Order is Preserved on Positive Reals by Squaring
Theorem
Let $x, y \in \R: x > 0, y >0$.
Then:
- $x < y \iff x^2 < y^2$
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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:
- $x < y$
$\blacksquare$