Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to write a proof that $a \vert b$ if and only if $da \vert db$. This is what I have so far:

$da \vert db$ if and only if $\lfloor db / da \rfloor = db / da$

$\frac{db}{da} = \frac{b}{a}$ and therefore $\lfloor b / a \rfloor = b / a$

consequently $a \vert b$

I feel like this addresses the if but not the and only if part of the initial statement. Where do I go from here?

share|cite|improve this question
What is the definition of "$a|b$" you are using? – Andrés E. Caicedo Dec 19 '12 at 8:05
@AndresCaicedo Fixed the typos. The definition I am using is that $a \vert b$ if $b \mod a = 0$ – jsj Dec 19 '12 at 8:09
Then you better use the definition in your argument, because at the moment, it looks like the definition is $\lfloor b/a\rfloor =b/a$. If this is something that has been previously established, you may want to mention that: "Recall that we previously showed ..." Anyway, you can reorganize your argument: – Andrés E. Caicedo Dec 19 '12 at 8:13
$db/da=b/a$, so also $\lfloor b/a\rfloor =\lfloor db/da\rfloor$. It follows that $da|db$ if and only if $\lfloor db/da\rfloor=db/da$ if and only if $\lfloor b/a\rfloor =b/a$ if and only if $a|b$. This presentation covers both directions of the equivalence directly. (It worries me that your approach assumes $a,d$ are different from zero.) – Andrés E. Caicedo Dec 19 '12 at 8:15
@AndresCaicedo My definition is based on the fact that $a \mod b = a-b \lfloor a/b \rfloor$. If $a-b \lfloor a/b \rfloor = 0$ then $\lfloor a/b \rfloor = a/b$... You are right though that that part of the proof definately needs to be in there. Thanks for your help. I might end up using Daniel's approach anyway, even though somehow the zero remainder definition makes more sense to me. – jsj Dec 19 '12 at 8:30
up vote 1 down vote accepted

For $a$ and $b \in \mathbb{Z}$ we have that $a \mid b$ if there is a $k \in \mathbb{Z}$ such that:

$$b = ak $$

Therefore, if $a \mid b$ then $b = ak$ and so $db = dak$. Thus, $da\mid db$.

If on the other hand, we begin with the assumption that $da \mid db$, that is:

$$db = dak$$

for some $k \in \mathbb{Z}$. Then the fact that $a \mid b$ follows by dividing by $d$ as long as $d \neq 0$.

Edit: The other direction is essentially your original proof backwards.

If $a \mid b$ then $\lfloor{db/da}\rfloor =\lfloor{b/a}\rfloor= b/a = db/da$ and thus $da \mid db$.

share|cite|improve this answer
that's probably a better approach, thanks – jsj Dec 19 '12 at 8:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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