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Our teacher went over the following proof but didn't post the notes on it. I'm trying to understand how it checks out, could someone explain it without using contraposition.

For any integers a,b,p, if p is prime and p ∣ ab, then p ∣ a or p ∣ b.

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If $p \mid a$ you are finished, otherwise $\gcd(p,a) = 1 $ and there are $x,y$ such that $px+ay = 1$. Then $pbx+aby = b$ and since $p \mid pbx$ and $p \mid (ab) y$, we see that $p \mid b$.

copper.hat
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