Prove $0\le a<b\:\text{implies}\:0\le a^2<b^2\:\text{and}\:0\le \sqrt{a}<\sqrt{b}$

Question

This proof can be done by using a contradiction. For proving a contradiction, if $0\le a<b$ implies $0>a^2\ge b^2\:\text{or}\:0>sqrt{a}\ge sqrt{b}$. If this is the right direction, then I would have to provide each case.

Answer 1

Hint:  $u^2-v^2=(u-v)(u+v)\,$, and if $u,v \ge 0$ then $u+v \ge 0$, so $u-v$ and $u^2-v^2$ have the same sign. Use that twice, once for $u=a, v=b$, then for $u=\sqrt{a}, b=\sqrt{b}\,$.

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[ EDIT ]  For example, with $u=a, v=b$ where $a,b \ge 0\,$, it follows from the above that:

$$ a-b \le 0 \iff a^2-b^2 \le 0 \quad\style{font-family:inherit}{\text{which is the same as}}\quad a \le b \iff a^2 \le b^2 $$

Answer 2

Assume $0\leq a<b$.

Suppose $a^2\geq b^2$, then $a^2\geq b^2=b\cdot b>ab>a\cdot a=a^2$, which is a contradiction in itself ($a^2<a^2$ is not possible), so $a^2<b^2$.

Suppose $\sqrt{a}\geq \sqrt{b}$, then $a=\sqrt{a}\cdot \sqrt{a}\geq \sqrt{b}\cdot\sqrt{a}\geq\sqrt{b}\cdot \sqrt{b}=b$, which is in contradiction with $a<b$, so $\sqrt{a}<\sqrt{b}$.