# 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:

\(\ds x < y\) | \(\implies\) | \(\ds x \times x < x \times y\) | 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 | |||||||||||

\(\ds \) | \(\leadsto\) | \(\ds x^2 < y^2\) | 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:

\(\ds x \ge y\) | \(\implies\) | \(\ds x \times x \ge x \times y\) | 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 | |||||||||||

\(\ds \) | \(\leadsto\) | \(\ds x^2 \ge y^2\) | 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

- 1977: K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*... (previous) ... (next): $\S 1$: Real Numbers: $\S 1.6$:*Example*