Prove that $0\le a^2 < b^2$ implies $a < b$ Q) $a$ and $b$ are positive real numbers and $a^2 < b^2$. Prove that $a < b$
My working so far: Assuming always that $a, b > 0$, the contrapositive of $$a^2 < b^2 \implies a < b$$ is 
$$a \ge b \implies a^2 \ge b^2\;.$$
Thanks

A more direct approach.
If $a,b>0$, then $a+b>0$, and hence $\frac{1}{a+b}>0$.
If $a^2<b^2$ then $$0 < b^2-a^2 = (b-a)(b+a)$$
Multiply both sides by $\frac{1}{a+b}$ to get $$0< b-a$$ so $a<b$.
This shows that you only really need $a+b>0$ and $b^2>a^2$ to show $b>a$.
