Square Root is Strictly Increasing

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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 $\sqrt x \ge \sqrt y$.

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

Thus a contradiction is created.

Therefore:

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

$\blacksquare$