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
 A: 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, $c\leqslant0$.
A: Going from $bc-ac\geq 0$ to $b\geq a$, you implicitly assume $c\geq 0$ which is incorrect. The correct way here would be: $$bc-ac=(b-a)c\geq 0\implies (b\geq a~\land~c\geq 0)~\lor~(a\geq b)~\land~c\leq 0)$$ and since you assumed $a\geq b$, we conclude $c\leq 0$

There is a more simper proof though:
$$ac\leq bc\iff (a-b)c\leq 0\tag 1$$
Since $a\gt b$, we have $a-b\gt 0$, so divide both sides of $(1)$ by $a-b$ to get $c\leq 0$
A: For an ordered field $\mathbb{F}$ the following order axioms hold 


*

*$\forall x,y\in \mathbb{F}$ exactly one is true $x>y, x=y, x<y$

*$\forall x,y,z\in \mathbb{F} \quad x<y \quad y<z\implies x<z$

*$\forall x,y,z\in \mathbb{F}\quad x>y \implies x+z>y+z$

*$\forall x,y,z\in \mathbb{F}\quad x>y \quad z>0 \implies xz>yz$
The property can be easily proved by contradiction by the last axiom.
A: So if you want to do a direct proof, not by contradiction:
$ac \le bc$ so $ac - bc = c(a-b) \le 0$.
$a > b$ so $a - b > 0$ so $(a-b)^{-1} > 0$.
So $c= c(a-b)(a-b)^{-1} \le 0 *(a-b)^{-1} = 0$
