inequality proof I have come across a problem:

Let $a,b$ and $c$ be real numbers where $a > b$. Prove that if $ac \leq bc$, then $c \leq 0$.

I tried using the Indirect proof
If $a > b$, and $c > 0$, then by the 4th axiom of Inequality, Multiplicativity, $ac > bc$. 
However, $ac \leq bc$, therefore $c \leq 0$.
I'm not sure if this is the correct way. 

Your proof is correct.
Here's an alternative proof.
$bc \geq ac \Longleftrightarrow c(b-a) \geq 0 \Longleftrightarrow -c(a-b) \geq 0$
Since $a > b \Longleftrightarrow (a-b) >0$, so $c \leq 0$.
