Prove that if $d|a$, then $d||a|$

Question

I have no idea where to take this. It says to consider both cases of $d|a$ and $d|-a$, but I don't how to prove that.

Answer 1

$|a| = a$ if $a\geq 0$. $|a| = -a$ if $a \lt 0$.

Now, is it true that if $d\mid a$ then $d \mid a$?

And is it true that if $d\mid a$, then $d\mid -a$?

If you can justify "yes" for both questions, you are done. Hint: use the definition of $a \mid b$: If $a\mid b$, then $b = ka$ for some integer $k$.

Answer 2

The definition of divisibility is that $d$ divides $a$ if there exists an integer $q$ such that $a = dq$.

For $|a|$, there are two cases: $|a| = a$ and $|a| = -a$. For each case, can you find an integer $p$ such that $|a| = d p$?