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|