$(a\mod m)/(b\mod m) = (a/b)\mod m$? b and m are relatively prime (m is prime and $b \in \mathbb Z_m^* $). In truth, I would like to be able to get to the following point (it is a simplified example):

$\frac{ab \mod m}{b \mod m} = a \mod m$

Is it possible? 



EDIT

My apologies. Thanks to Henning Malkholm I noticed I had used the wrong notation. I actually meant this:

$(ab \mod m)(b^{-1} \mod m) \equiv abb^{-1} \mod m$

$abb^{-1} \mod m \equiv (a \mod m)(bb^{-1} \mod m) \equiv (a \mod m)$

Is this valid? Again, I apologise if this seems too silly. I just don't feel secure enough to trust what I've found.
Let $a = 3, b = 4, m = 5$. Then
$$ab \mod m = 12 \mod 5 = 2$$
$$b \mod m = 4$$
$$a \mod m = 3$$
Thus, you can see that your assumption is not true.
It is possible, but it is not necessarily true.

For example, take $m=11$, $a=4$, $b=3$. Then $ab\bmod m=1$, $a\bmod m=4$ and $b\bmod m=3$, yet
$$ \frac{1}{3} \ne 4 $$



In order to extend division to modular arithmetic, what you need to do is find the modular inverse $b^{-1}$ of $b$, defined as the solution to
$$ xb \equiv 1 \pmod m$$
which is $3$.

Then you can let "$a$ divided by $b$, modulo $m$" mean $ab^{-1}\pmod m$.