Theorem:
For any integers $a, b, c$, with $c >0$,
$$a > b \iff ac > bc.$$
Definition:
For any integers $x, y$, $x > y$ means $x - y = k$, where $k$ is some positive number.
Distributive Law:
For any integer: $x(y + z) = xy + xz$.
Lemma:
For any integers, if $x > 0$, $y > 0$ then
- $xy > 0$
- $x/y > 0$
I know for bi conditionals you must prove the forward and backwards directions. The forward would be $a > b \implies ac > bc$ and the backward would just be flipped. Using contradiction for the forward direction you can assume $a > b \land \lnot(ac > bc)$.
How would I prove this in the forward direction?
Let $a, b$ be such that $a>b$ and let $c>0$. We wish to prove that $ca<cb$.
By definition, $$a-b=k$$ for some $k>0$. Multiplying both sides by $c$ we get $$ca-cb=ck$$ Since $c, k>0$ we know $ck>0$. Therefore, $ca<cb$ by definition.