Square Root is Strictly Increasing

Theorem

The positive square root function is strictly increasing, that is:

$\forall x, y \in \R_{>0}: x < y \implies \sqrt x < \sqrt y$

Proof

Let $x$ and $y$ be positive real numbers such that $x < y$.

Aiming for a contradiction, suppose $\sqrt x \ge \sqrt y$.

\(\text {(1)}: \quad\)

\(\ds \sqrt x\)

\(\ge\)

\(\ds \sqrt y\)

\(\text {(2)}: \quad\)

\(\ds \sqrt x\)

\(\ge\)

\(\ds \sqrt y\)

\(\ds x\)

\(\ge\)

\(\ds y\)

Real Number Axiom $\R \text O2$: Usual Ordering is Compatible with Multiplication: $(1) \times (2)$

Thus a contradiction is created.

Therefore:

$\forall x, y \in \R_{>0}: x < y \implies \sqrt x < \sqrt y$

$\blacksquare$