hw6sol - H om e w o r k 6 Sol u t ion P r o b l e m 1 Se c...

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Unformatted text preview: H om e w o r k 6 Sol u t ion P r o b l e m 1 Se c t io n 4.1, E x e r c ise 4, p age 279 a) b) Both sides of P(1) shown in part (a) equal 1. c) d) For each, k 1 that P(k) implies P(k+1) e) ! f) We have completed both the basis step and inductive step so by the principle of mathematical induction the statement is true for every positive integer n. P r o b l e m 2 Se c t io n 4.1, E x e r c ise 6, p age 280 Proof) Let P(n) be 1)Basic step: P(1) is true because 1 1!=1=(1+1)!-1 2) Inductive step: Assume that P(k) is true. Then, (1 1!+2 2!+ +k k!)+(k+1)(k+1)!=(k+1)!-1+(k+1)(k+1)!=(k+1)!(1+k+1)-1=(k+2)(k+1)!-1 =(k+2)!-1=P(k+1) We have completed both the basis step and inductive step so by the principle of mathematical induction the statement is true for every positive integer n. P r o b l e m 3 Se c t io n 4.1, E x e r c ise 14, p age 280 Proof) Let P(n) be 1) Basic step: P(1)=1 21=(1-1)21+1+2=2 2) Inductive step: Assume P(k) is true. Then, (1 21+2 22+ +k 2k)+(k+1) 2k+1=(k-1) 2k+1+2+(k+1) 2k+1 =2k+1 (k-1+k+1)+2=2k 2k+1 +2=k 2k+2 +2=P(k+1) We have completed both the basis step and inductive step so by the principle of mathematical induction the statement is true for every positive integer n. P r o b l e m 4 Se c t io n 4.2, E x e r c ise 4, p age 291 a) P(18) is true, because we can form 18 cents of postage with one 4-cent stamp and two 7-cent stamps. P(19) is true, because we can form 19 cents of postage with three 4-cent stamps and one 7-cent stamp. P(20) is true, because we can form 20 cents of postage with five 4-cent stamps. P(21) is true, because we can form 21 cents of postage with three 7-cent stamps. b) Inductive hypothesis: the statement that using 4-cent and 7-cent stamps we can form j cents postages for all j with 18 j k, where we assume k 21. c) Assuming the inductive hypothesis, we can form k+1 cents postage using just 4-cent and 7cent stamps. d) Because k 21, we know that P(k-3) is true, that is, that we can form k-3 cents of postage. Put one more 4-cent stamp on the envelope, and we have formed k+1 cents of postage. e) We have completed both basis step and the inductive step, so by the principle of strong induction, the statement is true for every integer n greater than or equal to 18. P r o b l e m 5 Se c t io n 4.2, E x e r c ise 12, p age 292 Proof) 1)Basic step: 1 = 20, 2 = 21, 3 = 21 + 20, 4 = 22, 5 = 22 + 20, and so on. 2) Inductive step: Assume the inductive hypothesis that every positive integer up to k can be written as a sum of distinct powers of 2. We must show that k + 1 can be written as a sum of distinct powers of 2. If k + 1 is odd, then k is even, so 20 added. If k + 1 is even, then (k + 1)/2 is a positive integer, so by the inductive hypothesis (k + 1)/2 can be written as a sum of distinct powers of 2. Increasing each exponent by 1 doubles the value and gives us the desired sum for k + 1. P r o b l e m 6 Se c t io n 4.2, E x e r c ise 30, p age 293 In the inductive step, when k=0, a0=1 for all nonzero a, but a1 is not one. In the denominator, we have , but -1 is negative, so it's not covered under our inductive assumption. Therefore, the inductive step fails. ...
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This note was uploaded on 11/05/2011 for the course CSCE 222 taught by Professor Math during the Spring '11 term at Texas A&M.

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