# Let a,b and c be real numbers where a>b. Prove that if ac<=bc, then c<=0.

Let $a,b$ and $c$ be real numbers where $a>b$. Prove that if $ac\le bc$, then $c\le 0$.
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Liam Everett
$bc\ge ac⟺c\left(b-a\right)\ge 0⟺-c\left(a-b\right)\ge 0$
Since $a>b⟺\left(a-b\right)>0$ , so $c\le 0$.