Instructor’s Solutions Manual for Elementary Number Theory and Its Applications

268 Pages·2011·2.03 MB·english

by  Kenneth H. Rosen

O
NLINE
I ’
NSTRUCTOR S
S M
OLUTIONS ANUAL
E N T
LEMENTARY UMBER HEORY
A
AND ITS PPLICATIONS
S E
IXTH DITION
Kenneth Rosen
Monmouth University
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ISBN-13: 978-0-321-53801-7
ISBN-10: 0-321-53801-3
Contents
1
Chapter1. TheIntegers 1
1.1. NumbersandSequences 1
1.2. SumsandProducts 7
1.3. MathematicalInduction 10
1.4. TheFibonacciNumbers 14
1.5. Divisibility 19
Chapter2. IntegerRepresentationsandOperations 25
2.1. RepresentationsofIntegers 25
2.2. ComputerOperationswithIntegers 29
2.3. ComplexityandIntegerOperations 31
Chapter3. PrimesandGreatestCommonDivisors 35
3.1. PrimeNumbers 35
3.2. TheDistributionofPrimes 38
3.3. GreatestCommonDivisorsandtheirProperties 44
3.4. TheEuclideanAlgorithm 49
3.5. TheFundamentalTheoremofArithmetic 53
3.6. FactorizationMethodsandtheFermatNumbers 63
3.7. LinearDiophantineEquations 66
Chapter4. Congruences 71
4.1. IntroductiontoCongruences 71
4.2. LinearCongruences 78
4.3. TheChineseRemainderTheorem 82
4.4. SolvingPolynomialCongruences 86
4.5. SystemsofLinearCongruences 89
4.6. FactoringUsingthePollardRhoMethod 91
Chapter5. ApplicationsofCongruences 93
5.1. DivisibilityTests 93
5.2. ThePerpetualCalendar 98
5.3. Round-RobinTournaments 101
5.4. HashingFunctions 103
5.5. CheckDigits 104
Chapter6. SomeSpecialCongruences 111
6.1. Wilson’sTheoremandFermat’sLittleTheorem 111
6.2. Pseudoprimes 115
6.3. Euler’sTheorem 117
Chapter7. MultiplicativeFunctions 121
7.1. TheEulerPhi-Function 121
7.2. TheSumandNumberofDivisors 127
7.3. PerfectNumbersandMersennePrimes 133
7.4. Mo¨biusInversion 137
7.5. Partitions 141
1
2
Chapter8. Cryptology 151
8.1. CharacterCiphers 151
8.2. BlockandStreamCiphers 152
8.3. ExponentiationCiphers 158
8.4. PublicKeyCryptography 159
8.5. KnapsackCiphers 161
8.6. CryptographicProtocolsandApplications 162
Chapter9. PrimitiveRoots 165
9.1. TheOrderofanIntegerandPrimitiveRoots 165
9.2. PrimitiveRootsforPrimes 168
9.3. TheExistenceofPrimitiveRoots 171
9.4. DiscreteLogarithmsandIndexArithmetic 173
9.5. PrimalityTestsUsingOrdersofIntegersandPrimitiveRoots 176
9.6. UniversalExponents 178
Chapter10. ApplicationsofPrimitiveRootsandtheOrderofanInteger 181
10.1. PseudorandomNumbers 181
10.2. TheElGamalCryptosystem 183
10.3. AnApplicationtotheSplicingofTelephoneCables 184
Chapter11. QuadraticResidues 187
11.1. QuadraticResiduesandNonresidues 187
11.2. TheLawofQuadraticReciprocity 194
11.3. TheJacobiSymbol 199
11.4. EulerPseudoprimes 202
11.5. Zero-KnowledgeProofs 203
Chapter12. DecimalFractionsandContinuedFractions 205
12.1. DecimalFractions 205
12.2. FiniteContinuedFractions 208
12.3. InfiniteContinuedFractions 212
12.4. PeriodicContinuedFractions 214
12.5. FactoringUsingContinuedFractions 218
Chapter13. SomeNonlinearDiophantineEquations 221
13.1. PythagoreanTriples 221
13.2. Fermat’sLastTheorem 224
13.3. SumsofSquares 227
13.4. Pell’sEquation 230
13.5. CongruentNumbers 231
Chapter14. TheGaussianIntegers 239
14.1. GaussianIntegersandGaussianPrimes 239
14.2. GreatestCommonDivisorsandUniqueFactorization 247
14.3. GaussianIntegersandSumsofSquares 256
AppendixA. AxiomsfortheSetofIntegers 261
AppendixB. BinomialCoefficients 263
CHAPTER 1
The Integers
1.1. NumbersandSequences
1.1.1.a. Thesetofintegersgreaterthan3iswell-ordered. Everysubsetofthissetisalsoasubsetoftheset
ofpositiveintegers,andhencemusthavealeastelement.
b. Thesetofevenpositiveintegersiswell-ordered. Everysubsetofthissetisalsoasubsetoftheset
ofpositiveintegers,andhencemusthavealeastelement.
c. The set of positive rational numbers is not well-ordered. This set does not have a least element.
If a/b were the least positive rational number then a/(b+a) would be a smaller positive rational
number,whichisacontradiction.
d. Thesetofpositiverationalnumbersoftheforma/2iswell-ordered. Considerasubsetofnumbers
of this form. The set of numerators of the numbers in this subset is a subset of the set of positive
integers,soitmusthavealeastelementb. Thenb/2istheleastelementofthesubset.
e. Thesetofnonnegativerationalnumbersisnotwell-ordered. Thesetofpositiverationalnumbers
isasubsetwithnoleastelement,asshowninpartc.
1.1.2. LetS bethesetofallpositiveintegersoftheforma−bk. S isnotemptybecausea−b(−1) = a+b
is a positive integer. Then the well-ordering principle implies that S has a least element, which is the
numberwe’relookingfor.
1.1.3. Supposethatxandyarerationalnumbers. Thenx=a/bandy =c/d,wherea,b,c,anddareintegers
withb(cid:1)=0andd(cid:1)=0. Thenxy =(a/b)·(c/d)=ac/bdandx+y =a/b+c/d=(ad+bc)/bdwherebd(cid:1)=
0. Becausebothx+yandxyareratiosofintegers,theyarebothrational.
1.1.4.a. Supposethatxisrationalandyisirrational.Thenthereexistintegersaandbsuchthatx= a where
b
aandbareintegerswithb(cid:1)=0. Supposethatx+yisrational. Thenthereexistintegerscanddwith
d(cid:1)=0suchthatx+y = c. Thisimpliesthaty =(x+y)−x=(a/b)−(c/d)=(ad−bc)/bd,which
d
meansthatyisrational,acontradiction. Hencex+yisirrational.
√ √
b. Thisisfalse. Acounterexampleisgivenby 2+(− 2)=0.
√
c. Thisisfalse. Acounterexampleisgivenby0· 2=0.
√ √
d. Thisisfalse. Acounterexampleisgivenby 2· 2=2.
√ √
1.1.5. Supposethat 3were√rational. The√ntherewouldexistpositiveintegersaandbwith √3=a/b. Con-
sequently,thesetS ={k 3|kandk 3arepositiveintegers}isn√onemptybecau√sea=b 3√. There√fore,
bythe√well-ordering√property,S hasasmallestelemen√t,says=t 3. W√ehaves 3−s=s 3−t 3=
(s−t) 3. Becauses 3 = 3tands√arebothint√egers,s 3−√s =(s−t) 3mustalsobeaninteger. F√ur-
th√ermore, it is p√ositive, because s 3−s = s( 3−1) and 3 > 1. It is less than s because s = t 3,
s 3√= 3t, and 3 < 3. ThiscontradictsthechoiceofsasthesmallestpositiveintegerinS. Itfollows
that 3isirrational.
1
2 Section1.1
1.1.6. LetS beasetofnegativeintegers. ThenthesetT = {−s : s ∈ S}isasetofpositiveintegers. Bythe
well-orderingprinciple,T hasaleastelementt0. Weprovethat−t0isagreatestelementofS. Firstnote
thatbecauset0 ∈ S,thent0 = −s0 forsomes0 ∈ S. Then−t0 = s0 ∈ S. Second,ifs ∈ S,then−s ∈ T,
sot0 ≤ −s. Multiplyingby−1yieldss ≤ −t0. Becausethechoiceofswasarbitrary,weseethat−t0 is
greaterthanorequaltoeveryelementofS.
1.1.7.a. Because0≤1/4<1,wehave[1/4]=0.
b. Because−1≤−3/4<0,wehave[−3/4]=−1.
c. Because3≤22/7<4,wehave[22/7]=3.
d. Because−2≤−2<−1,wehave[−2]=−2.
e. Wecompute[1/2+[1/2]]=[1/2+0]=[1/2]=0.
f. Wecompute[−3+[−1/2]]=[−3−1]=[−4]=−4.
1.1.8.a. Because−1≤−1/4<0,wehave[−1/4]=−1.
b. Because−4≤−22/7<−3,wehave[−22/7]=−4.
c. Because1≤5/4<2,wehave[5/4]=1.
d. Wecompute[[1/2]]=[0]=0.
e. Wecompute[[3/2]+[−3/2]]=[1+(−2)]=[−1]=−1.
f. Wecompute[3−[1/2]]=[3−0]=[3]=3.
1.1.9.a. Because[8/5]=1,wehave{8/5}=8/5−[8/5]=8/5−1=3/5.
b. Because[1/7]=0,wehave{1/7}=1/7−[1/7]=1/7−0=1/7.
c. Because[−11/4]=−3,wehave{−11/4}=−11/4−[−11/4]=−11/4−(−3)=1/4.
d. Because[7]=7,wehave{7}=7−[7]=7−7=0.
1.1.10.a. Because[−8/5]=−2,wehave{−8/5}=−8/5−[−8/5]=−8/5−(−2)=2/5.
b. Because[22/7]=3,wehave{22/7}=22/7−[22/7]=22/7−3=1/7.
c. Because[−1]=−1,wehave{−1}=−1−[−1]=−1−1=0.
d. Because[−1/3]=−1,wehave{−1/3}=−1/3−[−1/3]=−1/3−(−1)=2/3.
1.1.11. Ifxisaninteger,then[x]+[−x] = x−x = 0. Otherwise,x = z+r,wherez isanintegerandr isa
realnumberwith0<r <1. Inthiscase,[x]+[−x]=[z+r]+[−z−r]=z+(−z−1)=−1.
1.1.12. Let x = [x]+r where 0 ≤ r < 1. We consider two cases. First suppose that r < 1. Then x+ 1 =
2 2
[x]+(r+ 1) < [x]+1becauser+ 1 < 1. Itfollowsthat[x+ 1] = [x]. Also2x = 2[x]+2r < 2[x]+1
2 2 2
because2r <1. Hence[2x]=2[x]. Itfollowsthat[x]+[x+1]=[2x]. Nextsupposethat 1 ≤r <1. Then
2 2
[x]+1 ≤ x+(r+ 1) < [x]+2,sothat[x+ 1] = [x]+1. Also2[x]+1 ≤ 2[x]+2r = 2([x]+r) = 2x <
2 2
2[x]+2sothat[2x]=2[x]+1. Itfollowsthat[x]+[x+ 1]=[x]+[x]+1=2[x]+1=[2x].
2
Copyright(cid:5)c 2011PearsonEducation,Inc. PublishingasAddison-Wesley
Chapter1 3
1.1.13. Wehave[x] ≤ xand[y] ≤ y. Addingthesetwoinequalitiesgives[x]+[y] ≤ x+y. Hence[x+y] ≥
[[x]+[y]]=[x]+[y].
1.1.14. Letx=a+randy =b+s,whereaandbareintegersandrandsarerealnumberssuchthat0≤r,s<
1. ByExercise14,[2x]+[2y]=[x]+[x+ 1]+[y]+[y+ 1].Wenowneedtoshowthat[x+ 1]+[y+ 1]≥
2 2 2 2
[x+y]. Suppose0 ≤ r,s < 1. Then[x+ 1]+[y+ 1] = a+b+[r+ 1]+[s+ 1] = a+b,and[x+y] =
2 2 2 2 2
a+b+[r+s]=a+b,asdesired.Supposethat 1 ≤r,s<1.Then[x+1]+[y+1]=a+b+[r+1]+[s+1]=
2 2 2 2 2
a+b+2,and[x+y] = a+b+[r+s] = a+b+1,asdesired. Supposethat0 ≤ r < 1 ≤ s < 1. Then
2
[x+ 1]+[y+ 1]=a+b+1,and[x+y]≤a+b+1.
2 2
1.1.15. Letx = a+r andy = b+s, whereaandbareintegersandr andsarerealnumberssuchthat0 ≤
r,s<1. Then[xy]=[ab+as+br+sr]=ab+[as+br+sr],whereas[x][y]=ab. Thuswehave[xy]≥
[x][y]whenxandy arebothpositive. Ifxandy arebothnegative,then[xy] ≤ [x][y]. Ifoneofxandy
ispositiveandtheothernegative,thentheinequalitycouldgoeitherdirection. Forexamplestakex =
−1.5,y =5andx=−1,y =5.5. Inthefirstcasewehave[−1.5·5]=[−7.5]=−8>[−1.5][5]=−2·5=
−10. Inthesecondcasewehave[−1·5.5]=[−5.5]=−6<[−1][5.5]=−1·5=−5.
1.1.16. Ifxisanintegerthen−[−x] = −(−x) = x,whichcertainlyistheleastintegergreaterthanorequal
tox. Letx = a+r, whereaisanintegerand0 < r < 1. Then−[−x] = −[−a−r] = −(−a+[−r]) =
a−[−r]=a+1,asdesired.
1.1.17. Letx=[x]+r. Because0≤r <1,x+ 1 =[x]+r+ 1. Ifr < 1,then[x]istheintegernearesttoxand
2 2 2
[x+ 1] = [x]because[x] ≤ x+ 1 = [x]+r+ 1 < [x]+1. Ifr ≥ 1,then[x]+1istheintegernearestto
2 2 2 2
x(choosingthisintegerifxismidwaybetween[x]and[x+1])and[x+ 1] = [x]+1because[x]+1 ≤
2
x+r+ 1 <[x]+2.
2
1.1.18. Lety =x+n. Then[y]=[x]+n,becausenisaninteger. Thereforetheproblemisequivalenttoprov-
ingthat[y/m]=[[y]/m]whichwasdoneinExample1.34.
1.1.19. Letx=k+(cid:1)wherekisanintegerand0≤(cid:1)<1. Further,letk =a2+b,√whereaisthe(cid:1)largestin√teger
suchthata2 ≤k. Thena2 ≤k =a2+b≤x=a2+b+(cid:1)<(a+1)2.Then[ x]=aand[ [x]]=[ k]=
aalso,provingthetheorem.
1.1.20. Letx = k+(cid:1)wherek isanintegerand0 ≤ (cid:1) < 1. Choosewfrom0,1,2,...,m−1suchthatw/m ≤
(cid:1) < (w +1)/m. Then w ≤ m(cid:1) < w +1. Then [mx] = [mk +m(cid:1)] = mk +[m(cid:1)] = mk +w. On the
other hand, the same inequality gives us (w+j)/m ≤ (cid:1)+j/m < (w+1+j)/m, for any integer j =
0,1,2,...,m−1. Notethatthisimplies[(cid:1)+j/m]=[(w+j)/m]whichiseither0or1forjinthisrange.
Indeed,itequ(cid:2)als1preciselywhe(cid:2)nw+j ≥m,whichhap(cid:2)pensforexactlywvaluesof(cid:2)jinthisrange.Now
wecompute m−1[x+j/m]= m−1[k+(cid:1)+j/m]= m−1k+[(cid:1)+j/m]=mk+ m−1[(w+j)/m]=
(cid:2) j=0 j=0 j=0 j=0
mk+ m−1 1=mk+wwhichisthesameasthevalueabove.
j=m−w
1.1.21.a. Becausethedifferencebetweenanytwoconsecutivetermsofthissequenceis8,wemaycompute
thenthtermbyadding8tothefirsttermn−1times. Thatis,an =3+(n−1)8=8n−5.
b. Foreachn,wehavean−an−1 =2n−1,sowemaycomputethenthtermofthissequencebyadding
allthepowersof2,uptothe(n−1)th,tothefirstterm. Hencean =5+2+22+23+···+2n−1 =
5+2n−2=2n+3.
c. Thenthtermofthissequenceappears√tobez√ero,unlessnisaperfectsquare,inwhichcasetheterm
is1. Ifnisnotaperfectsquar√e,then√[ n]< n,whe√re[x√]representsthegreatestintegerfunction.
Ifnisaperfectsquare,then[ n]= n. Therefore,[[ n]/ n]equals1ifnisaperfectsquareand0
otherwise,asdesired.
d. ThisisaFibonacci-likesequence,withan =an−1+an−2,forn≥3,anda1 =1,anda2 =3.
Copyright(cid:5)c 2011PearsonEducation,Inc. PublishingasAddison-Wesley
4 Section1.1
1.1.22.a. Eachtermgivenis3timestheprecedingterm, soweconjecturethatthenthtermisthefirstterm
multipliedby3,n−1times. Soan =2·3n−1.
b. Inthissequence,an =0ifnisamultipleof3,andequals1otherwise. Let[x]representthegreatest
integer function. Because [n/3] < n/3 when n is not a multiple of 3 and [n/3] = n/3 when n is a
multipleof3,wehavethatan =1−[[n/3]/(n/3)].
c. If we look at the difference of successive terms, we have the sequence 1,1,2,2,3,3,.... So if n is
odd,sayn=2k+1,thenanisobtainedbyadding1+1+2+2+3+3+···+k+k =2tktothefirst
term,whichis1. (Heretk standsforthekthtriangularnumber.) Soifnisodd,thenan = 1+2tk
wherek =(n−1)/2.Ifniseven,sayn=2k,thenan =a2k+1−k =1−k+2tk.
d. ThisisaFibonacci-likesequence,withan =an−1+2an−2,forn≥3,anda1 =3,anda2 =5.
1.1.23. Threepossibleanswersarean =2n−1,an =(n2−n+2)/2,andan =an−1+2an−2.
1.1.24. Threepossibleanswersarean =an−1an−2,an =an−1+2n−3,andan =thenumberoflettersinthe
nthwordofthesentence“IfouransweriscorrectwewilljointheAntidisestablishmentarianismSociety
andboldlystatethat‘IfouransweriscorrectwewilljointheAntidisestablishmentarianismSocietyand
boldlystate....’ ”
1.1.25. Thissetisexactlythesequencean =n−100,andhenceiscountable.
1.1.26. Thefunctionf(n)=5nisaone-to-onecorrespondencebetweenthissetandthesetofintegers,which
isknowntobecountable.
1.1.27. One way to√show this is to imitate the proofthat the set of rat√ional numbers is countable, replacing
a/bwitha+b 2. Anotherwayistoconsiderthefunctionf(a+b 2)=2a3bwhichisaone-to-onemap
ofthissetintotherationalnumbers,whichisknowntobecountable.
1.1.28. LetAandB betwocountablesets. Ifoneorbothofthesetsarefinite,sayAisfinite,thenthelisting
a1,a2,...,an,b1,b2,...,whereanybi whichisalsoinAisdeletedfromthelist,demonstratesthecount-
abilityofA∪B. Ifbothsetsareinfinite,theneachcanberepresentedasasequence: A = {a1,a2,...},
andB = {b1,b2,...}. Considerthelistinga1,b1,a2,b2,a3,b3,...andformanewsequenceci asfollows.
Let c1 = a1. Given that cn is determined, let cn+1 be the next element in the listing which is different
fromeachciwithi=1,2,...,n. ThenthissequenceisexactlytheelementsofA∪B,whichistherefore
countable.
1.1.29. Suppose {Ai} is a countable collection of countable sets. Then each Ai can be represented by a se-
quence,asfollows:
A = a a a ...
1 11 12 13
A = a a a ...
2 21 22 23
A = a a a ...
3 31 32 33
.
.
.
Considerthelistinga11,a12,a21,a13,a22,a31,...,inwhichwefirstlisttheelementswithsubscripts
adding to 2, then the elements with subscripts adding to 3 and so on. Further, we order the elements
withsubscriptsaddingtok inorderofthefirstsubscript. Formanewsequenceci asfollows. Letc1 =
a1. Giventhatcn isdetermined,letcn+1 bethenextelementinthelistingwhichisdifferentfromeach
(cid:3)∞
ciwithi=1,2,...,n. Thenthissequenceisexactlytheelementsof Ai,whichisthereforecountable.
√ √ i=1
1.1.30.a. Notethat 2√≈ 1.4 = 7/5,sowemightguessthat 2−7/5 ≈ 0. Ifwemultiplythroughby5we
expectthat5 2−7shouldbesmall,anditsvalueisapproximately0.071whichismuchlessthan
1/8=0.125. Sowemaytakea=5≤8andb=7.
Copyright(cid:5)c 2011PearsonEducation,Inc. PublishingasAddison-Wesley
Chapter1 5
√ √
b. Asinparta.,notethat 32 = 1.2599... ≈ 1.25 = 5/4,soweinvestigate432−5 = 0.039... ≤ 1/8.
Sowemaytakea=4≤8andb=5.
c. Becauseweknowthatπ ≈ 22/7weinvestigate|7π−22| = 0.0088... ≤ 1/8. Sowemaytakea =
7≤8andb=22.
d. Becausee ≈ 2.75 = 11/4weinvestigate|4e−11| = 0.126...,whichistoolarge. Acloserapproxi-
mationtoeis2.718. Weconsiderthedecimalexpansionsofthemultiplesof1/7andfindthat5/7=
.714...,soe ≈ 19/7. Thereforeweinvestigate|7e−19| = 0.027 ≤ 1/8. Sowemaytakea = 7 ≤ 8
andb=19.
√ √
1.1.31.a. Notethat √3 = 1.73 ≈ 7/4,sowemightguessthat 3−7/4 ≈ 0. Ifwemultiplythroughby4we
findthat|4 3−7|=0.07...<1/10. Sowemaytakea=4≤10andb=7.
b. I√tishelpfultokeepthedecimalexpansionsofthemu√ltiplesof1/7inmindintheseexercises. Here
3√3 = 1.442...and3/7 = 0.428...sothatwehave 33 ≈ 10/7.Then, asinparta., weinvestigate
|733−10|=0.095...<1/10. Sowemaytakea=7≤10andb=10.
c. Because π2 = 9.869... and 6/7 = 0.857..., we have that π2 ≈ 69/7, so we compute |7π2 −69| =
0.087...<1/10. Sowemaytakea=7≤10andb=69.
d. Becausee3 =20.0855...wemaytakea=1andb=20toget|1e3−20|=0.855...<1/10.
1.1.32. For j = 0,1,2,...,n+1, consider the n+2 numbers {jα}, which all lie in the interval 0 ≤ {jα} <
1. We can partition this interval into the n+1 subintervals (k −1)/(n+1) ≤ x < k/(n+1) for k =
1,...,n+1. Becausewehaven+2numbersandonlyn+1intervals,bythepigeonholeprinciple,some
intervalmustcontainatleasttwoofthenumbers. Sothereexistintegersrandssuchthat0 ≤ r < s ≤
n+1and|{rα}−{sα}| ≤ 1/(n+1). Leta = s−randb = [sα]−[rα]. Because0 ≤ r < s ≤ n+1,we
have1≤a≤n. Also,|aα−b|=|(s−r)α−([sα]−[rα])|=|(sα−[sα])−(rα−[rα]a)|=|{sα}−{rα}|<
1/(n+1). Therefore,aandbhavethedesiredproperties.
1.1.33. Thenumberα mustlieinsomeintervaloftheformr/k ≤ α < (r+1)/k. Ifwedividethisinterval
intoequalhalves,thenαmustlieinoneofthehalves,soeitherr/k ≤α<(2r+1)/2kor(2r+1)/2k ≤
α < (r+1)/k. Inthefirstcasewehave|α−r/k| < 1/2k,sowetakeu = r. Inthesecondcasewehave
|α−(r+1)/k|<1/2k,sowetakeu=r+1.
1.1.34. Supposethatthereareonlyfinitelymanypositiveintegersq1,q2,...,qn withcorrespondingintegers
p1,p2,...,pn suchthat|α−pi/qi| < 1/qi2.Becauseαisirrational,|α−pi/qi|ispositiveforeveryi,and
so is |qiα−pi| so we may choose an integer N so large that |qiα−pi| > 1/N for all i. By Dirichlet’s
ApproximationTheorem,thereexistintegersr andswith1 ≤ s ≤ N suchthat|sα−r| < 1/N < 1/s,
sothat|α−r/s|<1/s2,andsisnotoneoftheqi. Therefore,wehaveanothersolutiontotheinequality.
Sonofinitelistofsolutionscanbecomplete,andweconcludethattheremustbeaninfinitenumberof
solutions.
√ √
1.1.35. Firstwehave| 2−1/1|=0.414...<1/12. Second,Exer√cise30,parta.,givesus| 2−7/5|<1/50<
1/52.Third,observingthat3/7 = 0.428...leadsustot√ry| 2−10/7| = 0.014... < 1/72 = 0.0204....
Fourth,observingthat5/12=0.4166...leadsustotry| 2−17/12|=0.00245...<1/122 =0.00694....
√ √
1.1.36.√Firstwehave|35−1/√1| = 0.7099... < 1/12. Second, |35−5/3| = 0.04... < 1/32. Third, because
35 =√1.7099..., wetry|35−17/10| = 0.0099... < 1/102.Likewise, wegetafourthrationalnu√mber
with |35−171/100| = 0.000024... < 1/1002. Fifth, consideration of multiples of 1/7 leads to |35−
12/7|=0.0043...<1/72.
1.1.37. We may assume that b and q are positive. Note that if q > b, we have |p/q −a/b| = |pb−aq|/qb ≥
1/qb > 1/q2. Therefore, solutions to the inequality must have 1 ≤ q ≤ b. For a given q, there can be
onlyfinitelymanypsuchthatthedistancebetweentherationalnumbersa/bandp/q islessthan1/q2
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6 Section1.1
(indeedthereisatmostone.) Thereforethereareonlyfinitelymanyp/qsatisfyingtheinequality.
1.1.38.a. Becausen2isanintegerforalln,sois[n2],sothefirsttentermsofthespectrumsequenceare2,4,
6,8,10,12,14,16,18,20.
√
b. The sequence for n 2, rounded, is 1.414, 2.828, 4.242, 5.656, 7.071, 8.485, 9.899, 11.314, 12.728,
14.142. Whenweapplythefloorfunctiontothesenumbersweget1,2,4,5,7,8,9,11,12,14for
thespectrumsequence.
√
c. Thesequenceforn(2+ 2), rounded, is3.414, 6.828, 10.24, 13.66, 17.07, 20.48, 23.90, 27.31, 30.73,
34.14. Whenweapplythefloorfunctiontothesenumbersweget3,6,10,13,17,20,23,27,30,34,
forthespectrumsequence.
d. The sequence for ne, rounded is 2.718, 5.436, 8.155, 10.87, 13.59, 16.31, 19.03, 21.75, 24.46, 27.18.
Whenweapplythefloorfunctiontothesenumbersweget2,5,8,10,13,16,19,21,24,27,forthe
spectrumsequence.
√
e. Thesequenceforn(1+ 5)/2,rounded,is1.618,3.236,4.854,6.472,8.090,9.708,11.33,12.94,14.56,
16.18. Whenweapplythefloorfunctiontothesenumbersweget1,3,4,6,8,9,11,12,14,16forthe
spectrumsequence.
1.1.39.a. Becausen3isanintegerforalln,sois[n3],sothefirsttentermsofthespectrumsequenceare3,6,
9,12,15,18,21,24,27,30.
√
b. Thesequenceforn 3,rounded,is1.732,3.464,5.196,6.928,8.660,10.39,12.12,13.86,15.59,17.32.
Whenweapplythefloorfunctiontothesenumbersweget1,3,5,6,8,10,12,13,15,17forthespec-
trumsequence.
√
c. Thesequenceforn(3+ 3)/2,rounded,is2.366,4.732,7.098,9.464,11.83,14.20,16.56,18.93,21.29,
23.66. Whenweapplythefloorfunctiontothesenumbersweget2,4,7,9,11,14,16,18,21,23for
thespectrumsequence.
d. The sequence for nπ, rounded is 3.142, 6.283, 9.425, 12.57, 15.71, 18.85, 21.99, 25.13, 28.27, 31.42.
Whenweapplythefloorfunctiontothesenumbersweget3,6,9,12,15,18,21,25,28,31,forthe
spectrumsequence.
1.1.40. Becauseα(cid:1)=β,theirdecimalexpansionsmustbedifferent. Iftheydifferindigitsthataretotheleftof
thedecimalpoint,then[α](cid:1)=[β],socertainlythespectrumsequencesaredifferent. Otherwise,suppose
thattheydifferinthekthpositiontotherightofthedecimal. Then[10kα](cid:1)=[10kβ],andsothespectrum
sequenceswillagaindiffer.
1.1.41. Assume that 1/α+1/β = 1. Note first that for all integers n and m, mα (cid:1)= nβ, for otherwise, we
solvetheequationsmα=nβ and1/α+1/β =1andgetrationalsolutionsforαandβ,acontradiction.
Thereforethesequencesmαandnβ aredisjoint.
Foraninteger k, defineN(k)tobethenumberofelementsofthesequences mα andnβ whichare
less than k. Now mα < k if and only if m < k/α, so there are exactly [k/α] members of the sequence
mαlessthank. Likewise, thereareexactly[k/β]membersofthesequencenβ lessthank. Sowehave
N(k)=[k/α]+[k/β]. Bydefinitionofthegreatestintegerfunction,wehavek/α−1<[k/α]<k/αand
k/β −1 < [k/β] < k/β,wheretheinequalitiesarestrictbecausethenumbersareirrational. Ifweadd
these inequalities we get k/α+k/β −2 < N(k) < k/α+k/β which simplifies to k−2 < N(k) < k.
BecauseN(k)isaninteger,weconcludethatN(k)=k−1. Thisshowsthatthereisexactlyonemember
oftheunionofthesequencesmαandnβineachintervaloftheformk−1≤x<k,andtherefore,when
weapplythefloorfunctiontoeachmember,exactlyonewilltakeonthevaluek.
Conversely,supposethatαandβ areirrationalnumberssuchthat1/α+1/β (cid:1)= 1. If1/α+1/γ = 1
thenweknowfromthefirstpartofthetheoremthatthespectrumsequencesforαandγ partitionthe
positiveintegers. ByExercise40,weknowthatthespectrumsequencesforβ andγ aredifferent,sothe
Copyright(cid:5)c 2011PearsonEducation,Inc. PublishingasAddison-Wesley