Showing $a^2 < b^2$, if $0 < a < b$ Lately, I've been stumbling with proofs of inequalities.
For example:

Given $0 < a < b$
  Show $a^2 < b^2$

The only thing I've been able to come up with so far:

$a^2 < b^2$
  $\sqrt{a^2} < \sqrt{b^2}$
  $a < b$

OR


$a < b$
  $a^2 < b^2$

However, neither of these solutions seem to be really "showing" that $a^2 < b^2$, assuming $0 < a < b$. I've tried some other things, but to no avail. Am I merely overthinking the problem when, in fact, these are actually acceptable solutions, or am I truly missing something here?

$$0<a<b\Rightarrow 0\cdot a<a\cdot a<b\cdot a<b\cdot b\Rightarrow
a^{2}<b^{2}.$$
