Changing modulus in modular arithmetic Is it true that
$$a\equiv b\pmod{m}\implies\frac{a}{n}\equiv\frac{b}{n} \pmod{\frac{m}{n}},$$ where $a, b, m, n, \frac{a}{n}, \frac{b}{n}, \frac{m}{n}\in\mathbb{N}$? If so, how do I prove it?

From the definition of modulo:
$$a \equiv b \pmod m \implies a=km+b \tag{1}$$
$$\frac{a}{n} \equiv {\frac{b}{n}} \pmod {\frac{m}{n}} \implies \frac{a}{n}=k \frac{m}{n} + \frac{b}{n} \tag{2}$$
Observe $(2)$. 
\begin{align}
\frac{a}{n}&=k \frac{m}{n} + \frac{b}{n} \tag {2} \\
\frac{a}{n} \cdot n&=k \frac{m}{n} \cdot n + \frac{b}{n} \cdot n \\
a & = km+b \tag{1}
\end{align}
By multiplying $(2)$ by $n$, we get $(1)$, so the expressions are the same. Hence $a \equiv b \pmod m \implies \frac{a}{n} \equiv \frac{b}{n} \pmod{ \frac{m}{n}}$
