Negative of Absolute Value

Theorem
Let $$x \in \R$$ 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

 * $$\left|{x}\right| < y \iff -y < x < y$$;
 * $$\left|{x}\right| \le y \iff -y \le x \le y$$.

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

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 $$-y < x < y$$.

Now suppose that $$\left|{x}\right| \le y$$.

If $$\left|{x}\right| < y$$ then $$-y < x < y$$ and so $$-y \le x \le y$$.

Otherwise, if $$\left|{x}\right| = y$$ then either $$x = y$$ or $$-x = y$$ and hence the result.


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

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

Now suppose that $$-y \le x \le y$$.

If $$-y < x < y$$ then $$\left|{x}\right| < y$$ and hence $$\left|{x}\right| \le y$$.

Else, if either $$-y = x$$ or $$x = y$$ then $$\left|{x}\right| = y$$ and hence the result.