Issuu

Problem-SolvingCorner:Quantiﬁers

1.ThestatementofExample1.6.6is

∀x∃y(x + y =0)

AswaspointedoutinExample1.6.6,thisstatementistrue.Now

∀x∀y(x + y =0) isfalse;acounterexampleis x = y =1.Also

∃x∀y(x + y =0) isfalsesince,givenany x,if y =1 x,then x + y =0.

2.Yes;thestatement ∀m∃n(m<n)withdomainofdiscourse Z × Z ofExample1.6.1alsosolvesproblems (a)and(b).

Section2.1

2.Forall x,forall y, x + y = y + x

3.An isoscelestrapezoid isatrapezoidwithequallegs.

5.Themediansofanytriangleintersectatasinglepoint.

6.If0 <x< 1and ε> 0,thereexistsapositiveinteger n satisfying xn <ε.

8.Let m and n beoddintegers.Thenthereexist k1 and k2 suchthat m =2k1 +1and

=2k2 +1.Now m + n =(2k1 +1)+(2k2 +1)=2(k1 + k2 +1)

Therefore, m + n iseven.

9.Let m and n beevenintegers.Thenthereexist k1 and k2 suchthat m =2k1 and n =2k2.Now mn =(2k1)(2k2)=2(2k1k2). Therefore, mn iseven.

11.Let m beanoddintegerand n beaneveninteger.Thenthereexist k1 and k2 suchthat m =2k1 +1 and n =2k2.Now mn =(2k1 +1)(2k2)=2(2k1k2 + k2) Therefore, mn iseven.

12.Let m and n beintegerssuchthat m and m + n areeven.Thenthereexist k1 and k2 suchthat m =2k1 and m + n =2k2.Now n =(m + n) m =2k2 2k1 =2(k2 k1) Therefore, n iseven.

14.Let x and y berationalnumbers.Thenthereexistintegers m1,n1,m2,n2 suchthat x = m1/n1 and y = m2/n2.Now xy =(m1m2)/(n1n2).Therefore xy isrational.

15.Let x beanonzerorationalnumber.Thenthereexistintegers m =0and n =0suchthat x = m/n Now1/x = n/m.Therefore1/x isrational.

17.Let m =3k1 +2and n =3k2 +2beintegersoftheprescribedfrom.Then mn =9k1k2 +6k1 +6k2 +4=3(3k1k2 +2k1 +2k2 +1)+1 isoftheform3k3 +1,where k3 =3k1k2 +2k1 +2k2 +1.

19. x 0+0= x 0because b +0= b forallrealnumbers b = x · (0+0)because b +0= b forallrealnumbers b = x · 0+ x · 0because a(b + c)= ab + ac forallrealnumbers a,b,c

Taking a = c = x · 0and b =0,theprecedingequationbecomes a + b = a + c;therefore,0= b = c = x · 0.

20.Wemusthave X = Y .Toprovethis,supposethat x ∈ X.Since Y isnonempty,choose y ∈ Y .Then (x,y) ∈ X × Y .Since X × Y = Y × X,(x,y) ∈ Y × X.Therefore x ∈ Y .Similarly,if x ∈ Y ,then x ∈ X.Thus X = Y

22.Let x ∈ X.Then x ∈ X ∪ Y .Therefore X ⊆ X ∪ Y

23.Let x ∈ X ∪ Z.Then x ∈ X or x ∈ Z.If x ∈ X,since X ⊆ Y , x ∈ Y .Therefore x ∈ Y ∪ Z.If x ∈ Z, then x ∈ Y ∪ Z.Ineithercase, x ∈ Y ∪ Z.Therefore X ∪ Z ⊆ Y ∪ Z

25.Let x ∈ Z Y .Then x ∈ Z and x/ ∈ Y .Now x cannotbein X,forif x ∈ X,since X ⊆ Y ,then x ∈ Y , whichisnotthecase.Since x ∈ Z and x/ ∈ X, x ∈ Z X.Therefore Z Y ⊆ Z X

26.Let x ∈ Y (Y X).Then x ∈ Y and x/ ∈ Y X.Since x ∈ Y ,wemusthave x ∈ X (if x/ ∈ X,we wouldhave x ∈ Y X).Therefore Y (Y X) ⊆ X

Nowlet x ∈ X.Then x/∈ Y X.Since X ⊆ Y , x ∈ Y .Thus x ∈ Y (Y X).Therefore X ⊆ Y (Y X).Wehaveshownthat Y (Y X)= X

28.Let Z ∈P(X) ∪P(Y ).Then Z ∈P(X)or Z ∈P(Y ).If Z ∈P(X),then Z isasubsetof X and,thus, Z isalsoasubsetof X ∪ Y .Therefore Z ∈P(X ∪ Y ).Similarly,if Z ∈P(Y ), Z ∈P(X ∪ Y ).Ineither case, Z ∈P(X ∪ Y ).Therefore P(X) ∪P(Y ) ⊆P(X ∪ Y ).

29.Let Z ∈P(X ∩ Y ).Then Z isasubsetof X ∩ Y .Therefore Z isasubsetof X andasubsetof Y .Thus Z ∈P(X) ∩P(Y ).Wehaveprovedthat P(X ∩ Y ) ⊆P(X) ∩P(Y ).

Let Z ∈P(X) ∩P(Y ).Then Z ∈P(X)and Z ∈P(Y ).Since Z ∈P(X), Z isasubsetof X.Since Z ∈P(Y ), Z isasubsetof Y .Since Z isasubsetof X and Y , Z isasubsetof X ∩Y .Thus Z ∈P(X ∩Y ). Therefore P(X) ∩P(Y ) ⊆P(X ∩ Y ).Itfollowsthat P(X ∩ Y )= P(X) ∩P(Y ).

31.Let X = {a} and Y = {b}.Then

P(X)= {∅, {a}}, P(Y )= {∅, {b}}, so

P(X) ∪P(Y )= {∅, {a}, {b}}

Since X ∪ Y = {a,b},

P(X ∪ Y )= {∅, {a}, {b}, {a,b}}

Now {a,b}∈P(X ∪ Y ),but {a,b} / ∈P(X) ∪P(Y ).Therefore P(X ∪ Y ) ⊆P(X) ∪P(Y )isfalsein general.

32.(X ∩ Y ) (X ∩ Z)=(X ∩ Y ) ∩ (X ∩ Z)[A B = A ∩ B] =(X ∩ Y ) ∩ (X ∪ Z)[DeMorgan’slaw; Theorem1.1.21,part(k)]

=((X ∩ Y ) ∩ X) ∪ ((X ∩ Y ) ∩ Z)[Distributivelaw; Theorem1.1.21,part(c)]

=((Y ∩ X) ∩ X) ∪ ((X ∩ Y ) ∩ Z)[Commutativelaw; Theorem1.1.21,part(b)] =(Y ∩ (X ∩ X)) ∪ (X ∩ (Y ∩ Z))[Associativelaw; Theorem1.1.21,part(a)] =(Y ∩ ∅) ∪ (X ∩ (Y ∩ Z))[Complementlaw; Theorem1.1.21,part(e)] = ∅ ∪ (X ∩ (Y ∩ Z))[Boundlaw; Theorem1.1.21,part(g)] =(X ∩ (Y ∩ Z)) ∪ ∅ [Commutativelaw; Theorem1.1.21,part(b)]

= X ∩ (Y ∩ Z)[Identitylaw; Theorem1.1.21,part(d)]

= X ∩ (Y Z)[A B = A ∩ B]

34.False.Let X = {a} and Y = Z = {b}.Then X ∪ (Y Z)= {a}, (X ∪ Y ) (X ∪ Z)= ∅

35.True. Y X = Y ∩ X = Y ∪ X = Y ∪ X = X ∪ Y.

36.False.Let X = {a} and Y = Z = {b}.Then X (Y ∪ Z)= {a}, (X Y ) ∪ Z = {a,b}

38.False.Let X = {a}, Y = {b},and U = {a,b}.Then X Y = {b}, Y X = {a}

40.True.Let x ∈ (X ∩ Y ) ∪ (Y X).Noweither x ∈ X ∩ Y or x ∈ Y X.Ineithercase, x ∈ Y .Therefore (X ∩ Y ) ∪ (Y X) ⊆ Y Nowsupposethat x ∈ Y .Either x ∈ X or x/ ∈ X.If x ∈ X,then x ∈ X ∩Y .Thus x ∈ (X ∩Y )∪(Y X). If x/∈ X,then x ∈ Y X.Again x ∈ (X ∩ Y ) ∪ (Y X).Thus Y ⊆ (X ∩ Y ) ∪ (Y X).Therefore (X ∩ Y ) ∪ (Y X)= Y

41.True.Let a ∈ X × (Y ∪ Z).Then a =(x,y)where x ∈ X and y ∈ Y ∪ Z.Now y ∈ Y or y ∈ Z.If y ∈ Y ,then a =(x,y) ∈ X × Y .Thus a ∈ (X × Y ) ∪ (X × Z).If y ∈ Z,then a =(x,y) ∈ X × Z.Again a ∈ (X × Y ) ∪ (X × Z).Therefore X × (Y ∪ Z) ⊆ (X × Y ) ∪ (X × Z).

Nowsupposethat a ∈ (X × Y ) ∪ (X × Z).Theneither a ∈ X × Y or a ∈ X × Z.If a ∈ X × Y ,then a =(x,y)where x ∈ X and y ∈ Y .Inparticular, y ∈ Y ∪ Z.Thus a =(x,y) ∈ X ×(Y ∪ Z).If a ∈ X ×Z, then a =(x,z)where x ∈ X and z ∈ Z.Inparticular, z ∈ Y ∪ Z.Thus a =(x,z) ∈ X × (Y ∪ Z). Therefore(X × Y ) ∪ (X × Z) ⊆ X × (Y ∪ Z).Wehaveprovedthat X × (Y ∪ Z)=(X × Y ) ∪ (X × Z).

43.True.Let a ∈ X × (Y Z).Then a =(x,y),where x ∈ X and y ∈ Y Z.Thus y ∈ Y and y/ ∈ Z and, so,(x,y) ∈ X × Y and(x,y) / ∈ X × Z.Therefore a =(x,y) ∈ (X × Y ) (X × Z).Wehaveshownthat X × (Y Z) ⊆ (X × Y ) (X × Z).

Nowsupposethat a ∈ (X × Y ) (X × Z).Then a ∈ X × Y and a/∈ X × Z.Thus a =(x,y), where x ∈ X, y ∈ Y ,and y/∈ Z.Therefore a =(x,y) ∈ X × (Y Z).Wehaveshownthat (X × Y ) (X × Z) ⊆ X × (Y Z).Itfollowsthat X × (Y Z)=(X × Y ) (X × Z).

44.False.Take X = {1, 2}, Y = {1}, Z = {2}.Then Y × Z = {(1, 2)},X Y = {2},X Z = {1}

Thus X (Y × Z)= {1, 2} and(X Y ) × (X Z)= {(2, 1)}.

47–56.Argueasintheproofgiveninthebookoftheﬁrstassociativelaw[Theorem1.1.21,part(a)].

58.Bydeﬁnition (A B) A =[(A B) ∪ A] [(A B) ∩ A].

Showthat (A B) ∪ A = A ∪ B and(A B) ∩ A = A ∩ B. Thestatementthenfollowseasily.

59.Thestatementistrue.Weﬁrstprovethat A ⊆ B.Let x ∈ A. Wedividetheproofintotwocases.First,weconsiderthecasethat x ∈ C.Then x ∈ A C.Therefore x ∈ B C.Thus x ∈ B (sinceif x ∈ B,then x ∈ B C). Next,weconsiderthecasethat x ∈ C.Then x ∈ A C.Therefore x ∈ B C.Thus x ∈ B. Ineithercase, x ∈ B,andso A ⊆ B.Similarly, B ⊆ A,andso A = B

61.Thestatementisfalse.Let A = {1, 2},B = {2, 3},C = {1, 3}

Since B ∩ C = {3}, A (B ∩ C)= {1, 2, 3}

Now A B = {1, 3} and A C = {2, 3}, thus (A B) ∩ (A C)= {3}.

62.Thestatementisfalse.Let A = {1, 2},B = {2, 3},C = {1, 3}

Since B C = {1, 2}, A ∪ (B C)= {1, 2}

Since A ∪ B = A ∪ C = {1, 2, 3}, (A ∪ B) (A ∪ C)= ∅

64.Yes, iscommutative:

65.Yes, isassociative.Weﬁrstprovethat (A B) C =(A ∩ B

)

) ∪ (A ∩

) ∪ (A ∩ B

C). (1) [Forthemotivationofthisformula,drawtheVenndiagramof (A B) C.]ByExercise57, (A B) C =[(A

) C] ∪ [C (A B)] AgainusingExercise57andthefactthat X Y = X ∩ Y ,wehave (A B) C =[(A B) ∪ (B A)] C =[(A ∩ B) ∪ (B ∩ A)] ∩ C.

Usingthedeﬁnitionof ,thefactthat X Y = X ∩ Y ,andDeMorgan’slaws,wehave

CombiningtheprecedingequationsandusingTheorem1.1.21,weobtainequation(1)

ByExercise64, iscommutative.Thus

Wecanobtainaformulafor(B C) A usingequation(1)with A replacedby B, B replacedby C, and C replacedby A.However,notingthattheright-handsideofequation(1)is symmetricin A, B, and C,weseethatthetwoexpressions

(A B) C and A (B C) areequal.Therefore, isassociative.

Section2.2

2.False; x = √2isacounterexample.

3.Weprovethecontrapositive:If x isrational,then x3 isrational. Supposethat x isrational.Thenthereexistintegers p and q suchthat x = p/q.Now x3 = p3/q3.Thus x3 isrational.

5.Suppose,bywayofcontradiction,that x< 1and y< 1and z< 1.Addingtheseinequalitiesgives x + y + z< 3,whichisacontradiction.

6.Suppose,bywayofcontradiction,that x> √2and y> √2.Multiplyingtheseinequalitiesgives xy> 2, whichisacontradiction.

8.Suppose,bywayofcontradiction,that x + y isrational.Since x and x + y arerational,thereexist integers p1,p2,q1,q2 suchthat x = p1/q1 and x + y = p2/q2.Now y =(x + y)

Therefore y isrational,whichisacontradiction.

9.False;acounterexampleis x =0,y = √2.

11.Sincetheintegersincreasewithoutbound,thereexists n ∈ Z suchthat √2/(b a) <n.Therefore √2/n<b a.Choose m ∈ Z aslargeaspossiblesatisfying m√2/n ≤ a.Then,bythechoiceof m, a< (m +1)√2/n.Also (m +1)√2 n = m√2 n + √2 n <a +(b a)= b.

Therefore x =(m +1)√2/n isanirrationalnumbersatisfying a<x<b.(If(m +1)√2/n isrational, say(m +1)√2/n = p/q where p and q areintegers,then √2= np/[(m +1)q]isrational,whichisnot thecase.)

12.If √2√2 isrational,thenwehavefoundirrationalnumbers a and b (namely a = b = √2)suchthat ab is rational.Supposethat √2√2 isirrational.Let a = √2√2 and b = √2.Now

isrational.Wehavefoundirrationalnumbers a and b suchthat ab isrational. Thisproofisnonconstructivesinceitdoesnotshowwhether thedesiredpairis a = b = √2or a = √2√2 ,b = √2.

14.Let a =2and b =1/2.Then a and b arerational.Now ab =21/2 = √2isirrational.Thisproofisa constructiveexistenceproof.

15.Suppose,bywayofcontradiction,that x>y.Let ε =(x y)/2.Then

whichisacontradiction.

17.Firstprovethatif b isarationalnumber,then bn isrationalforeverypositiveinteger n.Tothisend,let b = p/q,where p and q areintegers.Then bn = pn/qn.Since b isthequotientofintegers, b isrational. Nowsuppose,butwayofcontradiction,that ar isrationalforsomepositiverationalnumber r = p/q, where p and q arepositiveintegers.Because ar isrational,(ar)q isrationalbytheresultoftheﬁrst paragraph.Since(ar )q = ap , ap isrational.Thiscontradictsthehypothesisthat an isirrationalfor everypositiveinteger n

18.WeshowthatAbby,Cary,Dale,andEdiewenttotheconcert,butnotBosco.SupposethatBoscowent. ThenCaryandDalealsowent.SinceCarywent,Ediewent;andsinceDalewent,Abbywent.But thiscontradictsthehypothesisthatexactlyfourwenttotheconcert.Therefore,Boscodidnotgo.This meansAbby,Cary,Dale,andEdiewenttotheconcert.

20.Suppose,bywayofcontradiction,that X × ∅ isnotempty.Thenthereexists(x,y) ∈ X × ∅.Now y ∈ ∅,whichisacontradiction.

21.Supposethateveryboxcontainslessthan12balls.Theneachboxcontainsatmost11ballsandthe maximumnumberofballscontainedbythenineboxesis9 11=99.Contradiction.

23.Suppose,bywayofcontradiction,thateachoftheotherthreesuitscontainsatmostsixcards.Thenthese threesuitstogethercontainatmost3 · 6=18cards.Togetherwiththeothersuit,whichcontainsexactly sevencards,wecanaccountforatmost25cards.Since S contains26cards,wehaveacontradiction. Thereforethereisanothersuitinwhich S hasatleastsevencards.

24.Sincethereisasuitinwhich S1 hasatleastninecards, S2 hasatmostfourcardsinthissuit.Now suppose,bywayofcontradiction,that S2 hasatmostsevencardsintheotherthreesuits.Thesethree suitscontainatmost3 7=21cards.Togetherwithatmostfourcardsintheothersuit,wecanaccount foratmost25cardsin S2.Since S2 contains26cards,wehaveacontradiction.Thereforethere isasuit inwhich S2 hasatleasteightcards.

26.For n =3,wehave n2 > 2n

28.Thestatementisfalse.Let s1 = s2 =3.Then A =3.Forno i dowehave si >A.Theproofisby counterexample.

29.Thestatementistrueandweproveitusingproofbycontradiction.Supposethatforevery j, sj ≤ A Since sj ≤ A forall j and si <A,

s1 + + si + + sn <A + + A + + A = nA.

Dividingby n,weobtain

s1 + + sn n <A, whichisacontradiction.

31.Since si = sj ,either si = A or sj = A.Bychangingthenotation,ifnecessary,wemayassumethat si = A.Either si <A or si >A.If si >A,theproofiscomplete;soassumethat si <A.Weshowthat thereexists k suchthat sk >A.Suppose,bywayofcontradiction,that sm ≤ A forall m,thatis,

s1 ≤ A

s2 ≤ A

sn ≤ A.

Addingtheseinequalitiesyields

1 + s2 + ··· + si + ··· + sn <nA

since si <A.Dividingby n gives s1 + s2 + + sn n <A, whichisacontradiction.Thereforethereexists k suchthat sk >A.

33.If m and n arepositiveintegersand m> 3,then m3 +2n2 > 36.If m and n arepositiveintegersand n> 4,then m3 +2n2 > 36.Thusitsuﬃcestoconsiderthecases1 ≤ m ≤ 3and1 ≤ n ≤ 4.The followingtable,whichshowsthevaluesof m3 +2n2,showsthatthereisnosolutionto m3 +2n2 =36: m 123 1 31029 n 2 91635 3 192645 4 334059

34.Noticethat2m2 +4n2 1isoddand2(m + n)iseven.Therefore2m2 +4n2 1 =2(m + n)forall positiveintegers m and n

36.Weconsidertwocases: n iseven, n isodd.Firstsupposethat n iseven.ByExercise9,Section2.1, theproductofevenintegersiseven.Therefore n2 = n · n iseven.AgainbyExercise9,Section2.1, n3 = n2 · n iseven.ByExercise7,Section2.1,thesumofevenintegersiseven.Therefore n3 + n iseven. Nowsupposethat n isodd.ByExercise10,Section2.1,theproductofoddintegersisodd.Therefore n2 = n n isodd.AgainbyExercise10,Section2.1, n3 = n2 n isodd.ByExercise8,Section2.1,the sumofoddintegersiseven.Therefore n3 + n iseven.Ineithercase, n3 + n iseven.

38.First,notethatfromExercise37,forall x, |−x| = |( 1)x| = |−1||x| = |x|

Example2.2.7statesthatforall x, x ≤|x|.Usingtheseresults,weconsidertwocases: x + y ≥ 0and x + y< 0.If x + y ≥ 0,wehave |x + y| = x + y ≤|x| + |y|. If x + y< 0,wehave |x + y| = (x + y)= x + y ≤|−x| + |−y| = |x| + |y| Copyright c 2018PearsonEducation,Inc.

40.Supposethat xy> 0.Theneither x> 0and y> 0or x< 0and y< 0.If x> 0and y> 0,

sgn(xy)=1=1 · 1=sgn(x)sgn(y)

If x< 0and y< 0,

sgn(xy)=1= 1 1=sgn(x)sgn(y).

Next,supposethat xy =0.Theneither x =0or y =0.Thuseithersgn(x)=0orsgn(y)=0.Ineither case,sgn(x)sgn(y)=0.Therefore

sgn(xy)=0=sgn(x)sgn(y)

Finally,supposethat xy< 0.Theneither x> 0and y< 0or x< 0and y> 0.If x> 0and y< 0, sgn(xy)= 1=1 1=sgn(x)sgn(y)

If x< 0and y> 0,

sgn(xy)= 1= 1 1=sgn(x)sgn(y).

41. |xy| =sgn(xy)xy =sgn(x)sgn(y)xy =[sgn(x)x][sgn(y)y]= |x||y|

43.Supposethat x ≥ y.Then

Thus max

Theothercaseis x<y.Then

max{x,y} = x and |x y| = x y.

44.Supposethat x ≥ y.Then

{x,y} = y and |x y| = x y. Thus

Theothercaseis x<y.Then

45.max{x,y} +min{x,y} =

47.Supposethat n isodd.Then n =2k +1.Now n +2=(2k +1)+2=2(k +1)+1isodd.

Nowsupposethat n +2isodd.Then n +2=2k +1.Now n =(2k +1) 2=2(k 1)+1isodd. Therefore n isoddifandonlyif n +2isodd.

49.Supposethat A ⊆ C and B ⊆ C.Let x ∈ A ∪ B.Theneither x ∈ A or x ∈ B.If x ∈ A,since A ⊆ C, x ∈ C.If x ∈ B,since B ⊆ C, x ∈ C.Ineithercase, x ∈ C.Therefore A ∪ B ⊆ C

Nowsupposethat A ∪ B ⊆ C.Let x ∈ A.Then x ∈ A ∪ B.Since A ∪ B ⊆ C, x ∈ C.Therefore A ⊆ C Let x ∈ B.Then x ∈ A ∪ B.Since A ∪ B ⊆ C, x ∈ C.Therefore B ⊆ C.Weconcludethat A ⊆ C and B ⊆ C.Itfollowsthat A ⊆ C and B ⊆ C ifandonlyif A ∪ B ⊆ C

50.Supposethat C ⊆ A and C ⊆ B.Let x ∈ C.Since C ⊆ A, x ∈ A.Since C ⊆ B, x ∈ B.Since x ∈ A and x ∈ B, x ∈ A ∩ B.Therefore C ⊆ A ∩ B.

Nowsupposethat C ⊆ A ∩ B.Let x ∈ C.Then x ∈ A ∩ B.Inparticular, x ∈ A.Therefore C ⊆ A Againlet x ∈ C.Then x ∈ A ∩ B.Inparticular, x ∈ B.Therefore C ⊆ B.Thus C ⊆ A and C ⊆ B.It followsthat C ⊆ A and C ⊆ B ifandonlyif C ⊆ A ∩ B

53.[(a) → (b)]Weassumethat A ∩ B = ∅ andprovethat B ⊆ A.Let x ∈ B.If x ∈ A,weobtainthe contradiction A ∩ B = ∅.Thus x/ ∈ A.Hence x ∈ A.Therefore B ⊆ A

[(b) → (c)]Weassumethat B ⊆ A andprovethat A B = A ∪ B

Let x ∈ A B.Bydeﬁnition, A B =(A ∪ B) (A ∩ B),thus x ∈ A ∪ B.Therefore A B ⊆ A ∪ B

Let x ∈ A ∪ B.Weﬁrstprovethat x/ ∈ A ∩ B.Suppose,bywayofcontradiction,that x ∈ A ∩ B.Then x ∈ A and x ∈ B.Since B ⊆ A, x ∈ A,whichimpliesthat x/∈ A.Wehavethedesiredcontradiction. Therefore x/ ∈ A ∩ B.Now x ∈ (A ∪ B) (A ∩ B)= A B.Therefore A ∪ B ⊆ A B.Itfollowsthat A B = A ∪ B

[(c) → (a)]Weassumethat A B = A ∪ B andprovethat A ∩ B = ∅.Suppose,bywayofcontradiction, that A ∩ B isnotempty.Thenthereexists x ∈ A ∩ B.Then x ∈ A ∪ B.Thisimpliesthat x/ ∈ A B Since x ∈ A ∪ B, A B = A ∪ B,whichisacontradiction.Therefore A ∩ B = ∅

54.[(a) → (b)]Weassumethat A ∪ B = U andprovethat A ∩ B = ∅.Takingthecomplementofbothsides oftheequation A ∪ B = U andusingDeMorgan’slawandthe0/1law(Theorem1.1.22),we obtain A ∩ B = A ∪ B = U = ∅.

[(b) → (c)]Weassumethat A ∩ B = ∅ andprovethat A ⊆ B.Replace A by B and B by A inExercise 53(a)toobtain B ∩ A = ∅.SinceExercise53(a)isequivalenttoExercise53(b),weobtain A ⊆ B or A ⊆ B

[(c) → (a)]Weassumethat A ⊆ B andprovethat A∪B = U .Since U isauniversalset,weautomatically have A ∪ B ⊆ U .

Let x ∈ U .If x ∈ A,then x ∈ A ∪ B.If x/ ∈ A,then x ∈ A.Since A ⊆ B, x ∈ B.Again x ∈ A ∪ B Therefore U ⊆ A ∪ B.Itfollowsthat A ∪ B = U

Problem-SolvingCorner:Proofs

1.Theleastupperboundofanonemptyﬁnitesetofrealnumbersisthemaximumnumberintheset.

2.Callthegivenset X.Weprovethattheleastupperboundof X is1.Since 1 1 n < 1

forallpositiveintegers n,1isanupperboundof X.Let a beanupperboundfor X.Suppose,byway ofcontradiction,that a< 1.Sincetheintegersareunbounded,thereexistsapositive integer k suchthat 1 1 a <k.

Multiplyingby(1 a)/k gives 1 k < 1 a, which,inturn,isequivalentto a< 1 1 k

Thiscontradictsthefactthat a isanupperboundof X.Thus1 ≤ a and1istheleastupperboundof X

3.Let b betheleastupperboundof Y .If x ∈ X,then x ∈ Y and x ≤ b.Thus b isanupperboundof X If a istheleastupperboundof X, a ≤ b

4.0

5.Let Z = {x + y | x ∈ X and y ∈ Y } andlet z ∈ Z.Then z = x + y forsome x ∈ X,y ∈ Y .Now z = x + y ≤ a + b.Therefore Z isboundedaboveby a + b. Let c beanupperboundof Z.Suppose,bywayofcontradiction,that c<a + b.Let ε = a + b c.Now a ε/2isnotanupperboundof X sothereexists x ∈ X suchthat a ε 2 <x.

Similarly,thereexists y ∈ Y suchthat b ε 2 <y.

Addingthepreviousinequalitiesgives c = a + b ε<x + y,

whichcontradictsthefactthat c isanupperboundof Z.Therefore c ≥ a + b and a + b istheleastupper boundof Z

6.Since a isagreatestlowerboundfor X and b isalowerboundfor X, b ≤ a.Since b isagreatestlower boundfor X and a isalowerboundfor X, a ≤ b.Therefore a = b.

7.Let X beanonemptysetofrealnumbersboundedbelow.Let Y bethesetoflowerboundsof X.The set Y isnonemptysince X isboundedbelow.Let x beanelementof X.Forevery y ∈ Y ,wehave y ≤ x since y isalowerboundof X.Therefore Y isboundedaboveby x.Thus Y ishasaleastupperbound, say a.

Nextweshowthat a isalowerboundof X.Suppose,bywayofcontradiction,that a isnotalower boundof X.Thenthereexists x ∈ X suchthat x<a.Then x isnotanupperboundof Y .Therefore thereexists y ∈ Y suchthat x<y.Butthiscontradictsthefactthat y isalowerboundof X.Therefore a isalowerboundof X.

Finally,weshowthat a isthegreatestlowerboundof X.Let b bealowerboundof X.Then b ∈ Y Since a isanupperboundof Y , b ≤ a.Therefore a isthegreatestlowerboundof X

8.Since a + ε>a, a + ε isnotalowerboundof X.Thereforethereexists x ∈ X suchthat a + ε>x.Since a isalowerboundof X, x ≥ a.

9.Let tX denotetheset {tx | x ∈ X}

Wemustprovethat

(a) z ≥ ta forevery z ∈ tX (i.e., ta isanlowerboundfor tX), (b)if b isanlowerboundfor tX,then b ≤ ta (i.e., ta isthegreatestlowerboundfor tX).

Weﬁrstprovepart(a).Let z ∈ tX.Then z = tx forsome x ∈ X.Since a isanupperboundfor X, x ≤ a.Multiplyingby t andnotingthat t< 0,wehave z = tx ≥ ta.Therefore, z ≥ ta forevery z ∈ tX andtheproofofpart(a)iscomplete.

Nextweprovepart(b).Let b bealowerboundfor tX.Then tx ≥ b forevery x ∈ X.Dividingby t andnotingthat t< 0,wehave x ≤ b/t forevery x ∈ X.Therefore b/t isanupperboundfor X.Since a istheleastupperboundfor X, b/t ≥ a.Multiplyingby t andnotingagainthat t< 0,wehave b ≤ ta Therefore ta isthegreatestlowerboundfor tX.Theproofiscomplete.

Section2.3

3.1. ¬p ∨ r

2. ¬r ∨ q

3. p

4. ¬p ∨ q from1,2

5. q from3,4

4.1. ¬p ∨ t

2. ¬q ∨ s

3. ¬r ∨ s

4. ¬r ∨ t

5. p ∨ q ∨ r ∨ u

6. t ∨ q ∨ r ∨ u from1,5

7. s ∨ t ∨ r ∨ u from2,6

8. s ∨ t ∨ u from3,7

6.(p ↔ r) ≡ (p → r)(r → p) ≡ (¬p ∨ r)(¬r ∨ p)

1. ¬p ∨ r

2. ¬r ∨ p

3. r 4. p from2,3

8.1. a ∨¬b

2. a ∨ c

3. ¬a

4. ¬d

5. b negatedconclusion

6. ¬b from1,3 Now5and6combinetogiveacontradiction.

Section2.4

Insomeofthesesolutions,theBasisStepsareomitted.

12.ThesolutionissimilartothatforExercise11,whichisgiveninthebook.

14.Firstnotethat

c 2018PearsonEducation,Inc.

Thislastinequalityissuccessivelyequivalentto

Thislastinequalityistrueforall n ≥ 1.

15.2(

17.Bytheinductiveassumption,

Multiplyinginequalities(2)and(3),wehave

ApplyingtheBasisSteptothenumbers A and B,wehave

or,equivalently,

Combininginequalities(4)and(5),wehave

Takingthesquarerootofbothsidesofthelastinequalitygivesthedesiredresult. 18.(1+

20.Ifwesumthetermsinthediagonaldirection,weobtainone r,two r2’s,three r3’s,andsoon;thatis, weobtainthesum

MultiplyingtheinequalityofExercise19by r yields

Thus,thesumoftheentriesintheﬁrstcolumnislessthan r/(1 r).Similarly,thesumoftheentries inthesecondcolumnislessthan r2/(1 r),andsoon.Itfollowsfromtheprecedingdiscussionthat

Usinginequality(6),weobtainthedesiredresult

21.Take r =1/2inExercise20.

23.Assumethat11n 6isdivisibleby5.

whichisdivisibleby5.

24.Supposethat4divides6

.Now

Since4divides6

26.Weprovepart(a)only.TheBasisStepisimmediate.

Assumethat

Wemustprovethat

.Bytheinductiveassumption,

Bytheassociativelaw,

Bythedistributivelaw,

Therefore

andtheInductiveStepiscomplete.

29. n n +1

31.Weuseinductionon n,thenumberoflines,toprovetheresult.Ifthereisoneline,theresultiscertainly true.Supposethatthereare n> 1lines.Removeoneline.Bytheinductivehypothesis,theregionsthat resultmaybecoloredredandgreensothatnotworegionsthat shareanedgearethesamecolor.Now restoretheremovedline.Theregionsabove(or,ifthelineisvertical,totheleftof)therestoredline arecoloredredandgreensothatnotworegionsthatshareanedgearethesamecolor,andtheregions below(or,ifthelineisvertical,totherightof)therestoredlinearealsocoloredredandgreensothat notworegionsthatshareanedgearethesamecolor.Nowreversethecolorofeveryregionbelow(or, ifthelineisvertical,totherightof)therestoredline.Theregionsbelow(or,ifthelineisvertical,to therightof)therestoredlinearestillcoloredredandgreensothatnotworegionsthatshareanedge arethesamecolor.Sincethecolorsbelowtherestoredlinehavebeenreversed,regionsthatsharean edgethatispartoftherestoredlinedonothavethesamecolor.Thereforetheregionsmaybecolored redandgreensothatnotworegionsthatshareanedgearethesamecolor,andtheinductiveproofis complete.

32.Theproofisbyinductiononthenumber n ofzeroswiththeBasisStep,asusual,omitted.

Supposethattheresultistruefor n zeros,andwearegiven n +1zerosand n +1onesdistributedaround acircle.Findazerofollowed,inclockwiseorder,byaone.Temporarilyremovethesetwonumbers.By theinductiveassumption,itispossibletostartatsomenumberandproceedaroundthecircletothe originalstartingpositioninsuchawaythat,atanypointduringthecycle,onehasseenatleastasmany zerosasones.Noticethatthislaststatementremainstrueifwerestoretheremovedzeroandone.

34.Atrominocancoverthesquaretotheleftofthemissingsquareasshown

orinasymmetricfashionbyreversing“up”and“down.”Inthe ﬁrstcase,itisimpossibletocover thetwosquaresinthetoprowattheextremeleft.Inthesecondcase,itisimpossibletocoverthe twosquaresinthebottomrowattheextremeleft.Therefore, itisimpossibletotiletheboardwith trominoes.

35.Theonly5 × 5deﬁcientboardsthatcanbetiledwithtrominoesarethosewiththecornersquareor centersquaremissing.

36.Suchaboardcanbetiledwith ij 2 × 3rectanglesoftheform

38Bysymmetry,wemayassumethatthemissingsquareislocatedinthe7 × 7subboardshowninthe followingﬁgure.Exercise37showshowtotilethissubboard.Exercise34showsthatthetwo6 × 4 subboardscanbetiled.Exercise32showsthatthe5 × 5subboardwithacornersquarecanbetiled. Thusthedeﬁcient11 × 11boardcanbetiledwithtrominoes.

39. BasisStep (n =0).Inthiscase,the2n × 2n L-shapeisatrominoand,so,itistiled. InductiveStep.Assumethatwecantilea2n 1 × 2n 1 L-shapewithtrominoes.Givena2n × 2n L-shape,divideitintofour2n 1 × 2n 1 L-shapes:

Bytheinductiveassumption,wecantileeachofthefour2n 1 × 2n 1 L-shapeswithtrominoes.The inductivestepiscomplete.

42.ArguingasinthesolutiontoExercise41,thenumberings

showthattheonlypossibilityforthemissingsquareisthecentersquare.Thisboardcanbetiled:

43.AnargumentlikethoseinthesolutionstoExercises41and42showsthattheonlyboardthatcanbe tiledwithstraighttrominoesistheonewiththemissingsquareinrow3,column3(andthethreeboards symmetrictoit).

44.SeeD.W.Walkup,“Coveringarectanglewith T-tetrominoes,” Amer.Math.Mo.,(November1965) 986–988.

46. BasisStep (k = i +1)

Case1 a[k] < val. Now h becomes i +1andtheswapchangesnothing.Thus h ≤ k.Theonly p satisfying i<p ≤ h is p = i +1.Forthisvalueof p,wedohave a[p] < val .No p satisﬁes h<p ≤ k,so thethirdconditionisvacuouslysatisﬁed.

Case2 a[k] ≥ val. Atthebottomoftheloop,wehave k = i +1and h = i.Thus h ≤ k.Sinceno p satisﬁes i<p ≤ h,thesecondconditionisvacuouslysatisﬁed.Theonly p satisfying h<p ≤ k is p = i +1.Forthisvalueof p,wedohave a[p] ≥ val

InductiveStep

Bytheinductiveassumption,theinvariantholdsforthepreviousiterationoftheloop.Wedenotethese valuesof h and k as hold and kold.Thuswehave hold ≤ kold;forall p,i<p ≤ hold,a[p] < val ;andfor p,hold <p ≤ kold,a[p] ≥ val .Wedenotethenewvaluesof h and k as hnew and knew.Atthetopof theloop, k isincremented,sowehave knew = kold +1.

Case1 a[knew] ≥ val. Atthebottomoftheloop,wehave hnew = hold.Thus hnew = hold ≤ kold <knew, sotheﬁrstconditionholds.Supposethat i<p ≤ hnew.Since hnew = hold ,p satisﬁes i<p ≤ hold Therefore a[p] < val andthesecondconditionholds.Supposethat hnew <p ≤ knew.If hnew <p<knew, then hold <p ≤ kold.Inthiscase, a[p] < val .If p = knew,also a[p] ≥ val .Thereforethethirdcondition holds.

Case2 a[knew] < val. Firstsupposethat hold = kold.Then hnew = hold +1= kold +1= knew.Thus thefirstconditionholds.Theswapchangesnovalues.Supposethat i<p ≤ hnew.If p<hnew,then i<p ≤ hold,so a[p]= val .If p = hnew,a[p]= a[hnew]= a[knew] < val ,sothesecondconditionholds. Sinceno p satisfies hnew <p ≤ knew,thethirdconditionisvacuouslysatisfied.

Nowsupposethat hold = kold.Then hold <kold.After h and k areincremented, hnew <knew,sothe firstconditionissatisfied.Now hnew = hold +1.Since hold <hnew ≤ kold,a[hnew] ≥ val .Itfollows thataftertheswap, a[hnew] < val and a[knew] ≥ val .Supposethat i<p ≤ hnew.If i<p<hnew,then i<p ≤ hold ,so a[p] < val .If p = hnew,wealsohave a[p] < val .Thusthesecondconditionholds. Supposethat hnew <p ≤ knew.If hnew <p<knew,then hold <p ≤ kold and p = hnew,so a[p] ≥ val .If p = knew,a[p] ≥ val.Thusthethirdconditionholds.TheInductiveStepiscomplete. Beforethefinalswap,forall p,i<p ≤ h,a[p] < val andforall p,h<p ≤ j,a[p] ≥ val .Notethat a[i]= val and a[h] < val .Thus,afterswapping a[i]and a[h], a[h]= val ;forall p,i ≤ p<h,a[p] < val ; andforall p,h<p ≤ j,a[p] ≥ val

47.TheargumentisessentiallyidenticaltothatofExample 2.4.7thatshowsthatany2n ×2n deﬁcientboard canbetiledwithtrominoes.

48.Noticethat k3 1=(k 1)[(k 2)(k 4)+7(k 1)]. Since7divides k3 1,7divides k 1or(k 2)(k 4)+7(k 1).If7dividesthelatterexpression,7 alsodivides(k 2)(k 4).If7divides(k 2)(k 4),7divideseither k 2or k 4.

50.TheInductiveStepfailsifeither a or b is1.Inthiscase,theinductivehypothesisiserroneouslyapplied tothepair a 1,b 1,whichincludesanonpositiveinteger.

51.Toarguebycontradiction,onemustassumethatthepropositionfails forsome n ≥ 2.Theallegedproof assumesthatthepropositionfails forall n ≥ 2.

52.For n =2,theinequalitybecomes 1 2 + 2 3 < 4 3 ,whichistrue.ThustheBasisStepistrue. Assumethatthegivenstatementholdsfor n.Now 1 2 + 2 3 + ··· + n n +1 + n +1 n +2 < n2 n +1 + n +1 n +2

TheInductiveStepwillbeprovedprovided

Ifwemultiplythelastinequalityby(n +1)(n +2),weobtain

whichisreadilyveriﬁedastrue.

54.Inthefollowingﬁgure ab a and b arebothsurvivors.

55.Supposethattherearethreepersons.Thetwopersonsclosesttogetherthrowateachother,andthe thirdpersonthrowsatoneofthetwoclosest.Thereforethethirdpersonsurvives.Thiscompletethe BasisStep.

Supposethattheassertionistruefor n,andconsider n +2persons.Again,theclosestpairthrowsat eachother.Therearenowtwocasestoconsider.Iftheremaining n personsallthrowatoneanother,by theinductiveassumption,thereisasurvivor.Ifatleastoneoftheremaining n personsthrowsatoneof theclosestpair,amongtheremaining n persons,atmost n 1piesarethrownatoneanother.Inthis case,someonesurvivesbecausetherearenotenoughpiestogoaround.TheInductiveStepiscomplete.

57.Thestatementisfalse.Inthefollowingﬁgure a a throwsapiethegreatestdistance,butisnotasurvivor.

59.Let x1 beacommonpointof X2,X3,X4;let x2 beacommonpointof X1,X3,X4;let x3 beacommon pointof X1,X2,X4;andlet x4 beacommonpointof X1,X2,X3.Since x1,x2,x3 ∈ X4,thetriangle x1x2x3 (perimeterandinterior)isin X4.Similarly,thetriangle x1x2x4 isin X3;thetriangle x1x3x4 is in X2;andthetriangle x2x3x4 isin X1.Weconsidertwocases:

Case1:Oneofthepoints x1,x2,x3,x4 isinthetrianglewhoseverticesaretheotherthreepoints. For example,supposethat x1 isintriangle x2x3x4.Sincetriangle x2x3x4 isin X1, x1 ∈ X1.Bydeﬁnition, x1 ∈ X2 ∩ X3 ∩ X4.Therefore, x1 ∈ X1 ∩ X2 ∩ X3 ∩ X4

Case2:Noneofthepoints x1,x2,x3,x4 isinthetrianglewhoseverticesaretheotherthreepoints. In thiscase, x1,x2,x3,x4 aretheverticesofaconvexquadrilateral:

x4 x2 x3 x1 x

Nowtheintersection, x,ofthediagonalsofthisquadrilateralbelongstoeachofthetrianglesand,thus, toeachof X1,X2,X3,X4.

60.Theproofisbyinductionon n.TheBasisStepis n =4,whichisExercise59.

WeturntotheInductiveStep.Assumethatif X1,...,Xn areconvexsets,eachthreeofwhichhavea commonpoint,thenall n setshaveacommonpoint.

Let X1,...,Xn,Xn+1 beconvexsets,eachthreeofwhichhaveacommonpoint.Wemustshowthatall n +1setshaveacommonpoint.ByExercise55,

areconvexsets.Weclaimthatanythreeofthesetsin(7)have acommonpoint.Theclaimistrueby hypothesisifthethreesetsareanyof X1,...,Xn 1.Consider Xi,Xj,Xn ∩ Xn+1, i<j ≤ n 1.By hypothesis,anythreeof Xi,Xj,Xn,Xn+1 haveacommonpoint.ByExercise56, Xi,Xj ,Xn,Xn+1 have acommonpoint.Therefore, Xi,Xj,Xn ∩ Xn+1 haveacommonpoint.Thus,anythreeofthesetsin(7) haveacommonpoint.Bytheinductiveassumption,thesetsin (7)haveacommonpoint.TheInductive Stepiscomplete.

62.Weﬁrstprovetheresultfor n =3.Let A1,A2,A3 beopenintervalssuchthateachpairhasanonempty intersection.Choose x1 ∈ A

,x3 ∈ A2 ∩ A3.Notethatifanypair(x1,x2 or x1,x3 or x3,x3)isequal,itisin A1 ∩ A2 ∩ A3.Wemayassume x1 <x2.Weconsiderthreecases.Firstsuppose that x3 <x1.Since x2,x3 ∈ A3,[x3,x2] ⊆ A3.([a,b]isthesetofall x satisfying a ≤ x ≤ b.)Thus x1 ∈ A3.Therefore x1 ∈ A1 ∩ A2 ∩ A3.

Nextsupposethat x1 <x3 <x2.Since x1,x2 ∈ A1,[x1,x2] ⊆ A1.Thus x3 ∈ A1.Therefore x3 ∈ A1 ∩ A2 ∩ A3

Finallysupposethat x1 <x2 <x3.Since x1,x3 ∈ A2,[x1,x3] ⊆ A2.Thus x2 ∈ A2.Therefore x2 ∈ A1 ∩ A2 ∩ A3.Wehaveshownthatif A1,A2,A3 areopenintervalssuchthateachpairhasa nonemptyintersection,then A1 ∩ A2 ∩ A3 isnonempty.

Wenowprovethatgivenstatementusinginductionon n.TheBasisStep(n =2)istrivial.

Assumethatif I1,...,In isasetofopenintervalssuchthateachpairhasanonemptyintersection,then I1 ∩···∩ In isnonempty.Let I1,...,In+1 beasetofopenintervalssuchthateachpairhasanonempty intersection.Since In ∩ In+1 isnonempty,itisanopeninterval.Weclaimthat

I1,...,In 1,In ∩ In+1

isasetofopenintervalssuchthateachpairhasanonemptyintersection.Thisiscertainlytrueforpairs oftheform Ii,Ij,1 ≤ i<j ≤ n 1.Considerapairoftheform Ii, i ≤ n 1,and In ∩ In+1.Sinceeach pairamong Ii,In,In+1 hasnonemptyintersection,bythecase n =3provedpreviously, Ii ∩ In ∩ In+1 is nonempty.Therefore,

I1,...,In 1,In ∩ In+1 isasetofopenintervalssuchthateachpairhasanonemptyintersection.Bytheinductiveassumption

Ii ∩···∩ In 1 ∩ (In ∩ In+1) isnonempty.Theinductivestepiscomplete.

64.565.5

67.After j rounds,2, 4,..., 2j havebeeneliminated.Atthispoint,thereare2i persons.Thisisexactlythe Josephusproblemwhenthenumberofpersonsisapowerof2,exceptthattheroundbeginswithperson 2j +1,ratherthanwithperson1.ByExercise66,person2j +1isthesurvivor.

65.977

71.∆

bn =∆an.Then

ByExercise70,

Combiningthepreviousequations,weobtain

Solvingfor1+2+ + n,weobtain

72.Let an = n!.Then

bn =∆an.Then

ByExercise70,

Combiningthepreviousequations,weobtain 1(1!)+2(2!)+

74.Since p isdivisibleby k,thereexists t1 suchthat p = t1k.Since q isdivisibleby k,thereexists t2 such that p = t2k.Now

Therefore, p + q isdivisibleby k

Problem-SolvingCorner:MathematicalInduction

1.TheBasisStep(n =0)is H1 ≤ 1+0.Since H1 =1,theBasisStepistrue.

Nowassumethat H2n ≤ 1+ n.Then

TheInductiveStepiscomplete.

2.TheBasisStep(n =1)is H1 =2H1 1.Since H1 =1,theBasisStepistrue.

Nowassumethat

Then

TheInductiveStepiscomplete.

4.Weprovetheassertionbyinduction.TheBasisStepis n =1:

FortheInductiveStep,assumethattheassertionistruefor n.Now

5.Weprovetheassertionbyinduction.TheBasisStepis n =1:

WeproceedtotheInductiveStep.First,noticethat

Assumethattheassertionistruefor n.Then

Section2.5

2.Verifydirectlythecases n =24,..., 28.Assumethatthestatementistrueforpostage i satisfying 24 ≤ i<n.Wemustshowthatwecanmake n centspostageusingonly5-centand7-centstamps.We mayassumethat n> 28.Then n>n 5 ≥ 24.Bytheinductiveassumption,wecanmake n 5cents postageusing5-centand7-centstamps.Adda5-centstampto obtain n centspostage.

3.Verifydirectlythecases n =12, 13, 14 Assumethatthestatementistrueforpostage i satisfying 12 ≤ i<n.Wemustshowthatwecanmake n centspostageusingonly3-centand7-centstamps.We mayassumethat n> 14.Then n>n 3 ≥ 12.Bytheinductiveassumption,wecanmake n 3cents postageusing3-centand7-centstamps.Adda3-centstampto obtain n centspostage.

5.TheBasisStep(n =6)isprovedbyusingthree2-centstamps.Nowassumethatwe canmakepostage for n cents.Ifthereisatleastone7-centstamp,replaceitbyfour2-centstampstomake n +1cents postage.Ifthereareno7-centstamps,thereareatleastthree2-centstamps(because n ≥ 6).Replace three2-centstampsbyone7-centstamptomake n +1centspostage.TheInductiveStepiscomplete.

6.TheBasisStep(n =24)isprovedbyusingtwo5-centstampsandtwo7-centstamps.Nowassumethat wecanmakepostagefor n cents.Ifthereareatleasttwo7-centstamps,replacetwo7-centstampswith three5-centstampstomake n +1centspostage.Ifthereisexactlyone7-centstamp,thenthereare atleastfour5-centstamps(because n ≥ 24).Replaceone7-centstampandfour5-centstampswith four7-centstampstomake n +1centspostage.Ifthereareno7-centstamps,thenthereareatleast ﬁve5-centstamps(againbecause n ≥ 24).Replaceﬁve5-centstampswiththree7-centstampsandone 5-centtomake n +1centspostage.TheInductiveStepiscomplete.

8.Wemusthave4 ≤ n/2 .Sincethisinequalityfailsfor n =5, 6, 7,theBasisStepsare n =4, 5, 6, 7.

9.Wemusthave2 ≤ n/3 .Sincethisinequalityfailsfor n =3, 4, 5,theBasisStepsare n =2, 3, 4, 5.

11.WeomittheBasisStep.FortheInductiveStep,wehave

13.WeomittheBasisStep.FortheInductiveStep,wehave cn =4c n/2 + n ≤ 4[4( n/2 − 1)2]+ n ≤ 4[4(n/2 1)2]+ n =4n 2 15n +16 ≤ 4(n 1)2 Thelastinequalityreducesto12 ≤ 7n,whichistruesince n> 1.

14.WeomittheBasisSteps(n =2, 3).WeturntotheInductiveStep.Assumethat n ≥ 4.Then n/2 ≥ 2, so n/2 ≥ 2.Then

cn =4c n/2 + n> 4( n/2 +1)2/8+ n ≥ 4[(n 1)/2+1]2/8+ n =(n +1)2/8+ n > (n +1)2/8

Weusedthefactthat n/2 ≥ (n 1)/2forall n

16. BasisStep (n =1).Theﬁrstplayerremovesonecard(fromeitherpile).The secondplayerthen removesthelastcardandwinsthegame.

InductiveStep. Supposethat n> 1andwhenevertherearetwopilesof k<n cards,thesecond playercanalwayswinthegame.

Supposethattherearetwopilesof n cards.Thefirstplayerremoves i cardsfromoneofthepiles.If i = n (i.e.,thefirstplayerremovesallofthecardsfromonepile),thesecondplayercanwinbyremoving allofthecardsfromtheremainingpile.If i<n,thesecondplayerremoves i cardsfromtheotherpile leavingtwopileseachwith n i cards.Thegamethenresumeswiththefirstplayerfacingtwopiles eachwith k = n i<n cards.Bytheinductiveassumption,thesecondplayercanwinthegame.The inductiveproofiscomplete.

18. q = 6, r =719. q =0, r =721. q =0, r =022. q =1, r =0

24.If

where n1 <n2 <...<nk,anotherrepresentationis

25.(b)Since p/q< 1,n> 1.Since n isthesmallestpositiveintegersatisfying1/n ≤ p/q and n 1isa positiveintegerlessthan n, p/q< 1/(n 1). (d)Wehave

(9)

Since1/n<p/q,equation(9)showsthat

wehave

Thethirdinequalityisestablished.

Now

Inparticular, p1 q1 < 1

Wehaveestablishedthesecondinequality. Bytheinductiveassumption, p1/q1 canbeexpressedinEgyptianform.Thelastequationfollows. (e)See(10).

(f)Theequationistruebecauseof(d).Forany i =1,...,k,

Itfollowsthat n,n1,...,nk aredistinct.

26. 3 8 = 1 3 + 1 24 , 5 7 = 1 2 + 1 5 + 1 70 , 13 19 = 1 2 + 1 6 + 1 57

29.Enclosethemissingsquareinacorner(n 3) × (n 3)subboardasshowninthefollowingﬁgure.Since 3divides n2 1,3alsodivides(n 3)2 1.Now n 3isodd, n 3 > 5,and3divides(n 3)2 1,soby Exercise28,wemaytilethissubboard.Tilethetwo3 × (n 4)subboardsusingtheresultofExercise 34,Section2.4.Tilethedeﬁcient4 × 4subboardusingExample2.4.7.The n × n boardistiled. (n 3) × (n 3) 3 × (n 4) 4 × 4 3×(n 4) n n

30.SeeI.P.ChuandR.Johnsonbaugh,“Tilingboardswithtrominoes,” J.Recr.Math.,18(1985–86) 188–193.

32.

33.SeeC.Jepsen,“Ontilingdeﬁcientboardswithtrominoes,” J.Recr.Math.,27(1995)125–130.

34.If n =0, d 1= d> 0,and1isin X.If n> 0, d(2n)= n(2d) >n;thus2n isin X.Ineithercase X isnonempty.Since d> 0and n ≥ 0, X containsonlypositiveintegers.BytheWell-OrderingProperty, X containsaleastelement q > 0.Then dq >n.Let q = q 1.Wecannothave dq>n (forthen q wouldnotbethe least elementin X);therefore, dq ≤ n.Let r = n dq.Then r ≥ 0.Also

r = n dq = n d(q 1) <dq d(q 1)= d.

Therefore,wehavefound q and r satisfying

n = dq + r 0 ≤ r<d.

35.WeﬁrstproveTheorem2.5.6for n> 0.TheBasisStepis n =1.If d =1,wehave n = dq + r,where q = n and r =0,0 ≤ r<d.If d> 1,wehave n = dq + r,where q =0and r =1,0 ≤ r<d.Thus Theorem2.5.6istruefor n =1.

AssumethatTheorem2.5.6holdsfor n.Thenthereexists q and r suchthat

n = dq + r 0 ≤ r <d.

Now

n +1= dq +(r +1)

If r <d 1,then r +1 <d.Inthiscase,ifwetake q = q and r = r +1,wehave

n +1= dq + r 0 ≤ r<d.

If r = d 1,wehave

n +1= d(q +1).

Inthiscase,ifwetake q = q +1and r =0,wehave

n +1= dq + r 0 ≤ r<d.

TheInductiveStepiscomplete.Therefore,Theorem2.5.6is trueforall n> 0.

If n =0,wemaywrite

n = dq + r,

where q = r =0.Therefore,Theorem2.5.6istruefor n =0.

Finally,supposethat n< 0.Then n> 0,sobytheﬁrstpartoftheproof,thereexist q and r such that

n = dq + r 0 ≤ r <d.

If r =0,wemaytake q = q and r =0toobtain

n = dq +0.

If r > 0,wetake q = q 1and r = d r .Then0 <r<d and n = d( q ) r = d(q +1)+(r d)= dq + r.

Therefore,Theorem2.5.6istruefor n< 0.

37.Supposethatwehaveapropositionalfunction S(n)whosedomainofdiscourseisthesetofintegers greaterthanorequalto n0.Supposethat S(n0)istrueand,forall n>n0,if S(k)istrueforall k,n0 ≤ k<n,then S(n)istrue.Wemustprovethat S(n)istrueforeveryinteger n ≥ n0.Weﬁrst assumethat n0 ≥ 0.

Wearguebycontradiction.Soassumethat S(n)isfalseforsomeinteger n1 ≥ n0.Let X betheset ofnonnegativeintegersforwhich S(n)isfalse.Then X isnonempty.BytheWell-OrderingProperty, X hasaleastelement n2.Since S(n0)istrue, n2 >n0.Furthermore,forany k,n0 ≤ k<n2, S(k) istrue[otherwise n2 wouldnotbetheleastinteger n forwhich S(n)isfalse].Since S(k)istrueforall k,n0 ≤ k<n2,byhypothesis, S(n2)istrue.Contradiction. If n0 < 0,applythepreviousargumenttothepropositionalfunction S (n): S(n + n0) withdomainofdiscoursethesetofnonnegativeintegers.

38.ThestrongformofinductionclearlyimpliestheformofinductionwheretheInductiveStepis:“If S(n) istrue,then S(n +1)istrue.”Fortheconverse,useExercises36and37.