Order is Preserved on Positive Reals by Squaring

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Theorem

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


Then:

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



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Proof 1

Necessary Condition

Assume $x < y$.

Then:

\(\ds x < y\) \(\implies\) \(\ds x \times x < x \times y\) Real Number Ordering is Compatible with Multiplication
\(\ds x < y\) \(\implies\) \(\ds x \times y < y \times y\) Real Number Ordering is Compatible with Multiplication
\(\ds \) \(\leadsto\) \(\ds x^2 < y^2\) Transitive Law

So:

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

$\Box$


Sufficient Condition

Assume $x^2 < y^2$.

Then:

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

So:

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

$\blacksquare$


Proof 2

From Real Numbers form Totally Ordered Field, the real numbers form an ordered field.

The result follows from Order of Squares in Ordered Field.

$\blacksquare$


Proof 3

From Real Numbers form Totally Ordered Field, the real numbers form a totally ordered field.

By definition, a totally ordered field is a totally ordered ring without proper zero divisors.

The result follows from Order of Squares in Totally Ordered Ring without Proper Zero Divisors.

$\blacksquare$


Proof 4

Necessary Condition

Let $x < y$.

Then:

\(\ds x < y\) \(\implies\) \(\ds x \times x < x \times y\) Real Number Ordering is Compatible with Multiplication
\(\ds x < y\) \(\implies\) \(\ds x \times y < y \times y\) Real Number Ordering is Compatible with Multiplication
\(\ds \) \(\leadsto\) \(\ds x^2 < y^2\) Transitive Law

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:

\(\ds x \ge y\) \(\implies\) \(\ds x \times x \ge x \times y\) Real Number Ordering is Compatible with Multiplication
\(\ds x \ge y\) \(\implies\) \(\ds x \times y \ge y \times y\) Real Number Ordering is Compatible with Multiplication
\(\ds \) \(\leadsto\) \(\ds x^2 \ge y^2\) Transitive Law

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

Hence by Proof by Contradiction it follows that:

$x < y$

$\blacksquare$

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