Direct Proof - $ac\leq bc \Rightarrow c \leq 0$ Hello Mathematics Stackexchange Community,
currently I am trying to proof following by direct proof only:
Suppose $a,b$ and $c$ $\in$ $\mathbb{R}$ and $a>b$
Proof that if $ac \leq bc$ $\Rightarrow$ $c\leq 0$
$\\$
My attempt:
Suppose $ac \leq bc$ and $a>b$ 
It follows that: $bc-ac>0$
Therefore: $b>a$
But because we assumed that $a>b$ it follows that $ac>bc$
Therefore: $c>0$
So if $ac\leq bc$ $\Rightarrow c\leq 0$.
Somehow I have the feeling that my proof is wrong.
It would be nice if someone would verificate my proof and hint at errors and/or give me advice regarding my proof.
Sincerely,
Aquila

Your proof is wrong: how do you jump from $bc-ac>0$ to $b-a>0$?
Suppose that $a>b$ and that $ac\leqslant bc$. This is equivalent to asserting that $a-b>0$ and $ac-bc\leqslant0$. But $ac-bc=(a-b)c$. If $c>0$ then, since $a-b>0$, we would have $(a-b)c>0$. But $(a-b)c\leqslant0$. Therefore, $c\leqslant0$.
