Proof that $a|b \iff ac|bc$ I have to prove, using the concept of divisibility (and no division), that if $c\ne 0$ then $a\mid b \leftrightarrow ac\mid bc$. I first proved $a \mid b \to ac\mid bc$, then got stuck on the opposite process. Here's what I did on the first part (please assume all numbers to be integers):
$$a\mid b$$
$$b=ka$$
$$bc=kac$$
$$ac\mid bc$$
Is this right? 
As for $ac\mid bc \implies a\mid b$, I can't figure out a way to remove $c$ from $b$ without messing everything on $a$'s side. 
Googling for an answer, I only found solutions that used division itself, which was of no help. Thanks in advance for your help!
 A: Although this is a bit of a cheat (it's barely avoiding division), you could do the following:
Sketch:
Suppose that $ac\mid bc$.  Then there is some integer $k$ so that $ack=bc$.  Therefore, by factoring $c(ak-b)=0$.  Since $c\not=0$, you can conclude something about the other factor...
Can you take it from here?
A: Let $ ac | bc $. Then there is some $ k \in \mathbb{Z} $ such that $ ack = bc $. Then $ (ak - b)c = 0 $. As $ c \neq 0 $ by assumption, we have $ ak - b = 0 $, so $ ak = b $, so $ a | b $.
The property of $ \mathbb Z $ that $ ab = 0 $ implies $ a = 0 $ or $ b = 0 $ is essential to this result, so hopefully using this is allowed.
A: You want to go through your equivalences backwards and of course, you need to divide by $c$. So, the actual question is: How to write the division without calling it division?
$bc=kac \Leftrightarrow c(b-ka)=0$, so since $c$ is not zero you get $b=ka$. 
A: Strong Hint:$$\begin{align}ac&=bc\\ \Leftrightarrow\quad \bigg(\frac 1c\bigg)ac&= \bigg(\frac 1c\bigg)bc \\ \Leftrightarrow\space\, \bigg(\frac 1c\cdot c\bigg)a&= \bigg(\frac 1c\cdot c\bigg)b \\ \Leftrightarrow\quad\quad\:\, 1\cdot a&=1\cdot b \\ \Leftrightarrow\quad\quad\quad\,\,\, a&=b\end{align}$$ This is known as the $\color{green}{\text{Multiplicative Cancellation Law}}$.
