Best argument to prove $|x|\le a \iff -a\le x \le a$ $$|x|\le a \iff -a\le x \le a$$
I can only verify the integrity of this by talking about distances on the number line. But is there a algebraic argument that proves this?
 A: Since you have an absolute value, which is defined (for reals) as
$$|x|=\begin{cases}x && \text{if }x\geq 0 \\ -x && \text{if }x\leq 0\end{cases}$$
we are behooved to split into cases ourself to verify this. So, we have two cases:


*

*Suppose $x\geq 0$. Then $|x|=x$ and our inequality is $x\leq a$. This condition, in conjunction with the fact that $x$ must be non-negative for the equality $|x|=x$ to be valid yields that if $x$ is such that $0\leq x \leq a$, then $|x|\leq a$.

*Suppose $x\leq 0$. Then $|x|=-x$ and our inequality is $-x\leq a$ which rearranges to $x\geq -a$. However, since $x$ is negative, we receive $-a \leq x \leq 0$ as a set of solutions as well.


Any $|x|$ must fall into one of the cases above and hence we can state that $|x|\leq a$ is equivalent to saying that at least one of the following holds
$$-a \leq x \leq 0$$
$$0 \leq x \leq a.$$
It should be obvious that, since the intervals $[-a,0]$ and $[0,a]$ overlap, this is equivalent to
$$-a\leq x \leq a$$
A: Go back to the definition of $|x|$. 
If $x \geq 0$, then $|x|  = x$; hence in that case 
$$0 \leq x \leq a$$
If $x < 0$, then $|x| = -x$; in that case $-x \leq a$ or
$$-a \leq x < 0$$
As $x$ must be either positive, zero, or negative,
$$-a \leq x \leq a$$
A: With $a\geq 0$,
$$
|x|\leq a\iff x^2\leq a^2\iff(x-a)(x+a)\leq 0\iff x\in[-a,a].
$$
