Why does $-b < a < b \iff |a| < b$?

Question

I don't have any intuition for this because it's just something I memorized. I only understand that $|a| = a$ if $a$ is already positive (or $0$), and $|a| = -a$ if $a$ is negative since we want to turn it positive ($-(-a)$). So does that mean we will have a piecewise inequality also?

But how do we get absolute value from $-b < a < b$?

Answer 1

Note that $|a|=\max\{a,-a\}$. Hence $$\begin{align}|a|<b&\iff \max\{a,-a\}<b\\ &\iff a<b\land -a<b\\&\iff a<b\land a>-b\\&\iff -b<a<b.\end{align}$$

Answer 2

Make cases:
Let $|a|<b$, then $a < b$ if $a$ is positive. If $a$ is negative, $a<-a=|a|<b$ and $-a < b$ if $a$ is negative, so $a > -b$. If $a$ is positive, $-a < a = |a| < b$ so again $a > -b$.
This proves $|a| < b \Rightarrow -b<a<b$. The converse is analog (make cases for $a\ge 0, a<0$).

Answer 3

First of all you should remark that: $-b < a < b$ implies that $b \geq 0$. Then you get:

$$-b < a \Rightarrow -a < b$$

Because of the usual rules of inequalities.

You also have the inequality:

$$a < b$$

Which implies:

$$ max(-a, a) < b$$

And

$$|a| = max(a, -a)$$

So you can conclude from here.

The converse follow the same path of reasonning backward