5 Absolute Value of a Real Number

$\bullet\qquad$ The absolute value of $a\in \mathbb{R}$ is the number $|a|$ defined by

$|a| = \left\{\begin{matrix} a & \hbox{ if } \quad a\geqslant 0 \\ \\ - a & \hbox{ if } \quad a < 0\end{matrix}\right.$

$\bullet\qquad$ $|a-b|$ is the distance between the real numbers $a$ and $b$ .

$\bullet\qquad$ $| - a| = |a| \qquad \qquad \qquad \sqrt{a^2} = |a|\qquad \qquad \qquad | a \times b |= |a| \times |b|$

$\qquad| a+ b| \leqslant |a| + |b| \qquad \hbox{triangular inequality}$

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Exercise 1

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The distance between $a$ and $b$ is equal to $d(a,b) = |a-b|$ where

$| a - b | = a - b$ if $a\geq b$ and $| a- b | = - a + b$ if $a\leq b$ .

Exercise 2

Exercise 3

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For the last expression, we have

$\displaystyle{ |(-8)- 5| -2 |-3| + | 3(-4)| -\Big| \frac{10}{-2}\Big| - \frac{|-15|}{-3} + 0.5 + \frac{1}{4} |-2| }$

$\displaystyle{= |-13| -6+ | -12| -\Big|-5\Big| +5 + 0.5 + 0.25 (2) }$
$= 13 -6 +12 - 5 +5 + 1= 20$ .

Exercise 4

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* For $x\geq 5$ , we have $|x-5| = x-5$ , then

$|x-5| = x+5 \quad \Leftrightarrow \quad x-5 = x+5 \quad \Leftrightarrow \quad -5 = 5$

leads to a contradiction.

** For $x< 5$ , we have $|x-5| = -(x-5) = 5- x$ , then

$|x-5| = x+5 \quad \Leftrightarrow \quad -x+5 = x+5 \quad \Leftrightarrow \quad x = 0.$

*** The equality is true only for $x=0$ .

Exercise 5

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We have

$- |- 4 - x^2| = - |-(4+x^2)| = - |-1| |4 + x^2| = - (1) |4 + x^2|$ .

Since $4 + x^2\geq 4 >0$ , we deduce that $|4 + x^2| = 4 + x^2$ .

Thus, the statement is true.

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Guide to Precalculus Review Copyright © 2025 by Samia CHALLAL is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License, except where otherwise noted.

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