Order is Preserved on Positive Reals by Squaring/Proof 4

From ProofWiki
Jump to navigation Jump to search

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

Retrieved from "https://proofwiki.org/w/index.php?title=Order_is_Preserved_on_Positive_Reals_by_Squaring/Proof_4&oldid=406821"