Assignment 6

# Assignment 6 - Assignment#6 Section 2.4 9b Use PMI to prove...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

{[ snackBarMessage ]}

### Page1 / 4

Assignment 6 - Assignment#6 Section 2.4 9b Use PMI to prove...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online