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

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!

• What are you working in? $\mathbb{Z}$? Are $a$ and $b$ integers? – Zubin Mukerjee Sep 3 '15 at 17:51
• It doesn't matter, the statement is true in any ring with 1. – vadim123 Sep 3 '15 at 18:02
• write a = a_1 * a_2 * a_3 * ... * a_n. Since b|a, b = a_(i_1) * a_(i_2) * .. a_(i_k). Now write, a = (-1)*(-1)*(prime factors of a). Write -b = (-1)*(facors of b). Now, its clear I suppose. – user265328 Sep 3 '15 at 18:14

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.

• It helps a lot to show that for any $x \in \mathbb{Z}$, $-x = -1 \cdot x$ ... that also lets you show that $-(-x) = x$. EDIT: OP hasn't said that $a$ and $b$ are integers, but assuming they are, your answer is a good one, +1 – Zubin Mukerjee Sep 3 '15 at 17:53
• Yes, you make a good point. I made the assumption that OP is working with integers since that is the most common thing for such introductory courses. I edited the answer to make that assumption explicit. Thanks for the feedback! – 123 Sep 3 '15 at 18:04

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.