Vtyjkyuk

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 $sqrta^2 < sqrtb^2$ we get our original hypothesis $a < b$ . Thus if $0 < a < b$ then $a^2 < b^2$.

In the proof from the book you presented, you've assumed $0<a<b$ and deduced $a^2<b^2$.

In your presented proof, you've essentially assumed $a^2<b^2$ and deduced $a<b$. Why? The hypothesis $0<a<b$ is never used and by taking the square root of $a^2<b^2$, you implicitly assume that statement, deducing $a<b$ from it.

Thus, conceptually you have shown $BRightarrow A$ for the corresponding statements $A,B$, i.e. you have established $ALeftrightarrow B$ using $ARightarrow B$ from the previous proof.

EDIT: As discussed with Ennar in the comments, there are additionally some issues with deducing $a<b$ from $a^2<b^2$ as it requires the square root function to be monotone, a property that follows from the fact the the square function is monotone on $[0,infty)$. Then of course, you can not establish $BRightarrow A$ without first establishing $ARightarrow B$, i.e. there is another circularity of reasoning.

As you can see, there are even more implicit assumptions than I had initially pointed out.