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For every prime number $p$, either $p=2$ or $p$ is odd. We can rephrase this: for every prime number $p$, if $p \neq 2$, then $p$ is odd. Now try to prove it.
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Proof.

Let $p$ be an arbitrary prime number. Assume $p$ is not odd. So $p$ is divisible by 2 . Since $p$ is prime, it must have exactly two divisors, and it has 2 as a divisor, so $p$ must be divisible by only 1 and 2. Therefore $p=2$. This completes the proof (by contrapositive).

by Diamond (38,951 points)

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