Are these proofs correct? Answer either Yes or No for each. Claim: all natural numbers are either even or odd.Proof: By induction. Base case: 0 is even, since 0 = 2· 0.Now suppose that n – 1 is either even or odd, and we must show thatn is either even or odd.Case 1: If n – 1 is even, then n – 1= 2k for some k, so n = 2k + 1, showing n must be odd.Case 2: If n – l is odd, then n – 1 = 2k + 1 for some k, so n2k + 2 = 2(k +1), showing n must be even. Claim: all integers n > 4 can be written as a sum of two or more prime numbers (possibly repeated).Proof: By strong induction. Base case: for n = 4, the sum 2 + 2 works.For the inductive step, assume it's true for all k < n, and we'll show it holds for n. Consider k = n – 2. By the inductive hypothesis, this must equal a sum of primes pi +p2 +...+ Pm. Then we can add 2 to this to get a sum of primes that equals n.

Name: Are these proofs correct? Answer either Yes or No for each. Claim: all
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Description: Are these proofs correct? Answer either Yes or No for each. Claim: all natural numbers are either even or odd.Proof: By induction. Base case: 0 is even, since 0 = 2· 0.Now suppose that n – 1 is either even or odd, and we must show thatn is either eve

1) Yes. The proof of first claim is correct. Here use 1st principal of mathematical induction.

Math

Advanced Math

Claim: all natural numbers are either even or odd. Proof: By induction. Base case: 0 is even, since 0 = 2 · 0. Now suppose that n – 1 is either even or odd, and we must show that n is either even or odd. Case 1: If n – 1 is even, then n – 1 = 2k for some k, so n = 2k + 1, showing n must be odd. Case 2: If n – 1 is odd, then n – 1 = 2k + 1 for some k, son = 2k + 2 = 2(k + 1), showing n must be even.

Claim: all natural numbers are either even or odd. Proof: By induction. Base case: 0 is even, since 0 = 2 · 0. Now suppose that n – 1 is either even or odd, and we must show that n is either even or odd. Case 1: If n – 1 is even, then n – 1 = 2k for some k, so n = 2k + 1, showing n must be odd. Case 2: If n – 1 is odd, then n – 1 = 2k + 1 for some k, son = 2k + 2 = 2(k + 1), showing n must be even.

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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