I was trying to prove that given $a>0$:$\tag{1}|x|<a\implies -a<x<a$
First I wrote $\tag{2}|x|=\sqrt{x^2}\implies \sqrt{x^2}<a\implies x^2<a^2$
I got stuck here as I had never taken square root in inequalities before.
From searching the definition of square root I found $\sqrt{x}$ is not a function at all, but the positive part is taken as a function called 'primary sqrt.'
The solution to an equation like $x^2=a$ is both $+\sqrt{a}$ and $-\sqrt{a}$, but I don't know what to do in the inequality case.
One tricky proof to (1), I found was :
$|x|<a\implies -|x|>-a$
we know $-|x|\le x\le |x|$ and from above equation $-a<-|x|\le x\le |x|<a$
But this doesn't work for the case of $|x|>a\implies-x<-a$ and $x>a$…
My question is, is (2) valid? and is there a way to prove (1) from (2) or using that idea?
In the case of an equality $x^2=a^2$ gives $x$ can be $a$ or $-a$. Applying *sqrt. on both sides gives me 4 possible equations $x=a,x=-a,-x=a,-x=-a$, but 2 of them are equivalent and $x=\pm a$.
In the case of inequalities by a similar idea I could say $x^2<a^2$ also gives me 4 possible solutions $x<a,x<-a,-x<a,-x<-a$. But only $-a<x<a$ and $a<x\cap x<-a$ are non contradictory combinations. and only $-a<x<a$ has a non empty solution set.
Best Answer
Equation (2) is valid, from there you could get $$ x^2-a^2 <0,$$ which, being a difference of squares, is equivalent to $$(x-a)(x+a)<0 .$$ A product of two real numbers is negative precisely when one is negative and the other positive. This gives two possible cases
Clearly the second case is empty, leaving only the first.