Prove if $ac>bc$ and $c>0$, then $a>b$

Question

How can I prove that if $ac>bc$ and $c>0$, then $a>b$ without division? Or Should I prove it by contrapositive?

Answer 1

of $ac>bc$, it follows that $ac-bc>0$ and $(a-b).c>0$. Knowing that the data numbers $x,y \in \mathbb{Z}$, such that $x.y>0$, then $x$ and $y$ are either positive or negative, i.e, $x,y>0$ or $x,y<0$.

By hypothesis, $c>O$, and so $a-b>0$. Therefore, $a>b$, which solves the problem.

Answer 2

$$ ac> bc $$ $$ c(a-b) \gt 0 $$

$$ \text{two cases: } (c\gt 0 \quad \text{ and } \quad a-b\gt 0) \quad \text{ or } \quad (c \lt 0 \quad \text{ and } \quad a-b \lt 0) $$

and because $$c \gt 0$$ then $$a-b\gt 0$$