Proving an absolute value inequality

[cbarker1]

If $\left| a \right| \le b$, then $-b\le a\le b$.
Let $a,b \in\Bbb{R}$ The definition of the absolute value is $ \left| x \right|= x, x\ge 0$ and $\left| x \right|=-x, x< 0$, where x is some real number.

Case I:$a\ge 0$, $\left| a \right|=a>b$

Case II: a<0, $\left| a \right|=-a<b$the solution is $-b<0\le a\le b$

I work on a number line. yet I still have trouble with the proof.

 

[mathmari]

At which point of the proof are you facing difficulties?

 

[Evgeny.Makarov]

Well, I, for one, have trouble understanding the proof since I don't follow its logical structure, and not just because $a>b$ should be $a\le b$ in case I. Maybe someone can write a more coherent proof.