$(a\mod m)/(b\mod m) = (a/b)\mod m$? 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.
 A: 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.
A: 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 $ab^{-1}\pmod m$.
A: It depends what the symbols mean. Most commonly, the $\mod m$ notation is used together with the $\equiv$ symbol meaning congruence, as in $15\equiv1\mod 7$.
However, you can also note that congruence modulo $m$ is an equivalence relation, and define $\mathbb{Z}_m$ as the set of congruence classes modulo $m$. Algebraically speaking, you get a ring, and you might write $a\bmod m$ for the equivalence class of $a$ modulo $m$. In which case, it is true by definition that $(a\bmod m)(b\bmod m)=ab\bmod m$, and you can then get the identity you ask for by dividing by $b\bmod m$ – provided that this is an invertible element of $\mathbb{Z}_m$, which is equivalent to $\gcd(b,m)=1$.
Finally, and this is somewhat more low-brow, you can define $a\bmod m$ to be the particular member of the congruence class of $a$ which is in $\{0,1,\ldots,m-1\}$. In which case, your conjecture is dead wrong, as already indicated by others.
