Chapter IProblems from I 2.5I2.5.1. Userosternotationtodesignatethegivensubsetsof R.(a)A = {x : x21 = 0}isequivalenttoA = {1, 1}.(b)B= {x : (x 1)2= 0}isequivalenttoB= {1}.(c)C= {x : x + 8 = 9}isequivalenttoC= {1}.(d) Given D = {x : x32x2+x = 2} observe that x32x2+x2 = 0 factors to (x2)(x2+1) = 0.Sox 2 = 0orx2+ 1 = 0. Noticethatthesecondfactorhasnorealroots. ThereforeD = {2}.(e)E= {x : (x + 8)2= 92}isequivalenttoE= {17, 1}.(f )F= {x : (x2+ 16x)2= 172}isequivalenttoF= {17, 1, 8 47, 8 +47}.I2.5.2. ListalltheinclusionrelationsthatholdamongthesetsA, B, C, D, E, F.Wehavethefollowinginclusionrelationsamongthegivensets:B CandC B= B= CB= C AB= C EB= C FE FWealsohavethetrivialcaseswhereeachsetisasubsetofitself.I2.5.3. LetA = {1}andB= {1, 2}. Discussthevalidityofthegivenstatements.(a)ItistruethatA Bsincex A = x Bforeveryx A.(b)ItistruethatA Bforthesamereasonas(a).(c)Thisisfalsebecause {1}/ B.1(d)Itistruethat1 A.(e)Itisfalsethat1 Asince1isanelementofA,butnotasubsetofA.(f )Itisfalsethat1 Bforthesamereasonasin(e).I2.5.4. Repeatexercise3ifA = {1}andB= {{1}, 1}.(a)A B;True.(b)A B;True.(c)A B;True.(d)1 A;True.(e)1 A;False.(f )1 B;False.I2.5.5. GiventhesetS= {1, 2, 3, 4},displayallsubsetsofS.1. {1}2. {1, 2}3. {1, 2, 3}4. {1, 2, 3, 4}5. {2}6. {2, 3}7. {2, 3, 4}8. {3}9. {3, 4}10. {3, 1}11. {4}12. {4, 1}13. {4, 2}14. {4, 1, 3}15. {4, 1, 2}16. I 2.5.6. Let A={1, 2}, B={{1}, {2}}, C={{1}, {1, 2}}, D={{1}, {2}, {1, 2}}. Arethefollowingstatementstrueorfalse?(a)A = B;False. 1 Aand1/ B.2(b)A B;Falseforthesamereasonas(a).(c)A C;Falsesince1 Abut1/ C.(d)A C;True. {1, 2} C.(e)A D;Trueasin(d).(f )B C;False, {2} Bbut {2}/ C.(g)B D;True,eachelementofBisinD.(h)B D;False.(i)A D;True.I2.5.7. Provethefollowingpropertiesofsetequality:(a) {a, a} = {a};Proof. Observethat {a, a} {a}and {a} {a, a}. Therefore {a, a} = {a}. (b) {a, b} = {b, a};Proof. Notethatx {a, b}ix {b, a}. Therefore {a, b} {b, a}and {b, a} {a, b}. Hence{a, b} = {b, a}. (c) {a} = {b, c}ia = b = c.Proof. (=)Supposethat {a} = {b, c}. Thenx {a}ix {b, c}. Sinceaistheonlyelementof{a},wemusthavea {b, c}andanyelementx = aisnotin {a, b}. Hencea = b = c.(=) Supposethat a=b=c. Thencertainly {a}= {a}. By(a)wehave {a}= {a, a}. Byhypothesisweobtain {a} = {b, c}. I2.5.8. Provethecommutativelawsofsetrelations.i. A B= B A;Proof. ThisproofisprovidedasanexampleonPg16ofthetext. ii. A B= B A;Proof. Leta A B. Thena Aanda B. Clearlya B A. Theotherdirectionisentirelysimilar. HenceA B= B AsinceA B B AandB A A B. 3I2.5.9. Provetheassociativelaws:A (B C) = (A B) CandA (B C) = (A B) C.Proof. ToproveA (B C) = (A B) C,wemustshowthatA (B C) (A B) Cand(AB) C A(B C). Letx A(B C). Thenx Aandx B C= x A, B, C. Sox A Bandsincex Cwehavex (A B) C. ThereforeA (B C) (A B) C. Theproof of (AB) C A(BC) is entirely similar. We then have A(BC) = (AB) C. I2.5.10. Provethedistributivelaws:A (B C) = (A B) (A C)andA (B C) = (A B) (A C).Proof. We rst showthat A (B C) =(A B) (A C). Let a A (B C). Thena Aandeithera Bora C. Inotherwords, wehavea A Bora A C. HenceA (B C) (A B) (A C). Nowleta (A B) (A C). Thena A Bora A C.Inotherwords,weknowa AandthataisineitherBorC. Hencea A (B C). Therefore(A B) (A C) A (B C). Thuswehavethedesiredequality.Nowweshowthat A (B C) =(A B) (A C). Let a A (B C). Thena Aora B C. SoweknowthataisinbothBandCora A. Soa (A B) (A C). Nowsupposethata (A B) (A C). Thena A Banda A C. Soa AoraisinbothBandC. Hencea A (B C). Thisprovestheequality. I2.5.11. ProvethatA A = AandA A = A.Proof. Fortherstequality, wemustshowthatA A AandA A A. Fortheformer, leta A A. Thena Aanda Atriviallyimpliesthata Asotherstinclusionistrue. Nowleta A. Thencertainlya A A. Sotheinclusiongoesbothways,givingthedesiredequality.For the second equality, we must show that AA A and A AA. Let a AA. Then a A4ora Aimpliesthata A. SoA A A. Nowsupposethata A. Certainlya A A. ThusA A = A. I2.5.12. ProvethatA A BandA B A.Proof. WerstprovethatA A B. Leta A. Notethenthata A BsinceaisinoneofAorB. Hencetheinclusionholds. TheproofoftheotherinclusionisomittedsinceitmaybefoundasanexampleonPg.16ofthetext. I2.5.13. ProvethatA = AandA = .Proof. WerstshowthatA =A. Leta A . Thena Aora . Thelattercannotbetruesince hasnoelements. Hencea Awhenevera A . Nowsupposethata A. Clearlya A . Hencetheequalityholds.Wenowshowthat A = . For contradiction, supposethat A =BwhereB = . Leta B. Then a A. But this produces a contradiction since there is no element in . Hence wemusthaveA = . I2.5.14. ProvethatA (A B) = AandthatA (A B) = A.Proof. WerstshowthatA (A B)=A. Leta A (A B). Thena Aora A B.Clearlya A. Fortheotherdirection,supposethata A. Thena A DwhereDisanyset.Thereforea A (A B). Sotheequalityholds.WenowshowthatA (A B) = A. Supposethata A (A B). Thena Aanda A B.ThusA (A B) A. Nowsupposethata A. Thena A DforanysetD. Thereforea A (A B)andwehaveA (A B). Thustheequalityholds. I2.5.15. ProvethatifA CandB C,thenA B C.Proof. LetA CandB C. Thenx A = x C. Similarlyy B= y C. Soifa Aora Bthena CimpliesthatA B C. I2.5.16. ProvethatifC AandC B,thenC A B.5Proof. Suppose that C A and C B. Then c C =c A and c B. hence C AB. I2.5.17.(a)IfA BandB C,provethatA C.Proof. Suppose that A BandBC. Thena A = a B = a C. Hencea A=a CandwehaveA C. (b)IfA BandB C,provethatA C.Proof. Theproofisentirelysimilarto(a). (c)WhatcanyouconcludeifA BandB C?WecanonlyconcludethatA C.(d)Ifx AandA B,isitnecessarilytruethatx B?Yes,itisnecessarilytrue.(e)Ifx AandA B,isitnecessarilytruethatx B?It is not necessarilytrue. Let A= {0}andB= {1, {0}}. Here, wehave0 Abut 0 / B.I2.5.18. ProvethatA(B C) = (AB) (AC).Proof. It suces to show that A(BC) (AB)(AC) and (AB)(AC) A(BC).Letx A (B C). Thenx Aandx/ B C. Sox Aexactlywhenx/ Bandx/ C.Therefore x (AB)(AC) whenever x A(BC). Hence A(BC) (AB)(AC).Nowlet x (A B) (A C). Thenx (A B) or x (A C). Sox Asuchthatx / B C. Therefore xA (B C) =(A B) (A C) A (B C). ThusA(B C) = (AB) (AC). I2.5.19. LetFbeaclassofsets. ThenB _AFA =

AF(B A) and B

AFA =_AF(B A).6Proof. WerstshowthatB

AFA=

AF(B A). Leta B

AFA. Thena Bsuchthata/ AFA. Hence, a (B A)foreveryA F. Inotherwordsa AF(B A). Nowsupposethata

AF(B A). Soa (B A)foreveryA F. Inotherwords,a Bsuchthata/ AforanyA F. Hencea B

AFA. Thustheequalityholds.WenowshowthatB

AFA =

AF(B A). Supposethata B

AFA. Thena Bsuchthata/ AFA. Soa (B A)forsomeA F. Thusa AF(B A). Nowsupposethata

AF(BA). Then a is in at least one BA for A F. Then a cannot be a common elementtoeveryA F. Inotherwords, a Bsuchthata/ AFA. Soa B

AFA, provingtheequality. 20. (a) Prove that one of A(BC) = (AB) Cand A(BC) = (AB) Cis alwaysrightandtheotherissometimeswrong. (b)Stateanadditionalnecessaryandsucientconditionfortheformulawhichissometimesincorrecttobealwaysright.(a)Therstequalityisnotalwaystrue.Proof. LetA= {1, 2, 3}, B= {4, 5}, andC= {10, 11}. ThenA (B C)=AandA B=A.NoticethatA C= {1, 2, 3, 10, 11}= A = A(B C). ThenitmustbetruethatA(B C) = (AB) C.Proof. Let a A (B C). Thena (A B) (A C) byproblem19. It follows thata (AB)Csince this is also the complement of A with respect to Band C; this case just takesthecomplementA Brstandthenremovesanyremainingelementsof CfromA. Sosupposethata (AB) C. Thereversereasoningoftherstcaseshowsthata A(B C). (b)TheequalityholdsifandonlyifA B= andC AorA = B= C.7Problems from I 3.3I3.3.1. ProveTheoremsI.5throughI.15.ThmI.5:a(b c) = ab ac.Proof. Observethatb = b c +c = ab = a ((b c) +c)= ab = a(b c) +ac= ab ac = a(b c).

ThmI.6:0 a = a 0 = 0.Proof. Bycommutativity,weneedonlytoshowthata 0 = 0. Byaxiom51 1 = 0. Hencea 0 = a(1 1)= a 1 a 1 [byThmI.5]= a a= 0.

ThmI.7:Ifab = acanda = 0,thenb = c.Proof. Letab = acanda = 0. Byaxiom6 x Rs.t. xa = ax = 1. Thenwehave(xa)b = (xa)cwhichsimpliesto1 b = b = c = 1 c. ThmI.8:Givenaandbwitha = 0,thereisexactlyonexs.t. ax = b.Proof. Given b and a = 0, choose ys.t. ay= 1 and let x = yb. Then ax = a(yb) = (ay)b = 1 b = b.Thereforethereisatleastxs.t. ax = b. ButbyThmI.7thisxisunique. ThmI.9:Ifa = 0,thenba= ba1.Proof. ByThmI.8, !x Rs.t. ax=b. ByThmI.7a1ax=a1b. Byassociativity(a1a)x=a1b, whichgivesx=ba1byaxioms1,4,6. Similarly,1aax=1abbyThmI.7. Bythesamestepsasabove,weobtainx = b1a=ba. Sincexisunique,wehaveba= ba1. ThmI.10:Ifa = 0,then(a1)1= a.Proof. Trivially, a = a. By axiom 6 we have a = (aa1)a. By Thm I.7 we have 1 = aa1. By axioms2,6andThmI.7,1 (a1)1= a_a1(a1)1_= a 1. Axioms4and6giveus(a1)1= a. ThmI.11:Ifab = 0,thena = 0orb = 0.8Proof. Letab = 0. Byaxiom6 y Rs.t. by= 1. Thena = 1 a = a 1 = a(by) = (ab)y= 0 y= 0.Axiom6guaranteesb = 0. Similarly, x Rs.t. ax = 1fora = 0. Thenwehaveb = 1 b = (ax)b = (xa)b = x(ab) = x 0 = 0.Thereforeeitheraorbmustbe0. ThmI.12:Provethat(a)b = (ab)and(a)(b) = ab.Proof. Werst provethat (a)b = (ab). Observethat (a + a)b =(a)b + ab =0. Then(a)b = (ab).Nowweshowthat(a)(b) = ab.Observethat0 = (a) 0 = (a)(b +b) = (a)(b) + (a)b = (a)(b) (ab)wherethenalequalityfollowsfromourpreviousresult. Thenweobtain(a)(b) = ab. ThmI.13:Ifb, d = 0,then(a/b) + (c/d) = (ad +bc)/(bd).Proof. Letb, d = 0. Then1 = (bd)1(bd)andso(ad +bc) = (ad +bc)(bd)1(bd). Multiplyingbothsidesbyb1d1gives(b1d1)(ad +bc) = (ad +bc)(bd)1. Distributingonthelefthandsidegivesusab1+cd1= (ad +bc)(bd)1. ThmI.14:Ifb, d = 0,then(ab1)(cd1) = (ac)(bd)1.Proof. Suppose b, d=0. Certainly1 1 =1. Then(bb1)(dd1) =(bd)(bd)1. This impliesthat b1d1=(bd)1andso (ac)b1d1=(ac)(bd)1. Rearranging gives the desiredresult(ab1)(cd1) = (ac)(bd)1. ThmI.15:Ifb, c, d = 0,then(ab1)(cd1) = (ad)(bc)1.Proof. I3.3.2. Provethat 0 = 0.Proof.0 = 0 1 = (0 1) [ByThmI.12]= 0(1) [ByThmI.12]= 0.

9I3.3.3. Prove11= 1.Proof. Observethat1 = 1 (1 11)= (1 1) 11= 1 11= 11.

I3.3.4. Provethatzerohasnoreciprocal.Proof. For contradiction, suppose that zero does have a reciprocal. Then y R s.t. 0y= y0 = 1.ThiscontradictsThmI.6since0 a = a 0 = 0 a R. Thuszerohasnoreciprocal. I3.3.5. Prove (a +b) = a b.Proof. Observethat(a +b) = (a +b) 1= (a +b) (1) [ThmI.12]= a(1) +b(1)= (a 1) + ((b 1)) [ThmI.12]= a b.

I3.3.6. Prove (a b) = a +b.Proof.(a b) = (a b) 1= (a b) (1)= a(1) + (b)(1)= a +b [ThmI.12].

10I3.3.7. Prove(a b) + (b c) = a c.Proof. Observethat(a b) + (b c) = (a + (b) + (b + (c))= a + ((b) +b) + (c)= (a + 0) + (c)= a c.

I3.3.8. Provethatifa, b = 0,then(ab)1= a1b1.Proof. Supposea, b = 0. Then(ab)1 (ab) = 1 =(ab)1a(b b1) = 1 b1=(ab)1(a 1) = b1=a1((ab)1a) = a1b1=(a1a)(ab)1= a1b1=1 (ab)1= a1b1=(ab)1= a1b1.

I3.3.9. Provethatifb = 0then (ab1) = (a)b1= a(b1).Proof. Letb =0. Then(a)b1= (ab1)byThmI.12anda(b1)= (ab1)byThmI.12.Hence (ab1) = (a)b1= a(b1). I3.3.10. Provethatifb, d = 0,then(ab1) (cd1) = (ad bc)(bd)1.11Proof. Letb, d = 0. Thenobservethat(ad bc)(bd)1= (ad)(bd)1(bc)(bd)1= (ad)(b1d1) (bc)(b1d1) [by3.3.8]= (ab1)(dd1) (cd1)(bb1)= (ab1) 1 (cd1) 1= ab1cd1.

12Problems from I 3.5I3.5.1. ProveTheoremsI.22throughI.25.ThmI.24:Ifab > 0theneitherbotha, barepositiveorbotharenegative.Proof. Seekingacontradiction,supposethatab > 0anda, b Rs.t. aandbareofoppositesign.WLOG, leta0. Thena= xforsomex R+. Thenab= xb= (xb) >0byThmI.12. Byaxiom7, wehavexb R+. Then (xb) 0.Proof. Supposethataandbarenotbothzero. First, leta, b =0. Thena2>0andb2>0soa2+b2> 0. Sincea2> 0andb2> 0,thena2+b2> 0ifeither,butnotboth,ofa, bare0. I3.5.9. Provethereisnoa Rs.t. x a x R.Proof. Forcontradiction, supposethereissuchana R. Thenwemaychoose2a=xandhavex = 2a a. Thisimpliesthata a2whichisthedesiredcontradiction. I3.5.10. Provethatifxsatsies0 x < h h R+,thenx = 0.Proof. Forcontradiction,supposethatxsatises0 x < h h R+andx = 0. Thenwemayseth =x2andwehave0 x 1n. I3.12.4. Provethatifx R,thenthereisexactlyonen Zsatisfyingn x < n + 1.Proof. Letx Rbearbitrary. I3.12.7. Ifxisrationalandnonzeroandyisirrational,provethatx +y,x y,xy,x/y,andy/xareallirrational.Proof. Letx Qs.t. x = 0andletybeirrational. Wewillrstshowthatx +yisirrational. Forcontradiction, supposethatx + y Q. Thenx + y=abforsomea, b Zwhereb =0. Thenwehavey=ab xwhichwemayrewriteasy=ab cdforsomec, d Zwhered =0sincex Q.ByThmI.13wehavey=adbcbd. Thisisthedesiredcontradictionsincetherighthandsideisarationalnumber. Theproofofx ybeingirrationalfollowsfromthepreviousresultsinceitisthesamecasewithx + (y)and yisstillirrational.Nowweshowthatxyisalsoirrational. Again, supposethatxyisrational forcontradiction.Thenxy=abforsomea, b Z. Hencewehavey=xab. Sincex Qwehavey=acbdforsome15b, d Zwhered = 0. Thereforey Q,whichisthedesiredcontradiction. Theproofsofx/yandy/xfollowsincetheyarethesamecaseasabovewithx(y1)andy(x1)andx1 Qandy1isstillirrational. I3.12.8. Isthesumorproductoftwoirrationalnumbersalwaysirrational?Theanswerisnoineithercase. Observethat2 + (2) = 0and2 2 = 2.I3.12.9. If x, y R and x < y, prove that z RQ s.t. x < z< yand hence innitely manysuchzexist.Proof. Letx, y Rs.t. x < y. I3.12.10. Anintegernisevenifn = 2mforsomem Zandoddifn + 1iseven. Provethefollowingstatements:(a)Provethatanintegercannotbebothevenandodd.Proof. Forcontradiction,supposethataninteger,sayn,canbebothoddandeven. Thenn = 2kandn = 2l 1forsomek, l Z. Sowemusthave2k = 2l 1. Butthisgivesk = l 12. Thereforekcannotbeaninteger,givingthedesiredcontradiction. (b)Provethateveryintegeriseitherevenorodd.Proof. Thisfollowsfromthecontrapositiveof(a). (c)(i)Provethatthesumorproductoftwoevenintegersiseven.Proof. Seekingacontradiction,supposethatthesumandproductoftwoevenintegers,sayaandb,isodd. Byhypothesisa = 2kandb = 2lforsomek, l Z. Thena +b = 2k + 2l = 2(k +l)andab = (2k)(2l) = 2(2kl). Bothoftheseareeven,givingthedesiredcontradiction. (c)(ii)Whatcanyousayaboutthesumorproductoftwooddintegers?Theproductoftwooddintegersisodd. However,thesumoftwooddintegersiseven.Proof. Leta, b Zs.t. a = 2k 1andb = 2l 1forsomek, l Z. Observethatab = (2k1)(2l 1) = 2(2kl kl)+1. Hence ab is odd. Furthermore, a+b = (2k1)+(2l 1) =2(k +l 1)whichiseven. Hencethea +biseven. (d)(i)Provethatifn2iseven,thensoisn.Proof. Let n be odd. Then n = 2k 1) for some k Z. Therefore n2= (2k 1)2= 2(2k22k) +1.Son2isalsoodd. Itfollowsthatifn2iseven,thenniseven. (d)(ii)Provethatifa2= 2b2wherea, b Z,thenbothaandbareeven.16Proof. Let a2= 2b2for a, b Z. Then a2is even. By part (i), a is even. Therefore a = 2kfor somek Z. Thena2= (2k)2= 4k2= 2b2gives2k2= b2whichimpliesthatb2iseven. Bypart(i),biseven. (e)Provethateveryrational numbercanbeexpressedintheforma/b, wherea, b Z, atleastoneofwhichisodd.Proof. Bydenitionof Q,everyx Qcanbewrittenx =abforsomea, b Z. I3.12.11. Provethatthereisnorationalnumberwhosesquarerootis2.Proof. Forcontradiction,supposethereisarationalnumber(a/b)2= 2inwherea, b Zatleastoneofwhichisodd. Thena2=2b2impliesthata2=4k2forsomek Z. Then4k2=2b2gives2k2=b2implyingthatb2isalsoeven. Byhypothesis, atleastoneof aandbisodd. Buttheproductoftwooddnumbersisodd. Thereforethesquareofanoddnumbercannotbeeven. Thisisthedesiredcontradictionsinceourhypothesisleadsustoa2andb2beingeven. 17Problems from I 4.4I4.4.1. Provethefollowingformulasbyinduction:a. 1 + 2 + 3 + +n = n(n + 1)/2Proof. Weproceedbyinductiononn. As abasecase, let n=1. Observethat wethenhave1 = 1(1 +1)/2 = 1. Thisshowsthattheformulaistrueforn = 1. Nowsupposeitistrueforsomen = k Z+. Observethat1 + 2 + 3 + +k =k(k + 1)2byhypothesis,1 + 2 + +k + (k + 1) =k(k + 1)2+ (k + 1)byaddingk + 1toeachside,and1 + 2 + +k + (k + 1) =k2+ 3k + 22=(k + 1)(k + 2)2via simplication. Hence the formula is true for k +1 if kis true. By the principle of mathematicalinduction,theformulaistrue n Z+. (b)1 + 3 + 5 + + (2n 1) = n2.Proof. Refer to the given formula as the assertion A(n). Let n = 1. Then 1 = 12which is certainlytrue,soA(1)holds. NowsupposethatA(k)holdsforsomen = k Z+. Wethenhave1 + 3 + 5 + + (2k 1) = k2byourinductivehypothesis. Then1 + 3 + 5 + + (2k 1) + (2k + 1) = k2+ (2k + 1)= k2+ 2k + 1= (k + 1)2.HenceA(k) =A(k + 1). ThereforeA(n)holdsforalln Z+bytheprincipleofmathematicalinduction. (c)13+ 23+ 33+ +n3= (1 + 2 + 3 + +n)2.Proof. Letn = 1. Then13= 1 = 12. Nowsupposetheequalityholdsforsomen = k Z+. Thenwehave13+ 23+ +k3= (1 + 2 + +k)2(1)byhypothesis. Bypart(a),noticethattherighthandsideoftheequationgives(1 + 2 + +k)2=k2(k + 1)24.18Henceitsucestoshowthat13+ 23+ +k3+ (k + 1)3=(k + 1)2(k + 2)24.Startingfromtherighthandsideof(1),weobtain13+ 23+ +k3+ (k + 1)3= (1 + 2 + +k)2+ (k + 1)3=k2(k + 1)24+ (k + 1)3=k2(k + 1)2+ 4(k + 1)34=(k + 1)2(k2+ 4k + 4)4=(k + 1)2(k + 2)24Thisisthedesiredresult. Thustheequalityholdsforalln Z+. (d)Wewanttoshowthat13+ 23+ + (n 1)3