# 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 $\sqrt x \ge \sqrt y$.

 $\text {(1)}: \quad$ $\displaystyle \sqrt x$ $\ge$ $\displaystyle \sqrt y$ $\text {(2)}: \quad$ $\displaystyle \sqrt x$ $\ge$ $\displaystyle \sqrt y$ $\displaystyle x$ $\ge$ $\displaystyle y$ Real Number Axioms: $\R O2$: compatibility with multiplication, $(1) \times (2)$

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