Showing that the square root is monotone I've shown the existence of unique square roots of all positive rational numbers, so now I want to prove that the square root is monotone: 
$0<a<b$ if and only if $\sqrt{a} < \sqrt{b}$

Let $0\leq a\leq b$ and $\sqrt b <\sqrt a.$ Then there is an $ε>0$ such that $\sqrt a=\sqrt b + ε.$ So we have $a-b=(\sqrt a)^2-(\sqrt b)^2=(\sqrt b + ε)^2-(\sqrt b)^2=2ε\sqrt b+ε^2=ε(2\sqrt b+ε)=ε(\sqrt b+\sqrt a)>0$, which contradicts to our assumption. Thus $\sqrt a \leq\sqrt b.$
