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 $- \left|{x}\right| = x < 0 < \left|{x}\right|$.

$\blacksquare$

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.

$\blacksquare$

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 1.15$
• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 1.20 \ (1)$