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. This is also called a proof by contradiction since you assume that what you want to prove is false and then you derive a contradiction with what you had assumed in the problem.
So, if $a > b$ and $c> 0$, then $ac > ab$, but this contradicts the fact that $ac \leq ab$ and so the assumption that $c>0$ must be wrong. Hence $c \leq 0$.
