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.

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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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