Exercise 16.

What is wrong with the following “proof” of the “fact” that for all \(n \in \N\text{,}\) the number \(n^2 + n\) is odd?

Proof.

Let \(P(n)\) be the statement “\(n^2 + n\) is odd.” We will prove that \(P(n)\) is true for all \(n \in \N\text{.}\) Suppose for induction that \(P(k)\) is true, that is, that \(k^2 + k\) is odd. Now consider the statement \(P(k+1)\text{.}\) Now \((k+1)^2 + (k+1) = k^2 + 2k + 1 + k + 1 = k^2 + k + 2k + 2\text{.}\) By the inductive hypothesis, \(k^2 + k\) is odd, and of course \(2k + 2\) is even. An odd plus an even is always odd, so therefore \((k+1)^2 + (k+1)\) is odd. Therefore by the principle of mathematical induction, \(P(n)\) is true for all \(n \in \N\text{.}\)

Solution.
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