if $\displaystyle ac\equiv bc \ \mbox{(mod m)}$, then $\displaystyle a\equiv b \ \mbox{(mod m)}$. $\displaystyle a,b,c\in\mathbb{Z} \ \mbox{and} \ m\in\mathbb{Z}^+$

$\displaystyle m|(ac-bc)\rightarrow mx=(a-b)c\rightarrow m\left(\frac{x}{c}\right)=(a-b)$

This false and only true if $\displaystyle c\neq 0$, correct?