certain pairsofthementerANDgates,andtheoutputsoftheseentertheﬁnalORgate. p r q r

2. Therearetwocases.If p istrue,then ¬(¬p)isthenegationofafalseproposition,hencetrue.Similarly,if p isfalse, then ¬(¬p)isalsofalse.Thereforethetwopropositionsarelogicallyequivalent.

4.a) Weconstructtherelevanttruthtableandnotethattheﬁfthandseventhcolumnsareidentical.

b) Againweconstructtherelevanttruthtableandnotethattheﬁfthandseventhcolumnsareidentical.

6. Weseethatthefourthandseventhcolumnsareidentical.

8. Weneedtonegateeachpartandswap“and”with“or.”

a) Kwamewillnottakeajobinindustryandwillnotgotograduateschool.

b) YoshikodoesnotknowJavaordoesnotknowcalculus.

c) Jamesisnotyoung,orheisnotstrong.

d) RitawillnotmovetoOregonandwillnotmovetoWashington.

10. Weapplytheequivalence p → q ≡ ¬p ∨ q totheconditionalsintheoriginalstatements.

a) ¬p → ¬q ≡ p ∨¬q

b) (p ∨ q) → ¬p ≡ ¬(p ∨ q) ∨¬p bytheconditional-disjunctionequivalence

≡ (¬p ∧¬q) ∨¬p bythesecondDeMorgan’slaw ≡ ¬p bytheﬁrstabsorptionlaw

c) (p → ¬q) → (¬p → q) ≡ ¬(p → ¬q) ∨ (¬p → q)bytheconditional-disjunctionequivalence

≡ ¬(¬p ∨¬q) ∨ (¬¬p ∨ q)bytheconditional-disjunctionequivalence

≡ (p ∧ q) ∨ (p ∨ q)bythedoublenegationandDeMorgan’slaws

≡ (p ∧ q) ∨ p ∨ q bytheassociativelaw

≡ p ∨ q bytheabsorbtionlaw c McGrawHillLLC.Allrightsreserved.NoreproductionordistributionwithoutthepriorwrittenconsentofMcGrawHillLLC.

12. WeconstructatruthtableforeachconditionalstatementandnotethattherelevantcolumncontainsonlyT’s.Forpart (a)wehavethefollowingtable.

Forpart(b)wehavethefollowingtable.

Forpart(c)wehavethefollowingtable.

Forpart(d)wehavethefollowingtable.Wehaveomittedsomeintermediatestepstomakethetableﬁt.

14. Theinstructionsaretousethefactthattheonlywayaconditionalstatementcanbefalseisforthehypothesistobetrue andtheconclusiontobefalse;henceitissuﬃcienttoshowthatitcannotbethatthehypothesisistrueandtheconclusion isfalse.

a) Ifthiswerenotatautology,then ¬p ∧ (p ∨ q)wouldbetruebut q wouldbefalse.Thiscannothappen,becausethe truthof ¬p ∧ (p ∨ q)forces p tobefalse,which,whencombinedwiththetruthof p ∨ q ,inturnforces q tobetrue,a contradiction.

b) Ifthiswerenotatautology,then(p → q) ∧ (q → r)wouldbetruebut p → r wouldbefalse.For p → r tobefalseit mustbethat p istrueand r isfalse.But p beingtrueimpliesbytheﬁrstexpressionthat q isalsotrueandhencethat r isalsotrue,acontradiction.

c) Ifthiswerenotatautology,then p ∧ (p → q)wouldbetruebut q wouldbefalse.Thiscannothappen,becausethe ﬁrstexpressionforces q tobetrue.

d) Ifthiswerenotatautology,then(p ∨ q) ∧ (p → r) ∧ (q → r)wouldbetruebut r wouldbefalse.Theﬁrstpartofthe hypothesisforcesatleastoneof p and q tobetrueandthesecondandthirdpartsthenforce r tobetrue,acontradiction.

16. Thesolutionsprovidedhereusetheconditional-disjunctionequivalenceastheinitialstep,butotherapproachescanbe equallyeﬀective.Relevantequivalencesarelistedateachstep,althoughcommutativityandassociativityarefrequently usedwithoutcomment.

a) [¬p ∧ (p ∨ q)] → q ≡ ¬[¬p ∧ (p ∨ q)] ∨ q bytheconditional-disjunctionequivalence

≡ p ∨¬(p ∨ q) ∨ q byaDeMorgan’slaw

≡ (p ∨ q) ∨¬(p ∨ q)bycommutativityandassociativity ≡ �� byanegationlaw

b) [(p → q) ∧ (q → r)] → (p → r)

≡ ¬[(p → q) ∧ (q → r)] ∨ (p → r)bytheconditional-disjunctionequivalence

≡ ¬(p → q) ∨¬(q → r) ∨ (p → r)byaDeMorgan’slaw

≡ (p ∧¬q) ∨ (q ∧¬r) ∨¬p ∨ r bythenegationofconditionalsand theconditional-disjunctionequivalences

≡ [¬p ∨ (p ∧¬q)] ∨ [r ∨ (q ∧¬r)]byassociativity

≡ [(¬p ∨ p) ∧ (¬p ∨¬q)] ∨ [(r ∨ q) ∧ (r ∨¬r)]byadistributivelaw

≡ [�� ∧ (¬p ∨¬q)] ∨ [(r ∨ q) ∧ ��]byanegationlaw

≡ (¬p ∨¬q) ∨ (r ∨ q)byanidentitylaw

≡ (¬p ∨ r) ∨ (¬q ∨ q)byassociativity

≡ (¬p ∨ r) ∨ �� byanegationlaw

≡ �� byadominationlaw

c) [p ∧ (p → q)] → q ≡ ¬[p ∧ (p → q)] ∨ q bytheconditional-disjunctionequivalence

≡ ¬p ∨¬(p → q) ∨ q byaDeMorgan’slaw

≡ (¬p ∨ q) ∨¬(p → q)bycommutativityandassociativity

≡ (p → q) ∨¬(p → q)bytheconditional-disjunctionequivalence

≡ �� byanegationlaw

d) [(p ∨ q) ∧ (p → r) ∧ (q → r)] → r

≡ ¬[(p ∨ q) ∧ (p → r) ∧ (q → r)] ∨ r bytheconditional-disjunctionequivalence

≡ ¬(p ∨ q) ∨¬(p → r) ∨¬(q → r) ∨ r byaDeMorgan’slaw

≡ ¬(p ∨ q) ∨ [(p ∧¬r) ∨ r] ∨ [(q ∧¬r) ∨ r]byanidempotentlaw, commutativity,andassociativity

≡ ¬(p ∨ q) ∨ [(p ∨ r) ∧ (¬r ∨ r)] ∨ [(q ∨ r) ∧ (¬r ∨ r)]byadistributivelaw

≡ ¬(p ∨ q) ∨ [(p

]byanegationlaw

≡ ¬(p ∨ q) ∨ (p ∨ r) ∨ (q ∨ r)byanidentitylaw

≡ ¬(p ∨ q) ∨ (p ∨ q) ∨ r byassociativityandanidempotentlaw

≡ �� ∨ r byanegationlaw ≡ �� byadominationlaw

18. Thisisnotatautology.Itissayingthatknowingthatthehypothesisofanconditionalstatementisfalseallowsusto concludethattheconclusionisalsofalse,andweknowthatthisisnotvalidreasoning.Toshowthatitisnotatautology, weneedtoﬁndtruthassignmentsfor p and q thatmaketheentirepropositionfalse.Sincethisispossibleonlyifthe conclusioniffalse,wewanttolet q betrue;andsincewewantthehypothesistobetrue,wemustalsolet p befalse. c McGrawHillLLC.Allrightsreserved.NoreproductionordistributionwithoutthepriorwrittenconsentofMcGrawHillLLC.

Itiseasytocheckthatif,indeed, p isfalseand q istrue,thentheconditionalstatementisfalse.Thereforeitisnota tautology.

20. Theﬁrstofthesepropositionsistrueifandonlyif p and q havethesametruthvalue.Thesecondistrueifandonlyif either p and q arebothtrue,or p and q arebothfalse.Clearlythesetwoconditionsaresayingthesamething.

22. Itiseasytoseefromthedeﬁnitionsofconditionalstatementandnegationthateachofthesepropositionsisfalseinthe caseinwhich p istrueand q isfalse,andtrueintheotherthreecases.Thereforethetwopropositionsarelogically equivalent.

24. Itiseasytoseefromthedeﬁnitionsofthelogicaloperationsinvolvedherethateachofthesepropositionsistrueinthe casesinwhich p and q havethesametruthvalue,andfalseinthecasesinwhich p and q haveoppositetruthvalues. Thereforethetwopropositionsarelogicallyequivalent.

26. Supposethat(p → q) ∧ (p → r)istrue.Wewanttoshowthat p → (q ∧ r)istrue,whichmeansthatwewanttoshowthat q ∧ r istruewhenever p istrue.If p istrue,sinceweknowthatboth p → q and p → r aretruefromourassumption, wecanconcludethat q istrueandthat r istrue.Therefore q ∧ r istrue,asdesired.Conversely,supposethat p → (q ∧ r) istrue.Weneedtoshowthat p → q istrueandthat p → r istrue,whichmeansthatif p istrue,thensoare q and r Butthisfollowsfrom p → (q ∧ r).

28. WedetermineexactlywhichrowsofthetruthtablewillhaveTastheirentries.Note(p → q) ∨ (p → r)willbetruewhen eitheroftheconditionalstatementsistrue.Theconditionalstatementwillbetrueif p isfalse,orif q inonecaseor r intheothercaseistrue,i.e.,when q ∨ r istrue,whichispreciselywhen p → (q ∨ r)istrue.Sincethetwopropositions aretrueinexactlythesamesituations,theyarelogicallyequivalent.

30. ApplyingthethirdandﬁrstequivalencesinTable7,wehave ¬p → (q

r .Applying theﬁrstequivalenceinTable7to q → (p ∨ r)showsthat ¬q ∨ p ∨ r isequivalenttoit.Buttheseareequivalentbythe commutativeandassociativelaws.

32. Weknowthat p ↔ q istruepreciselywhen p and q havethesametruthvalue.Butthishappenspreciselywhen ¬p and ¬q havethesametruthvalue,thatis, ¬p ↔ ¬q .

34. Theconclusion q ∨ r willbetrueineverycaseexceptwhen q and r arebothfalse.Butif q and r arebothfalse,then oneof p ∨ q or ¬p ∨ r isfalse,becauseoneof p or ¬p isfalse.Thusinthiscasethehypothesis(p ∨ q) ∧ (¬p ∨ r)is false.Anconditionalstatementinwhichtheconclusionistrueorthehypothesisisfalseistrue,andthatcompletesthe argument.

36. Wejustneedtoﬁndanassignmentoftruthvaluesthatmakesoneofthesepropositionstrueandtheotherfalse.Wecan let p betrueandtheothertwovariablesbefalse.Thentheﬁrststatementwillbe �� → �� ,whichistrue,butthesecond willbe �� ∧ �� ,whichisfalse.

38. Weapplytherulesstatedinthepreamble.

40. If s hasanyoccurrencesof ∧ , ∨ , �� ,or �� ,thentheprocessofformingthedualwillchangeit.Therefore s∗ = s ifand onlyif s issimplyonepropositionalvariable(like p ).Amorediﬃcultquestionistodeterminewhen s∗ willbelogically equivalentto s .Forexample, p ∨ �� islogicallyequivalenttoitsdual p ∧ �� ,becausebotharelogicallyequivalentto p .

42. Thetableisinfactdisplayedsoastoexhibittheduality.Thetwoidentitylawsaredualsofeachother,thetwodomination lawsaredualsofeachother,etc.Theonlylawnotlistedwithanother,thedoublenegationlaw,isitsowndual,since therearenooccurrencesof ∧ , ∨ , T,or F toreplace.

44. Followingthehint,weeasilyseethattheansweris p ∧ q ∧¬r .

46. Thestatementoftheproblemisreallythesolution.Eachlineofthetruthtablecorrespondstoexactlyonecombination oftruthvaluesforthe n atomicpropositionsinvolved.Wecanwritedownaconjunctionthatistruepreciselyinthiscase, namelytheconjunctionofalltheatomicpropositionsthataretrueandthenegationsofalltheatomicpropositionsthat arefalse.Ifwedothisfor each lineofthetruthtableforwhichthevalueofthecompoundpropositionistobetrue,and takethedisjunctionoftheresultingpropositions,thenwehavethedesiredpropositioninitsdisjunctivenormalform.

48. Givenacompoundproposition p ,wecan,byExercise47,writedownaproposition q thatislogicallyequivalentto p andusesonly ¬ , ∧ ,and ∨ .NowbyDeMorgan’slawwecangetridofallthe ∨’sbyreplacingeachoccurrenceof p1 ∨ p2 ∨ ∨ pn with ¬(¬p1 ∧¬p2 ∧ ∧¬pn ).

50. Wewritedownthetruthtablecorrespondingtothedeﬁnition.

52. Wewritedownthetruthtablecorrespondingtothedeﬁnition. pqp ↓ q

54.a) Fromthedeﬁnition(orasseeninthetruthtableconstructedinExercise52), p ↓ p isfalsewhen p istrueandtrue when p isfalse,exactlyas ¬p is;thusthetwoarelogicallyequivalent.

b) Theproposition(p ↓ q) ↓ (p ↓ q)isequivalent,bypart(a),to ¬(p ↓ q),whichfromthedeﬁnition(ortruthtableor Exercise53)isequivalentto p ∨ q .

c) ByExercise49,everycompoundpropositionislogicallyequivalenttoonethatusesonly ¬ and ∨ .Butbyparts (a)and(b)ofthepresentexercise,wecangetridofallthenegationsanddisjunctionsbyusing NOR’s.Thusevery compoundpropositioncanbeconvertedintoalogicallyequivalentcompoundpropositioninvolvingonly NOR’s.

56. ThisexerciseissimilartoExercise54.Firstwecanseefromthetruthtablesthat(p ∣ p) ≡ (¬p)andthat((p ∣ p) ∣ (q ∣ q)) ≡ (p ∨ q).Thenweargueexactlyasinpart(c)ofExercise54:byExercise49,everycompoundpropositionis logicallyequivalenttoonethatusesonly ¬ and ∨ .Butbyourobservationsatthebeginningofthissolution,wecanget ridofallthenegationsanddisjunctionsbyusing NAND’s.Thuseverycompoundpropositioncanbeconvertedintoa logicallyequivalentcompoundpropositioninvolvingonly NAND’s.

58. Toshowthattheseare not logicallyequivalent,weneedonlyfindoneassignmentoftruthvaluesto p , q ,and r forwhich thetruthvaluesof p ∣ (q ∣ r)and(p ∣ q) ∣ r differ.OnesuchassignmentisTfor p andFfor q and r .Thencomputing fromthetruthtables(ordefinitions),weseethat p ∣ (q ∣ r)isfalseand(p ∣ q) ∣ r istrue.

60. Tosaythat p and q arelogicallyequivalentistosaythatthetruthtablesfor p and q areidentical;similarly,tosaythat q and r arelogicallyequivalentistosaythatthetruthtablesfor q and r areidentical.Clearlyifthetruthtablesfor p and q areidentical,andthetruthtablesfor q and r areidentical,thenthetruthtablesfor p and r areidentical(thisisa

fundamentalaxiomofthenotionofequality).Therefore p and r arelogicallyequivalent.(Weareassuming—andthere isnolossofgeneralityindoingso—thatthesameatomicvariablesappearinallthreepropositions.)

62. Ifwewanttheﬁrsttwoofthesetobetrue,then p and q musthavethesametruthvalue.If q istrue,thenthethirdand fourthexpressionswillbetrue,andif r isfalse,thelastexpressionwillbetrue.Soallﬁveofthesedisjunctionswillbe trueifweset p and q tobetrue,and r tobefalse.

64. Thesefollowdirectlyfromthedefinitions.Anunsatisfiablecompoundpropositionisonethatistruefornoassignment oftruthvaluestoitsvariables,whichisthesameassayingthatitisfalseforeveryassignmentoftruthvalues,whichis thesamesamesayingthatitsnegationistrueforeveryassignmentoftruthvalues.Thatisthedefinitionofatautology. Conversely,thenegationofatautology(i.e.,apropositionthatistrueforeveryassignmentoftruthvaluestoitsvariables) willbefalseforeveryassignmentoftruthvalues,andthereforewillbeunsatisfiable.

66. Ineachcasewehuntfortruthassignmentsthatmakeallthedisjunctionstrue.

a) Since p occursinfouroftheﬁvedisjunctions,wecanmake p true,andthenmake q false(andmake r and s anything weplease).Thusthispropositionissatisﬁable.

b) Thisissatisfiableby,forexample,setting p tobefalse(thattakescareofthefirst,second,andfourthdisjunctions), s tobefalse(forthethirdandsixthdisjunctions), q tobetrue(forthefifthdisjunction),and r tobeanything.

c) Itisnothardtoﬁndasatisfyingtruthassignment,suchas p , q ,and s true,and r false.

68. Addasixthclause, Q6 = ⋀⌊n∕2⌋ j=1 (¬p(1,2j)),whichassertsthatthereisnotaqueenintheevennumberedrowsofthe ﬁrstcolumn.

70. Recallthat p(i, j, n)assertsthatthecellinrow i ,column j containsthenumber n .Thus ⋁9 n=1 p(i, j, n)assertsthatthis cellcontainsatleastonenumber.Toassertthateverycellcontainsatleastonenumber,wetaketheconjunctionofthese statementsoverallcells: ⋀9 i=1 ⋀9 j=1 ⋁9 n=1 p(i, j, n).

72. Therearenineblocks,inthreerowsandthreecolumns.Let r and s indextherowandcolumnoftheblock,respectively, wherewestartcountingat0,sothat0 ≤ r ≤ 2and0 ≤ s ≤ 2.(Forexample, r = 0, s = 1correspondstotheblockin theﬁrstrowofblocksandsecondcolumnofblocks.)Thekeypointistonoticethattheblockcorrespondingtothepair (r, s)containsthecellsthatareinrows3r + 1,3r + 2,and3r + 3andcolumns3s + 1,3s + 2,and3s + 3.Therefore p(3r + i,3s + j, n)assertsthataparticularcellinthisblockcontainsthenumber n ,where1 ≤ i ≤ 3and1 ≤ j ≤ 3.If wetakethedisjunctionoverallthesevaluesof i and j ,thenweobtain ⋁3 i=1 ⋁3 j=1 p(3r + i,3s + j, n),assertingthatsome cellinthisblockcontainsthenumber n .Becausewewantthistobetrueforeverynumberandforeveryblock,weform thetriply-indexedconjunctiongiveninthetext.

SECTION1.4PredicatesandQuantiﬁers

2.a) Thisistrue,sincethereisan a in orange b) Thisisfalse,sincethereisno a in lemon c) Thisisfalse,sincethereisno a in true d) Thisistrue,sincethereisan a in false

4.a) Here x isstillequalto0,sincetheconditionisfalse.

b) Here x isstillequalto1,sincetheconditionisfalse.

c) Thistime x isequalto1attheend,sincetheconditionistrue,sothestatement x := 1isexecuted.

6. Theanswersgivenherearenotunique,butcaremustbetakennottoconfusenonequivalentsentences.Parts(c)and(f) areequivalent;andparts(d)and(e)areequivalent.Butthesetwopairsarenotequivalenttoeachother.

a) SomestudentintheschoolhasvisitedNorthDakota.(Alternatively,thereexistsastudentintheschoolwhohasvisited NorthDakota.)

b) EverystudentintheschoolhasvisitedNorthDakota.(Alternatively,allstudentsintheschoolhavevisitedNorth Dakota.)

c) Thisisthenegationofpart(a):NostudentintheschoolhasvisitedNorthDakota.(Alternatively,theredoesnotexist astudentintheschoolwhohasvisitedNorthDakota.)

d) SomestudentintheschoolhasnotvisitedNorthDakota.(Alternatively,thereexistsastudentintheschoolwhohas notvisitedNorthDakota.)

e) Thisisthenegationofpart(b):ItisnottruethateverystudentintheschoolhasvisitedNorthDakota.(Alternatively, notallstudentsintheschoolhavevisitedNorthDakota.)

f) AllstudentsintheschoolhavenotvisitedNorthDakota.(Thisistechnicallythecorrectanswer,althoughcommon Englishusagetakesthissentencetomean—incorrectly—theanswertopart(e).Tobeperfectlyclear,onecouldsaythat everystudentinthisschoolhasfailedtovisitNorthDakota,orsimplythatnostudenthasvisitedNorthDakota.)

8. Notethatpart(b)andpart(c)arenotthesortsofthingsonewouldnormallysay.

a) Ifananimalisarabbit,thenthatanimalhops.(Alternatively,everyrabbithops.)

b) Everyanimalisarabbitandhops.

c) Thereexistsananimalsuchthatifitisarabbit,thenithops.(Notethatthisistriviallytrue,satisﬁed,forexample,by lions,soitisnotthesortofthingonewouldsay.)

d) Thereexistsananimalthatisarabbitandhops.(Alternatively,somerabbitshop.Alternatively,somehoppinganimals arerabbits.)

10.a) Weassumethatthismeansthatonestudenthasallthreeanimals: ∃x(C(x) ∧ D(x) ∧ F (x)).

b) ∀x(C(x) ∨ D(x) ∨ F (x)) c) ∃x(C(x) ∧ F (x) ∧¬D(x))

d) Thisisthenegationofpart(a): ¬∃x(C(x) ∧ D(x) ∧ F (x)).

e) Heretheownersofthesepetscanbediﬀerent:(∃xC(x)) ∧ (∃xD(x)) ∧ (∃xF (x)).Thereisnoharminusingthesame dummyvariable,butthiscouldalsobewritten,forexample,as(∃xC(x)) ∧ (∃yD(y)) ∧ (∃zF (z)).

12.a) Since0 + 1 > 2 0,weknowthat Q(0)istrue.

b) Since( 1) + 1 > 2 ⋅ ( 1),weknowthat Q( 1)istrue.

c) Since1 + 1 ≯ 2 1,weknowthat Q(1)isfalse.

d) Frompart(a)weknowthatthereisatleastone x thatmakes Q(x)true,so ∃xQ(x)istrue.

e) Frompart(c)weknowthatthereisatleastone x thatmakes Q(x)false,so ∀xQ(x)isfalse.

f) Frompart(c)weknowthatthereisatleastone x thatmakes Q(x)false,so ∃x ¬Q(x)istrue.

g) Frompart(a)weknowthatthereisatleastone x thatmakes Q(x)true,so ∀x ¬Q(x)isfalse.

14.a) Since( 1)3 =−1,thisistrue.

b) Since( 1 2 )4 < ( 1 2 )2 ,thisistrue.

c) Since( x)2 = (( 1)x)2 = ( 1)2 x2 = x2 ,weknowthat ∀x(( x)2 = x2 )istrue.

d) Twiceapositivenumberislargerthanthenumber,butthisinequalityisnottruefornegativenumbersor0.Therefore ∀x(2x > x)isfalse.

16.a) True( x = √2) b) False( √ 1isnotarealnumber)

c) True(theleft-handsideisalwaysatleast2) d) False(nottruefor x = 1or x = 0)

18. Existentialquantiﬁersarelikedisjunctions,anduniversalquantiﬁersarelikeconjunctions.SeeExamples15and16.

a) Wewanttoassertthat P(x)istrueforsome x inthedomain,soeither P( 2)istrueor P( 1)istrueor P(0)istrue or P(1)istrueor P(2)istrue.Thustheansweris P( 2) ∨ P( 1) ∨ P(0) ∨ P(1) ∨ P(2).Theotherpartsofthisexercise aresimilar.NotethatbyDeMorgan’slaws,theexpressioninpart(c)islogicallyequivalenttotheexpressioninpart(f), andtheexpressioninpart(d)islogicallyequivalenttotheexpressioninpart(e).

b) P( 2) ∧ P( 1) ∧ P(0) ∧ P(1) ∧ P(2)

c) ¬P( 2) ∨¬P( 1) ∨¬P(0) ∨¬P(1) ∨¬P(2)

d) ¬P( 2) ∧¬P( 1) ∧¬P(0) ∧¬P(1) ∧¬P(2)

e) Thisisjustthenegationofpart(a): ¬(P( 2) ∨ P( 1) ∨ P(0) ∨ P(1) ∨ P(2))

f) Thisisjustthenegationofpart(b): ¬(P( 2) ∧ P( 1) ∧ P(0) ∧ P(1) ∧ P(2))

20. Existentialquantiﬁersarelikedisjunctions,anduniversalquantiﬁersarelikeconjunctions.SeeExamples15and16.

a) Wewanttoassertthat P(x)istrueforsome x inthedomain,soeither P( 5)istrueor P( 3)istrueor P( 1)istrue or P(1)istrueor P(3)istrueor P(5)istrue.Thustheansweris P( 5) ∨ P( 3) ∨ P( 1) ∨ P(1) ∨ P(3) ∨ P(5).

b) P( 5) ∧ P( 3) ∧ P( 1) ∧ P(1) ∧ P(3) ∧ P(5)

c) Theformaltranslationisasfollows:(( 5 ≠ 1) → P( 5)) ∧ (( 3 ≠ 1) → P( 3)) ∧ (( 1 ≠ 1) → P( 1)) ∧ ((1 ≠ 1) → P(1)) ∧ ((3 ≠ 1) → P(3)) ∧ ((5 ≠ 1) → P(5)).However,sincethehypothesis x ≠ 1isfalsewhen x is1andtruewhen x isanythingotherthan1,wehavemoresimply P( 5) ∧ P( 3) ∧ P( 1) ∧ P(3) ∧ P(5).

d) Theformaltranslationisasfollows:(( 5 ≥ 0) ∧ P( 5)) ∨ (( 3 ≥ 0) ∧ P( 3)) ∨ (( 1

( 1)) ∨ ((1

∧ P(1)) ∨ ((3 ≥ 0) ∧ P(3)) ∨ ((5 ≥ 0) ∧ P(5)).Sinceonlythreeofthe x’sinthedomainmeetthecondition,theansweris equivalentto P(1) ∨ P(3) ∨ P(5).

e) Forthesecondpartweagainrestrictthedomain:(¬P( 5) ∨¬P( 3) ∨¬P( 1) ∨¬

P( 3) ∧ P( 5)).Thisisequivalentto(¬P(1) ∨¬P(3)

22. Manyanswerarepossibleineachcase.

a) AdomainconsistingofafewadultsincertainpartsofIndiawouldmakethistrue.Ifthedomainwereallresidentsof theUnitedStates,thenthisiscertainlyfalse.

b) IfthedomainisallresidentsoftheUnitedStates,thenthisistrue.Ifthedomainisthesetofpupilsinaﬁrstgrade class,itisfalse.

c) IfthedomainconsistsofalltheUnitedStatesPresidentswhoselastnameisBush,thenthestatementistrue.Ifthe domainconsistsofallUnitedStatesPresidents,thenthestatementisfalse.

d) IfthedomainwereallresidentsoftheUnitedStates,thenthisiscertainlytrue.Ifthedomainconsistsofallbabies borninthelastﬁveminutes,onewouldexpectthestatementtobefalse(it’snotevenclearthatthesebabies“know”their mothersyet).

24. Inordertodothetranslationthesecondway,welet C(x)bethepropositionalfunction“ x isinyourclass.”Notethatfor thesecondway,wealwayswanttouseconditionalstatementswithuniversalquantiﬁersandconjunctionswithexistential quantiﬁers.

a) Let P(x)be“ x hasacellularphone.”Thenwehave ∀xP(x)theﬁrstway,or ∀x(C(x) → P(x))thesecondway.

b) Let F (x)be“ x hasseenaforeignmovie.”Thenwehave ∃xF (x)theﬁrstway,or ∃x(C(x) ∧ F (x))thesecondway.

c) Let S(x)be“ x canswim.”Thenwehave ∃x ¬S(x)theﬁrstway,or ∃x(C(x) ∧¬S(x))thesecondway.

d) Let Q(x)be“ x cansolvequadraticequations.”Thenwehave ∀xQ(x)theﬁrstway,or ∀x(C(x) → Q(x))the secondway.

e) Let R(x)be“ x wantstoberich.”Thenwehave ∃x ¬R(x)theﬁrstway,or ∃x(C(x) ∧¬R(x))thesecondway.

26. Inallofthese,wewilllet Y (x)bethepropositionalfunctionthat x isinyourschoolorclass,asappropriate.

a) Ifwelet U (x)be“ x hasvisitedUzbekistan,”thenwehave ∃xU (x)ifthedomainisjustyourschoolmates,or ∃x(Y (x) ∧ U (x))ifthedomainisallpeople.Ifwelet V (x, y)meanthatperson x hasvisitedcountry y ,thenwecanrewritethis lastoneas ∃x(Y (x) ∧ V (x,Uzbekistan)).

b) Ifwelet C(x)and P(x)bethepropositionalfunctionsassertingthat x hasstudiedcalculusandC++,respectively, thenwehave ∀x(C(x) ∧ P(x))ifthedomainisjustyourschoolmates,or ∀x(Y (x) → (C(x) ∧ P(x)))ifthedomainis allpeople.Ifwelet S(x, y)meanthatperson x hasstudiedsubject y ,thenwecanrewritethislastoneas ∀x(Y (x) → (S(x,calculus) ∧ S(x,C++))).

c) Ifwelet B(x)and M (x)bethepropositionalfunctionsassertingthat x ownsabicycleandamotorcycle,respectively, thenwehave ∀x(¬(B(x) ∧ M (x)))ifthedomainisjustyourschoolmates,or ∀x(Y (x) → ¬(B(x) ∧ M (x)))ifthedomainis allpeople.Notethat“noone”became“forall...not.”Ifwelet O(x, y)meanthatperson x ownsitem y ,thenwecan rewritethislastoneas ∀x(Y (x) → ¬(O(x,bicycle) ∧ O(x,motorcycle))).

d) Ifwelet H (x)be“ x ishappy,”thenwehave ∃x ¬H (x)ifthedomainisjustyourschoolmates,or ∃x(Y (x) ∧¬H (x)) ifthedomainisallpeople.Ifwelet E(x, y)meanthatperson x isinmentalstate y ,thenwecanrewritethislastoneas ∃x(Y (x) ∧¬E(x,happy)).

e) Ifwelet T (x)be“ x wasborninthetwentiethcentury,”thenwehave ∀xT (x)ifthedomainisjustyourschoolmates, or ∀x(Y (x) → T (x))ifthedomainisallpeople.Ifwelet B(x, y)meanthatperson x wasborninthe yth century,then wecanrewritethislastoneas ∀x(Y (x) → B(x,20)).

28. Let R(x)be“ x isinthecorrectplace,”let E(x)be“ x isinexcellentcondition,”let T (x)be“ x isa[oryour]tool,”and letthedomainofdiscoursebeallthings.

a) Thereexistssomethingnotinthecorrectplace: ∃x ¬R(x).

b) Ifsomethingisatool,thenitisinthecorrectplaceplaceandinexcellentcondition: ∀x (T (x) → (R(x) ∧ E(x))).

c) ∀x (R(x) ∧ E(x))

d) Thisissayingthateverythingfailstosatisfythecondition: ∀x ¬(R(x) ∧ E(x)).

e) Thereexistsatoolwiththisproperty: ∃x (T (x) ∧¬R(x) ∧ E(x)).

32. Ineachcaseweneedtospecifysomepropositionalfunctions(predicates)andidentifythedomainofdiscourse.

a) Let F (x)be“ x hasﬂeas,”andletthedomainofdiscoursebedogs.Ouroriginalstatementis ∀xF (x).Itsnegationis ∃x ¬F (x).InEnglishthisreads“Thereisadogthatdoesnothaveﬂeas.”

b) Let H (x)be“ x canadd,”wherethedomainofdiscourseishorses.Thenouroriginalstatementis ∃xH (x).Itsnegation is ∀x ¬H (x).InEnglishthisisrenderedmostsimplyas“Nohorsecanadd.”

c) Let C(x)be“ x canclimb,”andletthedomainofdiscoursebekoalas.Ouroriginalstatementis ∀xC(x).Itsnegation is ∃x ¬C(x).InEnglishthisreads“Thereisakoalathatcannotclimb.”

d) Let F (x)be“ x canspeakFrench,”andletthedomainofdiscoursebemonkeys.Ouroriginalstatementis ¬∃xF (x) or ∀x ¬F (x).Itsnegationis ∃xF (x).InEnglishthisreads“ThereisamonkeythatcanspeakFrench.”

e) Let S(x)be“ x canswim”andlet C(x)be“ x cancatchfish,”wherethedomainofdiscourseispigs.Thenouroriginal statementis ∃x (S(x) ∧ C(x)).Itsnegationis ∀x ¬(S(x) ∧ C(x)),whichcouldalsobewritten ∀x (¬S(x) ∨¬C(x))byDe Morgan’slaw.InEnglishthisis“Nopigcanbothswimandcatchfish,”or“Everypigeitherisunabletoswimoris unabletocatchfish.”

34.a) Let S(x)be“ x obeysthespeedlimit,”wherethedomainofdiscourseisdrivers.Theoriginalstatementis ∃x ¬S(x), thenegationis ∀xS(x),“Alldriversobeythespeedlimit.”

b) Let S(x)be“ x isserious,”wherethedomainofdiscourseisSwedishmovies.Theoriginalstatementis ∀xS(x),the negationis ∃x ¬S(x),“SomeSwedishmoviesarenotserious.”

c) Let S(x)be“ x cankeepasecret,”wherethedomainofdiscourseispeople.Theoriginalstatementis ¬∃xS(x),the negationis ∃xS(x),“Somepeoplecankeepasecret.”

d) Let A(x)be“ x hasagoodattitude,”wherethedomainofdiscourseispeopleinthisclass.Theoriginalstatementis ∃x ¬A(x),thenegationis ∀xA(x),“Everyoneinthisclasshasagoodattitude.”

36.a) ∃x((x ≤ 2) ∨ (x ≥ 3)) b) ∃x((x < 0) ∨ (x ≥ 5)) c) ∀x((x < 4) ∨ (x > 1))

d) ∀x((x ≤ 5) ∨ (x ≥ 1))

38.a) Since12 = 1,thisstatementisfalse; x = 1isacounterexample.Sois x = 0(thesearetheonlytwocounterexamples).

b) Therearetwocounterexamples: x = √2and x =−√2.

c) Thereisonecounterexample: x = 0.

40.a) Somesystemisopen. b) Everysystemiseithermalfunctioningorinadiagnosticstate.

c) Somesystemisopen,orsomesystemisinadiagnosticstate. d) Somesystemisunavailable.

e) Nosystemisworking.(Wecouldalsosay“Everysystemisnotworking,”aslongasweunderstoodthatthisis diﬀerentfrom“Noteverysystemisworking.”)

42. Therearemanywaystowritethese,dependingonwhatweuseforpredicates.

a) Let F (x)be“Thereislessthan x megabytesfreeontheharddisk,”withthedomainofdiscoursebeingpositive numbers,andlet W (x)be“User x issentawarningmessage.”Thenwehave F (30) → ∀xW (x).

b) Let O(x)be“Directory x canbeopened,”let C(x)be“File x canbeclosed,”andlet E betheproposition“System errorshavebeendetected.”Thenwehave E → ((∀x ¬O(x)) ∧ (∀x ¬C(x))).

c) Let B betheproposition“Theﬁlesystemcanbebackedup,”andlet L(x)be“User x iscurrentlyloggedon.”Then wehave(∃xL(x)) → ¬B

d) Let D(x)be“Product x canbedelivered,”andlet M (x)be“Thereareatleast x megabytesofmemoryavailable” and S(x)be“Theconnectionspeedisatleast x kilobitspersecond,”wherethedomainofdiscourseforthelasttwo propositionalfunctionsarepositivenumbers.Thenwehave(M (8) ∧ S(56)) → D(videoondemand).

44. Therearemanywaystowritethese,dependingonwhatweuseforpredicates.

a) Let A(x)be“User x hasaccesstoanelectronicmailbox.”Thenwehave ∀xA(x).

b) Let A(x, y)be“Groupmember x canaccessresource y ,”andlet S(x, y)be“System x isinstate y .”Thenwehave S(ﬁlesystem,locked) → ∀xA(x,systemmailbox).

c) Let S(x, y)be“System x isinstate y .”Recallingthat“onlyif”indicatesanecessarycondition,wehave S(ﬁrewall, diagnostic) → S(proxyserver,diagnostic).

d) Let T (x)be“Thethroughputisatleast x kbps,”wherethedomainofdiscourseispositivenumbers,let M (x, y) be“Resource x isinmode y ,”andlet S(x, y)be“Router x isinstate y .”Thenwehave(T (100) ∧¬T (500) ∧¬M (proxyserver,diagnostic)) → ∃xS(x,normal).

46. Wewantpropositionalfunctions P and Q thataresometimes,butnotalways,true(sothatthesecondbiconditionalis �� ↔ �� andhencetrue),butsuchthatthereisan x makingonetrueandtheotherfalse.Forexample,wecantake P(x) tomeanthat x isanevennumber(amultipleof2)and Q(x)tomeanthat x isamultipleof3.Thenanexamplelike x = 4or x = 9showsthat ∀x(P(x) ↔ Q(x))isfalse.

48.a) Therearetwocases.If A istrue,then(∀xP(x)) ∨ A istrue,andsince P(x) ∨ A istrueforall x , ∀x(P(x) ∨ A)isalso true.Thusbothsidesofthelogicalequivalencearetrue(henceequivalent).Nowsupposethat A isfalse.If P(x)istrue forall x ,thentheleft-handsideistrue.Furthermore,theright-handsideisalsotrue(since P(x) ∨ A istrueforall x ). Ontheotherhand,if P(x)isfalseforsome x ,thenbothsidesarefalse.Thereforeagainthetwosidesarelogically equivalent.

b) Therearetwocases.If A istrue,then(∃xP(x)) ∨ A istrue,andsince P(x) ∨ A istrueforsome(reallyall) x , ∃x(P(x) ∨ A)isalsotrue.Thusbothsidesofthelogicalequivalencearetrue(henceequivalent).Nowsupposethat A is false.If P(x)istrueforatleastone x ,thentheleft-handsideistrue.Furthermore,theright-handsideisalsotrue(since P(x) ∨ A istrueforthat x ).Ontheotherhand,if P(x)isfalseforall x ,thenbothsidesarefalse.Thereforeagainthe twosidesarelogicallyequivalent.

50.a) Therearetwocases.If A isfalse,thenbothsidesoftheequivalencearetrue,becauseaconditionalstatementwitha falsehypothesisistrue.If A istrue,then A → P(x)isequivalentto P(x)foreach x ,sotheleft-handsideisequivalent to ∀xP(x),whichisequivalenttotheright-handside.

b) Therearetwocases.If A isfalse,thenbothsidesoftheequivalencearetrue,becauseaconditionalstatementwitha falsehypothesisistrue(andweareassumingthatthedomainisnonempty).If A istrue,then A → P(x)isequivalentto P(x)foreach x ,sotheleft-handsideisequivalentto ∃xP(x),whichisequivalenttotheright-handside.

52. Itisenoughtofindacounterexample.Itisintuitivelyclearthatthefirstpropositionisassertingmuchmorethanthe second.Itissayingthatoneofthetwopredicates, P or Q ,isuniversallytrue;whereasthesecondpropositionissimply sayingthatforevery x either P(x)or Q(x)holds,butwhichitismaywelldependon x .Asasimplecounterexample, let P(x)bethestatementthat x isodd,andlet Q(x)bethestatementthat x iseven.Letthedomainofdiscoursebe thepositiveintegers.Thesecondpropositionistrue,sinceeverypositiveintegeriseitheroddoreven.Butthefirst propositionisfalse,sinceitisneitherthecasethatallpositiveintegersareoddnorthecasethatallofthemareeven.

54.a) Thisisfalse,sincetherearemanyvaluesof x thatmake x > 1true.

b) Thisisfalse,sincetherearetwovaluesof x thatmake x2 = 1true.

c) Thisistrue,sincebyalgebraweseethattheuniquesolutiontotheequationis x = 3.

d) Thisisfalse,sincetherearenovaluesof x thatmake x = x + 1true.

56. Thereareonlythreecasesinwhich ∃!xP(x)istrue,soweformthedisjunctionofthesethreecases.Theansweristhus

58. APrologqueryreturnsayes/noansweriftherearenovariablesinthequery,anditreturnsthevaluesthatmakethequery trueifthereare.

a) NoneofthefactswasthatKevinwasenrolledinEE222.Sotheresponseis no

b) OneofthefactswasthatKikowasenrolledinMath273.Sotheresponseis yes.

c) PrologreturnsthenamesofthecoursesforwhichGrossmanistheinstructor,namelyjust cs301

d) PrologreturnsthenamesoftheinstructorforCS301,namely grossman.

e) PrologreturnsthenamesoftheinstructorsteachinganycoursethatKevinisenrolledin,namely chan,sinceChanis theinstructorinMath273,theonlycourseKevinisenrolledin.

60. FollowingtheideaandsyntaxofExample28,wehavethefollowingrule: grandfather(X,Y):-father(X,Z),father(Z,Y);father(X,Z),mother(Z,Y). Notethatweusedthecommatomean“and”andthesemicolontomean“or.”For X tobethegrandfatherof Y, X must beeither Y’sfather’sfatheror Y’smother’sfather.

62.a)

d) Yes.Theunsatisfactoryexcuseguaranteedbypart(b)cannotbeaclearexplanationbypart(a).

64.a)

e) Yes.If x isoneofmypoultry,thenheisaduck(bypart(c)),hencenotwillingtowaltz(part(a)).Sinceoﬃcersare alwayswillingtowaltz(part(b)), x isnotanoﬃcer.

SECTION1.5NestedQuantiﬁers

2.a) Thereexistsarealnumber x suchthatforeveryrealnumber y , xy = y .Thisisassertingtheexistenceofamultiplicativeidentityfortherealnumbers,andthestatementistrue,sincewecantake x = 1.

b) Foreveryrealnumber x andrealnumber y ,if x isnonnegativeand y isnegative,thenthediﬀerence x y ispositive. Or,moresimply,anonnegativenumberminusanegativenumberispositive(whichistrue).

c) Foreveryrealnumber x andrealnumber y ,thereexistsarealnumber z suchthat x = y + z .Thisisatruestatement, sincewecantake z = x y ineachcase.

4.a) Somestudentinyourclasshastakensomecomputersciencecourse.

b) Thereisastudentinyourclasswhohastakeneverycomputersciencecourse.

c) Everystudentinyourclasshastakenatleastonecomputersciencecourse.

d) Thereisacomputersciencecoursethateverystudentinyourclasshastaken.

e) Everycomputersciencecoursehasbeentakenbyatleastonestudentinyourclass.

f) Everystudentinyourclasshastakeneverycomputersciencecourse.

6.a) RandyGoldbergisenrolledinCS252.

b) SomeoneisenrolledinMath695.

c) CarolSiteaisenrolledinsomecourse.

d) SomestudentisenrolledsimultaneouslyinMath222andCS252.

e) Thereexisttwodistinctpeople,thesecondofwhomisenrolledineverycoursethattheﬁrstisenrolledin.

f) Thereexisttwodistinctpeopleenrolledinexactlythesamecourses.

8.a) ∃x∃yQ(x, y)

b) Thisisthenegationofpart(a),andsocouldbewritteneither ¬∃x∃yQ(x, y)or ∀x∀y¬Q(x, y).

c) Weassumefromthewordingthatthestatementmeansthatthesamepersonappearedonbothshows: ∃x(Q(x,Jeopardy!) ∧ Q(x,WheelofFortune))

d) ∀y∃xQ(x, y) e) ∃x1 ∃x2 (Q(x1 ,Jeopardy!) ∧ Q(x2 ,Jeopardy!) ∧ x1 ≠ x2 )

10.a) ∀xF (x,Fred) b) ∀yF (Evelyn, y)

∀x

yF (x, y)

¬∃x

yF (x, y)

∀y∃xF (x, y) f) ¬∃x(F (x,Fred) ∧ F (x,Jerry)) g)

y2 (F (Nancy, y1 )

(Nancy, y

= y2 ))) h)

y(∀xF (x, y) ∧∀z(∀xF (x, z) → z = y))

))(Wedonotassumethatthissentenceisassertingthatthis personcanorcannotfoolher/himself.)

12. Theanswerstothisexercisearenotunique;therearemanywaysofexpressingthesamepropositionssymbolically.Note that C(x, y)and C(y, x)saythesamething.

a) ¬I (Jerry) b) ¬C(Rachel,Chelsea) c) ¬C(Jan,Sharon) d) ¬∃xC(x,Bob)

e) ∀x(x ≠ Joseph ↔ C(x,Sanjay)) f) ∃x ¬I (

14. Theanswerstothisexercisearenotunique;therearemanywaysofexpressingthesamepropositionssymbolically.Our domainofdiscourseforpersonshereconsistsofpeopleinthisclass.Weneedtomakeupapredicateineachcase.

a) Let S(x, y)meanthatperson x canspeaklanguage y .Thenourstatementis ∃xS(x,Hindi).

b) Let P(x, y)meanthatperson x playssport y .Thenourstatementis ∀x∃yP(x, y).

c) Let V (x, y)meanthatperson x hasvisitedstate y .Thenourstatementis ∃x(V (x,Alaska) ∧¬V (x,Hawaii)).

d) Let L(x, y)meanthatperson x haslearnedprogramminglanguage y .Thenourstatementis ∀x∃yL(x, y).

e) Let T (x, y)meanthatperson x hastakencourse y ,andlet O(y, z)meanthatcourse y isoﬀeredbydepartment z Thenourstatementis ∃x∃z∀y(O(y, z) → T (x, y)).

f) Let G(x, y)meanthatpersons x and y grewupinthesametown.Thenourstatementis ∃x

∀z(G(x, z) → (x = y ∨ x = z))).

)

g) Let C(x, y, z)meanthatpersons x and y havechattedwitheachotherinchatgroup z .Thenourstatementis ∀x∃y∃z(x ≠ y ∧ C(x, y, z)).

16. Welet P(s, c, m)bethestatementthatstudent s hasclassstanding c andismajoringin m .Thevariable s rangesover studentsintheclass,thevariable c rangesoverthefourclassstandings,andthevariable m rangesoverallpossible majors.

a) Thepropositionis ∃s∃mP(s,junior, m).Itistruefromthegiveninformation.

b) Thepropositionis ∀s∃cP(s, c,computerscience).Thisisfalse,sincetherearesomemathematicsmajors.

c) Thepropositionis ∃s∃c∃m(P(s, c, m) ∧ (c ≠ junior) ∧ (m ≠ mathematics)) .Thisistrue,sincethereisasophomore majoringincomputerscience.

d) Thepropositionis ∀s(∃cP(s, c,computerscience) ∨∃mP(s,sophomore, m)) .Thisisfalse,sincethereisafreshman mathematicsmajor.

e) Thepropositionis ∃m∀c∃sP(s, c, m).Thisisfalse.Itcannotbethat m ismathematics,sincethereisnosenior mathematicsmajor,anditcannotbethat m iscomputerscience,sincethereisnofreshmancomputersciencemajor. Nor,ofcourse,can m beanyothermajor.

18.a) ∀f (H (f ) → ∃cA(c)),where A(x)meansthatconsole x isaccessible,and H (x)meansthatfaultcondition x is happening

b) (∀u∃m (A(m) ∧ S(u, m))) → ∀uR(u),where A(x)meansthatthearchivecontainsmessage x , S(x, y)meansthatuser x sentmessage y ,and R(x)meansthattheemailaddressofuser x canberetrieved c McGrawHillLLC.Allrightsreserved.NoreproductionordistributionwithoutthepriorwrittenconsentofMcGrawHillLLC.

c) (∀b∃mD(m, b)) ↔ ∃p ¬C(p),where D(x, y)meansthatmechanism x candetectbreach y ,and C(x)meansthat process x hasbeencompromised

d) ∀x∀y (x ≠ y →

C(q, x, y))),where C(p, x, y)meansthatpath p connectsendpoint x to endpoint y

e) ∀x ((∀uK (x, u)) ↔ x = SysAdm),where K (x, y)meansthatperson x knowsthepasswordofuser y

20.a)

c) Whatdoes“necessarily”meaninthiscontext?Thebestexplanationistoassertthatacertainuniversalconditional statementisnottrue.Sowehave ¬∀

y < 0)).Notethatwedonotwanttoputthenegation symbolinside(itisnottruethatthediﬀerenceoftwonegativeintegersisnevernegative),nordowewanttonegatejust theconclusion(itisnottruethatthesumisalwaysnonnegative).Wecouldrewriteoursolutionbypassingthenegation inside,obtaining ∃x∃y((x < 0)

0)).

),wherethedomainofdiscourseconsistsofallintegers

24.a) Thereexistsanadditiveidentityfortherealnumbers—anumberthatwhenaddedtoeverynumberdoesnotchange itsvalue.

b) Anonnegativenumberminusanegativenumberispositive.

c) Thediﬀerenceoftwononpositivenumbersisnotnecessarilynonpositive.

d) Theproductoftwonumbersisnonzeroifandonlyifbothfactorsarenonzero.

26.a) Thisisfalse,since1 + 1 ≠ 1 1. b) Thisistrue,since2 + 0 = 2 0.

c) Thisisfalse,sincetherearemanyvaluesof y forwhich1 + y ≠ 1 y

d) Thisisfalse,sincetheequation x + 2 = x 2hasnosolution.

e) Thisistrue,sincewecantake x = y = 0. f) Thisistrue,sincewecantake y = 0foreach x .

g) Thisistrue,sincewecantake y = 0. h) Thisisfalse,sincepart(d)wasfalse.

i) Thisiscertainlyfalse.

28.a) Thisistrue,sinceforagivenreal x ,let y = x2

b) Thisisfalse,sincenosuch y existsif x isnegative.

c) Thisistrue,sincewecanset x = 0.

d) Thisisfalse,sincethecommutativelawforadditionalwaysholds.

e) Thisistrue,sincewecantake y = 1∕x .

f) Thisisfalse,sincethereciprocalof y dependson y —thereisnotone x thatworksforall y

g) Thisistrue,sincewecanlet y = 1 x

h) Thisisfalse,sincethissystemofequationsisinconsistent.

i) Thisisfalse,sincethissystemhasonlyonesolution.If x = 0,forexample,thenno y satisﬁes y = 2 ∧−y = 1.

j) Thisistrue,sincewecanlet z = (x + y)∕2).

30. WeneedtousethetransformationsshowninTable2ofSection1.4,replacing ¬∀ by ∃¬ ,andreplacing ¬∃ by ∀¬ . Inotherwords,wepushallthenegationsymbolsinsidethequantiﬁers,changingthesenseofthequantiﬁersaswedo so,becauseoftheequivalencesinTable2ofSection1.4.Inaddition,weneedtouseDeMorgan’slaws(inTable6of Section1.3)tochangethenegationofaconjunctiontothedisjunctionofthenegationsandtochangethenegationofa disjunctiontotheconjunctionofthenegations.Wealsousethefactthat

32. Aswepushthenegationsymboltowardtheinside,eachquantiﬁeritpassesmustchangeitstype.Forlogicalconnectives weeitheruseDeMorgan’slawsorrecallthat ¬(

→ q) ≡ p ∧¬q (Table7inSection1.3)andthat ¬(p ↔ q) ≡ ¬p ↔ q (Exercise25inSection1.3).

a) ¬∃

34. Thelogicalexpressionisassertingthatthedomainconsistsofatmosttwomembers.(Itissayingthatwheneveryou havetwounequalobjects,anyobjecthastobeoneofthosetwo.Notethatthisisvacuouslytruefordomainswithone element.)Thereforeanydomainhavingoneortwomemberswillmakeittrue(suchasthefemalemembersoftheUnited StatesSupremeCourtin2005),andanydomainwithmorethantwomemberswillmakeitfalse(suchasallmembersof theUnitedStatesSupremeCourtin2005).

36. Ineachcaseweneedtospecifysomepredicatesandidentifythedomainofdiscourse.

a) Let L(x, y)meanthatperson x haslost y dollarsplayingthelottery.Theoriginalstatementisthen ¬∃x∃y(y > 1000 ∧ L(x, y)).Itsnegationofcourseis ∃x∃y(y > 1000 ∧ L(x, y));someonehaslostmorethan$1000playingthelottery.

b) Let C(x, y)meanthatperson x haschattedwithperson y .Thegivenstatementis ∃x∃y(y

C(x, z)))).Thenegationistherefore ∀x

x ∧¬(z = y ↔ C(x, z)))).InEnglish,everybodyinthisclass haseitherchattedwithnooneelseorhaschattedwithtwoormoreothers.

c) Let E(x, y)meanthatperson x hassentemailtoperson y .Thegivenstatementis ¬∃

z ∧∀w(w ≠ x → (E(x, w) ↔ (w

y ∨ w = z)))).Thenegationisobviously

(E(x, w) ↔ (w = y ∨ w = z)))).InEnglish,somestudentinthisclasshassentemailtoexactlytwootherstudentsinthis class.

d) Let S(x, y)meanthatstudent x hassolvedexercise y .Thestatementis ∃x∀yS(x, y).Thenegationis ∀x∃y ¬S(x, y).In English,foreverystudentinthisclass,thereissomeexercisethatheorshehasnotsolved.(Onecouldalsointerpretthe givenstatementasassertingthatforeveryexercise,thereexistsastudent—perhapsadiﬀerentoneforeachexercise— whohassolvedit.Inthatcasetheorderofthequantiﬁerswouldbereversed.WordorderinEnglishsometimesmakes foralittleambiguity.)

e) Let S(x, y)meanthatstudent x hassolvedexercise y ,andlet B(y, z)meanthatexercise y isinsection z ofthebook. Thestatementis ¬∃x∀z∃y(B(y, z) ∧ S(x, y)).Thenegationisofcourse ∃x∀z∃y(B(y, z) ∧ S(x, y)).InEnglish,somestudent hassolvedatleastoneexerciseineverysectionofthisbook.

38.a) InEnglish,thenegationis“Somestudentinthisclassdoesnotlikemathematics.”Withtheobviouspropositional function,thisis ∃x¬L(x).

b) InEnglish,thenegationis“Everystudentinthisclasshasseenacomputer.”Withtheobviouspropositionalfunction, thisis ∀xS(x).

c) InEnglish,thenegationis“Foreverystudentinthisclass,thereisamathematicscoursethatthisstudenthasnot taken.”Withtheobviouspropositionalfunction,thisis ∀x∃c¬T (x, c).

d) Asinpart(f)ofExercise15,let P(z, y)be“Room z isinbuilding y ,”andlet Q(x, z)be“Student x hasbeenin room z .”Thentheoriginalstatementis ∃x∀y∃z(P(z, y) ∧ Q(x, z)) .Toformthenegation,wechangeallthequantiﬁers andputthenegationontheinside,thenapplyDeMorgan’slaw.Thenegationistherefore

)) , whichisalsoequivalentto ∀x∃y∀z(P(z, y) → ¬Q(x, z)) .InEnglish,thiscouldberead,“Foreverystudentthereisa buildingsuchthatforeveryroominthatbuilding,thestudenthasnotbeeninthatroom.”

40.a) Therearemanycounterexamples.If x = 2,thenthereisno y amongtheintegerssuchthat2 = 1∕y ,sincetheonly solutionofthisequationis y = 1∕2.Evenifwewereworkinginthedomainofrealnumbers, x = 0wouldprovidea counterexample,since0 = 1∕y fornorealnumber y

b) Wecanrewrite y2 x < 100as y2 < 100 + x .Sincesquarescanneverbenegative,nosuch y existsif x is,say, 200.This x providesacounterexample.

c) Thisisnottrue,sincesixthpowersarebothsquaresandcubes.Trivialcounterexampleswouldinclude x = y = 0and x = y = 1,butwecanalsotakesomethinglike x = 27and y = 9,since272 = 36 = 93

42. Thedistributivelawisjustthestatementthat x(y + z) = xy + xz forallrealnumbers.Thereforetheexpressionwewant is ∀x∀y∀z (x(y + z) = xy + xz),wherethequantiﬁersareassumedtorangeover(i.e.,thedomainofdiscourseis)thereal numbers.

44. Wewanttosaythatforeachtripleofcoeﬃcients(the a , b ,and c intheexpression ax2 + bx + c ,whereweinsistthat a ≠ 0sothatthisactuallyisquadratic),thereareatmosttwovaluesof x makingthatexpressionequalto0.Thedomain hereisallrealnumbers.Wewrite ∀

0) → (x1 = x2 ∨ x1 = x3 ∨ x2 = x3 )).

46. Thisstatementsaysthatthereisanumberthatislessthanorequaltoallsquares.

a) Thisisfalse,sincenomatterhowsmallapositivenumber x wemightchoose,ifwelet y = √x∕2,then x = 2y2 ,and itwillnotbetruethat x ≤ y2 .

b) Thisistrue,sincewecantake x =−1,forexample.

c) Thisistrue,sincewecantake x =−1,forexample.

48. Weneedtoshowthateachofthesepropositionsimpliestheother.Supposethat ∀xP(x) ∨∀xQ(x)istrue.Wewantto showthat ∀x∀y(P(x) ∨ Q(y))istrue.Byourhypothesis,oneoftwothingsmustbetrue.Either P isuniversallytrue,or Q isuniversallytrue.Inthefirstcase, ∀x∀y(P(x) ∨ Q(y))istrue,sincethefirstexpressioninthedisjunctionistrue,no matterwhat x and y are;andinthesecondcase, ∀x∀y(P(x) ∨ Q(y))isalsotrue,sincenowthesecondexpressioninthe disjunctionistrue,nomatterwhat x and y are.Nextweneedtoprovetheconverse.Sosupposethat ∀x∀y(P(x) ∨ Q(y)) istrue.Wewanttoshowthat ∀xP(x) ∨∀xQ(x)istrue.If ∀xP(x)istrue,thenwearedone.Otherwise, P(x0 )mustbe falseforsome x0 inthedomainofdiscourse.Forthis x0 ,then,thehypothesistellsusthat P(x0 ) ∨ Q(y)istrue,nomatter what y is.Since P(x0 )isfalse,itmustbethecasethat Q(y)istrueforeach y .Inotherwords, ∀yQ(y)istrue,or,to changethenameofthemeaninglessquantifiedvariable, ∀xQ(x)istrue.Thiscertainlyimpliesthat ∀xP(x) ∨∀xQ(x)is true,asdesired.

50.a) ByExercises47and48binSection1.4,wecansimplybringtheexistentialquantiﬁeroutside: ∃x(P(x) ∨ Q(x) ∨ A). b) ByExercise48ofthecurrentsection,theexpressioninsidetheparenthesesislogicallyequivalentto ∀x∀y(P(x) ∨ Q(y)). Applyingthenegationoperation,weobtain ∃x∃y¬(P(x) ∨ Q(y)).

c) FirstwerewritethisusingTable7inSection1.3as ∃xQ(x) ∨¬∃xP(x),whichisequivalentto ∃xQ(x) ∨∀x¬P(x). Tocombinetheexistentialanduniversalstatementsweusepart(b)ofExercise49ofthecurrentsection,obtaining ∀x∃y(¬P(x) ∨ Q(y)),whichisinprenexnormalform.

52. Wesimplywanttosaythatthereexistsan x suchthat P(x)holds,andthatevery y suchthat P(y)holdsmustbethis same x .Thuswewrite ∃x(P(x) ∧∀y(P(y) → y = x)) .Evenmorecompactly,wecanwrite ∃x∀y(P(y) ↔ y = x).

SECTION1.6RulesofInference

2. Thisismodustollens.Theﬁrststatementis p → q ,where p is“Georgedoesnothaveeightlegs”and q is“Georgeis notaspider.”Thesecondstatementis ¬q .Thethirdis ¬p .Modustollensisvalid.Wecanthereforeconcludethatthe conclusionoftheargument(thirdstatement)istrue,giventhatthehypotheses(theﬁrsttwostatements)aretrue.

4.a) Wehavetakentheconjunctionoftwopropositionsandassertedoneofthem.Thisis,accordingtoTable1, simpliﬁcation.

b) Wehavetakenthedisjunctionoftwopropositionsandthenegationofoneofthem,andassertedtheother.Thisis, accordingtoTable1,disjunctivesyllogism.SeeTable1fortheotherpartsofthisexerciseaswell.

c) modusponens d) addition e) hypotheticalsyllogism

6. Let r betheproposition“Itrains,”let f betheproposition“Itisfoggy,”let s betheproposition“Thesailingracewill beheld,”let l betheproposition“Thelifesavingdemonstrationwillgoon,”andlet t betheproposition“Thetrophy willbeawarded.”Wearegivenpremises(¬r ∨¬f ) → (s ∧ l), s → t ,and ¬t .Wewanttoconclude r .Wesetupthe proofintwocolumns,withreasons,asinExample6.Notethatitisvalidtoreplacesubexpressionsbyotherexpressions logicallyequivalenttothem.

StepReason

1. ¬t Hypothesis

2. s → t Hypothesis

3. ¬s Modustollensusing(1)and(2)

4.(¬r ∨¬f ) → (s ∧ l)Hypothesis

5.(¬(s ∧ l)) → ¬(¬r ∨¬f )Contrapositiveof(4)

6.(¬s ∨¬l) → (r ∧ f )DeMorgan’slawanddoublenegative

7. ¬s ∨¬l Addition,using(3)

8. r ∧ f Modusponensusing(6)and(7)

9. r Simpliﬁcationusing(8)

8. Firstweuseuniversalinstantiationtoconcludefrom“Forall x ,if x isaman,then x isnotanisland”thespecialcaseof interest,“IfManhattanisaman,thenManhattanisnotanisland.”Thenweformthecontrapositive(usingalsodouble negative):“IfManhattanisanisland,thenManhattanisnotaman.”Finallyweusemodusponenstoconcludethat Manhattanisnotaman.Alternatively,wecouldapplymodustollens.

10.a) Ifweusemodustollensstartingfromtheback,thenweconcludethatIamnotsore.Anotherapplicationofmodus tollensthentellsusthatIdidnotplayhockey.

b) Wereallycan’tconcludeanythingspeciﬁchere.

c) Byuniversalinstantiation,weconcludefromtheﬁrstconditionalstatementbymodusponensthatdragonﬂieshave sixlegs,andweconcludebymodustollensthatspidersarenotinsects.Wecouldsayusingexistentialgeneralization that,forexample,thereexistsanon-six-leggedcreaturethateatsasix-leggedcreature,andthatthereexistsanon-insect thateatsaninsect.

d) WecanapplyuniversalinstantiationtotheconditionalstatementandconcludethatifHomer(respectively,Maggie) isastudent,thenhe(she)hasanInternetaccount.NowmodustollenstellsusthatHomerisnotastudent.Thereareno conclusionstobedrawnaboutMaggie.

e) Theﬁrstconditionalstatementisthatif x ishealthytoeat,then x doesnottastegood.Universalinstantiationand modusponensthereforetellusthattofudoesnottastegood.Thethirdsentencesaysthatifyoueat x ,then x tastesgood. Thereforethefourthhypothesisalreadyfollows(bymodustollens)fromtheﬁrstthree.Noconclusionscanbedrawn aboutcheeseburgersfromthesestatements.

f) Bydisjunctivesyllogism,theﬁrsttwohypothesesallowustoconcludethatIamhallucinating.Thereforebymodus ponensweknowthatIseeelephantsrunningdowntheroad.

12. ApplyingExercise11,wewanttoshowthattheconclusion r followsfromthefivepremises(p ∧ t) → (r ∨ s), q → (u ∧ t), u → p , ¬s ,and q .From q and q → (u ∧ t)weget u ∧ t bymodusponens.Fromtherewegetboth u and t by simplification(andthecommutativelaw).From u and u → p weget p bymodusponens.From p and t weget p ∧ t byconjunction.Fromthatand(p ∧ t) → (r ∨ s)weget r ∨ s bymodusponens.Fromthatand ¬s wefinallyget r by disjunctivesyllogism.

14. Ineachcasewesetuptheproofintwocolumns,withreasons,asinExample6.

a) Let c(x)be“ x isinthisclass,”let r(x)be“ x ownsaredconvertible,”andlet t(x)be“ x hasgottenaspeedingticket.” Wearegivenpremises c(Linda), r(Linda), ∀x(r(x) → t(x)),andwewanttoconclude ∃x(c(x) ∧ t(x)).

StepReason

1. ∀x(r(x) → t(x))Hypothesis

2. r(Linda) → t(Linda)Universalinstantiationusing(1)

3. r(Linda)Hypothesis

4. t(Linda)Modusponensusing(2)and(3)

5. c(Linda)Hypothesis

6. c(Linda) ∧ t(Linda)Conjunctionusing(4)and(5)

7. ∃x(c(x) ∧ t(x))Existentialgeneralizationusing(6)

b) Let r(x)be“ r isoneoftheﬁveroommateslisted,”let d (x)be“ x hastakenacourseindiscretemathematics,”andlet a(x)be“ x cantakeacourseinalgorithms.”Wearegivenpremises ∀x(r(x) → d (x))and ∀x(d (x) → a(x)),andwewant toconclude ∀x(r(x) → a(x)).Inwhatfollows y representsanarbitraryperson.

StepReason

1. ∀x(r(x) → d (x))Hypothesis

2. r(y) → d (y)Universalinstantiationusing(1)

3. ∀x(d (x) → a(x))Hypothesis

4. d (y) → a(y)Universalinstantiationusing(3)

5. r(y) → a(y)Hypotheticalsyllogismusing(2)and(4)

6. ∀x(r(x) → a(x))Universalgeneralizationusing(5)

c) Let s(x)be“ x isamovieproducedbySayles,”let c(x)be“ x isamovieaboutcoalminers,”andlet w(x)be“movie x iswonderful.”Wearegivenpremises ∀x(s(x) → w(x))and ∃x(s(x) ∧ c(x)),andwewanttoconclude ∃x(c(x) ∧ w(x)). Inourproof, y representsanunspeciﬁedparticularmovie.

StepReason

1. ∃x(s(x) ∧ c(x))Hypothesis

2. s(y) ∧ c(y)Existentialinstantiationusing(1)

3. s(y)Simpliﬁcationusing(2)

4. ∀x(s(x) → w(x))Hypothesis

5. s(y) → w(y)Universalinstantiationusing(4)

6. w(y)Modusponensusing(3)and(5)

7. c(y)Simpliﬁcationusing(2)

8. w(y) ∧ c(y)Conjunctionusing(6)and(7)

9. ∃x(c(x) ∧ w(x))Existentialgeneralizationusing(8)

d) Let c(x)be“ x isinthisclass,”let f (x)be“ x hasbeentoFrance,”andlet l(x)be“ x hasvisitedtheLouvre.”Weare givenpremises ∃x(c(x) ∧ f (x)), ∀x(f (x) → l(x)),andwewanttoconclude ∃x(c(x) ∧ l(x)).Inourproof, y representsan unspeciﬁedparticularperson.

StepReason

1. ∃x(c(x) ∧ f (x))Hypothesis

2. c(y) ∧ f (y)Existentialinstantiationusing(1)

3. f (y)Simpliﬁcationusing(2)

4. c(y)Simpliﬁcationusing(2)

5. ∀x(f (x) → l(x))Hypothesis

6. f (y) → l(y)Universalinstantiationusing(5)

7. l(y)Modusponensusing(3)and(6)

8. c(y) ∧ l(y)Conjunctionusing(4)and(7)

9. ∃x(c(x) ∧ l(x))Existentialgeneralizationusing(8)

16.a) Thisiscorrect,usinguniversalinstantiationandmodustollens.

b) Thisisnotcorrect.Afterapplyinguniversalinstantiation,itcontainsthefallacyofdenyingthehypothesis.

c) Afterapplyinguniversalinstantiation,itcontainsthefallacyofaﬃrmingtheconclusion.

d) Thisiscorrect,usinguniversalinstantiationandmodusponens.

18. Weknowthat somes existsthatmakes S(s,Max)true,butwecannotconcludethatMaxisonesuch s .Thereforethis ﬁrststepisinvalid.

20.a) Thisisinvalid.Itisthefallacyofaﬃrmingtheconclusion.Letting a =−2providesacounterexample.

b) Thisisvalid;itismodusponens.

22. Wewillgiveanargumentestablishingtheconclusion.Wewanttoshowthatallhummingbirdsaresmall.LetTweetybe anarbitraryhummingbird.WemustshowthatTweetyissmall.TheﬁrstpremiseimpliesthatifTweetyisahummingbird, thenTweetyisrichlycolored.Thereforeby(universal)modusponenswecanconcludethatTweetyisrichlycolored. ThethirdpremiseimpliesthatifTweetydoesnotliveonhoney,thenTweetyisnotrichlycolored.Thereforeby(universal) modustollenswecannowconcludethatTweetydoesliveonhoney.Finally,thesecondpremiseimpliesthatifTweety isalargebird,thenTweetydoesnotliveonhoney.Thereforeagainby(universal)modustollenswecannowconclude thatTweetyisnotalargebird,i.e.,thatTweetyissmall,asdesired.Noticethatweinvokeuniversalgeneralizationas thelaststep.

24. Steps3and5areincorrect;simpliﬁcationappliestoconjunctions,notdisjunctions.

26. Wewanttoshowthattheconditionalstatement P(a) → R(a)istrueforall a inthedomain;thedesiredconclusionthen followsbyuniversalgeneralization.Thuswewanttoshowthatif P(a)istrueforaparticular a ,then R(a)isalsotrue. Forsuchan a ,byuniversalmodusponensfromtheﬁrstpremisewehave Q(a),andthenbyuniversalmodusponens fromthesecondpremisewehave R(a),asdesired.

28. Wewanttoshowthattheconditionalstatement ¬R(a) → P(a)istrueforall a inthedomain;thedesiredconclusion thenfollowsbyuniversalgeneralization.Thuswewanttoshowthatif ¬R(a)istrueforaparticular a ,then P(a)isalso true.Forsuchan a ,universalmodustollensappliedtothesecondpremisegivesus ¬(¬P(a) ∧ Q(a)).Byrulesfrom propositionallogic,thisgivesus P(a) ∨¬Q(a).Byuniversalgeneralizationfromtheﬁrstpremise,wehave P(a) ∨ Q(a). Nowbyresolutionwecanconclude P(a) ∨ P(a),whichislogicallyequivalentto P(a),asdesired.

30. Let a be“Allenisagoodboy,”let h be“Hillaryisagoodgirl,”andlet d be“Davidishappy.”Thenourassumptions are ¬a ∨ h and a ∨ d .Usingresolutiongivesus h ∨ d ,asdesired.

c McGrawHillLLC.Allrightsreserved.NoreproductionordistributionwithoutthepriorwrittenconsentofMcGrawHillLLC.

32. Weapplyresolutiontogivethetautology(p ∨ ��) ∧ (¬p ∨ ��) → (�� ∨ ��).Theleft-handsideisequivalentto p ∧¬p ,since p ∨ �� isequivalentto p ,and ¬p ∨ �� isequivalentto ¬p .Theright-handsideisequivalentto �� .Sincetheconditional statementistrue,andtheconclusionisfalse,itfollowsthatthehypothesis, p ∧¬p ,isfalse,asdesired.

34. Letususethefollowingletterstostandfortherelevantpropositions: d for“logicisdifficult,” s for“manystudentslike logic,”and e for“mathematicsiseasy.”Thentheassumptionsare d ∨¬s and e → ¬d .Notethatthefirstofthese isequivalentto s → d ,sincebothformsarefalseifandonlyif s istrueand d isfalse.Inaddition,letusnotethat thesecondassumptionisequivalenttoitscontrapositive, d → ¬e .Andfinally,bycombiningthesetwoconditional statements,weseethat s → ¬e alsofollowsfromourassumptions.

a) Hereweareaskedwhetherwecanconcludethat s → ¬e .Aswenotedabove,theanswerisyes,thisconclusionis valid.

b) Thequestionconcerns ¬e → ¬s .Thisisequivalenttoitscontrapositive, s → e .Thatdoesn’tseemtofollowfromour assumptions,solet’sﬁndacaseinwhichtheassumptionsholdbutthisconditionalstatementdoesnot.Thisconditional statementfailsinthecaseinwhich s istrueand e isfalse.Ifwetake d tobetrueaswell,thenbothofourassumptions aretrue.Thereforethisconclusionisnotvalid.

c) Theissueis ¬e ∨ d ,whichisequivalenttotheconditionalstatement e → d .Thisdoes not followfromourassumptions.Ifwetake d tobefalse, e tobetrue,and s tobefalse,thenthispropositionisfalsebutourassumptionsaretrue.

d) Theissueis ¬d ∨¬e ,whichisequivalenttotheconditionalstatement d → ¬e .Wenotedabovethatthisvalidly followsfromourassumptions.

e) Thissentencesays ¬s → (¬e ∨¬d ).Theonlycaseinwhichthisisfalseiswhen s isfalseandboth e and d aretrue. Butinthiscase,ourassumption e → ¬d isalsoviolated.Therefore,inallcasesinwhichtheassumptionshold,this statementholdsaswell,soit is avalidconclusion.

SECTION1.7IntroductiontoProofs

2. Wemustshowthatwheneverwehavetwoevenintegers,theirsumiseven.Supposethat a and b aretwoevenintegers. Thenthereexistintegers s and t suchthat a = 2s and b = 2t .Adding,weobtain a + b = 2s + 2t = 2(s + t).Sincethis represents a + b as2timestheinteger s + t ,weconcludethat a + b iseven,asdesired.

4. Wemustshowthatwheneverwehaveaneveninteger,itsnegativeiseven.Supposethat a isaneveninteger.Then thereexistsaninteger s suchthat a = 2s .Itsadditiveinverseis 2s ,whichbyrulesofarithmeticandalgebra(see Appendix1)equals2( s).Sincethisis2timestheinteger s ,itiseven,asdesired.

6. Anoddnumberisoneoftheform2n + 1,where n isaninteger.Wearegiventwooddnumbers,say2a + 1and2b + 1. Theirproductis(2a + 1)(2b + 1) = 4ab + 2a + 2b + 1 = 2(2ab + a + b) + 1.Thislastexpressionshowsthattheproduct isodd,sinceitisoftheform2n + 1,with n = 2ab + a + b .

8. Let n = m2 .If m = 0,then n + 2 = 2,whichisnotaperfectsquare,sowecanassumethat m ≥ 1.Thesmallestperfect squaregreaterthan n is(m + 1)2 ,andwehave(m + 1)2 = m2 + 2m + 1 = n + 2m + 1 > n + 2 1 + 1 > n + 2.Therefore n + 2cannotbeaperfectsquare.

10. Arationalnumberisanumberthatcanbewrittenintheform x∕y where x and y areintegersand y ≠ 0.Supposethat wehavetworationalnumbers,say a∕b and c∕d .Thentheirproductis,bytheusualrulesformultiplicationoffractions, (ac)∕(bd ).Notethatboththenumeratorandthedenominatorareintegers,andthat bd ≠ 0since b and d wereboth nonzero.Thereforetheproductis,bydeﬁnition,arationalnumber.

12. Thisistrue.Supposethat a∕b isanonzerorationalnumberandthat x isanirrationalnumber.Wemustprovethatthe product xa∕b isalsoirrational.Wegiveaproofbycontradiction.Supposethat xa∕b wererational.Since a∕b ≠ 0, weknowthat a ≠ 0,so b∕a isalsoarationalnumber.Letusmultiplythisrationalnumber b∕a bytheassumed rationalnumber xa∕b .ByExercise10,theproductisrational.Buttheproductis(b∕a)(xa∕b) = x ,whichisirrational byhypothesis.Thisisacontradiction,soinfact xa∕b mustbeirrational,asdesired.

14. If x isrationalandnotzero,thenbydeﬁnitionwecanwrite x = p∕q ,where p and q arenonzerointegers.Since1∕x is then q∕p and p ≠ 0,wecanconcludethat1∕x isrational.

16. Assumetothecontrarythat x , y ,and z arealleven.Thenthereexistintegers a , b ,and c suchthat x = 2a , y = 2b , and z = 2c .Butthen x + y + z = 2a + 2b + 2c = 2(a + b + c)isevenbydeﬁnition.Thiscontradictsthehypothesisthat x + y + z isodd.Thereforetheassumptionwaswrong,andatleastoneof x , y ,and z isodd.

18. Wegiveaproofbycontraposition.Ifitisnottruethan m isevenor n iseven,then m and n arebothodd.ByExercise6, thistellsusthat mn isodd,andourproofiscomplete.

20.a) Wemustprovethecontrapositive:If n isodd,then3n + 2isodd.Assumethat n isodd.Thenwecanwrite n = 2k + 1 forsomeinteger k .Then3n + 2 = 3(2k + 1) + 2 = 6k + 5 = 2(3k + 2) + 1.Thus3n + 2istwotimessomeinteger plus1,soitisodd.

b) Supposethat3n + 2isevenandthat n isodd.Since3n + 2iseven,sois3n .Ifweaddsubtractanoddnumberfrom anevennumber,wegetanoddnumber,so3n n = 2n isodd.Butthisisobviouslynottrue.Thereforeoursupposition waswrong,andtheproofbycontradictioniscomplete.

22. Weneedtoprovetheproposition“If1isapositiveinteger,then12 ≥ 1.”Theconclusionisthetruestatement1 ≥ 1. Thereforetheconditionalstatementistrue.Thisisanexampleofatrivialproof,sincewemerelyshowedthatthe conclusionwastrue.

24. Wegiveaproofbycontradiction.Supposethatwedon’tgetapairofbluesocksorapairofblacksocks.Thenwedrew atmostoneofeachcolor.Thisaccountsforonlytwosocks.Butwearedrawingthreesocks.Thereforeoursupposition thatwedidnotgetapairofbluesocksorapairofblacksocksisincorrect,andourproofiscomplete.

26. Wegiveaproofbycontradiction.Iftherewereatmosttwodaysfallinginthesamemonth,thenwecouldhaveatmost 2 12 = 24days,sincethereare12months.Sincewehavechosen25days,atleastthreeofthemmustfallinthe samemonth.

28. Weneedtoprovetwothings,sincethisisan“ifandonlyif”statement.Firstletusprovedirectlythatif n iseventhen 7n + 4iseven.Since n iseven,itcanbewrittenas2k forsomeinteger k .Then7n + 4 = 14k + 4 = 2(7k + 2).This is2timesaninteger,soitiseven,asdesired.Nextwegiveaproofbycontrapositionthatif7n + 4iseventhen n is even.Sosupposethat n isnoteven,i.e.,that n isodd.Then n canbewrittenas2k + 1forsomeinteger k .Thus 7n + 4 = 14k + 11 = 2(7k + 5) + 1.Thisis1morethan2timesaninteger,soitisodd.Thatcompletestheproofby contraposition.

30. Therearetwothingstoprove.Forthe“if”part,therearetwocases.If m = n ,thenofcourse m2 = n2 ;if m =−n , then m2 = ( n)2 = ( 1)2 n2 = n2 .Forthe“onlyif”part,wesupposethat m2 = n2 .Puttingeverythingontheleftand factoring,wehave(m + n)(m n) = 0.Nowtheonlywaythataproductoftwonumberscanbezeroisifoneofthemis zero.Thereforeweconcludethateither m + n = 0(inwhichcase m =−n ),orelse m n = 0(inwhichcase m = n ), andourproofiscomplete.

32. Wewritetheseinsymbols: a < b ,(a + b)∕2 > a ,and(a + b)∕2 < b .Thelattertwoareequivalentto a + b > 2a and a + b < 2b ,respectively,andtheseareinturnequivalentto b > a and a < b ,respectively.Itisnowclearthatallthree statementsareequivalent.

34. Wegivedirectproofsthat(i)implies(ii),that(ii)implies(iii),andthat(iii)implies(i).Thatwillsuﬃce.Fortheﬁrst, supposethat x = p∕q where p and q areintegerswith q ≠ 0.Then x∕2 = p∕(2q),andthisisrational,since p and 2q areintegerswith2q ≠ 0.Forthesecond,supposethat x∕2 = p∕q where p and q areintegerswith q ≠ 0.Then x = (2p)∕q ,so3x 1 = (6p)∕q 1 = (6p q)∕q andthisisrational,since6p q and q areintegerswith q ≠ 0.For thelast,supposethat3x 1 = p∕q where p and q areintegerswith q ≠ 0.Then x = (p∕q + 1)∕3 = (p + q)∕(3q),and thisisrational,since p + q and3q areintegerswith3q ≠ 0.

36. No.Thislineofreasoningshowsthat if √2x2 1 = x ,thenwemusthave x = 1or x =−1.Thesearethereforethe onlypossiblesolutions,butwehavenoguaranteethatthey are solutions,sincenotallofourstepswerereversible(in particular,squaringbothsides).Thereforewe must substitutethesevaluesbackintotheoriginalequationtodetermine whethertheydoindeedsatisfyit.

38. Theonlyconditionalstatementsnotshowndirectlyare p1 ↔ p2 , p2 ↔ p4 ,and p3 ↔ p4 .Buttheseeachfollowwith oneormoreintermediatesteps: p

,since

(justestablished)and p1 ↔ p4 ;and p3 ↔ p4 ,since p3 ↔

and p1 ↔ p4

40. Wemustﬁndanumberthatcannotbewrittenasthesumofthesquaresofthreeintegers.Weclaimthat7issucha number(infact,itisthesmallestsuchnumber).Theonlysquaresthatcanbeusedtocontributetothesumare0,1, and4.Wecannotusetwo4’s,becausetheirsumexceeds7.Thereforewecanuseatmostone4,whichmeansthatwe mustget3usingjust0’sand1’s.Clearlythree1’sarerequiredforthis,bringingthetotalnumberofsquaresusedto four.Thus7cannotbewrittenasthesumofthreesquares.

42. Supposethatwelookatthetengroupsofintegersinthreeconsecutivelocationsaroundthecircle(first-second-third, second-third-fourth, ...,eighth-ninth-tenth,ninth-tenth-first,andtenth-first-second).Sinceeachnumberfrom1to10 getsusedthreetimesinthesegroups,thesumofthesumsofthetengroupsmustequalthreetimesthesumofthenumbers from1to10,namely3 55 = 165.Thereforetheaveragesumis165∕10 = 16 5.ByExercise41,atleastoneofthe sumsmustbegreaterthanorequalto16.5,andsincethesumsarewholenumbers,thismeansthatatleastoneofthe sumsmustbegreaterthanorequalto17.

44. Weshowthateachoftheseisequivalenttothestatement(v) n isodd,say n = 2k + 1.Example1showedthat(v) implies(i),andExample9showedthat(i)implies(v).For(v) → (ii)weseethat1 n = 1 (2k + 1) = 2( k)is even.Conversely,if n wereeven,say n = 2m ,thenwewouldhave1 n = 1 2m = 2( m) + 1,so1 n would beodd,andthiscompletestheproofbycontrapositionthat(ii) → (v).For(v) → (iii),weseethat n3 = (2k + 1)3 = 8k3 + 12k2 + 6k + 1 = 2(4k3 + 6k2 + 3k) + 1isodd.Conversely,if n wereeven,say n = 2m ,thenwewouldhave n3 = 2(4m3 ),so n3 wouldbeeven,andthiscompletestheproofbycontrapositionthat(iii) → (v).Finally,for(v) → (iv),weseethat n2 + 1 = (2k + 1)2 + 1 = 4k2 + 4k + 2 = 2(2k2 + 2k + 1)iseven.Conversely,if n wereeven,say n = 2m , thenwewouldhave n2 + 1 = 2(2m2 ) + 1,so n2 + 1wouldbeodd,andthiscompletestheproofbycontrapositionthat (iv) → (v).

SECTION1.8ProofMethodsandStrategy

2. Wemustshowthatforallpositiveintegers x ,itisnottruethat x2 = 10.Considerthetwocasesdescribedinthehint. Case(i):If1 ≤ x ≤ 3,then x2 ≤ 9,so x2 ≠ 10.Case(ii):If x ≥ 4,then x2 ≥ 16,so x2 ≠ 10.Thetwocasesrepresent allpossiblevaluesof x ,andinneithercaseis x2 = 10,so10isnotthesquareofaninteger.

4. Thecubesthatmightgointothesumare1,8,27,64,125,216,343,512,and729.Wemustshowthatnotwoof thesesumtoanumberonthislist.Ifwetrythe45combinations(1 + 1,1 + 8,...,1 + 729,8 + 8,8 + 27,...8 + 729, ...,729 + 729),weseethatnoneofthemworks.Havingexhaustedthepossibilities,weconcludethatnocubelessthan 1000isthesumoftwocubes.

6. Therearethreemaincases,dependingonwhichofthethreenumbersissmallest.If a issmallest(ortiedforsmallest), thenclearly a ≤ min(b, c),andsotheleft-handsideequals a .Ontheotherhand,fortheright-handsidewehave min(a, c) = a aswell.Inthesecondcase, b issmallest(ortiedforsmallest).Thesamereasoningshowsusthatthe right-handsideequals b ;andtheleft-handsideis min(a, b) = b aswell.Intheﬁnalcase,inwhich c issmallest(ortied forsmallest),theleft-handsideis min(a, c) = c ,whereastheright-handsideisclearlyalso c .Sinceoneofthethreehas tobesmallestwehavetakencareofallthecases.

8. Because x and y areofoppositeparities,wecanassume,withoutlossofgenerality,that x isevenand y isodd.This tellsusthat x = 2m forsomeinteger m and y = 2n + 1forsomeinteger n .Then5x + 5y = 5(2m) + 5(2n + 1) = 10m + 10n + 1 = 10(m + n) + 1 = 2 5(m + n) + 1,whichsatisﬁesthedeﬁnitionofbeinganoddnumber.

10. Thenumber1hasthisproperty,sincetheonlypositiveintegernotexceeding1is1itself,andthereforethesumis1. Thisisaconstructiveproof.

12. Theonlyperfectsquaresthatdiﬀerby1are0and1.Thereforethesetwoconsecutiveintegerscannotbothbeperfect squares.Thisisanonconstructiveproof—wedonotknowwhichofthemmeetstherequirement.(Infact,acomputer algebrasystemwilltellusthatneitherofthemisaperfectsquare.)

14. Ofthesethreenumbers,atleasttwomusthavethesamesign(bothpositiveorbothnegative),sincethereareonlytwo signs.(Itisconceivablethatsomeofthemarezero,butweviewzeroaspositiveforthepurposesofthisproblem.)The productoftwowiththesamesignisnonnegative.Thiswasanonconstructiveproof,sincewehavenotidentiﬁedwhich productisnonnegative.(Infact,acomputeralgebrasystemwilltellusthatallthreearepositive,soallthreeproducts arepositive.)

16. Anassertionlikethisoneisimplicitlyuniversallyquantiﬁed—itmeansthat forall rationalnumbers a and b , ab is rational.Todisprovesuchastatementitsuﬃcestoprovideonecounterexample.Take a = 2and b = 1∕2.Then ab = 21∕2 = √2,andweknowfromExample11inSection1.7that √2isnotrational.

18. Weknowfromalgebrathatthefollowingequationsareequivalent: ax + b = c , ax = c b x = (c b)∕a .Thisshows, constructively,whattheuniquesolutionofthegivenequationis.

20. Given r ,let a betheclosestintegerto r lessthan r ,andlet b betheclosestintegerto r greaterthan r .Inthenotation tobeintroducedinSection2.3, a = ⌊r ⌋ and b = ⌈r ⌉ .Infact, b = a + 1.Clearlythedistancebetween r andanyinteger otherthan a or b isgreaterthan1socannotbelessthan1∕2.Furthermore,since r isirrational,itcannotbeexactly half-waybetween a and b ,soexactlyoneof r a < 1∕2and b r < 1∕2holds.

22. Given x ,let n bethegreatestintegerlessthanorequalto x ,andlet �� = x n .Inthenotationtobeintroducedin Section2.3, n = ⌊x⌋ .Clearly0 ≤ ��< 1,and �� isuniqueforthis n .Anyotherchoiceof n wouldcausetherequired �� tobelessthan0orgreaterthanorequalto1,so n isuniqueaswell.

24. Wefollowthehint.Thesquareofeveryrealnumberisnonnegative,so(x 1∕x)2 ≥ 0.Multiplyingthisoutand simplifying,weobtain x2 2 + 1∕x2 ≥ 0,so x2 + 1∕x2 ≥ 2,asdesired.

26. Let x = 1and y = 10.Thentheirarithmeticmeanis5.5andtheirquadraticmeanis √50.5 ≈ 7.11.Similarly,if x = 5and y = 8,thenthearithmeticmeanis(5 + 8)∕2 = 6 5andthequadraticmeanis √(52 + 82 )∕2 ≈ 6 67.Sowe conjecturethatthequadraticmeanisalwaysgreaterthanorequaltothearithmeticmean.Thuswewanttoprovethat

forallpositiverealnumbers x and y .Doingsomealgebra,weﬁndthatthisinequalityisequivalenttothetruestatement that(x y)2 ≥ 0:

Infact,ourargumentalsoshowsthatequalityholdsifandonlyif x = y

28. Ifweweretoendupwithnine0’s,theninthestepbeforethiswemusthavehadeithernine0’sornine1’s,sinceeach adjacentpairofbitsmusthavebeenequalandthereforeallthebitsmusthavebeenthesame.Thusifwearetostartwith somethingotherthannine0’sandyetendupwithnine0’s,wemusthavehadnine1’satsomepoint.Butinthestep beforethateachadjacentpairofbitsmusthavebeendiﬀerent;inotherwords,theymusthavealternated0,1,0,1,and soon.Thisisimpossiblewithanoddnumberofbits.Thiscontradictionshowsthatwecannevergetnine0’s.

30. Clearlyonlythelasttwodigitsof n contributetothelasttwodigitsof n2 .Sowecancompute02 ,12 ,22 ,32 ,...,992 , andrecordthelasttwodigits,omittingrepetitions.Weobtain00,01,04,09,16,25,36,49,64,81,21,44,69,96, 56,89,24,61,41,84,29,76.Fromthatpointon,thelistrepeatsinreverseorder(aswetakethesquaresfrom252 to 492 ,andthenitallrepeatsagainaswetakethesquaresfrom502 to992 ).Thereasonfortheselasttwostatementsare that(50 n)2 = 2500 100n + n2 ,so(50 n)2 and n2 havethesametwoﬁnaldigits,and(50 + n)2 = 2500 + 100n + n2 , so(50 + n)2 and n2 havethesametwoﬁnaldigits.Thusourlist(whichcontains22numbers)iscomplete.

32. If |y| ≥ 2,then2x2 + 5y2 ≥ 2x2 + 20 ≥ 20,sotheonlypossiblevaluesof y totryare0and ±1.Intheformercasewe wouldbelookingforsolutionsto2x2 = 14andinthelattercaseto2x2 = 9.Clearlytherearenointegersolutionsto theseequations,sotherearenosolutionstotheoriginalequation.

34. Followingthehint,welet x = m2 n2 , y = 2mn ,and z = m2 + n2 .Then x2 + y2 = (m2 n2 )2

= m4 2

)2 = z2 .Thuswehavefoundinﬁnitelymanysolutions,since m and n canbearbitrarilylarge.

36. Oneproofthat 3 √2isirrationalissimilartotheproofthat √2isirrational,giveninExample11inSection1.7.Itisa proofbycontradiction.Supposethat21∕3 (or 3 √2,whichisthesamething)istherationalnumber p∕q ,where p and q arepositiveintegerswithnocommonfactors(thefractionisinlowestterms).Cubing,weseethat2 = p3 ∕q3 ,or, equivalently, p3 = 2q3 .Thus p3 iseven.Sincetheproductofoddnumbersisodd,thismeansthat p iseven,sowecan write p = 2s .Substitutingintotheequation p3 = 2q3 ,weobtain8s3 = 2q3 ,whichsimpliﬁesto4s3 = q3

Nowweplaythesamegamewith q .Since q3 iseven, q mustbeeven.Wehavenowconcludedthat p and q arebotheven,thatis,that2isacommondivisorof p and q .Thiscontradictsthechoiceof p∕q tobeinlowestterms. Thereforeouroriginalassumption—that 3 √2isrational—isinerror,sowehaveprovedthat 3 √2isirrational.

38. Theaverageoftwodiﬀerentnumbersiscertainlyalwaysbetweenthetwonumbers.Furthermore,theaverage a of rationalnumber x andirrationalnumber y mustbeirrational,becausetheequation a = (x + y)∕2leadsto y = 2a x , whichwouldberationalif a wererational.

40. Thesolutionisnotunique,buthereisonewaytomeasureoutfourgallons.Fillthe5-gallonjugfromthe8-gallonjug, leavingthecontents(3,5,0),whereweareusingtheorderedtripletorecordtheamountofwaterinthe8-gallonjug,the 5-gallonjug,andthe3-gallonjug,respectively.Nextfillthe3-gallonjugfromthe5-gallonjug,leaving(3,2,3).Pour thecontentsofthe3-gallonjugbackintothe8-gallonjug,leaving(6,2,0).Emptythe5-gallonjug’scontentsintothe 3-gallonjug,leaving(6,0,2),andthenfillthe5-gallonjugfromthe8-gallonjug,producing(1,5,2).Finally,topoff the3-gallonjugfromthe5-gallonjug,andwe’llhave(1,4,3),withfourgallonsinthe5-gallonjug.

42.a) 16 → 8 → 4 → 2 → 1

44. Thisiseasilydone,bylayingthedominoeshorizontally,threeinthefirstandlastrowsandfourineachoftheothersixrows.

46. Withoutlossofgenerality,wenumberthesquaresfrom1to25,startinginthetoprowandproceedinglefttorightin eachrow;andweassumethatsquares5(upperrightcorner),21(lowerleftcorner),and25(lowerrightcorner)arethe missingones.Wearguethatthereisnowaytocovertheremainingsquareswithdominoes.

Bysymmetrywecanassumethatthereisadominoplacedin1-2(usingtheobviousnotation).Ifsquare3iscovered by3-8,thenthefollowingdominoesareforcedinturn:4-9,10-15,19-20,23-24,17-22,and13-18,andnownodomino cancoversquare14.Thereforewemustuse3-4alongwith1-2.Ifweuseallof17-22,18-23,and19-24,thenweare againquicklyforcedintoasequenceofplacementsthatleadtoacontradiction.Thereforewithoutlossofgenerality,we canassumethatweuse22-23,whichthenforces19-24,15-20,9-10,13-14,7-8,6-11,and12-17,andwearestuckonce again.Thiscompletestheproofbycontradictionthatnoplacementispossible.

48. Thebarriersshowninthediagramsplittheboardintoonecontinuousclosedpathof64squares,eachadjacenttothe next(forexample,startattheupperleftcorner,goallthewaytotheright,thenallthewaydown,thenallthewaytothe left,andthenweaveyourwaybackuptothestartingpoint).Becauseeachsquareinthepathisadjacenttoitsneighbors, thecolorsalternate.Therefore,ifweremoveoneblacksquareandonewhitesquare,thisclosedpathdecomposesinto twopaths,eachofwhichstartsinonecolorandendsintheothercolor(andthereforehasevenlength).Clearlyeach suchpathcanbecoveredbydominoesbystartingatoneend.Thiscompletestheproof.

50. IfwestudyFigure7,weseethatbyrotatingorreﬂectingtheboard,wecanmakeanysquarewewishnonwhite,withthe exceptionofthesquareswithcoordinates(3,3),(3,6),(6,3),and(6,6).Thereforethesameargumentaswasusedin Example22showsthatwecannottiletheboardusingstraighttriominoesifanyoneofthoseother60squaresisremoved. Thefollowingdrawing(rotatedasnecessary)showsthatwecantiletheboardusingstraighttriominoesifoneofthose foursquaresisremoved.

52. Wewilluseacoloringofthe10 × 10boardwithfourcolorsasthebasisforaproofbycontradictionshowingthatno suchtilingexists.Assumethat25straighttetrominoescancovertheboard.Somewillbeplacedhorizontallyandsome vertically.Becausethereisanoddnumberoftiles,thenumberplacedhorizontallyandthenumberplacedvertically cannotbothbeodd,soassumewithoutlossofgeneralitythatanevennumberoftilesareplacedhorizontally.Color thesquaresinorderusingthecolorsred,blue,green,yellowinthatorderrepeatedly,startingintheupperleftcorner andproceedingrowbyrow,fromlefttorightineachrow.Thenitisclearthateveryhorizontallyplacedtilecoversone squareofeachcolorandeachverticallyplacedtilecoverseitherzeroortwosquaresofeachcolor.Itfollowsthatinthis tilinganevennumberofsquaresofeachcolorarecovered.Butthiscontradictsthefactthatthereare25squaresofeach color.Thereforenosuchcoloringexists.

SUPPLEMENTARYEXERCISESFORCHAPTER1

2. Thetruthtableisasfollows.

4.a) Theconverseis“IfIdrivetoworktoday,thenitwillrain.”Thecontrapositiveis“IfIdonotdrivetoworktoday,then itwillnotrain.”Theinverseis“Ifitdoesnotraintoday,thenIwillnotdrivetowork.”

b) Theconverseis“If x ≥ 0then |x| = x .”Thecontrapositiveis“If x < 0then |x| ≠ x .”Theinverseis“If |x| ≠ x , then x < 0.”

c) Theconverseis“If n2 isgreaterthan9,then n isgreaterthan3.”Thecontrapositiveis“If n2 isnotgreaterthan9, then n isnotgreaterthan3.”Theinverseis“If n isnotgreaterthan3,then n2 isnotgreaterthan9.”

6. Theinverseof p → q is ¬p → ¬q .Thereforetheinverseoftheinverseis ¬¬p → ¬¬q ,whichisequivalentto p → q (theoriginalproposition).Theconverseof p → q is q → p .Thereforetheinverseoftheconverseis ¬q → ¬p ,which isthecontrapositiveoftheoriginalproposition.Theinverseofthecontrapositiveis q → p ,whichisthesameasthe converseoftheoriginalstatement.

8. Let t be“Sergeitakesthejoboﬀer,”let b be“Sergeigetsasigningbonus,”andlet h be“Sergeiwillreceiveahigher salary.”Thegivenstatementsare t → b , t → h , b → ¬h ,and t .Bymodusponenswecanconclude b and h fromthe ﬁrsttwoconditionalstatements,andthereforewecanconclude ¬h fromthethirdconditionalstatement.Wenowhave thecontradiction h ∧¬h ,sothesestatementsareinconsistent.

10. Wemakeatableoftheeightpossibilitiesfor p , q ,and r ,showingthetruthvaluesofthethreepropositions.

pqrp → q ¬(p ∨ r) ∨ qq

Ifwelookattheﬁrstrowofthetable,weseethatifthestudentacceptsallthreepropositions,thentheresultingcommitmentsareconsistent,becausethepropositionsarealltrueinthiscaseinwhich p , q ,and r arealltrue.Similarly,looking atthesixthrowofthetable,where p and r arefalsebut q istrue,weseethatastudentwhoacceptstheﬁrsttwopropositionsandrejectsthethirdalsowins.Scanningtheentiretable,weseethatthewinninganswersareaccept-accept-accept, reject-reject-accept,accept-accept-reject,andaccept-reject-reject.

12. Aswesawfromtheexamplesinthepreviousexercises,onewinningstrategyisjusttoassumethatallthevariablesare trueandanswer“accept”or“reject”accordingtowhetherthegivenpropositionistrueorfalse.

14. Aknightwouldneverclaimthatsheisaknave,soweknowthatAnitaisaknave.Becausesheislyingandtheﬁrst partofherconjunctionistrue,itmustbethesecondpartthatisfalse,andsoBohanmustbeaknave.IfCarmenwerea knight,thenBohan’sstatementwouldbetrue;becauseBohanisaknave,weknowthatthatcannotbe,soweconclude thatCarmenisalsoaknave.

16. If S isaproposition,thenitiseithertrueorfalse.If S isfalse,thenthestatement“If S istrue,thenunicornslive”is vacuouslytrue;butthisstatement isS ,sowewouldhaveacontradiction.Therefore S istrue,sothestatement“If S is true,thenunicornslive”istrueandhasatruehypothesis.Henceithasatrueconclusion(modusponens),andsounicorns live.Butweknowthatunicornsdonotlive.Itfollowsthat S cannotbeaproposition.

18. Fromthegiveninformationweknowthat p1 , p3 , p5 ,...aretrueand p2 , p4 , p6 ,...arefalse.Therefore pi ∧ pi+1 is alwaysfalse,andsothedisjunction ⋁100 i=1 (pi ∧ pi+1 )isalsofalse.Ontheotherhand, pi ∨ pi+1 isalwaystrue,andsothe conjunction ⋀100 i=1 (pi ∨ pi+1 )isalsotrue.

20.a) Theansweris ∃xP(x)ifwedonotreadanysigniﬁcanceintotheuseoftheplural,and,ifwedo,theanswerwouldbe ∃x∃y(P(x) ∧ P(y) ∧ x ≠ y).

b) ¬∀xP(x),or,equivalently, ∃x¬P(x) c) ∀yQ(y)

d) ∀xP(x)(theclasshasnothingtodowithit) e) ∃y¬Q(y)

22. Thegivenstatementtellsusthatthereareexactlytwoelementsinthedomain.Thereforethestatementwillbetrueas longaswechoosethedomaintobeanythingwithsize2,suchastheUnitedStatespresidentsnamedBush.

24. Wewanttosaythatforevery y ,theredonotexistfourdiﬀerentpeopleeachofwhomisthegrandmotherof y .Thuswe have ∀x¬∃

26.a) Sincethereisnorealnumberwhosesquareis 1,itistruethatthereexistexactly0valuesof x suchthat x2 =−1. b) Thisistrue,because0istheoneandonlyvalueof x suchthat |x| = 0. c) Thisistrue,because √2and √2aretheonlyvaluesof x suchthat x2 = 2. d) Thisisfalse,becausetherearemorethanthreevaluesof x suchthat x = |x| ,namelyallpositiverealnumbers.

28. Letusassumethehypothesis.Thismeansthatthereissome x0 suchthat P(x0 , y)holdsforall y .Thenitiscertainly truethatforall y thereexistsan x suchthat P(x, y)istrue,sinceineachcasewecantake x = x0 .Notethattheconverse isnotalwaysatautology,sincethe x in ∀y∃xP(x, y)candependon y

30. No.Hereisanexample.Let P(x, y)be x > y ,wherewearetalkingaboutintegers.Thenforevery y theredoesexist an x suchthat x > y ;wecouldtake x = y + 1,forexample.However,theredoesnotexistan x suchthatfor everyy , x > y ;inotherwords,thereisnosuperlargeinteger(iffornootherreasonthanthatnointegercanbelargerthanitself).

32.a) Itwillsnowtoday,butIwillnotgoskiingtomorrow.

b) Somepersoninthisclassdoesnotunderstandmathematicalinduction.

c) Allstudentsinthisclasslikediscretemathematics.

d) Thereissomemathematicsclassinwhichallthestudentsstayawakeduringlectures.

34. Let W (r)meansthatroom r ispaintedwhite.Let I (r, b)meanthatroom r isinbuilding b .Let L(b, u)meanthatbuilding b isonthecampusofUnitedStatesuniversity u .Thenthestatementisthatthereissomeuniversity u andsomebuilding onthecampusof u suchthateveryroomin b ispaintedwhite.Insymbolsthisis ∃

W (r))).

36. Tosaythatthereareexactlytwoelementsthatmakethestatementtrueistosaythattwoelementsexistthatmakethe statementtrue,andthateveryelementthatmakesthestatementtrueisoneofthesetwoelements.Morecompactly,we canphrasethelastpartbysayingthatanelementmakesthestatementtrueifandonlyifitisoneofthesetwoelements. Insymbolsthisis ∃x∃y(x ≠ y ∧∀z(P(z) ↔ (z = x ∨ z = y))).InEnglishwemightexpresstheruleasfollows.The hypothesesarethat P(x)and P(y)arebothtrue,that x ≠ y ,andthatevery z thatsatisﬁes P(z)mustbeeither x or y Theconclusionisthatthereareexactlytwoelementsthatmake P true.

38. Wegiveaproofbycontraposition.If x isrational,then x = p∕q forsomeintegers p and q with q ≠ 0.Then x3 = p3 ∕q3 , andwehaveexpressed x3 asthequotientoftwointegers,thesecondofwhichisnotzero.Thisbydeﬁnitionmeansthat x3 isrational,andthatcompletestheproofofthecontrapositiveoftheoriginalstatement.

40. Let m bethesquarerootof n ,roundeddownifitisnotawholenumber.(InthenotationtobeintroducedinSection2.3, weareletting m = ⌊√n⌋ .)Wecanseethatthisistheuniquesolutioninacoupleofways.First,clearlythediﬀerent choicesof m correspondtoapartitionof N ,namelyinto {0} , {1,2,3} , {4,5,6,7,8} , {9,10,11,12,13,14,15} ,.... Soevery n isinexactlyoneofthesesets.Alternatively,takethesquarerootofthegiveninequalitiestogive m ≤ √n < m + 1.That m isthentheﬂoorof √n (andthat m isunique)followsfromstatement(1a)ofTable1inSection2.3.

42. Aconstructiveproofseemsindicated.Wecanlookforexamplesbyhandorwithacomputerprogram.Thesmallest onestobefoundare50 = 52 + 52 = 12 + 72 and65 = 42 + 72 = 12 + 82 .

44. Weclaimthatthenumber7isnotthesumofatmosttwosquaresandacube.Theﬁrsttwopositivesquaresare1and4, andtheﬁrstpositivecubeis1,andthesearetheonlynumbersthatcouldbeusedinformingthesum.Clearlynosumof threeorfeweroftheseis7.Thiscounterexampledisprovesthestatement.

46. Wegiveaproofbycontradiction.If √2 + √3wererational,thensowouldbeitssquare,whichis5 + 2√6.Subtracting 5anddividingby2thenshowsthat √6isrational,butthiscontradictsthetheoremwearetoldtoassume.

Discrete Mathematics And Its Applications 2025 Release 1St Rosen Solutions Manual

Solutions Manual for Discrete

and Its Applications 2025

1st Edition by Rosen

CHAPTER1

TheFoundations:LogicandProofs

SECTION1.2ApplicationsofPropositionalLogic

SECTION1.4PredicatesandQuantiﬁers

SECTION1.5NestedQuantiﬁers

SECTION1.6RulesofInference

StepReason

StepReason

StepReason

StepReason

SECTION1.7IntroductiontoProofs

SECTION1.8ProofMethodsandStrategy

SUPPLEMENTARYEXERCISESFORCHAPTER1