Why does $y<\pm\sqrt x$ mean $-\sqrt x<y<\sqrt x$

Question

If you have
$$y^2<x \iff \sqrt y^2<\sqrt x \iff |y|<\sqrt x$$

I understand that its possible to derive $-\sqrt x<y<\sqrt x$, by considering the cases where $y<0$ and $y\ge 0$.

But is it possible to prove it from $y<\pm\sqrt 2$, by considering the cases $y<\sqrt 2$ and $y<-\sqrt 2$?

Answer 1

We should avoid this notation $y<\pm\sqrt 2$, as noticed we can use instead the compact

$$|y|<\sqrt 2 \iff y>-\sqrt 2 \;\;\land \;\; y<\sqrt 2\iff -\sqrt 2<y<\sqrt 2$$

which is widely used and accepted.

Note that this one $y<\pm\sqrt 2$ could be interpreted instead as

$$y<\pm\sqrt 2 \iff y<-\sqrt 2 \;\;\land \;\; y<\sqrt 2 \iff y<-\sqrt 2$$

which is another good reason to avoid that notation.