Order is Preserved on Positive Reals by Squaring/Proof 1

Theorem
Let $x, y \in \R: x > 0, y >0$.

Then:
 * $x < y \iff x^2 < y^2$

Necessary Condition
Assume $x < y$.

Then:

So $x < y \implies x^2 < y^2$.

Sufficient Condition
Assume $x^2 < y^2$.

Then:

So $x^2 < y^2 \implies x < y$.

An alternative approach is to assume that $x^2 < y^2$ but $x \ge y$ and obtain a contradiction.