332 Mathematical Induction Solutions for Chapter 14. for every positive integer n. Proof. We will prove this with mathematical induction.

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1 33 Mathematcal Inducton. Solutons for Chapter. Prove that 3 n n n for every postve nteger n. Proof. We wll prove ths wth mathematcal nducton. Observe that f n, ths statement s, whch s obvously true. Consder any nteger k. We must show that S k mples S k. In other words, we must show that f 3 k k k s true, then 3 k k k k s also true. We use drect proof. Suppose k and 3 k k k. Observe that 3 k k 3 k k k k k k k k k k k k k. Therefore we have shown that 3 k k k k. 3. Prove that n 3 n n for every postve nteger n. Proof. We wll prove ths wth mathematcal nducton. When n the statement s 3, whch s true. Now assume the statement s true for some nteger n k, that s assume k 3 k k. Observe that ths mples the statement s true for n k k 3 k k 3 k 3 k k k 3 k k k 3 k k k 3 k k k k k k

2 Solutons for Chapter 333 k k k k Therefore k 3 k 3 kk, whch means the statement s true for n k. 5. If n N, then 3 n n. When n, ths statement s, or, whch s true. Now assume the statement s true for some nteger n k, that s assume 3 k k. Observe ths mples that the statement s true for n k, as follows: 3 k k 3 k k k k k k k Thus we have 3 k k k, so the statement s true for n k. Thus the result follows by mathematcal nducton. nn n 7 7. If n N, then nn. When n, we have 3 7, whch s the true statement 3 8. Now assume the statement s true for some nteger n k, that s assume kk kkk7. Now observe that kk k k kk k k kk k 7 k k kk k 7k k 3 kk k 7 k k 3 k kk 7 k 3 k k 3k 8

3 33 Mathematcal Inducton k k k 9 k k k 7 Thus we have kkkk kkk7, and ths means the statement s true for n k. Thus the result follows by mathematcal nducton. 9. Prove that 5 n for every nteger n 0. For n 0, the statement s 5 0. Ths s 0, whch s true. Now assume the statement s true for some nteger n k, that s assume 5 k. Ths means 5 k a for some nteger a, and from ths we get 5 k a. Now observe that 5 k 5 k 5 5 k 5 a 5a 5 a 5 5a. Ths shows 5 k 5a, whch means 5 k. Ths completes the proof by mathematcal nducton.. Prove that 3 n 3 5n for every nteger n 0. When n 0, the statement s , or 3, whch s true. Now assume the statement s true for some nteger n k 0, that s assume 3 k 3 5k. Ths means k 3 5k 3a for some nteger a. We need to show that 3 k 3 5k. Observe that k 3 5k k 3 3k 3k 5k 5 k 3 5k 3k 3k 3a 3k 3k 3a k k. Thus we have deduced k 3 k 3a k k. Snce a k k s an nteger, t follows that 3 k 3 5k. It follows by mathematcal nducton that 3 n 3 5n for every n 0.

4 Solutons for Chapter Prove that n 3 n for every nteger n 0. When n 0, the statement s 0 3 0, or 0, whch s true. Now assume the statement s true for some nteger n k 0, that s, assume k 3 k. Ths means k 3 k a for some nteger a. We need to show that k 3 k. Observe that k 3 k k 3 3k 3k k k 3 k 3k 3k a 3k 3k a 3kk. Thus we have deduced k 3 k a 3kk. Snce one of k or k must be even, t follows that kk s even, so kk b for some nteger b. Consequently k 3 k a 3kk a 3b a b. Snce k 3 k a b t follows that k 3 k. Thus the result follows by mathematcal nducton. 5. If n N, then nn n. When n, the statement s, whch smplfes to. Now assume the statement s true for some nteger n k, that s assume kk k. Next we show that the statement for n k s true. Observe that kk k k kk k k k k k k k k k k k k k k k k k k. Ths establshes kk that the statement s true for n k. k, whch s to say

5 33 Mathematcal Inducton Ths completes the proof by mathematcal nducton. 7. Suppose A, A,... A n are sets n some unversal set U, and n. Prove that A A A n A A A n. Proof. The proof s by strong nducton. When n the statement s A A A A. Ths s not an entrely obvous statement, so we have to prove t. Observe that A A {x : x U x A A } defnton of complement {x : x U x A A } {x : x U x A x A } defnton of {x : x U x A x A } DeMorgan {x : x U x A x A } {x : x U x A x U x A } dstrbutve prop. {x : x U x A } {x : x U x A } def. of A A defnton of complement Let k. Assume the statement s true f t nvolves k or fewer sets. Then A A A k A k A k A A A k A k A k A A A k A k A k A A A k A k A k Thus the statement s true when t nvolves k sets. Ths completes the proof by strong nducton. 9. Prove n k /k /n for every n. Proof. Ths clearly holds for n. Assume t holds for some n. Then n k /k /n /n n n nn /n. The proof s complete.. If n N, then 3 n n. Proof. If n, the result s obvous. Assume the proposton holds for some n >. Then 3 n 3 n n n n n 3 n n n 3 n Now, the sum n n n 3 on the rght has n n n terms, n all greater than or equal to, so the sum s greater than n n n. Therefore. n

6 Solutons for Chapter 337 we get 3 n n n n n 3 n n n. Ths means the result s true for n, so the theorem s proved. 3. Use nducton to prove the bnomal theorem x y n n n 0 x n y. Proof. Notce that when n, the formula s x y 0 x y 0 x 0 y x y, whch s true. Now assume the theorem s true for some n >. We wll show that ths mples that t s true for the power n. Just observe that x y n x yx y n n n x y x n y 0 n n n n x n y x n y 0 0 [ ] n n n x n y y n 0 n n n x n y y n 0 n n n x n y. 0 Ths shows that the formula s true for x y n, so the theorem s proved. 5. Concernng the Fbonacc sequence, prove that F F F 3 F...F n F n. Proof. The proof s by nducton. When n the statement s F F F 3, whch s true. Also when n the statement s F F F F 3, whch s true, as F F. Now assume k and F F F 3 F... F k F k. We need to show F F F 3 F... F k F k F k3. Observe that F F F 3 F... F k F k F F F 3 F... F k F k F k F k F k F k F k3. Ths completes the proof by nducton. 7. Concernng the Fbonacc sequence, prove that F F 3 F n F n.

7 338 Mathematcal Inducton Proof. If n, the result s mmedate. Assume for some n > we have n F F n. Then n F F n n F F n F n F n F n as desred. 9. Prove that n 0 n n n 3 3 n 0 n Fn. Proof. Strong Inducton For n ths s F F. Thus the asserton s true when n. Now fx n and assume that k 0 k k k 3 3 k 0 k Fk whenever k < n. In what follows we use the dentty n k n k n k. We also often use 0 whenever t s untrue that 0 b a. a b n n n 0 0 n n n n n 0 n n n n n n 3 n n n n n n n 3 n n n [ ] [ ] n n 0 n n n 0 n F n F n F n Ths completes the proof. 3. Prove that n k k0 r n r, where r N. Hnt: Use nducton on the nteger n. After dong the bass step, break up the expresson k r as k r k r k r. Then regroup, use the nducton hypothess, and recombne usng the above dentty. 33. Suppose that n nfntely long straght lnes le on the plane n such a way that no two are parallel, and no three ntersect at a sngle pont. Show that ths arrangement dvdes the plane nto n n regons. Proof. The proof s by nducton. For the bass step, suppose n. Then there s one lne, and t clearly dvdes the plane nto regons, one on ether sde of the lne. As n n, the formula s correct when n. Now suppose there are n lnes on the plane, and that the formula s correct for when there are n lnes on the plane. Sngle out one of the n lnes on the plane, and call t l. Remove lne l, so that there are now n lnes on the plane.

8 Solutons for Chapter 339 By the nducton hypothess, these n lnes dvde the plane nto n n l regons. Now add 5 lne l back. Dong ths adds an addtonal n regons. The dagram llustrates the 3 case where n 5. Wthout l, there are n lnes. Addng l back produces n 5 new regons. Thus, wth n lnes there are all together n n n regons. Observe n n n n n n n n. Thus, wth n lnes, we have n n regons, whch means that the formula s true for when there are n lnes. We have shown that f the formula s true for n lnes, t s also true for n lnes. Ths completes the proof by nducton. 35. If n, k N, and n s even and k s odd, then n k s even. Proof. Notce that f k s not a value between 0 and n, then n k 0 s even; thus from here on we can assume that 0 < k < n. We wll use strong nducton. For the bass case, notce that the asserton s true for the even values n and n : ; ; 3 even n each case. Now fx and even n assume that m k s even whenever m s even, k s odd, and m < n. Usng the dentty n k n k n k three tmes, we get n k n n k k n n n n k k k k n n n. k k k Now, n s even, and k and k are odd. By the nductve hypothess, the outer terms of the above expresson are even, and the mddle s clearly even; thus we have expressed n k as the sum of three even ntegers, so t s even.