I'm learning some basic number theory from Strayer's 'Elementary Number Theory.' I've arrived at what seems to be a very basic problem, albeit complete with a nasty twist:
Let $a,b,c \in \mathbb Z$ with $c \neq 0$. Prove that $a\mid b$ if and only if $ac\mid bc$.
The proof is trivial when $c>0$.
What has me confused is the case when $c<0$. If $a\mid b$, then $b=aq$ where $q \in \mathbb Z$ and $a>0$. Multiplication by $c$ on $b=aq$ yields $bc=(ac)q$, where clearly $ac<0$.
This is my problem. The 'division algorithm,' as it's been taught in the early stages of this book (and number theory in general) doesn't allow for the divisor to be negative. If we fix $c$ to $q$ by saying $bc=a(cq)$, we avoid the issue of a negative divisor, but then we can only say $a|bc$. Similarly, if we negate the negative quality of $c$, perhaps by saying $bc=(-ac)(-q)$, then still we can only say $-ac\mid bc$.
What am I missing here? Surely Strayer meant what he said - that is, $c \neq 0$ - and not something that would make more sense like $c>0$. Any help would be greatly appreciated!