# Order is Preserved on Positive Reals by Squaring

## 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

The result follows from Order of Squares in Ordered Field.

$\blacksquare$

## Proof 3

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:

$x < y$

$\blacksquare$