How would I prove that for all $a,b$ and $c<0$ we have $a<b \implies ca>cb$, using that $(-c)>0$?

Question

I found this proof already, but in my text book there is a lead saying I should use that $(-c)>0$. How would I do this? I keep running into the problem that I sooner or later need to multiply or divide by a negative number.

Edit: I should add that this is a very basic introductory one-dimensional analysis course and you should assume I know nothing basically.

Answer 1

Using $-c>0$, $$\begin{align}a<b&\implies (-c)a<(-c)b\\&\implies -ca<-cb\\&\implies -ca+(cb+ca)<-cb+(cb+ca)\\&\implies cb<ca\\&\implies\boxed{ca>cb}\end{align}$$

Answer 2

Let $x=-a>0$, $y=-b>0$ and $z=-c>0$ thus we need to prove that

$$-x<-y \iff x>y\implies xz>yz$$

which is trivially true.

Answer 3

$-c>0$ and $b-a>0$.

Thus, $$(-c)(b-a)>0$$ or $$ca-cb>0,$$ which is $$ca>cb.$$