Order is Preserved on Positive Reals by Squaring/Proof 4

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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:

$x < y$

$\blacksquare$


Sources