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Unformatted text preview: Assignment #6 Section 2.4 9b) Use PMI to prove that 2^n > n^2 for all n > 4. Proof: Define S = {n in natural number  2^n > n^2 and n > 4} Since 32 > 25, 2^5 > 5^2. Thus, 5 is in S. Suppose k is in S. Then 2^k > k^2 and k > 4. Since k > 4, k^2 > 4k = 2k + 2k > 2k + 1. Then, 2^(k+1) = 2^k + 2^k > k^2 + k^2 > k^2 + 2k + 1 = (k+1)^2. Since k+1 > 4 and 2^(k+1) > (k+1)^2, k+1 is in S. Since k was arbitrary, (\forall natural number k > 4)(k is in S => k+1 is in S). Therefore, by PMI, S = N  {1,2,3,4} Finally, we conclude that 2^n > n^2 for all n>4. 12) a)Proof by PMI Base Case: n=2, 2 > 1 ==> a decreasing sequence of 2 natural number. Induction Step: Assume the statement is true for n=k, that means there is a decreasing sequence of k natural number(a1, a2,...,ak). Now consider the case where n=k+1, Since k+1 > k, then k+1 > a1 > a2 >... > ak ===>a decreasing sequence of k+1 natural numbers. DONE. 15) a) F. Fails for n+1=2. b) F. Never proves statement to be true for (n+1) c) F. n=1 is odd, but 1^1+1 is even. d) F. The induction step assumes n+1 is in S, but you must assume n is in S and then show n+1 is too. e) F. Factorization of xy + 1 is incorrect, and there is no reason to assume that x + 1 or y + 1 is prime. Section 2.5 1) Proof by PCI Base case: n = 4: 4=2(2)+5(0) n = 5: 5=2(0)+5(1) Induction Step: Assume the statement is true for n=1,...,k.>k = 2x+5y for some integers x,y....
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 Winter '10
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