Prove that $m\mid a$ if and only if $a \equiv 0 \pmod{m}$.

Question

Prove that $m\mid a$ if and only if $a \equiv 0 \pmod{m}$.

This is what I have thus far:

Proof: Let $a$ be in the set of integers such that $a=0+km$, where $k$ is an integer. Then by the definition of divides, $m\mid a$. Since $m\mid a$, by the definition of congruency $a\equiv 0 \pmod{m}$.

I feel like I am missing something…

Answer 1

We have $$\begin{align} m\mid a &\iff a=nm \text{ for some }n\in \Bbb Z \\ &\iff a-0=nm\text{ for some } n\in\Bbb Z \\ &\iff m\mid (a-0) \\ &\iff a\equiv 0\pmod{m}. \end{align}$$