Show that: $a·c \equiv b·c\ (\text{mod }m)$ with $a, b, c$ and $m$ integers with $m \ge 2$ does not imply $a \equiv b\ (\text{mod }m)$ Show that $a·c\equiv b·c\ (\text{mod }m)$ with $a, b, c$ and $m$ integers with $m \ge 2$ does not imply $a \equiv b\ (\text{mod }m)$

I've seen many similar examples, but can't seem to find a step by step explanation or solution to any of them (like an actual proof).
Here is a counter example. $a=b+1$ , $c=m$
$2\cdot2\equiv 2\cdot5(mod\ 2)$ but $2\not\equiv 5(mod\ 2)$