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
\[ \frac{{\left| {x + y} \right|}}{{1 + \left| {x + y} \right|}} = 1 – \frac{1}{{1 + \left| {x + y} \right|}} \leqslant 1 – \frac{1}{{1 + \left| x \right| + \left| y \right|}} \]
\[ \leqslant \frac{{\left| x \right| + \left| y \right|}}{{1 + \left| x \right| + \left| y \right|}} \]
\[ \leqslant \frac{{\left| x \right|}}{{1 + \left| x \right|}} + \frac{{\left| y \right|}}{{1 + \left| y \right|}};{\text{ as }}\left| {x + y} \right| \leqslant \left| x \right| + \left| y \right|\]