Prove that $|x|\le a$

Question

How can I prove this?

$|x|\le a$, $\space \space$ $\:-a\le x\le a\:, $ $\space \space$ $a\in\mathbb{R}$.

Thanks!

Answer 1

If $|x|\le a$

If $$x\le 0, ~then~ -x \le a \implies x \ge -a \implies x\in [-a, 0]$$

If $$x\ge 0, ~then~ x \le a \implies x \in [0, a] $$

The final result is union of these two intervals: $x\in [-a,a]$

Answer 2

Note that $$|x|=x, x\ge0$$ $$|x|=-x, x\le 0$$ Now suppose $x \in [0,a]$, $x \le a \implies |x| \le a$ and so if $x \in [-a,0]$, $x \ge -a \implies -x \le a \implies |x| \le a$.