If $a \ge 0$ and $b \ge 0$, then prove that $a \le b \Leftrightarrow a^2 \le b^2 \Leftrightarrow \sqrt{a} \le \sqrt{b}$ Assumptions: Field Axioms and Order Axioms of $\Bbb R$ hold. Also assume that square root of positive real numbers exist.
Problem: I know how to prove this:
$$a \lt b \Leftrightarrow a^2 \lt b^2 \Leftrightarrow \sqrt{a} \lt \sqrt{b}$$
The equality is tripping me.
 A: Because $x\geq y\Leftrightarrow x-y\geq0$ and $$a^2-b^2=(a-b)(a+b)=(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})(a+b).$$
A: You in fact already have a proof because if and only if is preserved under negation. You proved 
 $$a < b \Leftrightarrow a^2 < b^2 \Leftrightarrow \sqrt{a} < \sqrt{b}$$
so that
 $$ \neg(a < b )\Leftrightarrow \neg( a^2 < b^2) \Leftrightarrow \neg(\sqrt{a} < \sqrt{b})$$
which is exactly
$$b \le a \Leftrightarrow b^2 \le a^2 \Leftrightarrow \sqrt{b} \le \sqrt{a}.$$
A: "The equality is tripping me"
Then you are thinking too hard
1) $a = b \implies a^2 = b^2$ and $\sqrt a = \sqrt b$.
2) $a^2 = b^2 \implies \sqrt{a^2} = \sqrt{b^2}$ and as positive square roots  are unique, then $a = \sqrt {a^2}$ and $b=\sqrt{b^2}$.
3) $\sqrt{a} = \sqrt{b} \implies a = \sqrt{a}^2 = \sqrt{b}^2 =b$.
That's all.
...or....
2) $a \ne b \implies $ either i) $a < b \implies a^2 < b^2$ and $\sqrt a < \sqrt b$ or ii) $b < a \implies b^2 < a^2$ and $\sqrt b < \sqrt a$.
You've done the heart part, but have gotten tripped up on the utterly trivial.
=====
To be perfectly honest, had someone wrote.  "In the case of $a = b$ the result is obvious." I'd have accepted it.
Or if someone had simply used $\le $ signs in the proof (as in $0 \le a; 0 \le b$ so if $a \le b$ then $a*a \le a*b$ and $a*b \le b*b$ and so $a^2 \le a*b \le b^2$) I'd have accepted it without thought.
.... I probably shouldn't.... but I would.
A: Both $f(x) = x$, $f(x) = \sqrt(x)$ and $f(x) = x^2$ are all strictly monotonic functions, therefore you can use the definition of strict monotonicity.
$f$ is monotonic in $A \subset \mathbb{R}$ iff
$$
x < y \Leftrightarrow f(x) < f(y)
$$
For all $x,y$ in $A$
