The question:

Show that if $\displaystyle 0 <= a <= b$, then $\displaystyle \sqrt{a} <= \sqrt{b}$

I did:

$\displaystyle \sqrt{0} <= \sqrt{a} <= \sqrt{b}$

$\displaystyle 0 <= \sqrt{a} <= \sqrt{b}$

Therefore:

$\displaystyle \sqrt{a} <= \sqrt{b}$

My textbook doesn't have answers for these types of questions, so I can't check.