# Absolute Value of a Number

Sometimes it is useful to restrict our attention to non-negative real numbers only. For this purpose, we define a numerical or non-negative value of a real number which we call an absolute value or modulus of the real number.

Absolute Value: The absolute value of a real number, $a$ denoted by $\left| a \right|$, is the real number $a,{\text{ }} – a,{\text{ or }}0$ as a positive, negative, or zero, i.e.

$\left| a \right| = \left\{ {\begin{array}{*{20}{c}} a&{,{\rm{ }}a > 0}\\ { – a}&{,{\rm{ }}a < 0}\\ 0&{,{\rm{ }}a = 0} \end{array}} \right.$

From the definition of the absolute value of a real number, we have the examples below.

Example:

(1)   $\left| a \right| = {\text{max}}\left\{ {a, – a} \right\}$,
(2)   $– \left| a \right| = {\text{min}}\left\{ {a, – a} \right\}$,
(3)   $\left| a \right| \geqslant a \geqslant – \left| a \right|$.

Example:

(1) $\left| {ab} \right| = \left| a \right| \cdot \left| b \right|$,
(2) $\left| {\frac{a}{b}} \right| = \frac{{\left| a \right|}}{{\left| b \right|}},{\text{ }}\left( {b \ne 0} \right)$.

Example:

(1) $\left| a \right| + \left| b \right| \geqslant \left| {a + b} \right|$,
(2) $\left| {a – b} \right| \geqslant \left| {\left| a \right| – \left| b \right|} \right|$.

Example:

If $\varepsilon > 0$, then $\left| {a – b} \right| < \varepsilon \Leftrightarrow b – \varepsilon < a < b + \varepsilon$.

Solution:

We have
$\left| {a – b} \right| = {\text{max}}\left\{ {\left( {a, – b} \right), – \left( {a, – b} \right)} \right\} < \varepsilon$
$\Leftrightarrow \left( {a – b} \right) < \varepsilon \wedge – \left( {a – b} \right) < \varepsilon$
$\Leftrightarrow a < b + \varepsilon \wedge \left( { – a + b} \right) < \varepsilon$
$\Leftrightarrow a < b + \varepsilon \wedge \left( {b – \varepsilon } \right) < a$
$\Leftrightarrow b – \varepsilon < a < b + \varepsilon$

Example:

For all real numbers, $x$ and $y$, $\frac{{\left| {x + y} \right|}}{{1 + \left| {x + y} \right|}} \leqslant \frac{{\left| x \right|}}{{1 + \left| x \right|}} + \frac{{\left| y \right|}}{{1 + \left| y \right|}}$

Solution:

Here, for all real numbers $x$ and $y$, we have
$\frac{{\left| {x + y} \right|}}{{1 + \left| {x + y} \right|}} = 1 – \frac{1}{{1 + \left| {x + y} \right|}} \leqslant 1 – \frac{1}{{1 + \left| x \right| + \left| y \right|}}$
$\leqslant \frac{{\left| x \right| + \left| y \right|}}{{1 + \left| x \right| + \left| y \right|}}$
$\leqslant \frac{{\left| x \right|}}{{1 + \left| x \right|}} + \frac{{\left| y \right|}}{{1 + \left| y \right|}};{\text{ as }}\left| {x + y} \right| \leqslant \left| x \right| + \left| y \right|$