Negative of Absolute Value

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

Let $$\left|{x}\right|$$ be the absolute value of $$x$$.

Then $$- \left|{x}\right| \le x \le \left|{x}\right|$$.

Corollary
$$\left|{x}\right| < y \iff -y < x < y$$.

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


 * If $$x \ge 0$$, then $$- \left|{x}\right| \le 0 \le x = \left|{x}\right|$$.


 * If $$x < 0$$, then $$- \left|{x}\right| = x < 0 < \left|{x}\right|$$.

Proof of Corollary

 * First we show that $$\left|{x}\right| < y \Longrightarrow -y < x < y$$.

Suppose $$\left|{x}\right| < y$$.

Then from the above, $$x \le \left|{x}\right|$$ and $$\left|{x}\right| \ge -x$$.

So $$x < y$$ and $$-x < y$$, and so $$x > -y$$ from Ordering of Inverses.

It follows that $$-y < x < y$$.


 * Next we show that $$-y < x < y \Longrightarrow \left|{x}\right| < y$$.

Suppose $$-y < x < y$$.

Then $$x < y$$ and $$-x < y$$.

For all $$x$$, $$\left|{x}\right| = x$$ or $$\left|{x}\right| = -x$$.

Thus it follows that $$\left|{x}\right| < y$$.