Proving the contrapositive statements

Question

I have successfully proved $a^2 < b^2$ if $0 \leq a<b$, where a,b are real numbers

But I am struggling to prove the contrapositive of this statement i.e. $|a|<|b|$ , if and only if $a^2<b^2$

Thanks in advance!

Answer 1

HINTS

Assume that $a^2<b^2.$ There are three possibilities:

$$ |a|>|b|\\ |a|=|b|\\ |a|<|b| $$ If you assume either of the first two, you will get at contradiction, so the third must be the case.

Answer 2

$P\implies Q$ will be true if and only if $\lnot Q \implies \lnot P$.

So to prove $|a| < |b| \implies a^2 < b^2$ is straightforward and you claimed you did it.

I'd do it as $0 \le |a| < |b|$ and if $|a| = 0$ this is trivial. $|b^2| > 0$ so $|b^2| > 0 =|a^2|$. And if $0 < |a| < |b|$ then $a^2 = |a|*|a| < |b|*|a|$ and $|b|*|a|< |b|*|b|=b^2$ so $a^2 < b^2$.

To prove $a^2 < b^2 \implies |a| < |b|$ it'd suffice to prove:

$\lnot (|a| < |b|) \implies \lnot(a^2 < b^2)$ or in other words:

$|a| \ge |b| \implies b^2 \ge a^2$ or in still other words $|b|\le |a| \implies b^2 \le a^2$.

But you pretty much did that already. If $0 \le |b| < |a| $ then $b^2 < a^2$ as we proved above. And if $|b| =|a|$ then $b^2 = a^2$. So $b^2 \le a^2$.

And that's it. We're done.