I'm reading the book 'How to prove it' from Daniel Velleman which he presents a proof for the following; if $0 < a < b$ then $a^2 < b^2$ as;
Proof. Suppose $0 < a < b$. Multiplying the inequality $a < b$ by the positive number $a$ we can conclude that $a^2 < ab$, and similarly multiplying by $b$ we get $ab < b^2$. Therfore $a^2 < ab < b^2$, as required. Thus if $0 < a < b$ then $a^2 < b^2$.
However, I was also wondering if the statement could be proved using the following method.
Proof. Suppose that $0 < a < b$. Taking the square root of both sides of the inequality $\sqrt{a^2} < \sqrt{b^2}$ we get our original hypothesis $a < b$ . Thus if $0 < a < b$ then $a^2 < b^2$.