TMUA-P2-PRF3-009 — Paper P2 · Proof by contradiction · Level 2

Options

1. A
   Suppose $n$ is even. Then $n = 2k$ and $n^2 = 4k^2 = 2(2k^2)$ is even, as required.
2. B
   Suppose, for contradiction, that $n$ is odd. Then $n = 2k+1$ and $n^2 = 2(2k^2+2k)+1$ is odd, contradicting the hypothesis that $n^2$ is even. Hence $n$ is even.
3. C
   Suppose, for contradiction, that $n^2$ is odd. Then the claim holds vacuously.
4. D
   Check $n = 4$: $n^2 = 16$ is even and $n$ is even, so the claim holds generally.

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