Let $a \geq 0$. $|x| \leq a$ iff $-a \leq x \leq a$.

I am trying to prove the theorem on the title where the universe is the set of all real numbers. I got a result, but I think I have missed something. I need help confirming that I got the desired result or I have missed something (and not both).

In my attempted proof, I have used the following definition.

Definition 5.2

\begin{align} |x| = \begin{cases} x &\text{if $x \geq 0$}\\ -x &\text{if $x <0$}\end{cases} \end{align}

Let $a \geq 0$. $|x| \leq a$ iff $-a \leq x \leq a$.

Proof. Suppose $|x| \leq a$. By Definition 5.2, $x \leq a$ for $x \geq 0$ and $x \geq -a$ for $x<0$. Since $-a \leq x \leq a$, the implication have been proved.

On the other hand, suppose $-a \leq x \leq a$. From the inequality $x \geq -a$, by multiplying both sides by $-1$, one obtains $-x \leq a$. Because $x \leq a$ and $-x \leq a$, by Definition 5.2 $|x| \leq a$ and the converse have been proved.

Since the implication and the converse of the statement have been proved, $|x| \leq a$ iff $-a \leq x \leq a$ for each $a \geq 0$.

It needs a bit more justification. For example, for the forward direction, we should break it into two cases:

Case 1: $x\geq 0$. In this case, $0\leq x\leq a$. Since $a\geq0$, we have $-a\leq 0$, so $$-a\leq 0\leq x\leq a\implies -a\leq x\leq a$$

Case 2: $x<0$. In this case, $-a\leq x<0$. Since $x<0$, we have $x\leq 0$. Since $a\geq 0$, we have

$$-a\leq x\leq 0\leq a\implies -a\leq x\leq a.$$

For the converse, it does not follow directly from the definition that $|x|\leq a$; that is, this needs more justification as well. We break it into two cases:

Case 1: $x\geq 0$. In this case, $|x|=x$, so $x\leq a\implies |x|\leq a$.

Case 2: $x<0$. In this case, $|x|=-x$, so $-x\leq a\implies |x|\leq a.$