Order Preserved on Positive Reals by Squaring

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Theorem

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

Then:

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

Assume $x < y$.

Then:

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

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

$\Box$

Assume $x^2 < y^2$.

Then:

$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle x^2$	$<$	$\displaystyle y^2$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle \implies$	$\displaystyle $	$\displaystyle 0$	$<$	$\displaystyle y^2 - x^2$	$\displaystyle $	$\displaystyle $	$\displaystyle $	Real Number Ordering is Compatible with Addition
$\displaystyle $	$\displaystyle \implies$	$\displaystyle $	$\displaystyle \left({y-x}\right) \left({y+x}\right)$	$>$	$\displaystyle 0$	$\displaystyle $	$\displaystyle $	$\displaystyle $	Difference of Two Squares
$\displaystyle $	$\displaystyle \implies$	$\displaystyle $	$\displaystyle \left({y-x}\right) \left({y+x}\right) \left({y+x}\right)^{-1}$	$>$	$\displaystyle 0 \times \left({y+x}\right)^{-1}$	$\displaystyle $	$\displaystyle $	$\displaystyle $	(as $x + y > 0$)
$\displaystyle $	$\displaystyle \implies$	$\displaystyle $	$\displaystyle y - x$	$>$	$\displaystyle 0$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle \implies$	$\displaystyle $	$\displaystyle x$	$<$	$\displaystyle y$	$\displaystyle $	$\displaystyle $	$\displaystyle $

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

$\blacksquare$

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

\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle x < y\)	\(\implies\)	\(\displaystyle x x < x y\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Real Number Ordering is Compatible with Multiplication
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle x < y\)	\(\implies\)	\(\displaystyle x y < y y\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Real Number Ordering is Compatible with Multiplication
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\implies\)	\(\displaystyle x^2 < y^2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Ordering is Transitive

\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle x^2\)	\(<\)	\(\displaystyle y^2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle 0\)	\(<\)	\(\displaystyle y^2 - x^2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Real Number Ordering is Compatible with Addition
\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \left({y-x}\right) \left({y+x}\right)\)	\(>\)	\(\displaystyle 0\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Difference of Two Squares
\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \left({y-x}\right) \left({y+x}\right) \left({y+x}\right)^{-1}\)	\(>\)	\(\displaystyle 0 \times \left({y+x}\right)^{-1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	(as $x + y > 0$)
\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle y - x\)	\(>\)	\(\displaystyle 0\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x\)	\(<\)	\(\displaystyle y\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)