How would you go about proving this without using its contrapositive?

Question

Let a be an integer. Prove that if 3|2a , then 3|a.

I'm trying to think of a way to solve this with a direct proof, or at least without using a contrapositive which would be too easy. I considered trying a proof by cases where a was even and a was odd and that didn't work at all. Would a proof by cases where 3|a, 3|a+1, and 3|a+2 work? I could show when 3|a+1, and 3|a+2, it doesn't divide 2a.

Also I'm curious, is there a good way to introduce a lemma to solve this?

Answer 1

Many ways to do this. One approach: If ${3|2a}$ then $2a=3x$ where $x$ is some integer.(Factor-Multiple relationship) Actually, $x$ is not just some integer, here it comes...an EVEN integer. Do you know why even? Ok, and if $x$ is even then we can write $x=2t$ where $t$ is any integer. Conclusion?

Answer 2

Assume $3\mid 2a$.   That is equivalently that $\exists k\in\Bbb N:3k=2a$.   For that $k$, then $a= k+k/2$ and because $a\in\Bbb N$, therefore ...rhubarb rhubarb rhubarb.... and thus $3\mid a$.

Therefore $\forall a\in\Bbb N:[(3\mid 2a) \to (3\mid a)]$