Prove by contraditcion that $a\mid b$ then $ac\mid bc$

Question

Prove by contradiction if $a\mid b$ then $ac\mid bc $ for $a,b,c\in\Bbb Z$

I am having considerable difficulty with this problem. Could someone give me a step-by-step solution? Even though the problem seems simple, I cannot solve it. I am new to discrete maths: I only know direct proof and contraposition (I am trying to work on contradiction). All I know is that $a\mid b$ if for some other number $c$, we have $b = a\times c$.

Answer 1

Start with not ac | bc. We wish to show not a | b.
Assume a|b. Easily prove ac | bc, a contradiction
from the starting premise. Thus not a|b.

It is pointless to prove this by contradiction
as a direct proof is easy and simple.

Answer 2

Assume $c>0 $, suppose on contrary $\forall k\in \mathbb Z$ $$ac\neq k bc.$$ Let $ac>kbc$. Then $a>kb$. similarly for $ac<kbc$. consider the other case also $c<0$. In each case we get $a $ not divides $b$.