Q:

Michael Spivak's Calculus - Chapter 1 problem 5.10

The problem says
If $a,b \ge 0$ and $a^2 \lt b^2$, then $a \lt b$. (Hint use previous problem backwards)
The previous problem proved if 0 $\le$ a$\lt$b then $a^2 \lt b^2$
When I see this problem I don't understand how you can turn the problem around and it always holds. I'd think that you'd have to prove that if AND ONLY IF $0 \le a \lt b$ will $a^2 \lt b^2$.
Is this train of thought correct? The hint makes it seem this is a simple reference to the previous problem, like all you'd have to say is "because of previous problem" you'd only need one or two lines. 

A:

$0 < b^2 - a^2 = (b - a)(b + a)$
Prove b + a is positive.  Thus b - a is positive.

