INSTRUCTOR S SOLUTIONS MANUAL ELEMENTARY NUMBER THEORY. Bart Goddard Kenneth H. Rosen AT&T Labs AND ITS APPLICATIONS FIFTH EDITION.

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1 INSTRUCTOR S SOLUTIONS MANUAL to accomay ELEMENTARY NUMBER THEORY AND ITS APPLICATIONS FIFTH EDITION Bart Goddard Keeth H. Rose AT&T Labs

2 This work is rotected by Uited States coyright laws ad is rovided solely for the use of istructors i teachig their courses ad assessig studet learig. Dissemiatio or sale of ay art of this work icludig o the World Wide Web will destroy the itegrity of the work ad is ot ermitted. The work ad materials from it should ever be made available to studets excet by istructors usig the accomayig text i their classes. All reciiets of this work are exected to abide by these restrictios ad to hoor the iteded edagogical uroses ad the eeds of other istructors who rely o these materials. Reroduced by Pearso Addiso-Wesley from electroic files sulied by the author. Coyright 2005 Pearso Educatio, Ic. Publishig as Pearso Addiso-Wesley, 75 Arligto Street, Bosto, MA All rights reserved. No art of this ublicatio may be reroduced, stored i a retrieval system, or trasmitted, i ay form or by ay meas, electroic, mechaical, hotocoyig, recordig, or otherwise, without the rior writte ermissio of the ublisher. Prited i the Uited States of America. ISBN XXX

3 CHAPTER 1 The Itegers 1.1. Numbers, Sequeces, ad Sums a. The set of itegers greater tha 3 is well-ordered. Every subset of this set is also a subset of the set of ositive itegers, ad hece must have a least elemet. b. The set of eve ositive itegers is well-ordered. Every subset of this set is also a subset of the set of ositive itegers, ad hece must have a least elemet. c. The set of ositive ratioal umbers is ot well-ordered. This set does ot have a least elemet. If a/b were the least ositive ratioal umber the a/b + a would be a smaller ositive ratioal umber, which is a cotradictio. d. The set of ositive ratioal umbers of the form a/2 is well-ordered. Cosider a subset of umbers of this form. The set of umerators of the umbers i this subset is a subset of the set of ositive itegers, so it must have a least elemet b. The b/2 is the least elemet of the subset. e. The set of oegative ratioal umbers is ot well-ordered. The set of ositive ratioal umbers is a subset with o least elemet, as show i art c Let S be the set of all ositive itegers of the form a bk. S is ot emty sice a b 1 a + b is a ositive iteger. The the well-orderig ricile imlies that S has a least elemet, which is the umber we re lookig for Suose that x ad y are ratioal umbers. The x a/b ad y c/d, where a, b, c, ad d are itegers with b 0 ad d 0. The xy a/b c/d ac/bd ad x + y a/b + c/d ad + bc/bd where bd 0. Sice both x + y ad xy are ratios of itegers, they are both ratioal a. Suose that x is ratioal ad y is irratioal. The there exist itegers a ad b such that x a b where a ad b are itegers with b 0. Suose that x + y is ratioal. The there exist itegers c ad d with d 0 such that x + y c d. This imlies that y x + y x a/b c/d ad bc/bd, which meas that y is ratioal, a cotradictio. Hece x + y is irratioal. b. This is false. A couterexamle is give by c. This is false. A couterexamle is give by d. This is false. A couterexamle is give by Suose that 3 were ratioal. The there would exist ositive itegers a ad b with 3 a/b. Cosequetly, the set S {k 3 k ad k 3 are ositive itegers} is oemty sice a b 3. Therefore, by the well-orderig roerty, S has a smallest elemet, say s t 3. We have s 3 s s 3 t 3 s t 3. Sice s 3 3t ad s are both itegers, s 3 s s t 3 must also be a iteger. Furthermore, it is ositive, sice s 3 s s 3 1 ad 3 > 1. It is less tha s sice s t 3, s 3 3t, ad 3 < 3. This cotradicts the choice of s as the smallest ositive iteger i S. It follows that 3 is irratioal Let S be a set of egative itegers. The the set T { s : s S} is a set of ositive itegers. By the well-orderig ricile, T has a least elemet t 0. We rove that t 0 is a greatest elemet of S. First ote

4 2 1. THE INTEGERS that sice t 0 S, the t 0 s 0 for some s 0 S. The t 0 s 0 S. Secod, if s S, the s T, so t 0 s. Multilyig by 1 yields s t 0. Sice the choice of s was arbitrary, we see that t 0 is greater tha or equal to every elemet of S a. Sice 0 1/4 < 1, we have [1/4] 0. b. Sice 1 3/4 < 0, we have [ 3/4] 1. c. Sice 3 22/7 < 4, we have [22/7] 3. d. Sice 2 2 < 1, we have [ 2] 2. e. We comute [1/2 + [1/2]] [1/2 + 0] [1/2] 0. f. We comute [ 3 + [ 1/2]] [ 3 1] [ 4] a. Sice 1 1/4 < 0, we have [ 1/4] 1. b. Sice 4 22/7 < 3, we have [ 22/7] 4. c. Sice 1 5/4 < 2, we have [5/4] 1. d. We comute [[1/2]] [0] 0. e. We comute [[3/2] + [ 3/2]] [1 + 2] [ 1] 1. f. We comute [3 [1/2]] [3 0] [3] a. Sice [8/5] 1, we have {8/5} 8/5 [8/5] 8/5 1 3/5. b. Sice [1/7] 0, we have {1/7} 1/7 [1/7] 1/7 0 1/7. c. Sice [ 11/4] 3, we have { 11/4} 11/4 [ 11/4] 11/4 3 1/4. d. Sice [7] 7, we have {7} 7 [7] a. Sice [ 8/5] 2, we have { 8/5} 8/5 [ 8/5] 8/5 2 2/5. b. Sice [22/7] 3, we have {22/7} 22/7 [22/7] 22/7 3 1/7. c. Sice [ 1] 1, we have { 1} 1 [ 1] d. Sice [ 1/3] 1, we have { 1/3} 1/3 [ 1/3] 1/3 1 2/ If x is a iteger, the [x] + [ x] x x 0. Otherwise, x z + r, where z is a iteger ad r is a real umber with 0 < r < 1. I this case, [x] + [ x] [z + r] + [ z r] z + z Let x [x] + r where 0 r < 1. We cosider two cases. First suose that r < 1 2. The x [x] + r < [x] + 1 sice r < 1. It follows that [x ] [x]. Also 2x 2[x] + 2r < 2[x] + 1 sice 2r < 1. Hece [2x] 2[x]. It follows that [x] + [x ] [2x]. Next suose that 1 2 r < 1. The [x] + 1 x + r < [x] + 2, so that [x ] [x] + 1. Also 2[x] + 1 2[x] + 2r 2[x] + r 2x < 2[x] + 2 so that [2x] 2[x] + 1. It follows that [x] + [x ] [x] + [x] + 1 2[x] + 1 [2x] We have [x] x ad [y] y. Addig these two iequalities gives [x] + [y] x + y. Hece [x + y] [[x] + [y]] [x] + [y].

5 1.1. NUMBERS, SEQUENCES, AND SUMS Let x a+r ad y b+s, where a ad b are itegers ad r ad s are real umbers such that 0 r, s < 1. By Exercise 14, [2x] + [2y] [x] + [x ] + [y] + [y ]. We ow eed to show that [x ] + [y ] [x + y]. Suose 0 r, s < 1 2. The [x ] + [y ] a + b + [r ] + [s ] a + b, ad [x + y] a+b+[r+s] a+b, as desired. Suose that 1 2 r, s < 1. The [x+ 1 2 ]+[y+ 1 2 ] a+b+[r+ 1 2 ]+[s+ 1 2 ] a + b + 2, ad [x + y] a + b + [r + s] a + b + 1, as desired. Suose that 0 r < 1 2 s < 1. The [x ] + [y ] a + b + 1, ad [x + y] a + b Let x a + r ad y b + s, where a ad b are itegers ad r ad s are real umbers such that 0 r, s < 1. The [xy] [ab + as + br + sr] ab + [as + br + sr], whereas [x][y] ab. Thus we have [xy] [x][y]. If x ad y are both egative, the [xy] [x][y]. If oe of x ad y is ositive ad the other egative, the the iequality could go either directio. For examles take x 1.5, y 5 ad x 1, y 5.5. I the first case we have [ 1.5 5] [ 7.5] 8 > [ 1.5][5] I the secod case we have [ 1 5.5] [ 5.5] 6 < [ 1][5.5] If x is a iteger the [ x] x x, which certaily is the least iteger greater tha or equal to x. Let x a + r, where a is a iteger ad 0 < r < 1. The [ x] [ a r] a + [ r] a [ r] a + 1, as desired Let x [x] + r. Sice 0 r < 1, x [x] + r If r < 1 2, the [x] is the iteger earest to x ad [x+ 1 2 ] [x] sice [x] x+ 1 2 [x]+r+ 1 2 < [x]+1. If r 1 2, the [x]+1 is the iteger earest to x choosig this iteger if x is midway betwee [x] ad [x+1] ad [x+ 1 2 ] [x]+1 sice [x]+1 x+r+ 1 2 < [x] Let y x +. The [y] [x] +, sice is a iteger. Therefore the roblem is equivalet to rovig that [y/m] [[y]/m] which was doe i Examle Let x k + ɛ where k is a iteger ad 0 ɛ < 1. Further, let k a 2 + b, where a is the largest iteger such that a 2 k. The a 2 k a 2 + b x a 2 + b + ɛ < a The [ x] a ad [ [x]] [ k] a also, rovig the theorem Let x k + ɛ where k is a iteger ad 0 ɛ < 1. Choose w from 0, 1, 2,..., m 1 such that w/m ɛ < w + 1/m. The w mɛ < w + 1. The [mx] [mk + mɛ] mk + [mɛ] mk + w. O the other had, the same iequality gives us w + j/m ɛ + j/m < w j/m, for ay iteger j 0, 1, 2,..., m 1. Note that this imlies [ɛ + j/m] [w + j/m] which is either 0 or 1 for j i this rage. Ideed, it equals 1 recisely whe w+j m, which haes for exactly w values of j i this rage. Now we comute m 1 j0 [x + j/m] m 1 j0 [k + ɛ + j/m] m 1 j0 k + [ɛ + j/m] mk + m 1 j0 [w + j/m] mk + m 1 jm w 1 mk + w which is the same as the value above a. Sice the differece betwee ay two cosecutive terms of this sequece is 8, we may comute the th term by addig 8 to the first term 1 times. That is, a b. For each, we have a a 1 2 1, so we may comute the th term of this sequece by addig all the owers of 2, u to the 1th, to the first term. Hece a c. The th term of this sequece aears to be zero, uless is a erfect square, i which case the term is 1. If is ot a erfect square, the [ ] <, where [x] reresets the greatest iteger fuctio. If is a erfect square, the [ ]. Therefore, [[ ]/ ] equals 1 if is a erfect square ad 0 otherwise, as desired. d. This is a Fiboacci-like sequece, with a a 1 + a 2, for 3, ad a 1 1, ad a a. Each term give is 3 times the recedig term, so we cojecture that the th term is the first term multilied by 3, 1 times. So a

6 4 1. THE INTEGERS b. I this sequece, a 0 if is a multile of 3, ad equals 1 otherwise. Let [x] rereset the greatest iteger fuctio. Sice [/3] < /3 whe is ot a multile of 3 ad [/3] /3 whe is a multile of 3, we have that a 1 [[/3]//3]. c. If we look at the differece of successive terms, we have the sequece 1, 1, 2, 2, 3, 3,.... So if is odd, say 2k + 1, the a is obtaied by addig k + k 2t k to the first term, which is 1. Here t k stads for the kth triagular umber. So if is odd, the a 1 + 2t k where k 1/2. If is eve, say 2k, the a a 2k+1 k 1 k + 2t k. d. This is a Fiboacci-like sequece, with a a 1 + 2a 2, for 3, ad a 1 3, ad a Three ossible aswers are a 2 1, a 2 + 2/2, ad a a 1 + 2a Three ossible aswers are a a 1 a 2, a a , ad a the umber of letters i the th word of the setece If our aswer is correct we will joi the Atidisestablishmetariaism Society ad boldly state that If our aswer is correct we will joi the Atidisestablishmetariaism Society ad boldly state This set is exactly the sequece a 100, ad hece is coutable The fuctio f 5 is a oe-to-oe corresodece betwee this set ad the set of itegers, which is kow to be coutable Oe way to show this is to imitate the roof that the set of ratioal umbers is coutable, relacig a/b with a + b 2. Aother way is to cosider the fuctio fa + b 2 2 a 3 b which is a oe-to-oe ma of this set ito the ratioal umbers, which is kow to be coutable Let A ad B be two coutable sets. If oe or both of the sets are fiite, say A is fiite, the the listig a 1, a 2,..., a, b 1, b 2,..., where ay b i which is also i A is deleted from the list, demostrates the coutability of A B. If both sets are ifiite, the each ca be rereseted as a sequece: A {a 1, a 2,...}, ad B {b 1, b 2,...}. Cosider the listig a 1, b 1, a 2, b 2, a 3, b 3,... ad form a ew sequece c i as follows. Let c 1 a 1. Give that c is determied, let c +1 be the ext elemet i the listig which is differet from each c i with i 1, 2,...,. The this sequece is exactly the elemets of A B, which is therefore coutable Suose {A i } is a coutable collectio of coutable sets. The each A i ca be rereseted by a sequece, as follows: A 1 a 11 a 12 a A 2 a 21 a 22 a A 3 a 31 a 32 a Cosider the listig a 11, a 12, a 21, a 13, a 22, a 31,..., i which we first list the elemets with subscrits addig to 2, the the elemets with subscrits addig to 3 ad so o. Further, we order the elemets with subscrits addig to k i order of the first subscrit. Form a ew sequece c i as follows. Let c 1 a 1. Give that c is determied, let c +1 be the ext elemet i the listig which is differet from each c i with i 1, 2,...,. The this sequece is exactly the elemets of A i, which is therefore coutable a. Note that /5, so we might guess that 2 7/5 0. If we multily through by 5 we exect that should be small, ad its value is aroximately which is much less tha 1/ So we may take a 5 8 ad b 7. b. As i art a., ote that /4, so we ivestigate /8. So we may take a 4 8 ad b 5. i1

7 1.1. NUMBERS, SEQUENCES, AND SUMS 5 c. Sice we kow that π 22/7 we ivestigate 7π /8. So we may take a 7 8 ad b 22. d. Sice e /4 we ivestigate 4e , which is too large. A closer aroximatio to e is We cosider the decimal exasios of the multiles of 1/7 ad fid that 5/ , so e 19/7. Therefore we ivestigate 7e /8. So we may take a 7 8 ad b a. Note that /4, so we might guess that 3 7/4 0. If we multily through by 4 we fid that < 1/10. So we may take a 4 10 ad b 7. b. It is helful to kee the decimal exasios of the multiles of 1/7 i mid i these exercises. Here ad 3/ so that we have /7. The as i art a., we ivestigate < 1/10. So we may take a 7 10 ad b 10. c. Sice π ad 6/ , we have that π 2 69/7, so we comute 7π < 1/10. So we may take a 7 10 ad b 69. d. Sice e we may take a 1 ad b 20 to get 1e < 1/ For j 0, 1, 2,..., + 1, cosider the + 2 umbers {jα}, which all lie i the iterval 0 {jα} < 1. We ca artitio this iterval ito the + 1 subitervals k 1/ + 1 x < k/ + 1 for k 1,..., + 1. Sice we have + 2 umbers ad oly + 1 itervals, by the igeohole ricile, some iterval must cotai at least two of the umbers. So there exist itegers r ad s such that 0 r < s + 1 ad {rα} {sα} 1/ + 1. Let a s r ad b [sα] [rα]. Sice 0 r < s + 1, we have 1 a. Also, aα b s rα [sα] [rα] sα [sα] rα [rα]a {sα} {rα} < 1/ + 1. Therefore, a ad b have the desired roerties The umber α must lie i some iterval of the form r/k α < r + 1/k. If we divide this iterval ito equal halves, the α must lie i oe of the halves, so either r/k α < 2r + 1/2k or 2r + 1/2k α < r + 1/k. I the first case we have α r/k < 1/2k, so we take u r. I the secod case we have α r + 1/k < 1/2k, so we take u r Suose that there are oly fiitely may ositive itegers q 1, q 2,..., q with corresodig itegers 1, 2,..., such that α i /q i < 1/q 2 i. Sice α is irratioal, α i/q i is ositive for every i, ad so is q i α i so we may choose a iteger N so large that q i α i > 1/N for all i. By Dirichlet s Aroximatio Theorem, there exist itegers r ad s with 1 s N such that sα r < 1/N < 1/s, so that α r/s < 1/s 2, ad s is ot oe of the q i. Therefore, we have aother solutio to the iequality. So o fiite list of solutios ca be comlete, ad we coclude that there must be a ifiite umber of solutios First we have 2 1/ < 1/1 2. Secod, Exercise 30, art a., gives us 2 7/5 < 1/50 < 1/5 2. Third, observig that 3/ leads us to try 2 10/ < 1/ Fourth, observig that 5/ leads us to try 2 17/ < 1/ First we have 3 5 1/ < 1/1 2. Secod, 3 5 5/ < 1/3 2. Third, sice , we try / < 1/10 2. Likewise, we get a fourth ratioal umber with / < 1/ Fifth, cosideratio of multiles of 1/7 leads to / < 1/ We may assume that b ad q are ositive. Note that if q > b, we have /q a/b b aq /qb 1/qb > 1/q 2. Therefore, solutios to the iequality must have 1 q b. For a give q, there ca be oly fiitely may such that the distace betwee the ratioal umbers a/b ad /q is less tha 1/q 2 ideed there is at most oe. Therefore there are oly fiitely may /q satisfyig the iequality.

8 6 1. THE INTEGERS a. Sice 2 is a iteger for all, so is [2], so the first te terms of the sectrum sequece are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20. b. The sequece for 2, rouded, is 1.414, 2.828, 4.242, 5.656, 7.071, 8.485, 9.899, , , Whe we aly the floor fuctio to these umbers we get 1, 2, 4, 5, 7, 8, 9, 11, 12, 14 for the sectrum sequece. c. The sequece for 2 + 2, rouded, is 3.414, 6.828, 10.24, 13.66, 17.07, 20.48, 23.90, 27.31, 30.73, Whe we aly the floor fuctio to these umbers we get 3, 6, 10, 13, 17, 20, 23, 27, 30, 34, for the sectrum sequece. d. The sequece for e, rouded is 2.718, 5.436, 8.155, 10.87, 13.59, 16.31, 19.03, 21.75, 24.46, Whe we aly the floor fuctio to these umbers we get 2, 5, 8, 10, 13, 16, 19, 21, 24, 27, for the sectrum sequece. e. The sequece for 1 + 5/2, rouded, is 1.618, 3.236, 4.854, 6.472, 8.090, 9.708, 11.33, 12.94, 14.56, Whe we aly the floor fuctio to these umbers we get 1, 3, 4, 6, 8, 9, 11, 12, 14, 16 for the sectrum sequece a. Sice 3 is a iteger for all, so is [3], so the first te terms of the sectrum sequece are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30. b. The sequece for 3, rouded, is 1.732, 3.464, 5.196, 6.928, 8.660, 10.39, 12.12, 13.86, 15.59, Whe we aly the floor fuctio to these umbers we get 1, 3, 5, 6, 8, 10, 12, 13, 15, 17 for the sectrum sequece. c. The sequece for 3 + 3/2, rouded, is 2.366, 4.732, 7.098, 9.464, 11.83, 14.20, 16.56, 18.93, 21.29, Whe we aly the floor fuctio to these umbers we get 2, 4, 7, 9, 11, 14, 16, 18, 21, 23 for the sectrum sequece. d. The sequece for π, rouded is 3.142, 6.283, 9.425, 12.57, 15.71, 18.85, 21.99, 25.13, 28.27, Whe we aly the floor fuctio to these umbers we get 3, 6, 9, 12, 15, 18, 21, 25, 28, 31, for the sectrum sequece Sice α β, their decimal exasios must be differet. If they differ i digits that are to the left of the decimal oit, the [α] [β], so certaily the sectrum sequeces are differet. Otherwise, suose that they differ i the kth ositio to the right of the decimal. The [10 k α] [10 k β], ad so the sectrum sequeces will agai differ Assume that 1/α + 1/β 1. Note first that for all itegers ad m, mα β, for otherwise, we solve the equatios mα β ad 1/α + 1/β 1 ad get ratioal solutios for α ad β, a cotradictio. Therefore the sequeces mα ad β are disjoit. For a iteger k, defie Nk to be the umber of elemets of the sequeces mα ad β which are less tha k. Now mα < k if ad oly if m < k/α, so there are exactly [k/α] members of the sequece mα less tha k. Likewise, there are exactly [k/β] members of the sequece β less tha k. So we have Nk [k/α] + [k/β]. By defiitio of the greatest iteger fuctio, we have k/α 1 < [k/α] < k/α ad k/β 1 < [k/β] < k/β, where the iequalities are strict because the umbers are irratioal. If we add these iequalities we get k/α + k/β 2 < Nk < k/α + k/β which simlifies to k 2 < Nk < k. Sice Nk is a iteger, we coclude that Nk k 1. This shows that there is exactly oe member of the uio of the sequeces mα ad β i each iterval of the form k 1 x < k, ad therefore, whe we aly the floor fuctio to each member, exactly oe will take o the value k. Coversely, suose that α ad β are irratioal umbers such that 1/α + 1/β 1. If 1/α + 1/γ 1 the we kow from the first art of the theorem that the sectrum sequeces for α ad γ artitio the ositive itegers. By Exercise 40., we kow that the sectrum sequeces for β ad γ are differet, so the sequeces for α ad β ca ot artitio the ositive itegers.

9 1.2. SUMS AND PRODUCTS The first two Ulam umbers are 1 ad 2. Sice , it is the third Ulam umber ad sice , it is the fourth Ulam umber. Note that 5 is ot a Ulam umber sice The fifth Ulam umber is 6 sice ad o other two Ulam umbers have 6 as their sum. We have , so 7 is ot a Ulam umber. The sixth Ulam umber is Note that ad so either 9 or 10 is a Ulam umber. The seveth Ulam umber is 11 sice is the uique way to write 11 as the sum of two distict Ulam umbers. Next ote that so that 12 is ot a Ulam umber. Note that is the uique way to write 13 as the eighth Ulam umber. We see that ad , so that either 14 or 15 are Ulam umbers. We ote that is the uique way to write 16 as the sum of two Ulam umbers, so that the ith Ulam umber is 16. Note that so that 17 is ot a Ulam umber. Note that is the uique way to write 18 as the sum of two Ulam umbers so that 18 is the teth Ulam umber. I summary, the first te Ulam umbers are: 1, 2, 3, 4, 6, 8, 11, 13, 16, Assume that there are oly fiitely may Ulam umbers. Let the two largest Ulam umbers be u 1 ad u. The the iteger u + u 1 is a Ulam umber larger tha u. It is the uique sum of two Ulam umbers sice u i + u j < u + u 1 if j < or j ad i < Suose that e is ratioal so that e a/b where a ad b are itegers ad b 0. Let k b be a iteger ad set c k!e 1 1/1! 1/2! 1/3! 1/k!. Sice every deomiator i the exressio divides evely ito k!, we see that c is a iteger. Sice e 1 + 1/1! + 1/2! +, we have 0 < c k!1/k + 1! + 1/k + 2! + 1/k /k + 1k < 1/k /k This last geometric series is equal to 1/k, so we have that 0 < c < 1/k, which is imossible sice c is a iteger. Therefore e must be irratioal a. We have b. We have c. We have a. We have b. We have c. We have 1.2. Sums ad Products 5 j j j1 5 1/j + 1 1/2 + 1/3 + 1/4 + 1/5 + 1/6 29/20. j j0 4 j j0 4 j + 1/j + 2 1/2 + 2/3 + 3/4 + 4/5 + 5/6 71/20. j a. We use the formula from Examle 1.15 as follows. We evaluate the sum Examle The we have j1 8 2 j j0 8 2 j b. We could roceed as i art a, or we may do the followig: j0 8 2 j as i j j j j. We may aly the formula i Examle 1.15 to this last sum, with a 15, j0 7 ad r 3, to get the sum equal to j0

10 8 1. THE INTEGERS c. We maiulate the sum as i art b., so we ca aly the formula from Examle /2 j+1 j0 j0 7 3/2 1/2 j 3/2 1/28 3/2 1/2 1 j a. We have 8 3 j ad r /2 j , usig the formula from Examle 1.15 with a 8, 10 j0 j b. We have 2 j j , usig the formula from Ex- 2 1 amle 1.15 with a 2, 10 ad r 2. c. We have 10 j0 10 ad r 1/3. 1/3 j 1/ / , usig the formula from Examle 1.15 with a 1, The sum k1 [ k] couts 1 for every value of k with k 1. There are such values of k i the rage k 1, 2, 3,...,. It couts aother 1 for every value of k with k 2. There are 3 such values i the rage. The sum couts aother 1 for each value of k with k 3. There are 8 such values i the rage. I geeral, for m 1, 2, 3,..., [ ] the sum couts a 1 for each value of k with k m, ad there are m 2 1 values i the rage. Therefore k1 [ k] [ ] m1 m2 1 [ ] + 1 [ ] m1 m2 [ ] + 1 [ ][ ] + 12[ ] + 1/ We see that t j1 j, ad t 1 1 j1 j 1 j1 j. Now, t 1 + t 1 j1 j + j We see that t j1 j j1 j + 1. Thus, 2t j1 j + j1 j + 1 j It is clear that 1 1. Suose we kow k 1. To comute k we cosider k ested etagos as i the figure. Note that k k 1 couts the umber of dots o three sides of the outer etago. Each side cosists of k dots, but two of the dots belog to two sides. Therefore k k 1 3k 2, which is the formula desired. The k1 3k From Exercise 8, we have k1 3k 2 3 k1 k 2 k / /2. O the other had, t / /2, which is the same as above a. Cosider a regular hexago which we border successively by hexagos with 3, 4, 5,... o each side. Defie the hexagoal umber h k to be the umber of dots cotaied i the k ested hexagos. b. First ote that h 1 1. To get a recursive relatioshi we cosider h k h k 1, which couts the dots added to the k 1st hexago to obtai the kth hexago. To do this, we must add 4 sides of k dots each, but 3 of the dots belog to two sides. Therefore h k h k 1 4k 3. A closed formula is the give by addig these differeces together: h k k i1 4i 3 4t k 3k 4kk + 1/2 3k 2k 2 k a. Cosider a regular hetago which we border successively by hetagos with 3, 4, 5,... o each side. Defie the hetagoal umbers s 1, s 2, s 3,..., s k,... to be the umber of dots cotaied i the k ested hetagos. j1

11 1.2. SUMS AND PRODUCTS 9 b. First ote that s 1 1. To get a recursive relatioshi we cosider s k s k 1, which couts the dots added to the k 1st hetago to obtai the kth hetago. To do this, we must add 5 sides of k dots each, but 4 of the dots belog to two sides. Therefore s k s k 1 5k 4. A closed formula is the give by addig these differeces together: s k k i1 5i 4 5t k 4k 5kk + 1/2 4k 5k 2 3k/ First cosider the differece T k T k 1. This couts the umber of dots o oe face of the kth tetrahedro. But this is simly the kth ested triagle used to defie the triagular umbers. Therefore, T k T k 1 t k. Hece, sice T 1 t 1 1, it follows that T k1 t k We cotiue with the formula from Exercise 12. T k1 t k k1 kk + 1/2. Exloitig the same techique as i Examle 1.19, we cosider k k 3 3k 2 + 3k + 1 3k 2 + k + 1 ad solve for k 2 + k to get k 2 + k k k 3 /3 1/3. The T 1/2 k1 kk + 1 1/6 k1 k k 3 1/6 k1 1. The first sum is telescoig ad the secod sum is trivial, so we have T 1/ / / Usig the fact! 1!, we fid that 1! 1, 2! 2, 3! 6, 4! 24, 5! 120, 6! 720, 7! 5040, 8! 40320, 9! , ad 10! Each of these four quatities are roducts of 100 itegers. The largest roduct is , sice it is the roduct of 100 factors of 100. The secod largest is 100! which is the roduct of the itegers 1, 2,..., 100, ad each of these terms is less or equal to 100. The third largest is 50! 2 which is the roduct of 1 2, 2 2,..., 50 2, ad each of these factors j 2 is less tha j50 + j, whose roduct is 100!. The smallest is which is the roduct of s a b. c. ka i k i1 i1 i1 a i. ia i a 1 2a 2 a 1 2 a 1 a 2 a! a i. k a k i a i. i1 i1 1 kk + 1 k1 k1 i1 1 k 1. Let a j 1/j + 1. Notice that this is a telescoig sum, ad k a j 1 a j kk + 1 usig the otatio i the text recedig Examle 1.15, we have a a 0 1 1/ + 1 / k k k k k k 1 k + 1 k2 k2 1 k 1 1 k k 1 k k2 k2 k We sum both sides of the idetity k+1 3 k 3 3k 2 +3k+1 from k 1 to k. k1 k+13 k , sice the sum is telescoig. k1 3k2 + 3k k1 k2 + 3 k1 k + k1 1 3 k1 k /2 +. As these two exressios are equal, solvig for k1 k2, we fid that k1 k2 / We sum both sides of the idetity k k 4 4k 3 + 6k 2 + 4k + 1 from k 1 to k. Usig Exercise 19 we fid that k1 k /4. k1 j a. 10! 7! !720 7!6!.

12 10 1. THE INTEGERS b. 10! 7!6! 7!5! 6 7!5!3!. c. 16! 14! !240 14!5!2!. d. 9! 7!8 9 7! !3!3!2! Sice c a 1!a 2! a! ad b a 1!a 2! a! 1, it follows that c! c c 1! c b! a 1!a 2! a! b! Assume that x y. The z! x! + y! y! + y! 2y!. Sice z > y we have z! y + 1y!. This imlies that y Hece the oly solutio with x, y, ad z ositive itegers is x y 1 ad z a. b. 1 1 j 1 1/21 1/3 1 1/ j /j 1 + 1/j j2 j2 j2 j Mathematical Iductio For 1 we have 1 < This is the basis case. Now assume < 2. We the have + 1 < < This comletes the iductive ste ad the roof by mathematical iductio We have 2 2, , , , ad We cojecture that j1 2j + 1 sice this formula holds for small values of. To rove this by mathematical iductio we have 1 j1 2j so the result is true for 1. Now assume that the formula holds for. The +1 j1 2j j1 2j This comletes the roof For the basis ste we have The, 1 k 2 k1 k1 1 k1 1 k For the iductive ste, we assume that 1 k1 1 k 2 1 k by the iductio hyothesis. This is less tha , as desired For the basis ste, we have kk For the iductive ste, we assume that 2 k1 k+1 1 kk The, 1 kk kk , as desired. + 2 k1 k We see that A, A 2 1 2, A 3 A A ad so o. We cojecture that A 1. To rove this by mathematical iductio we first ote that the basis ste follows sice A. Next, we assume that A 1. The A +1 A A The basis ste holds sice /2. For the iductive ste assume that j1 j + 1/2. It follows that +1 j j1 j j

13 1.3. MATHEMATICAL INDUCTION 11 This fiishes the iductive roof For the basis ste, we have 1 j1 j /6. For the iductive ste, we assume that j1 j /6. The, +1 j1 j2 j1 j / / / [ ]/6, as desired For the basis ste, we have 1 j1 j3 1, ad 11+1/2 2 1 also. For the iductive ste, we assume that j1 j3 + 1/2 2. The, +1 j1 j3 j1 j / /2 2, as desired For the basis ste, we have 1 j1 jj /3. Assume it is true for. The +1 j1 jj / / / For the basis ste, we have 1 j1 1j 1 j /2. For the iductive ste, we assume that j1 1j 1 j /2. The, j1 1j 1 j 2 j1 1j 1 j / [2 + 1 ] /2, as desired We have j1 2 j 2 P j1 j 2 +1/2 sice j j We use mathematical iductio. For 1 we have 1 j1 j j! 1 1! ! 1 1. Now assume that j1 j j! + 1! 1. The +1 j1 j j! + 1! ! + 1! ! 1. This comletes the roof We will rove this usig mathematical iductio. We see that Now assume that ostage of cets ca be formed, with 4a + 5b, where a ad b are oegative itegers. To form + 1 cets ostage, if a > 0 we ca relace a 4-cet stam with a 5-cet stam; that is, + 1 4a 1 + 5b + 1. If o 4-cet stams are reset, the all 5-cet stams were used. It follows that there must be at least three 5-cet stams ad these ca be relaced by four 4-cet stams; that is, + 1 4a b We rove this usig mathematical iductio. We see that Now assume that ostage of cets ca be formed, with 10a + 7b, where a ad b are ositive itegers. To form + 1 cets ostage, if a > 1 we ca relace 2 te-cet stams with 3 seve-cet stams, that is, a 2 + 7b + 3. If a < 2, the otice that b 7. We ca relace 7 seve-cet stams with 5 te-cet stams, that is, a b We use mathematical iductio. The iequality is true for 0 sice H 2 0 H /2. Now assume that the iequality is true for, that is, H /2. The H j1 1/j j2 +1 1/j H j2 +1 1/ / / /2 + 1/ /2. This comletes the iductive roof For the basis ste, we have H 2 0 H For the iductive ste, we assume that H The, as desired. H 2 +1 H j j < , For the basis ste, we have 2 1! 2 < ! 2 4. For the iductive ste, we assume that 2! < 2 2! 2. The [2 + 1]! 2! < 2 2! < 2 2! [ + 1!] 2, as desired.

14 12 1. THE INTEGERS We will use the secod ricile of mathematical iductio to rove this. For the basis ste, we have x y is a factor of x 1 y 1. For the iductive ste, we assume that x y is a factor of x y ad x 1 y 1. The, x +1 y +1 x y x + y + xyx 1 y 1. Sice x y is a factor of both x y x + y ad xyx 1 y 1, it is a factor of x +1 y Let A be such a set. Defie B as B {x k + 1 x A ad x k}. Sice x k, B is a set of ositive itegers. Sice k A ad k k, k k is i B. Sice + 1 is i A wheever is, + 1 k + 1 is i B wheever k + 1 is. Thus B satisfies the hyothesis for mathematical iductio, i.e. B is the set of ositive itegers. Maig B back to A i the atural maer, we fid that A cotais the set of itegers greater tha or equal to k The basis ste holds sice < 4! 24. Now assume that 2 <!. The < 2! < + 1! + 1! For the basis ste, we have < 24 4!. For the iductive ste, we assume that 2 <!. The, <! <! + 3 <! +! 2! < + 1! + 1!, as desired The basis ste is clear whe 0. For the iductive ste, we assume that 1 + h 1 + h. The, 1 + h h 1 + h 1 + h1 + h 1 + h + h + h h + 1 sice h 2 is ositive. This last iequality roves the iductio hyothesis We use the secod ricile of mathematical iductio. For the basis ste, if the uzzle has oly oe iece, the it is assembled with exactly 0 moves. For the iductio ste, assume that all uzzles with k ieces require k 1 moves to assemble. Suose it takes m moves to assemble a uzzle with + 1 ieces. The the m move cosists of joiig two blocks of size a ad b, resectively, with a + b + 1. But by the iductio hyothesis, it requires exactly a 1 ad b 1 moves to assemble each of these blocks. Thus, m a 1 + b a + b This comletes the iductio The 2 case does ot follow from the 1 case, sice, whe 2, the set of horses labelled 1 to 1 which is just the set cotaiig horse 1 does ot have ay commo elemets with the set of horses labelled from 2 to which is just the set cotaiig horse Suose that f is defied recursively by secifyig the value of f1 ad a rule for fidig f + 1 from f. We will rove by mathematical iductio that such a fuctio is well-defied. First, ote that f1 is well-defied sice this value is exlicitly stated. Now assume that f is well-defied. The f + 1 also is well-defied sice a rule is give for determiig this value from f The fuctio is f 2. For the basis ste, we have f For the iductive ste, we assume that f 2. The, f + 1 2f , as desired We have g1 2, g2 2 g1 4, g3 2 g , ad g4 2 g The basis ste is give. For the iductive ste, we assume that the value of f at the first ositive itegers are uiquely determied. The f + 1 is uiquely determied from the rule. Therefore, by mathematical iductio, f is determied for every ositive iteger We use the secod ricile of mathematical iductio. The basis ste cosists of verifyig the formula for 1 ad 2. For 1 we have f ad for 2 we have f Now assume that fk 2 k + 1 k for all ositive itegers k with k < where > 2. By the iductio hyothesis it follows that f f 1 + 2f This fiishes the roof Sice > , the basis ste holds. Assume that 2 > 2. Note that for > 4, > > The we have < 2 2 < , which comletes the iductio.

15 1.3. MATHEMATICAL INDUCTION We use the secod ricile of mathematical iductio. We see that a , a i 3, ad a These are the basis cases. Now assume that a k 3 k for all itegers k with 0 k <. It follows that a a 1 + a 2 + a < The iductio argumet is comlete a. For the basis ste otice that for 1 rig oly, moves are eeded. For the iductive ste we assume that it takes 2 1 stes to trasfer rigs. To make the iductive ste, first trasfer of + 1 rigs to the third eg. This takes 2 1 stes. Now trasfer the bottom rig to the secod eg. This is oe ste. The trasfer the rigs o the third eg to the secod eg. This is 2 1 more stes. Altogether, this takes stes. b. The world will last, accordig to this leged, , 446, 744, 073, 709, 551, 615 secods miutes hours days years, that is more tha 580 billio years Let P be the statemet for. The P 2 is true, sice we have a 1 +a 2 /2 2 a 1 a 2 a 1 a 2 / Assume P is true. The by P 2, for 2 ositive real umbers a 1,..., a 2 we have a 1 + +a 2 2 a 1 a 2 + a3 a a 2 1 a 2. Aly P to this last exressio to get a a 2 2a 1 a 2 a 2 1/2 which establishes P for 2 k for all k. Agai, assume P is true. Let g a 1 a 2 a 1 1/ 1. Alyig P, we have a 1 + a a 1 + g a 1 a 2 a 1 g 1/ g 1 g 1/ g. Therefore, a 1 + a a 1 1g which establishes P 1. Thus P 2 k is true ad P imlies P 1. This establishes P for all There are four 2 2 chess boards with oe square missig. Each ca be covered with exactly oe L-shaed iece. This is the basis ste. Now assume that ay 2 2 chess board ca be covered with L- shaed ieces. Cosider a chess board with oe square missig. Slit this ito four 2 2 chess boards three of which cotai every square ad the fourth has oe square missig. By the iductive hyothesis we ca cover the fourth 2 2 chess board because it is missig oe square. Now use oe L-shaed iece to cover the three squares i the other three chess boards that touch at the ceter of the larger chess board. What is left to cover is all the rest of the squares i each of the three 2 2 chess boards. The iductive hyothesis says that we ca cover all the remaiig squares i each of these chess boards. This comletes the roof Note that sice 0 < < q we have 0 < /q < 1. The roositio is trivially true if 1. We roceed by strog iductio o. Let ad q be give ad assume the roositio is true for all ratioal umbers betwee 0 ad 1 with umerators less tha. To aly the algorithm, we fid the uit fractio 1/s such that 1/s 1 > /q > 1/s. Whe we subtract, the remaiig fractio is /q 1/s s q/qs. O the other had, if we multily the first iequality by qs 1 we have q > s 1 which leads to > s q, which shows that the umerator of /q is strictly greater tha the umerator of the remaider s q/qs after oe ste of the algorithm. By the iductio hyothesis, this remaider is exressible as a sum of uit fractios, 1/u /u k. Therefore /q 1/s + 1/u /u k which comletes the iductio ste a. Sice 1/2 < 2/3, we subtract to get 2/3 1/2 + 1/6. b. Sice 1/2 < 5/8, we subtract to get 5/8 1/2 + 1/8. c. Sice 1/2 < 11/17 we subtract to get 11/17 1/2 + 5/34. The largest uit fractio less tha 5/34 is 1/7 so we subtract to get 11/17 1/2 + 1/7 + 1/238. d. The largest uit fractio less tha 44/101 is 1/3 so we subtract ad get 44/101 1/3 + 31/303. The largest uit fractio less tha 31/303 is 1/10, so we subtract to get 44/101 1/3 + 1/10 + 7/3030. The largest uit fractio less tha 7/3030 is 1/433, so we subtract to get 44/101 1/3 + 1/10 + 1/ / Note that this is the result of the greedy algorithm. Other reresetatios are ossible, such as 44/101 1/3 + 1/10 + 1/ /26664.

16 14 1. THE INTEGERS 1.4. The Fiboacci Numbers a. We have f 1 1, f 2 1, ad f f 1 + f 2 for 3. Hece f 3 f 2 + f , f 4 f 3 + f , f , f , f , f , f , ad f b. We cotiue beyod art a fidig that f 11 f 10 + f , f , ad f c. We cotiue beyod art b fidig that f 14 f 13 + f , ad f d. We cotiue beyod art c fidig that f , f , ad f e. We cotiue beyod art d fidig that f , f f. We cotiue beyod art e fidig that f , f , f , f , ad f a. We cotiue from Exercise 1 art a, fidig that f ad f b. We cotiue from Exercise 1 art c, fidig that f c. We comuted f i Exercise 1 art f. d. We cotiue from Exercise 1 art f, fidig that f , f , f , f , ad f e. We cotiue from art d, fidig f , ad f f. We cotiue from art e, fidig f , f , f ad f Note that from the Fiboacci idetity, wheever is a ositive iteger, f +2 f f +1. The we have 2f +2 f f +2 + f +2 f f +2 + f +1 f +3. If we add f to both sides of this equatio, we have the desired idetity Assumig is a ositive iteger, we have comute 2f +1 + f f +1 + f +1 + f f +1 + f +2 f +3. If we subtract f from both sides of this equatio, we have the desired idetity For 1 we have f f f 0 f 1, ad for 2, we have f f f 1 f 2. So the basis ste holds for strog iductio. Assume, the that f 2 4 f 2 2+2f 3 f 2 ad f 2 2 f f 2 f 1. Now comute f 2 f 2 1 +f 2 2 2f 2 2 +f 2 3 3f 2 2 f 2 4. Now we may substitute i our iductio hyotheses to set this last exressio equal to 3f f 2 f 1 f 2 2 2f 3 f 2 3f f f 1 f 1 f f 1 2 2f 1 f 2 f f 1 2f f f 1 f 2 + 2f f f 1 2f 1 f f 1 f 2 + 2f 1 f which comletes the iductio ste For a ositive iteger greater tha 1, we have f +2 f +1 + f f + f 1 + f f + f f 2 + f 3f f 2. Addig f 2 to both sides yields the desired idetity Note that f 1 1 f 2, f 1 + f 3 3 f 4, ad f 1 + f 3 + f 5 8 f 6 so we cojecture that f 1 + f 3 + f f 2 1 f 2. We rove this by iductio. The basis ste is checked above. Assume that our formula is true for, ad cosider f 1 + f 3 + f f f 2+1 f 2 + f 2+1 f 2+2, which is

17 1.4. THE FIBONACCI NUMBERS 15 the iductio ste. Therefore the formula is correct Note that f 2 1 f 3 1, f 2 + f 4 4 f 5 1, ad f 2 + f 4 + f 6 12 f 7 1, so we cojecture that f 2 +f 4 +f 6 + +f 2 f We rove this by iductio. The basis ste is checked above. Assume that our formula is true for, ad cosider f 2 + f 4 + f f 2 + f 2+2 f f 2+2 f 2+3 1, which is the iductio ste. Therefore the formula is correct. Aother solutio is to subtract the formula i Exercise 7 from the formula i Examle 1.27, as follows: i1 f 2i 2 i1 f i i1 f 2i 1 f f 2 f First suose 2k is eve. The f f f 1 f 2k + f 2k f 1 2f 2k 1 + f 2k f 1 f 2k+2 1 2f 2k by the formulas i Examle 1.23 ad Exercise 3. This last equals f 2k+2 f 2k f 2k 1 f 2k+1 f 2k 1 f 2k 1 1 f 1 1. Now suose 2k + 1 is odd. The f f f 2k+1 f 2k f 2k f 1 f 2k+1 f 2k 1 1 by the formula just roved for the eve case. This last equals f 2k+1 f 2k f 2k + 1 f We ca uite the formulas for the odd ad eve cases by writig the formula as f For 1 we have f 3 2 f2 2 +f Ad whe 2 we have f f3 2 +f2 2, so the basis stes hold for mathematical iductio. Now assume, for the strog form of iductio, that the idetity holds for all values of u to k. The f 2k 3 fk f k 2 2 ad f 2k 1 fk 2 + f k 1 2. Now we calculate f 2k+1 f 2k + f 2k 1 f 2k 1 + f 2k 2 + f 2k 1 2f 2k 1 + f 2k 1 f 2k 3 3f 2k 1 f 2k 3. Now substitutig i the iductio hyothesis, makes this last exressio equal to 3fk 2 +f k 1 2 f k 1 2 f k 2 2 3fk 2 + 2f k 2 2 f k f k 1 2 2fk 2 + f k f kf k 1 2fk 2 + f k+1 f k 2 + 2f k f k+1 f k fk f k 2, which comletes the iductio ste We ca costruct a iductio roof similar to the oes i Exercises 5 ad 10, or we may roceed as follows. From Exercise 5, we have f 2 f 2 + 2f 1 f f f + f 1 + f 1 f +1 f 1 f +1 + f 1 f 2 +1 f 2 1, which is the desired idetity Let S f + f 1 + f 2 + 2f f f 1. We roceed by iductio. If 3 we have S 3 f 3 +f 2 +f , ad whe 4 we have S 4 f 4 +f 3 +f 2 +2f , so the basis stes hold. Now assume the idetity holds for all values less or equal to ad cosider S +1 f +1 +f +f 1 +2f 2 +4f f f f 1. We use the Fiboacci idetity to exad every term excet the last two to get S +1 f + f 1 + f 1 + f 2 + f 2 + f 3 +2f 3 +f 4 +4f 4 +f f 2 +f f f 1. Next we regrou, takig the first term from each set of aretheses, lus the secod last term together i oe grou, the last term from each set of aretheses together i aother grou, ad leavig the last term by itself to get S +1 f + f 1 + f 2 + 2f 3 + 4f f f 2 + f 1 + f 2 + f 3 + 2f 4 + 4f f f 1. The first grou is see to be equal to S whe we realize that the last f 2 f 1. The secod grou is equal to S 1, so we have S +1 S + S by the iductio hyothesis. Therefore, by mathematical iductio, the roositio is roved We roceed by mathematical iductio. For the basis ste, 1 j1 f j 2 f 1 2 f 1 f 2. To make the iductive ste we assume that j1 f j 2 f f +1. The +1 j1 f j 2 j1 f j 2 + f +1 2 f f +1 + f+1 2 f +1 f We use mathematical iductio. We will use the recursive defiitio f f 1 + f 2, with f 0 0 ad f 1 1. For 1 we have f 2 f 0 f Hece the basis ste holds. Now assume that f +1 f 1 f 2 1. The f +2 f f 2 +1 f +1 + f f f +1 f + f 1 f 2 f +1 f This comletes the roof From Exercise 13, we have f +1 f f 1 f 2 f f 2 f f 2 2 f 2 + f 2 1. The idetity i Exercise 10 shows that this is equal to f 2 1 whe is a ositive iteger, ad i articular whe is greater tha Sice f 1 f f 2 2, the basis ste holds. By the iductio hyothesis we have f 1 f f 2 1 f 2 +f 2 f 2+1 +f 2+1 f 2+1 f 2 2 +f 2 f 2+1 +f 2+1 f 2+1 f 2 f 2 +f 2+1 +f 2+1 f 2+1

18 16 1. THE INTEGERS f 2 f f 2+1 f 2+1 f 2 + f 2+1 f 2+1 f For fixed m, we roceed by iductio o. The basis ste is f m+1 f m f 2 + f m 1 f 1 f m 1 + f m 1 1 which is true. Assume the idetity holds for 1, 2,..., k. The f m+k f m f k+1 + f m 1 f k ad f m+k 1 f m f k + f m 1 f k 1. Addig these equatios gives us f m+k + f m+k 1 f m f k+1 + f k + f m 1 f k + f k 1. Alyig the recursive defiitio yields f m+k+1 f m f k+2 + f m 1 f k+1, which is recisely the idetity We re give that L 1 1 ad L 2 3. Addig each cosecutive air to geerate the ext Lucas umber yields the sequece 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, A few trial cases lead us to cojecture that i1 L i L We rove that this formula is correct by iductio. The basis ste is L 1 1 ad L , which checks. Assume that the formula holds for ad comute +1 i1 L i i1 L i + L +1 L L +1 by the iductio hyothesis. This last equals L +2 + L +1 3 L +3 3, which comletes the iductio ste A few trial cases lead us to cojecture that i1 L 2i 1 L 2 2. We rove that this formula is correct by iductio. The basis ste is L 1 1 L 2 2. Assume that the formula holds for ad comute +1 i1 L 2i 1 i1 L 2i 1 + L 2+1 L L 2+1 L 2+2 2, which comletes the iductio ste A few trial cases lead us to cojecture that i1 L 2i L We rove that this formula is correct by iductio. The basis ste is L 2 3 L 3 1. Assume that the formula holds for ad comute +1 i1 L 2i i1 L 2i + L 2+2 L L 2+2 L 2+3 1, which comletes the iductio ste We roceed by iductio. The basis ste is whe 2, ad we have L 2 2 L 3 L Now assume the idetity holds for. The for + 1 we have L 2 +1 L +2 L L + L 1 L +1 L +1 +L L L L +1 +L 1 L +1 L +1 L L 2 L 2 L 1 L , where we aly the iductio hyothesis at the eultimate ste We roceed by iductio. The basis ste is L L 1 L Assume the formula holds for ad cosider +1 i1 L2 i i1 L2 i + L2 +1 L L L 2 +1 L +1 L + L +1 2 L +1 L +2 2, which comletes the iductio ste For 2, we have L f 1 + f 3. For 3, we have L f 2 + f 4. This serves as the basis ste. Now assume that the statemet is true for k 2, 3, 4,...,. The L +1 L + L 1 f +1 + f 1 + f + f 2 f +1 + f + f 1 + f 2 f +2 + f, which comletes the iductio For the basis ste, we check that L 1 f f 2 ad L 2 f f 4. Assume the idetity is true for all ositive itegers u to. The we have f +1 L +1 f +2 f f +2 f from Exercise 16. This equals f 2 +2 f 2 f +1 + f 2 f 1 + f 2 2 f f +1 f + f 2 f 2 1 2f 1 f 2 f 2 2 f 2 +1 f f 2 f f +1 f f 1 f 2 f +1 f 1 f +1 + f 1 + f f 2 f + f 2 + 2f 2 1, where the last arethetical exressio is obtaied from Exercise 8. This equals f L + f 1 L 1 + 2f 2 1. Alyig the iductio hyothesis yields f 2 + f f 2 1 f 2 + f f f 2 2 f f 2 f 2+2, which comletes the iductio For the basis ste, we check that whe 1, 5f L 1 + L 3 ad whe 2, 5f L 2 + L 4. Now assume the idetity holds for itegers less tha, ad comute 5f +1 5f + 5f 1 L 1 + L +1 + L 2 + L L 1 + L 2 + L +1 + L L + L +1, which comletes the iductio ste We rove this by iductio o. Fix m a ositive iteger. If 2, the for the basis ste we eed to show that L m+2 f m+1 L 2 + f m L 1 3f m+1 + f m, for which we will use iductio o m. For m 1 we have L f 2 + f 1 ad for m 2 we have L f 3 + f 2, so the basis ste for m holds. Now assume that the basis ste for holds for all values of m less tha ad equal to m. The L m+3 L m+2 + L m+1 3f m+1 + f m + 3f m + f m 1 3f m+2 + f m+1, which comletes the iductio ste o m ad roves the basis ste for. To rove the iductio ste o, we comute L m++1 L m+ +

19 1.4. THE FIBONACCI NUMBERS 17 L m+ 1 f m+1 L + f m L 1 + f m+1 L 1 + f m L 2 f m+1 L + L 1 + f m L 1 + L 2 f m+1 L +1 + f m L, which comletes the iductio o ad roves the idetity First check that α 2 α + 1 ad β 2 β + 1. We roceed by iductio. The basis stes are α + β 1 + 5/ /2 1 L 1 ad α 2 + β α β 2 + L 1 3 L 2. Assume the idetity is true for all ositive itegers u to. The L +1 L + L 1 α + β + α 1 + β 1 α 1 α β 1 β + 1 α 1 α 2 + β 1 β 2 α +1 + β +1, which comletes the iductio We fid that f 9 + f 7 + f 4, f 10 + f 8 + f 6 + f 2, f 11 + f 8 ad f 12 + f 10 + f 2. I each case, we used the greedy algorithm, always subtractig the largest ossible Fiboacci umber from the remaider Suose there is a ositive iteger that has o Zeckedorf reresetatio. The by the well-orderig roerty, there is a smallest such iteger,. Let f k be the largest Fiboacci umber less tha or equal to. Note that if f k, the has a Zeckedorf reresetatio, cotrary to our assumtio. The f k is a ositive iteger less tha, so it has a Zeckedorf reresetatio f k m i1 f a i. Sice has o Zeckedorf reresetatio, it must be that oe of the f ai s is equal to or cosecutive to f k. That is, oe of f k 1, f k, or f k+1 aears i the summatio for f k. The m i1 f a i + f k f k 1 + f k f k+1. But this cotradicts the choice of f k as the largest Fiboacci umber less tha. This establishes existece. To establish uiqueess of the Zeckedorf reresetatio, suose that there is a ositive iteger that has two distict reresetatios. The the well-orderig roerty gives us a smallest such iteger,. Suose m i1 f a i l j1 f b l are two distict reresetatios for. The o f ai f bj, else we could cacel this term from each side ad have a smaller iteger with two distict reresetatios. Without loss of geerality, assume that f a1 > f a2 > > f am ad f b1 > f b2 > > f bl ad that f a1 > f b1. If b 1 is eve, we comute l i1 f b i f b1 + f b1 2 + f b f 2 f b1+1 1 by Exercise 4. But this last is less tha or equal to f a1 1 <, a cotradictio. If b 1 is odd, we comute, ow usig Exercise 3, l i1 f b i f b1 + f b1 2 + f b f 3 f b1 +1 f 1 f a1 1 <, which is also a cotradictio. This roves uiqueess We roceed by mathematical iductio. The basis stes 2 ad 3 are easily see to hold. For the iductive ste, we assume that f α 1 ad f 1 α 2. Now, f +1 f + f 1 α 1 + α 2 α, sice α satisfies α α 1 + α We roceed by the secod ricile of mathematical iductio o. For the basis ste, we observe that 0 0 f For the iductive ste, we assume that f+1, ad that f. Now, [ ] + [ ] f+1 + f f Usig Theorem 1.3 ad the otatio therei, we have α 2 α + 1 ad β 2 β + 1, sice they are roots of x 2 x 1 0. The we have f 2 α 2 β 2 / 5 1/ 5α + 1 β + 1 1/ 5 j0 j α j j0 j β j 1/ 5 j0 j α j β j j1 j fj sice the first term is zero i the eultimate sum We rove this usig mathematical iductio. For 1 we have F f2 f f 1 f 0 where f 0 0. Now assume that this formula is true for. The F +1 F f+1 f F 1 1 f f f+1 + f f +1 f + f 1 f f+2 f +1. f +1 f O oe had, detf detf 1. O the other had, f+1 f det f f f +1 f 1 f 2 1.