Prove that if $d|a$, then $-d|a$ and $d|-a$ 
Prove that if $d|a$, then $-d|a$ and $d|-a$
Additionally $d| |a|$.

I wouldn't mind someone going over the correctness of my proof if they have time please.
If $d|a$, then there is some integer $k$, such that $dk=a$
Also $-d(-k)=a$ so $-d|a$
Also $d(-k)=-a$ so $d|-a$
$d|0$, so that $d||a|$.
Apart from potentially missing something in logic, I'm sure this could be written in correct mathematical english a little better if someone has some input that would be appreciated.
Thanks
 A: Your argument is just fine (most of the way). There's no neater way, no better "mathematical English", than your case by case treatment.
For $d | |a|$ you should do cases too: the absolute value of $a$ is always either $a$ or $-a$ so your first argument covers it.
No need to treat $0$ as a special case. Indeed, everything divides $0$.
A: Your first three are absolutely perfect. However your last $d|0$ therefore $d||a|$ is utterly incomprehensible and makes no sense to me whatsoever. What does dividing into zero (which every number does) have to do with dividing into the absolute value of a. As every number divides into zero you can not use it to prove anything that isn't true for every number and not all numbers divide $|a|$ so $d|0$ can't prove $d||a|$ any more than $1847|0$ can be used to prove $1847|256$.
Instead simply say, As $|a| = \pm a$ and we have been given $d|a$ and we have proven $d|-a$ therefore $d||a|$ no matter whether $|a| = a$ or $|a| = -a$.
.....
Oh... I see.  You were assuming it was clear you were doing three cases.  If $a > 0$ then then $d|a \implies d||a|$.  If $a < 0$ then $d|-a \implies d||a|$ and if $a = 0$ then $d|0 \implies d||a|$.  That would have been fine except it wasn't clear that that was your intent.  
At least,  I completely missed it.
But you don't need to specifically show the case for $a=0$ because if $a = 0$ then $|a| = a$ and that was covered in the given case.
