$b$ divides $a \Leftrightarrow -b$ divides $a$

Question

Prove that $b$ divides $a$ if and only if $-b$ divides $a$.

I'm thinking something like $a = bp$ and $b = aq$, then go on from there? It seems simple enough but thanks for the help in advance!

Answer 1

Hint:

I assume you're working in $\mathbb{Z}$

You must prove both directions. I suggest starting with: $$b|a \to -b|a$$

next you must show $$-b|a \to b|a$$

for example,

Proof:

Let $a,b,\phi \in \mathbb{Z} s.t. b|a$ Then we have that: $$a=b \phi$$

Then we keep in mind that if: $$\frac{a}{\phi} \in \mathbb{Z}$$

then $$-\frac{a}{\phi} \in \mathbb{Z}$$

That should be enough to see how to finish this direction. The other direction is roughly the same.

Answer 2

In all integral domain A, if $b|a$ then $a\in bA$(principal ideal generated by b) and since $bA= (-b)A$ it is clear that $(-b)|a$ too. With $-(-b)=b$ we finish.