Order is Preserved on Positive Reals by Squaring/Proof 1

Theorem

$x < y \iff x^2 < y^2$

Proof

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

\(\ds \)

\(\leadsto\)

\(\ds x^2 < y^2\)

Real Number Ordering is Transitive

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:

$x^2 < y^2 \implies x < y$

$\blacksquare$

Sources

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