-8 and 5-5 problem sheet Solutons to the followng seven exercses and optonal bonus problem are to be submtted through gradescope by :0PM on Wednesday th September 08. There are also some practce problems, not to be turned n, for those seekng more practce and also for revew pror to the exam. Problem Usng nducton and wthout usng the formulae for and prove that n nn + n + + = for all n N =0 Soluton. We proceed by nducton on n. Let Pn= n nn+n+ =0 + =. Base case. When n = 0, the left-hand sde s 0 = 0 and the rght hand sde s = 0. These are equal, so P0 s true. 0 = 0 Inducton step. Fx n N and suppose that Pn s true. That s to say, n nn + n + + =. We need to prove that Well, n+ + = = = n+ + = = n + + n + n + = n + n + n +. by defnton of summaton nn + n + = + n + n + by the nducton hypothess nn + n + + n + n + = n + n + n + = by dstrbutvty Hence we have proved what we wanted to prove.
The clam follows for all natural numbers, by nducton. Problem Prove that for all n N +. n = + = n n + Soluton. We proceed by nducton on n. Let Pn= n = + = n n+. Base case. When n =, the left-hand sde s 4 = 4, and the rght-hand sde s 4. These are equal, so P s true. Inducton step. Fx n N + and suppose that Pn s true. That s to say, We need to prove that Well, n+ = n+ = n = n + = = + = n n +. + = n + n + +. + + n + n + 4 = n n + + n + n + 4 nn + 4 + = n + n + 4 = n + 4n + n + n + 4 n + n + = n + n + 4 = n + n + 4 Snce n + 4 = n + +, Pn+ s true. The clam follows for all postve ntegers, by nducton. by defnton of summaton by the nducton hypothess by combnng fractons by dstrbutvty by factorzaton by cancellaton
Problem For ntegers n fnd and prove a formula for n. = How to approach ths problem Ths part s meant to help you thnk about how to fnd the formula. In partcular, ths part should only appear on scratch paper but not on your answer sheet. Usually, when you have a sum or product of lots of thngs, the natural way to fnd the closed-form formula s to try some small n s to see how ths thng behaves. Therefore, When n = 4, When n = 5, When n =, = = 4 = = 5 = = = + = 4 4 4 = + + = 4 5 4 4 6 5 = 5 It seems that most terms n the mddle wll cancel out. More specfcally, when n s even, only the frst term remans, and when n s odd, only the frst and last term, namely n+ and n remans. Therefore, we may guess that: n { = n s even n s odd = Now, what we are left to do s to formally prove ths s the correct formula, whch wll be presented n the next part. We ll use nducton to prove ths. n+ n How to how to formally prove ths problem Ths part s the formal proof, whch should be on your answer sheet.
Let Pn = n { = n s even n+ n n s odd = Proof Because the cases splt by party, we rewrte n as ether k or k + to ndcate the party. And we wll nduct on k N and k. Base Case. When k =, we have + = = = = = and Inducton Step. Assume for some k and k N, k = Then we have k+ = = = = k+ and k+ = k + k + + = 4 = = + = = k + k + k + k + = k + = k + k + = k + k + IH k + 4
and k++ = = = k+ = + k + = k + 4 k + k + + = k + + k + whch also comples wth the formula we clamed. Therefore, by nducton, we have proved that for all ntegers n n { = n s even n s odd = n+ n Problem 4 Startng wth a sngle stack of 00 cons, two players take alternate turns; n each turn, a player removes,, or 4 cons from the stack. The player who removes the last con wns. Prove that the second player has a strategy to wn regardless of what strategy the frst player uses. Hnt: prove somethng more general Soluton. An teraton of the game conssts of a move by the frst player, followed by a move by the second player. Let pn be the logcal formula, If the game begns wth 5n cons, then the second player has a wnnng strategy. We wll use nducton to show that pn s true for all postve ntegers n. Note that the truth of p0 s exactly what we were asked to show. Base case. p s true, because f the frst player removes cons, then the second player can remove 5 cons e {5 : [4]} = [4]. Inducton Step. Now suppose that k s a postve nteger and that pk s true. Consder a game that begns wth 5k + 5 cons. As n the base case, f the frst player removes cons, then the second player can remove 5 cons, leavng 5k cons. By the truth of pk, and snce t wll be the frst player s turn, the second player has a wnnng strategy from ths pont. 5
Therefore, by the prncple of mathematcal nducton, we have shown that pn s true for all postve ntegers n, and the result follows. Problem 5 Suppose that a football team can score ponts or 7 ponts on any scorng opportunty e do not consder, 6 and 8 pont optons. Fnd, wth proof, the largest score that s not possble for the team to acheve. Soluton. We clam that s the largest unachevable score. Frst note that s, ndeed, not achevable. To acheve a score of, there would have to be ether zero or one 7 pont score. Snce s not a multple, zero 7 pont scores s mpossble, and snce 4 s not a multple of, one 7 pont score s mpossble. Now we ll prove that f n then n can be expressed as a nonnegatve multple of plus a nonnegatve multple of 7 by strong nducton on n. Let pn = n = + 7j for some natural numbers and j. Base case. Note that so p, p and p4 are all true. = 4 = 7 + 4 = 7 Inducton step. Fx n 4 and suppose, for all k N wth k n, that pk s true, e k can be expressed as a nonnegatve multple of plus a nonnegatve multple of 7. Snce n 4 we have n n. Hence, by our nducton hypothess, n = a + 7b for some a, b 0. Thus n + = a + + 7b and n + can be expressed as a nonnegatve multple of plus a nonnegatve multple of 7. Combnng the above arguments shows that s the largest unachevable score. Problem 6 Let a n n N be a sequence satsfyng a n = a n + a n for all n 6
a Gven that a 0, a are odd, prove that a n s odd for all n N. b Gven that a 0 = a =, prove that a n = n n+ for all n N. Soluton. a Suppose a 0 and a are odd. We ll prove that a n s odd for all n N by strong nducton on n. Let pn = a n s odd. For the base case, p0 and p are true, snce we re assumng that a 0, a are odd. For the nducton step, fx n and suppose that for all k n pk s true, e a k s odd. Then, n partcular, a n and a n are odd, so a n = p and a n = q for some p, q Z. Now snce n +, we have a n+ = a n + a n = p + q = p + q so a n+ s odd. Hence, by nducton, a n s odd for all n N. b We ll prove that a n = n n+ for all n N by strong nducton on n. Let pn = a n = n n+. Base case. The truth of p0 and p follow by pluggng n values: 0 = + = = a 0 = = = a Inducton step. Fx n > and suppose that for all 0 < n p s true, e a = +. Snce n > t follows that n, so we can apply the formula to obtan a n = a n + a n Snce 0 n < n and 0 n < n, the nducton hypothess yelds a n = n n + n n = n + n n n = n + n + n n = n n = n n+ Ths s what we needed to prove. By strong nducton, t follows that a n = n n+ for all n N. 7
Problem 7 Let f n n N be the sequence satsfyng f 0 = 0, f = and f n = f n + f n for all n. Show, by nducton, that f n + f n+ = f n+ and f n+ f n = f n+ for every n N. Soluton. For all natural numbers n, let P n denote the proposton that f n + f n+ = f n+ and f n+ f n = f n+. We wll show that P n s true for all n 0 by nducton. Base case. Compute f = f + f 0 =. Ths means that f 0 + f = 0 + = = f 0+ and f f 0 = 0 = = f 0+. Thus both equaltes are true when n = 0, mplyng that P 0 s true. Inducton step. Assume now that n N and P n s true, meanng that fn + fn+ = f n+ and fn+ fn = f n+. Our goal s to show that P n P n +,.e. that fn+ + fn+ = f n+ and fn+ fn+ = f n+4. To do ths, frst remark that the equaltes n P n can be rewrtten as Thus fn+ + fn+ IH = f n+ f n f n+ = f n+ f n and f n+ = f n + f n+. = f n+ + f n+ = f n+. + Ths proves the frst equalty n P n +. f n+ + f n 8
The second equalty s trcker; whle t s possble to go through the algebrac manpulatons drectly, they are somewhat unmotvated, and so we nstead choose to show a less computatonal soluton. Note that for any real numbers a and b, a + b + a b = a + ab + b + a ab + b = a + b. In partcular, settng a = f n+ and b = f n+ yelds Ths rearranges now to f n+ + f n+ = f n+ + f n+ + f n+ f n+ = f n+ + f n. f n+ f n+ = f n+ + f n+ f n = f n+ + f n+ + f n+ f n IH = fn+ + f n+ = f n+4. Note that n the step n whch we nvoke the nductve hypothess we also make use of the fact that f n+ + f n+ = f n+, whch we proved n the prevous paragraph. Ths proves the second equalty. Thus by the prncpal of mathematcal nducton, we are done. Bonus Problem ponts On the sland of Perfect Reasonng, k > 0 people dress unfashonably. Clve Newstead arrves on a boat, somewhat dstressed at the stereotypcal unfashonablty of the logcally adept, and announces, At least one of you s dressed unfashonably. No unfashonably dressed person realzes that they are dressed unfashonably, but everyone else on the sland realzes t. In n days from now f you have concluded that your apparel s unfashonable, you wll come to the beach at noon that day to denounce your apparel. The slanders gather at noon every day thereafter. When wll the unfashonable apparel be denounced, and how wll the wearers reason? Soluton. Frst consder the case when k =. Then, ths person wll mmedately know that they are dressed unfashonably, and the unfashonable apparel wll be denounced on day. Although not strctly necessary, nvestgatng the k = case s enlghtenng. Say the unfashonable people on the sland are p, p. From the perspectve of p, f he s not unfashonable, then p wll know that p s not unfashonable and p wll denounce hs fashon on day. Thus, f p does not denounce hs fashon on day, p knows that he must be unfashonable, and so he wll denounce hs fashon on day. A key aspect s that p wll thnk exactly the same way as p, and 9
so p wll do the same thng, and p, p wll both denounce ther fashon on day. Now, let P k be the proposton that for k > 0 unfashonable people, t wll take k days for the unfashonable apparel to be denounced; furthermore, all k unfashonable people on the sland wll denounce ther unfashonable apparel on the k th day. We have establshed that the base case P s true. We proceed va nducton. Consder the case where we have k + unfashonable people, say p, p,..., p k+. From the perspectve of p k+, f he s not unfashonable, then by the nductve hypothess p, p,..., p k wll all denounce ther unfashonable apparel on day k. Therefore, f p k+ does not see anybody denounce ther fashon on day k, then he wll conclude that he must be unfashonable and announce ths on day k +. Every unfashonable person p on the sland wll reason the same way, and so all k + unfashonable people wll denounce ther fashon on the k + th day. As P k P k +, and P s true, we have by the prncple of nducton that P k s true for every k. Extra Problem Let P n be a mathematcal statement dependng on an nteger n. Suppose that: P 0 s true; and Gven n Z, f P n s true, then at least one of P n + or P n s true. For whch n Z must P n be true? Soluton. P n needs to be true when n = 0; but gven any n 0, P n need not be true. P 0 must be true by. Let P n be the statement n 0. Then P 0 s the asserton 0 0, whch s true. Gven n N, suppose P n s true. Then n 0, so that n 0. Hence P n s true. Hence P n satsfes and, provng that P n need not be true for any n > 0. Now take P n to be the asserton n 0. The same argument as above yelds that P n satsfes and, and yet s false for all n < 0. 0
Extra Problem Let n be a postve nteger. For each sum below, wrte t n summaton notaton and fnd and prove a formula n terms of n: a + 7 + + + 4n ; b + 5 + 9 + + 4n + ; c + + 4 n + n; d + 5 7 + + 4n 4n. Soluton. a The sum s n = 4. Snce = nn+, t s equal to nn + n, whch s n turn equal to nn +. b The sum s n =0 4 +. Lke wth, ths s equal to nn + + n +, whch s equal to n + n +. c The sum s n =0 + here we group each par of terms,.e. + for each n. Hence the value of the sum s n. Ths could also be proved by nducton usng the expresson n =, but ths s more nvolved and s unnecessary. d Lke n c, the sum s n =, snce the pars of terms are all of the form +, whch s equal to. Hence the sum evaluates to 4n. Ths could also be proved by nducton usng the expresson n = +. Extra Problem Let a R wth a 0. Fnd the flaw n the followng proof that a n = for all n 0: We prove that a n = for all n 0 by nducton on n. Base case. a 0 = snce any nonzero real number rased to the power 0 s equal to. Inducton step. Let n N and assume the nducton hypothess. Then a n+ = an a n a n = =
By nducton, a n = for all n 0. Soluton. The error s n the nducton step. The nducton hypothess s not clearly stated; f t were, then t would be more obvous that t only assumes that a n =, not that a n = as well. To fx ths, we would need to use strong nducton; but then we d realze the clam fals n the nducton step, snce a = a, whch need not be equal to. Extra Problem 4 Show that for all postve ntegers n, the number 4 n + 5n s dvsble by 9. Soluton. Let P n denote the proposton that 4 n + 5n s dvsble by 9. We prove that P n s true for all n va nducton. BC: Note that for n =, we have 4 n + 5n = 4 + 5 = 8. Snce 8 s dvsble by 9, we see that P s true. IS: Now assume that P n s true for some n, so that 4 n + 5n s dvsble by 9. Then 4 n+ + 5n + = 4 4 n + 5n + 4 = 44 n + 5n + 8 45n. Note that by the Inducton Hypothess, 4 n + 5n s dvsble by 9, so there exsts k Z wth 4 n +5n = 9k. Thus, we have 4 n+ +5n+ = 4 9k +8 45n = 9 4k + 5n, so 4 n+ + 5n + s dvsble by 9. Thus P n P n +, as desred. Combnng both of the above bullet ponts gves that P n s true for all nteger n, as desred.