Absolute Value of a Number

Sometimes, it is useful to restrict our attention over non-negative real numbers only. For this purpose, we define 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 according 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 absolute value of a real number, we have

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 number, 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|

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