Order is Preserved on Positive Reals by Squaring/Proof 1

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Theorem

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


Proof

Necessary Condition

Assume $x < y$.

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

Assume $x^2 < y^2$.

Then:

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

So:

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

$\blacksquare$


Sources