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 $x < y$.

Suppose $\sqrt x \ge \sqrt y$.

Thus a contradiction is created.

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