Table Of Content“Appendix-1” — 2018/11/8 — 15:50 — page 1 — #1
Appendix: Solution Sketches for
Mostly Odd-Numbered Exercises
Thesolutionoftheexercisesisthemathematicsversionofusingourvocabularyandgram-
mar. If this is your first time writing proofs, DO NOT PANIC. It just takes practice. Also,
youhavemorehelpthanIdidasastudent.Atleastsolutionstoodd-numberexercisesare
to be found. And there is Google of course. Do not worry too much if your solutions are
differentfrommine.Therearemanypathstothemountaintop.
Chapter 1
Exercise 1.1.1. In Mathematica, I wrote Table[FactorInteger[x^2-x+41],
{x,0,41}].
∑ ( )
Exercise 1.1.3. Using Exercise 1.4.7 0:9999(cid:1)(cid:1)(cid:1)= 9 + 9 + 9 +(cid:1)(cid:1)(cid:1)= 9 1 1 n =
10 100 1000 10 n=0 10
9 1 =1.
101(cid:0)1
10
Exercise1.2.1.Thepointofbothoftheseexercisesisthatthenegationof(PandQaretrue)
is((Pfalse)or(Qfalse)),fortwopropositionsPandQ.Similarly,thenegationof(PorQ)is
((notP)and(notQ)).
(a) FirstweshowthatA(cid:0)(B[C)(cid:26)(A(cid:0)B)\(A(cid:0)C).Ifx2Aandx2=(B[C),thenx2A
and(x2=Bandx2=C).Thisimpliesx2(A(cid:0)B)\(A(cid:0)C).
On the other hand, to see that A(cid:0)(B[C)(cid:27)(A(cid:0)B)\(A(cid:0)C), note that if x2
(A(cid:0)B)\(A(cid:0)C),then(x2Aandx2=Bandx2=C). Itfollowsthatx2Aandx2=B[C.
Thusx2A(cid:0)(B[C).
(b) To cheat a bit, we combine the two inclusions instead of writing the same thing twice
witharrowsreversed.Myhighschoolteacherwouldneverforgivemeforthis.
x2A(cid:0)(B\C)
()x2Aandx2=(B\C)
()x2Aand(x2=Borx2=C)
()x2A(cid:0)(B[C).
Exercise1.2.3.Againwecheatabitandcombinethings.Weusethefactthatforproposi-
tionsP;Q;R,thestatement(Ptrue)and(QorRtrue)isthesameasthestatement(PandQ
true)or(PandRtrue).
x2A\(B[C)()x2Aand(x2Borx2C)
()(x2Aandx2B)or(x2Aandx2C)
()x2(A\B)[(A\C).
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2 Appendix: SolutionSketchesforMostlyOdd-NumberedExercises
Exercise1.2.5.
(a) true,asorderisirrelevantforsetmembership;
(b) false,asordermattersfororderedpairs;
f g
(c) false,as 0 isasetwithoneelement–nottheemptyset.
Exercise1.3.1.
′
(a) Supposethat0and0 arebothidentitiesforaddition.Then,byR1andR3,
0=0′+0=0+0′=0′. Thus0′=0.
′
Similarlyif1and1 areidentitiesformultiplication,byR1andR3,wehave
1=1′(cid:1)1=1(cid:1)1′=1′. Thus1′=1.
(b) Supposethatm+x=m+y=0.ThenusingR1,R2,andR3,wefindthat
y=0+y=(m+x)+y=(x+m)+y=x+(m+y)=x+0=x:
Exercise 1.3.3. Using R5, we see that m+((cid:0)1)m=1(cid:1)m+((cid:0)1)(cid:1)m=(1+((cid:0)1))(cid:1)m=
0(cid:1)m=0bytheprecedingexercise.Itfollowsthat(cid:0)m=((cid:0)1)m,usingExercise1.3.1.Since
((cid:0)m)+m=0,itfollowsthatm=(cid:0)((cid:0)m),usingExercise1.3.1.
Exercise 1.3.5. Using R1 and R5, we see that (m+n)(cid:1)k=k(cid:1)(m+n)=k(cid:1)m+k(cid:1)n=
m(cid:1)k+n(cid:1)k.
Exercise1.3.7.Fact2.ByO1,eitherx(cid:0)y2Porx(cid:0)y=0or(cid:0)(x(cid:0)y)=y(cid:0)x2P.Incase1,
wehavex>y.Incase2,wehavex=y.Incase3,wehavey>x.
Fact4.Ifc>0andx<y,thency(cid:0)cx=c(y(cid:0)x)2Pandcx<cy.
Fact5.Ifc<0andx<y,thencx(cid:0)cy=c(x(cid:0)y)=((cid:0)c)(y(cid:0)x)2P.Thuscx>cy.
Exercise1.3.9.
(a) SupposethatRisanorderedintegraldomainwithtwoelements.ByR3,R=f0;1g.We
know0<1.Bythethirdfactandfirstfactaboutorder,wehave1+1>1+0=1>0.
ThiswouldgiveathirdelementofR.Contradiction!
(b) Suppose R has n elements. We know that n>2 by part (a) and the fact that 0̸=1 is
always assumed. By R3, R must contain 0 and 1. Moreover it must contain 1+1=2,
2+1=3;::::;1+(cid:1)(cid:1)(cid:1)1=n. By fact 3 about inequalities, we know that 0<1<2<
3<(cid:1)(cid:1)(cid:1)<n. Butthenn<n+1andn+1mustbeinRasRisclosedunderaddition–
contradictingthefactthatRhasnelements.
Exerpcise 1.3.11.pNote that the facts about inequalities hold for the real numbers too. Set
x= aandy= b.Then,sincethesquareofanyrealnumberis(cid:21)0,weseethat
0(cid:20)(x(cid:0)y)2=x2(cid:0)2xy+y2;
whichimplies(byadding2xytobothsides)that
2xy(cid:20)x2+y2:
Then(dividingby2andgoingbacktotheoriginalvariablesaandb),weobtain:
p
x2+y2 a+b
xy(cid:20) whichmeans ab(cid:20) :
2 2
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Chapter1 3
Exercise1.3.13
(a) If there exists a2Z+ such that 0<a<1, then 0<a2<a by fact 4 about inequalities.
Butthenthenon-emptysetS=fanjn2Z+gisasubsetofZ+ withoutaleastelement–
contradictingthewellorderingprinciple.
(b) Iftherewereanintegerasuchthat1<a<2,thenbyfact3aboutorder(subtracting1
from all parts of the inequalities), we would know that 0=1(cid:0)1<a(cid:0)1<2(cid:0)1=1,
contradicting part (a). Similarly if a were an integer in Z+ such that n<a<n+1 for
someintegern,thenbyfact3(subtractingnfromallpartsoftheinequalities)wehave
0<a(cid:0)n<1, contradicting part (a). So the only integers in Z+ are those we know
about–namely1;2;3;4;:::
Exercise1.4.1.Firstweprovethecasen=1.Usingourknowledgeoffractions,wehave
1(cid:1)(1+1)(cid:1)(2+1) 2(cid:1)3
12= = =1:
6 6
Fortheinductionstepweassumetheresultforntermsontheleft,whichis
n(n+1)(2n+1)
12+22+(cid:1)(cid:1)(cid:1)+n2= : (A.1)
6
Thenweadd(n+1)2 tobothsidestoobtain
n(n+1)(2n+1)
12+22+(cid:1)(cid:1)(cid:1)+n2+(n+1)2= +(n+1)2:
6
Now we must simplify the right-hand side of the equation to see that it is (A.1) with n
replacedbyn+1.SoagainwemesswithfractionsusingthedistributivelawonQandthe
definitionofadditionoffractions:
{ }
n(n+1)(2n+1) n(2n+1)
+(n+1)2=(n+1) +(n+1)
6 6
{ }
n(2n+1)+6(n+1)
=(n+1)
6
{ }
2n2+7n+6
=(n+1)
6
(n+1)(n+2)(2n+3)
= ;
6
whichisequation(A.1)withnreplacedbyn+1.
Exercise1.4.3.SupposewehavealistofstatementsT ,n2Z+,suchthatweknowT istrue
n 1
andweknowthatTntrueimpliesTn+1trueforalln2Z+.DefineS=fn2Z+jTn isfalseg.If
Swerenon-empty,thenthewell-orderingprinciplewouldimplythatShasaleastelement
a. Since T is true, we know that a>1. By the definition of S and a, we know that T is
1 k
trueforallk=1;:::;a(cid:0)1. ButthenweknowthatT istrue–contradictinga2Sandthe
a
definitionofS.
Exercise1.4.5.Thecasen=1isclearasA\B =A\B .Thecasen=2wasExercise1.2.3.
1 1
(cid:21)
Sonowassumetheresultforn 2:namely,assumethat
A\(B [(cid:1)(cid:1)(cid:1)[B )=(A\B )[(cid:1)(cid:1)(cid:1)[(A\B ):
1 n 1 n
Herewealsoassumetheassociativityofunionsofarbitrarynumbersofsets,whichfollows
fromExercise1.2.2.
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4 Appendix: SolutionSketchesforMostlyOdd-NumberedExercises
Sowefindthat,usingthecasen=2,
A\((B1[(cid:1)(cid:1)(cid:1)[Bn)[Bn+1)=(A\(B1[(cid:1)(cid:1)(cid:1)[Bn))[(A\Bn+1):
Thenbytheinductionhypothesis,weget
A\(B1[(cid:1)(cid:1)(cid:1)[Bn+1)=(A\B1)[(cid:1)(cid:1)(cid:1)[(A\Bn)[(A\Bn+1):
Exercise1.4.7.
(a) Thecasen=0saysx0=1=x1(cid:0)1,whichiscertainlytrue.
x(cid:0)1
For the induction step, we assume the nth formula and add the (n+1)th term to it.
Thisgives
xn+1(cid:0)1
nthformula: 1+x+x2+(cid:1)(cid:1)(cid:1)xn=
(cid:0)
x 1
nextterm: xn+1=xn+1
xn+1(cid:0)1
(n+1)thformula:1+x+x2+(cid:1)(cid:1)(cid:1)xn+xn+1= +xn+1:
(cid:0)
x 1
Noweverythingridesonourabilitytosimplifytheright-handsideofthelastformula.
Weusetherulesforaddingfractionstoget
xn+1(cid:0)1 xn+1(cid:0)1+xn+1(x(cid:0)1)
+xn+1=
(cid:0) (cid:0)
x 1 x 1
xn+2(cid:0)xn+1+xn+1(cid:0)1 xn+2(cid:0)1
= = :
(cid:0) (cid:0)
x 1 x 1
Thisisexactlywhatweneededtofinishtheinductionstep.
(b) Forthisweneedtoremembersomethingfromcalculus,namelythefactthatifjxj<1,
then limn!1xn=0. Hopefully you believe this. Then you need to believe the various
factsaboutlimitsofdifferences,quotients,:::toseethat
lim xn+1(cid:0)1=nl!im1xn+1(cid:0)1=0(cid:0)1= 1 :
n!1 x(cid:0)1 x(cid:0)1 x(cid:0)1 1(cid:0)x
Exercise 1.4.9. We use mathematical induction I. Step 1 says 1+ 1(cid:21)1+ 1. That is
2 2
clearlytrue.
Theinductionstepassumesthenthformulaandaddstheterms
1 1 1
+ +(cid:1)(cid:1)(cid:1)+
2n+1 2n+2 2n+1
tobothsides.Thusweneedtoshowthat
n 1 1 1 n+1
1+ + + +(cid:1)(cid:1)(cid:1)+ (cid:21)1+ :
2 2n+1 2n+2 2n+1 2
Thisisequivalenttoshowingthat
1 1 1 1
+ +(cid:1)(cid:1)(cid:1)+ (cid:21) :
2n+1 2n+2 2n+1 2
(cid:21)
Theleft-handsideis
2n+1(cid:0)2n 1
= :
2n+1 2
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Chapter1 5
Exercise1.5.1.Ifaorbequals0,thenbothare0anda=(cid:6)b.Soassumeneitheranorbis0.
Thenajbmeansb=ac,forsomenonzerointegerc.Andbjameansa=bd,forsomenonzero
integer d. Thus, substituting the formula for a into b=ac, we get b=bdc, which implies,
since b̸=0, cd=1. If c>1, then d>0 and d(cid:21)1, which implies cd>1, a contradiction.
Thus c(cid:20)0. Of course then if c̸=(cid:0)1, this means c<(cid:0)1 and then also d(cid:20)(cid:0)1, implying
that cd>1, again a contradiction. Here we have the facts about inequalities, of course.
Theconclusionisthatc=d=(cid:6)1.Againa=(cid:6)b.
(cid:20)
Exercise1.5.3.Toshow(withoutacomputer)thattheprecedingexampleoftheprimes 31
is correct, the most efficient method is to use the Sieve of Erastothenes. We know that 2
is prime, since any divisor d>1 of 2 would satisfy 1<d(cid:20)2. There are no integers d that
satisfytheseinequalitiesexceptd=2.Therefore2isprime.Sonextwelisttheoddintegers
between 3 and 31. The smallest number on this list is 3, which is prime, by the same sort
ofreasoningthattoldus2isprime.Nextwecrossoutallnumbersnonthelistwhichare
multiplesof3suchthatn>3.Thatleaves:5;7;11;13;17;19;23;25;29;31.Again,5must
beprime,sinceithasnoproperdivisors.Weshouldnextcrossoutanymultiplesof5thatare
largerthan5.Thatmeanswecrossout25.Weclaimthattheremainingnumbersareprime.
Why? Otherwise the number n on the list haspa divisorpd – meaning that n=dk. Suppose
1<d(cid:20)k<n.Thismeansd2(cid:20)n.Butthend(cid:20) n.Now 31(cid:24)=5:5678.Thismeansthatthe
numbers n2f5;7;11;13;17;19;23;29;31g cannot have a prime divisor d<n. Thus, the
numbersonourlistareallprime.
f 2Zj (cid:21)(cid:0) g
Exercise1.5.5.Theset n n x isnon-emptyandthushasaleastelementmbythe
well-orderingaxiom.Then(cid:0)m=⌊x⌋.
⌊ ⌋
Exercise 1.5.7. Set q= a , Then set r= a(cid:0)bq. We must show that 0(cid:20)r<b. First q(cid:20)a
b b
impliesqb(cid:20)a,usingthefactsaboutinequalitiesandb>0.Therefore0(cid:20)a(cid:0)qb=r,again
usingthefactsaboutinequalities.Ifr(cid:21)b,thena(cid:0)qb(cid:0)b=a(cid:0)(q+1)b(cid:21)0.Thisimplies
a(cid:21)(q+1),contradictingthedefinitionofqasthelargestinteger(cid:20)a.
b b
Exercise 1.5.9. gcd(13;169)=gcd(13;132)=13. gcd(11;1793)=gcd(11;11(cid:1)163)=11.
HereIamusingthefactthat11and13areeachprime.
Exercise1.5.11.Weperformthedivisions
a=bq +r ; where 0(cid:20)r <b
1 1 1
b=r q +r ; where 0(cid:20)r <r
1 2 2 2 1
r =r q +r ; where 0(cid:20)r <r
1 2 3 3 3 2
. .
. .
. .
rn(cid:0)2=rn(cid:0)1qn+rn; where 0(cid:20)rn<rn(cid:0)1
rn(cid:0)1=rnqn+1+0; sothatrn=lastnonzeroremainder:
The fact that there is a last nonzero remainder comes from the well-ordering principle
as any set of non-negative integers must have a least element. So then why must r =
n
gcd(a;b)? First we see that rn divides rn(cid:0)1 and then rn(cid:0)2 and so on – reading the table
j j
frombottomtotop–endingwithr bandthusr a.Nextweneedtoseethatanycommon
n n
divisor c of a and b must divide r . For this read the table from top to bottom, to see that
n
j j j j j
cr1 andthencr2 andsoon–endingwithcrn(cid:0)2 andcrn(cid:0)1 andthencrn.
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6 Appendix: SolutionSketchesforMostlyOdd-NumberedExercises
Exercise1.5.13
1001=163(cid:1)6+23
1141=163(cid:1)7
163=23(cid:1)7+2
23=2(cid:1)11+1
gcd(1141;163)=gcd(163(cid:1)7;163)=163:
so gcd(1001;163)=1
1=23(cid:0)2(cid:1)11=23(cid:0)(163(cid:0)23(cid:1)7)(cid:1)11
=23(cid:1)78(cid:0)163(cid:1)11=(1001(cid:0)163(cid:1)6)(cid:1)78(cid:0)163(cid:1)11 163=1(cid:1)163+0(cid:1)1141
=78(cid:1)1001(cid:0)479(cid:1)163
Exercise 1.5.15. To prove the existence of the factorization, use the second induction
principle and induct on n2Z, n(cid:21)2. If n=2, then n is itself prime and we are done.
Assume the existence of the factorization for integers between 2 and n. We must prove
the result for n+1. If n+1 is prime, we are done. Otherwise n+1=ab, with 1<a;b<
n+1. By the induction assumption, both a and b are products of primes and thus so is
ab=n+1.QED.
Exercise1.5.17.Thepositivedivisorsof24are1;2;3;4;6;8;12;24.
Thepositivedivisorsof36are1;2;3;4;6;9;12;18;36.
Thepositivedivisorsof81are1;3;9;27;81.
p
Exercise 1.5.19. Suppose 2=m=n, where m;n2Z+ and gcd(m;n)=1. We can make
this assumption since we can always divide out any common factors in the numera-
tor and denominator of a fraction. Then upon squaring both sides, we have 2=m2=n2.
Next multiply both sides by n2 to see that 2n2=m2. This implies that m is even since
the square of an odd number is odd. To see this, write m=2k+1, for some k2Z. Then
m2=4k2+4k+1=2(2k2+2k)+1,whichisodd.Thusm=2kforsomek2Z.Substitute
thisinto2n2=m2toseethat2n2=4k2,whichimpliesn2isevenandniseven,contradicting
gcd(m;n)=1. p
Assume k(cid:21)3. Suppose that k2=m=n, where m;n2Z+ and gcd(m;n)=1. Then, as
before, if we take the kth power of the equality, we get 2=mk=nk. Then, multiplying by
nk, we find that 2nk=mk. This means that if we factor m as a product of primes, 2 must
appear. Moreover, by the uniqueness of factorization into primes, 2dk must be the power
of 2 in mk, where d(cid:21)1. But then, again using uniqueness of factorization into primes, if
2e is the power of 2 in n, we have 1+ke=dk. This says k divides 1, which is impossible
(cid:21)
fork 3.
Exercise 1.5.21. For the y-axis, we are seeing x=0 and gcd(x;y)=y. So the color at the
point(x;y)correspondstothesizeofy.Theanaloghappensonthex-axis.Whenx=y,we
have gcd(x;y)=x. The color at the point (x;y) corresponds to the size of x on the x-axis
(oryonthey-axis).
When y=kx, for k2Z+, we have gcd(x;y)=gcd(x;kx)=x. So, on the line y=kx, the
colorat(x;y)correspondstothesizeofx.
If x=ky, for k2Z+, then we have gcd(x;y)=gcd(ky;y)=y. In this case, the size of y
correspondstothecoloratthepoint(x;y).
Asthenumbersx;yincreaseweseeafullcolorspectrumred,blue,green,yellow,orange.
Ontheotherhandwhengcd(x;y)=1,weseered.
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Chapter1 7
∏ ∏ ∏
k k k
Exercise 1.5.23. Set m= pei and n= pfi, and d= pgi; whereg =minfe;fg. First
i i i i i i
i=1 i=1 i=1
∏
k
notethatddividesbothmandn.Nextletr= psi beadivisorofm.Itfollowsthats (cid:20)e,
i i i
i=1
(cid:20)
foralli.Similarlyifrdividesn,weknowthats f foralli.Thusifrisacommondivisor
i i
(cid:20)
of m and n, we see that s g for all i. This means that r divides d. Therefore d is indeed
i i
thegreatestcommondivisorofmandn.
Exercise1.6.1.12o’clocka.m.midnight.
Exercise 1.6.3. Suppose that a(cid:17)b(mod m). This means a(cid:0)b=mc for some integer c.
Then divide both a and b by m. We get a=mq+r and b=mq′+r′ with 0(cid:20)r;r′<m.
By interchange of a and b, if necessary, we may suppose that r(cid:0)r′(cid:21)0. Then r(cid:0)r′=
a(cid:0)mq(cid:0)(b(cid:0)mq′)=a(cid:0)b(cid:0)m(q(cid:0)q′)=mc(cid:0)m(q(cid:0)q′)=m(c(cid:0)q+q′). So mj(r(cid:0)r′).
But0(cid:20)r(cid:0)r′<mandthisimpliesr(cid:0)r′=0.
Conversely,supposea=mq+randb=mq′+rwith0(cid:20)r<m.Thisimpliesthata(cid:0)b=
m(q(cid:0)q′)andthusa(cid:17)b(modm).
Exercise1.6.5.
(a) Assumea(cid:17)b(modn).Thusa(cid:0)b=nqforsomeintegerq.Wemustshowthata+c(cid:17)
b+c(modn).Well(a+c)(cid:0)(b+c)=a(cid:0)b=nq.Thedesiredcongruencefollows.
(b) Wemustshowthata(cid:17)b(modn)andc(cid:17)d(modn)implythata+c(cid:17)b+d(modn).
We know that a+c(cid:17)b+c (mod n) by part (a). Also we know that b+c(cid:17)b+
d(modn),againbypart(a).Itfollowsthata+c(cid:17)b+d(modn).
Forthislaststepweareusingthefactthatcongruencehasthetransitivepropertyofan
equivalencerelation(asdefinedinSection1.6).Thatis,wemustshowthatifx(cid:17)y(modn)
and y(cid:17)z (mod n), then x(cid:17)z (mod n). Our hypotheses say that x(cid:0)y=un and y(cid:0)z=vn
for some integers u;v. This means that x(cid:0)z=(x(cid:0)y)+(y(cid:0)z)=un+vn=(u+v)n and
thusthatx(cid:17)z(modn).
Exercise1.6.7.
5100007(cid:17)1100007(cid:17)1(mod4):
3100007(cid:17)((cid:0)1)100007(cid:17)(cid:0)1(cid:17)3(mod4):
Exercise 1.6.9. We want to know when we can solve the congruence ax(cid:17)b (mod n) for x,
giventheintegersa;b;n,withn>0.Firstweassumethatd=gcd(a;n)dividesb.Weneed
to use Theorem 1.5.2. This theorem says d=gcd(a;n)=ra+sn, for some integers r;s. So
ifddividesb,wehaveb=dq=(ra+sn)q=arq+nsqandwecantakex=rq(modn).
On the other hand if a=da and n=dn , with gcd(a ;n )=1; and there is an x such
1 1 1 1
that ax(cid:17)b (mod n), then we have da x(cid:0)b=dn q, for some integer q. But this says
1 1
b=d(a x(cid:0)n q)whichimpliesthatddividesb.
1 1
Exercise1.6.11.Firstgcd(5;163)=1.Then(usingExercise1.6.9)weseethat
163=5(cid:1)32+3
5=3(cid:1)1+2
3=2(cid:1)1+1
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8 Appendix: SolutionSketchesforMostlyOdd-NumberedExercises
So
1=3(cid:0)2(cid:1)1
1=3(cid:0)(5(cid:0)3(cid:1)1)
1=3(cid:1)(1+1)(cid:0)5
1=(163(cid:0)5(cid:1)32)(cid:1)2(cid:0)5
1=163(cid:1)2(cid:0)5(cid:1)65
Sotosolve5x(cid:17)1(mod163),weneed1=5x+163yandthismeans
x(cid:17)(cid:0)65(mod163) or x(cid:17)98(mod163):
Exercise1.6.13.
( ) ( ) ( )
(cid:0)1 (cid:0)
1 2 1 2 1 5
= = mod7
0 1 0 1 0 1
( ) ( ) ( ) ( )
1 2 (cid:0)1 1 4 (cid:0)2 1 4 5 (cid:0)4 4 5
= = = mod7
3 4 4(cid:0)6 (cid:0)3 1 (cid:0)2 4 1 1 4 1
( ) ( ) ( )
3 4 5 12 15 5 1
= = = mod7
1 4 1 12 3 5 3
Exercise1.6.15 Solve2x(cid:17)1(mod5)bytrialanderror.Weseethatx(cid:17)3(mod5)works.
This means x=3+5y. Plug that into the second congruence 3x(cid:17)2 (mod 7). So we need
tosolve3(3+5y)(cid:17)2(mod7).Simplifytoseethatweneed15y(cid:17)2(cid:0)9(mod7).Thuswe
mustsolvey(cid:17)0(mod7).Thatgivesx=3(mod35).Checkthis!
Exercise 1.7.1. In Figure 1.10 only one line is visible, y=x. On the white side is y>x.
Onthepurplesideisy<x.
In Figure 1.11 we see many white lines. The line y=x comes from the fact that yjy, for
integers y. The line y=x=2 comes from the fact that nj2n with x=n and y=2n, n2Z.
Similarlyweseetheliney=x=3asnj3nwithx=n,y=3n; n2Z.Andsoweseethelines
y=x=k,k=4;5;:::
In Figure 1.12 we see the white lines y=x, y=x+5, y=x(cid:0)5;y=x+10;y=x(cid:0)10,
andgenerallyy=x(cid:6)5n,n=0;1;2;3;:::.
Exercise1.7.3.Itisnotsymmetricasajbdoesnotimplybjaunlessa=(cid:6)b.Youcanseethis
fromFigure1.11–asthelinesy=x=k,k=1;2;3;4;:::arevisiblebutnotthelinesx=y=k,
k=2;3;4;:::
Exercise1.7.5.Firstcheckthepropertiesofanequivalencerelation.
reflexive:a(cid:0)a=02Z.
symmetric:a(cid:0)b2Zimplies(cid:0)(a(cid:0)b)=b(cid:0)a2Z.
transitive:a(cid:0)bandb(cid:0)c2Zimplya(cid:0)c=(a(cid:0)b)+(b(cid:0)c)2Z.
Then note that a set of representatives for R=(cid:24) is the interval [0;1). We exclude 1 as
1(cid:24)0. To see no two elements a;b2[0;1) can be equivalent under this relation, we can
assumea(cid:21)b.Thena(cid:0)b2[0;1).However,[0;1)\Z=f0gimpliesthattheonlywaya(cid:24)b
would be a(cid:0)b=0 and a=b. Also, given any real number x, we can translate x to [0;1)
byaninteger,sincex(cid:0)⌊x⌋2[0;1),where⌊x⌋isthegreatestinteger(cid:20)x.
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Chapter1 9
Exercise 1.7.7. The set of positive divisors of 30 is f1;2;3;5;6;10;15;30g. The poset
diagramisbelow.
FigureA.1 Posetdiagramforthepositivedivisorsof30
30
6 15
10
2 5
3
1
Exercise1.8.1.
(a) f(A[B)=f(A)[f(B)istrue.y2f(A[B) ()y=f(x)forsomex2A[B()either
y2f(A)ory2f(B)()y2f(A)[f(B).
(b) f(A\B)=f(A)\f(B) is false, in general. We need only give a counterexample. Let
f:R!R+[f0gbedefinedbyf(x)=x2.LetA=R+=(0;1)andB=(cid:0)R+=((cid:0)1;0).
Thenf(A)\f(B)=R+ whileA\B=∅sothatf(A\B)=∅.
Exercise1.8.3.Ifg(f(x))=g(f(u)),thensincegis1–1,weknowf(x)=f(u).Butthen,since
fis1–1,wehavex=u.
Givenc2C,sincegisonto,9b2Bsuchthatc=g(b).Then,sincefisonto,9a2Asuch
thatb=f(a).Itfollowsthatc=g(f(a))andg◦fisonto.
Exercise1.8.5.
(a) f:Z!Zgivenbyf(x)=x6 isnot1–1norisitonto.Forf(x)=f((cid:0)x)andmostintegers
arenot6thpowers:forexample,2or3.
(b) f:Z!Z given by f(x)=3x is 1–1 but not onto. For 3x=3y implies x=y by the
cancellationlaw.However,mostintegersarenotdivisibleby3:forexample,2.
(c) f:Z!Z given by f(x)=x(cid:0)3 is 1–1 and onto. For x(cid:0)3=u(cid:0)3 implies x=u, after
adding3tobothsides.Also,giveny2Z,wecansolvey=x(cid:0)3withx=y+3.
Exercise1.8.7.LetS=fs ;:::;s gandT=ft ;:::;t g.Then
1 m 1 n
∪n
S[T= (S(cid:2)ftg):
i
i=1
NowS(cid:2)ftg=f(s ;t);:::;(s ;t)g.ItfollowsthatjS(cid:2)ftgj=jSj.Thusfromthepreced-
i 1 i m i i
ingexerciseweseethatjS(cid:2)Tj=jSjjTj.
Exercise 1.8.9. If A and B are finite sets, each with the same number n of elements, then
f:A!Bis1–1ifffisonto.
Suppose A=fa ;:::;a g and B=fb ;:::;b g. If f:A!B is 1–1, then f:A!f(A) is
1 n 1 n
1–1 and onto – which implies jf(A)j=jAj=jBj. If f(A) does not equal B, there is some
elementofBwhichisnotinf(A).Thatwouldmeanthatjf(A)j<jBj.Contradiction.
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10 Appendix: SolutionSketchesforMostlyOdd-NumberedExercises
Next suppose f:A!B is onto. If f were not 1–1, we could leave some element a out of
(cid:0)f g
A and still have an onto map from A a onto B. That would say that there is a proper
subset S of A such that f:S!B is 1–1 and onto – implying jSj=jBj. But jSj<jAj would
contradictjAj=jBj.
Exercise1.8.11.
( ) ( )
n n n! n! k+n(cid:0)k+1
(a) + = + =n!
k(cid:0)1 k (k(cid:0)1)!(n(cid:0)k+1)! k!(n(cid:0)k)! k!(n(cid:0)k+1)!
( )
n+1 (n+1)! n+1
=n! = = .
k!(n(cid:0)k+1)! k!(n+1(cid:0)k)! k
( ) ( )
n n! n! n
(b) = = = .
n(cid:0)k (n(cid:0)k)!(n(cid:0)(n(cid:0)k))! (n(cid:0)k)!k! k
Exercise1.8.13.Thecasen=1is∑(fg)′=(f′)g+fg′
Assumethenthcase:(fg)(n)= n n f(k)g(n(cid:0)k).
k=0 k
Differentiatethisusingthecasen=1toget:
∑n ( )( )
(fg)(n+1)= n f(k+1)g(n(cid:0)k)+f(k)g(n+1(cid:0)k)
k
k=0 ( ) ( )
∑n ∑n
=f(n+1)g(0)+ n f(r)g(n+1(cid:0)r)+ n f(r)g(n+1(cid:0)r)+g(n+1)f(0)
(cid:0)
r 1 r
r=1 r=1
( )
=∑n+1 n+1 f(t)g(n+1(cid:0)t):
t
t=0
Here in the second line we changed variables in the first part of the sum – setting r=
k+1,whileinthesecondpartofthesumwejustsetr=k.InthethirdlineweusedExercise
1.8.13.Thustheformulaisprovedbyinduction.
(cid:21)
Exercise1.8.15.Wewanttoprovethefollowingforx 0:
i
x +(cid:1)(cid:1)(cid:1)+x p
1 n (cid:21) nx (cid:1)(cid:1)(cid:1)x :
1 n
n
We use Jensen’s inequality from Exercise 1.8.16. Set f(x)=(cid:0)logx on (0;1). Then
f′(x)=(cid:0)1=x and f′′(x)=x(cid:0)2>0, for x>0. So f is convex. This is proved in most cal-
culus books. When I looked at my favorite book Purcell [87, pp. 185ff], however, I found
thatitsaid“concaveupward”ratherthan“convex.”ThusterminologyhaschangedsinceI
wasateachingassistantattheUniversityofIllinois.ThenIlookedatmyfavoriteadvanced
calculus book Lang [63, pp. 72–73] and he used the terminology “convex upward” rather
than“convex.”Soreaderbeware.
∑
n
WeapplyJensenforx >0,i=1;:::;n,andweightsa 2(0;1)suchthat a =1.Then
i i i
i=1
Jensensays
(cid:0)log(a x +(cid:1)(cid:1)(cid:1)+a x )(cid:20)(cid:0)a logx (cid:0)(cid:1)(cid:1)(cid:1)(cid:0)a logx :
1 1 n n 1 1 n n
Exponentiate this – using the fact that ex is a monotone strictly increasing function. This
gives
a x +(cid:1)(cid:1)(cid:1)+a x (cid:21)xa1(cid:1)(cid:1)(cid:1)xan:
1 1 n n 1 n
Thenseta =a =(cid:1)(cid:1)(cid:1)=a =1.Theresultisthenthearithmetic–geometricmeaninequality.
1 2 n n
