How can I square $-1 < x < 1$? If I square $-1 < x < 1$, I get $1 < x^2 < 1$ which doesn't make any sense. What additional algebraic steps do I need to apply in order to get the proper inequality $0 < x^2 < 1$?
And if we are going backwards, how do I algebraically solve $x^2 < 1$ to obtain $-1 < x < 1$? If I take the square root of both sides, I get $x < \pm 1$ which would mean that $x < 1$ and $x < -1$ so that's just $x < -1$ which is obviously not true. The solution would be to flip the inequality for the negative sign, but how do you know when to flip the inequality sign or not besides when you divide/multiply by opposite signs?
 A: You can square inequalities if all numbers are non negative. You also can square them, but have to reverse the inequalities  if all numbers are non positive. So in this (mixed) case, you 'll to slit your inequalities:


*

*$0\le x<1$ will give $0\le x^2<1$,

*$-1<x\le 0$ will produce $1>x^2\ge 0$.


So globally you get $$0\le x^2<1.$$
Of course all this can be visualised on the graph of the function $x\mapsto x^2$.
A: Recall $\forall x\in \mathbb{R}$, $|x^2|=x^2=|x|^2$ 


*

*$-1<x<1\implies 0\le|x|<1 \implies 0\le|x|^2<1 \implies 0\le x^2<1$

*$x^2<1\implies |x|^2<1 \implies |x|<1 \implies -1<x<1$ 
A: In the domain $-1 < x < 0$, $x$ takes a negative value, so squaring it in this domain is equivalent to multiplying by a negative number. What do we do with inequalities when multiplying by a negative number?
In a more general sense, when confronted with inequalities, it often makes sense to identify "breakpoints" and treat them by cases. So for $x^2 < 1$, we might ask, "what if $x$ is positive? What if $x$ is negative?" The final result is the combination of those answers.
A: To algebraically solve $x^2<1$ to obtain $-1<x<1$, just recall that $\sqrt{x^2}=|x|$, so $|x|<1$ gives $-1<x<1$.
