Order is Preserved on Positive Reals by Squaring

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

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

Necessary Condition
Assume $$x < y$$.

$$ $$ $$

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

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

$$ $$ $$ $$ $$ $$

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.