If $a \ge 0$ and $b \ge 0$, then prove that $a \le b \Leftrightarrow a^2 \le b^2 \Leftrightarrow \sqrt{a} \le \sqrt{b}$

Question

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.

Answer 1

"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.

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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.

Answer 2

Because $x\geq y\Leftrightarrow x-y\geq0$ and $$a^2-b^2=(a-b)(a+b)=(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})(a+b).$$

Answer 3

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 $$b \le a \Leftrightarrow b^2 \le a^2 \Leftrightarrow \sqrt{b} \le \sqrt{a}.$$

Answer 4

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 $A$