Let a,b,c be integers. Prove that if a|c and b|c, then either a|b or b|a.

Question

Let a,b,c be integers. Prove that if a|c and b|c, then either a|b or b|a.

Any ideas? (Suggested proof by contradiction). Not really sure how to go about this.

Answer 1

The statement is, in fact, false.

To prove a statement is false, you need only provide one counter-example. (One counter-example suffices, because the claim is being made about any/all integers $a, b, c$ such that $a\mid c$ and $b\mid c$. So if you find any particular triplet such that $a, b, c$ are integers and $a\mid c$ and $b\mid c$, but it is not the case that $a\mid b$ or $b\mid a$, the claim is in fact false.)

E.g., Suppose $a = 3, b=4, c= 12$. Then indeed, $3\mid 12\, $ and $\,4\mid 12,\,$ but $3$ does not divide $4$, nor does $4$ divide $3$.