Prove that $(a \cdot b) \text{ mod } n = ((a \text{ mod } n) \cdot (b \text{ mod } n)) \text{ mod } n$ 
Prove for all $n \in \mathbb{N}, n >1$ and $a,b \in \mathbb{Z}$ that
  $$(a \cdot b) \text{ mod } n = ((a \text{ mod } n) \cdot (b \text{ mod
} n)) \text{ mod } n$$

I'm not sure if this is a proof but I tried another way this time:
Let $x=a \text{ mod } n$, let $y= b \text{ mod } n$ where $x,y \in \mathbb{Z}$
$\Rightarrow$
$(x \cdot y) \text{ mod } n= (a \cdot b) \text{ mod } n$
Create a "matching" $p,q \in \mathbb{Z}$ such that:
$a=x+pn$ and $b=y+qn$
$\Rightarrow ((x+pn)(y+qn)) \text{ mod } n= (xy+xqn+ypn+pqn^{2}) \text{ mod } n=(xy+n(xq+yp+pqn)) \text{ mod } n = (x \cdot y) \text{ mod } n$

I hope this is correct and counts as proof?

Suppose that:
$$\begin{cases}
a \mod n = r_a\\
b \mod n = r_b\\
(a \cdot b) \mod n = r\\
\end{cases}$$
or equivalently
$$\begin{cases}
a = q_a \cdot n + r_a\\
b = q_b \cdot n + r_b\\
a \cdot b = q \cdot n + r\\
\end{cases}.$$
Then:
$$r_a \cdot r_b = (a-q_a \cdot n)\cdot (b-q_b \cdot n) = \\
= a \cdot b - a \cdot q_b \cdot n - b \cdot q_a \cdot n + q_a \cdot q_b \cdot n^2 = \\
= q \cdot n + r - a \cdot q_b \cdot n - b \cdot q_a \cdot n + q_a \cdot q_b \cdot n^2 = \\
= (q + q_a \cdot q_b \cdot n - a \cdot q_b - b \cdot q_a) \cdot n + r.$$
In other words:
$$r_a \cdot r_b = k \cdot n + r \Rightarrow \\
(r_a \cdot r_b) \mod n =  r \Rightarrow \\
r = (a \cdot b) \text{ mod } n = ((a \text{ mod } n) \cdot (b \text{ mod
} n)) \text{ mod } n .$$
