# Order is Preserved on Positive Reals by Squaring

## Contents

## Theorem

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

Then:

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

## Proof 1

### 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:

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

$\blacksquare$

## Proof 2

From Real Numbers form Totally Ordered Field, the real numbers form a totally 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:

\(\displaystyle x < y\) | \(\implies\) | \(\displaystyle x \times x < x \times y\) | 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:

- $x < y$

$\blacksquare$