Prove that $a \mid b \iff ac \mid bc$, provided $c \neq 0$.
We prove two statements:
(i) $a \mid b \implies ac \mid bc$, provided $c \neq 0$
(ii) $ac \mid bc \implies a \mid b$, provided $c \neq 0$
(i)
$a \mid b$ means there exists an integer $k$ such that
$$b = ak$$
Multiplying both sides by $c$ gives us
$$bc = ack$$
Which means $ac \mid bc$.
(ii)
$ac \mid bc$ means there exists an integer $k$ such that
$$bc = ack$$
Dividing both sides by $c$, which can't be 0, gives us
$$b = ak$$
Which means $a \mid b$.
Both results (i) and (ii) gives us $a \mid b \iff ac \mid bc$, provided $c \neq 0$.