# How can I improve this proof?

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?

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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

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 $da \mid db$.

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that's probably a better approach, thanks – jsj Dec 19 '12 at 8:11