# Order is Preserved on Positive Reals by Squaring/Proof 4

## Theorem

$x < y \iff x^2 < y^2$

## Proof

### Necessary Condition

Let $x < y$.

Then:

 $\displaystyle x < y$ $\implies$ $\displaystyle x \times x < x \times y$ Real Number Ordering is Compatible with Multiplication $\displaystyle x < y$ $\implies$ $\displaystyle x \times y < y \times y$ Real Number Ordering is Compatible with Multiplication $\displaystyle$ $\leadsto$ $\displaystyle x^2 < y^2$ Real Number Ordering is Transitive

So:

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

$\Box$

### Sufficient Condition

Let $x^2 < y^2$.

Aiming for a contradiction, suppose $x \ge y$.

Then:

 $\displaystyle x \ge y$ $\implies$ $\displaystyle x \times x \ge x \times y$ Real Number Ordering is Compatible with Multiplication $\displaystyle x \ge y$ $\implies$ $\displaystyle x \times y \ge y \times y$ Real Number Ordering is Compatible with Multiplication $\displaystyle$ $\leadsto$ $\displaystyle x^2 \ge y^2$ Real Number Ordering is Transitive

But this contradicts our assertion that $x^2 < y^2$.

Hence by Proof by Contradiction it follows that:

$x < y$

$\blacksquare$