What is going on here? $(-1<x^2<3 \Leftrightarrow -\sqrt{3} < x < \sqrt{3})$?

Question

Can someone explain how is this possible (see image)? Or is it wrong? Shouldn't it be $(-1<x^2<3 \Leftrightarrow \sqrt{-1} < x < \sqrt{3})$?

Answer 1

The implicit assumption in your example is that $f$ is a real-valued function of real variable. The alternative is to assume that you are working with complex numbers, but that isn't the case here. The obvious clue is that numbers are compared with $<$. If we were working with complex numbers, we would have to specify what we mean by $a < b$.

If $x$ is a real number, $x^2$ is greater than or equal to $0$. Therefore $x^2 > -1$ is always true and can be ignored. We are left with $x^2 < 3$. Every $x$ such that $-\sqrt{3} < x < \sqrt{x}$ is such that $x^2 < 3$, which is what the final line of the derivation shows. Moreover, every real number $x$ such that either $x \leq -\sqrt{3}$ or $x \geq \sqrt{3}$ is such that $x^2 \geq 3$. Hence

$$ -\sqrt{3} < x < \sqrt{3} $$

describes all values of $x$ for which $|f(x)| < 1$ holds.