0
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ What are you working in? $\mathbb{Z}$? Are $a$ and $b$ integers? $\endgroup$ – Zubin Mukerjee Sep 3 '15 at 17:51
  • $\begingroup$ It doesn't matter, the statement is true in any ring with 1. $\endgroup$ – vadim123 Sep 3 '15 at 18:02
  • $\begingroup$ 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. $\endgroup$ – user265328 Sep 3 '15 at 18:14
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – Zubin Mukerjee Sep 3 '15 at 17:53
  • $\begingroup$ 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! $\endgroup$ – 123 Sep 3 '15 at 18:04
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.