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