I am trying to find a way to prove that $b|a$ if and only if the remainder of $a$ divided by $b$ is $0$.
So far this is what I have:
If $r$ is the remainder of $a$ divided by $b$, this means $a=bq+r$. Subsequently, if $r=0$ then $a=bq$ which means $b|a$ by definition.
What I can't figure out is how to prove the if and only if part. I'm not sure if its better to start from $b|a$ and prove that $r=0$ or try to prove that if $r>0$ then $b \nmid a$. Either way I have had trouble with the steps to make the proper conclusion.
In the other direction, you have $b|a$, and you want to prove that the reminder $r$ is 0.
$b|a$, therefore $a = bq$ for some $q$. On the other hand, $a = bp + r$ for some $p$ and some $0 \le r < b$.
Therefore, $bq = bp + r$, so $b(q-p) = r$. If $q < p$, then $r < 0$, contradiction. On the other hand, if $q > p$, then $r > b$, again contradiction. So we have $q=p$, therefore $r=0$.
Also, if you have already proven the uniqueness of the reminder (which goes pretty much the same as what I have above), you can simply show that 0 works as reminder, so that must be it.