Negative of Absolute Value

Theorem
Let $x \in \R$ be a real number.

Let $\size x$ denote the absolute value of $x$.

Then:
 * $-\size x \le x \le \size x$

Proof
Either $x \ge 0$ or $x < 0$.


 * If $x \ge 0$, then $-\size x \le 0 \le x = \size x$.


 * If $x < 0$, then $-\size x = x < 0 < \size x$.