# Absolute value of a real number

Mathinker.com

## Definition

For any real number $x$ the absolute value
or modulus of $x$ is denoted by $|x|$ and is defined as

$$\bbox[8pt]{\Large{|x|=\left\vert \begin{array}{rcl} x &\text{if}& x\geqslant 0\\ -x&\text{if}& x < 0 \end{array} \right.}}$$

it means

if $x\geqslant 0$ then $|x| = x$

if $x < 0$ then $|x| = {\color{red}{-1}}\cdot x$

### Using Definition

$$|x|=\left\vert \begin{array}{rcl} x &\text{if}& x\geqslant 0\\ -x &\text{if}& x < 0 \end{array} \right.$$

find absolute value of an expression

ExpressionAnalysisCalculation
$|3|$$3\geqslant 0 \Rightarrow$$|3| = 3$
$|-5|$$-5<0 \Rightarrow$$|-5|={\color{red}{-1}}\cdot (-5) = 5$
$|12|$$12\geqslant 0 \Rightarrow$$|12|=12$
$|-7|$$-7<0 \Rightarrow$$|-7|={\color{red}{-1}}\cdot (-7) = 7$

### Find absolute value

ExpressionAnalysisCalculation
$|5-3|$$5-3\geqslant 0 \Rightarrow$$|5-3| = 5-3$
$|3-5|$$3-5<0 \Rightarrow$$|3-5|={\color{red}{-1}}\cdot (3-5)$
${\color{red}{-1}}\cdot (3-5) = -3+5$
$|\pi + 1|$$\pi + 1\geqslant 0 \Rightarrow$$|\pi + 1|=\pi +1$
$|3-\pi|$$3-\pi<0 \Rightarrow$$|3-\pi|={\color{red}{-1}}\cdot (3-\pi)$
${\color{red}{-1}}\cdot (3-\pi) = \pi - 3$
$|1-7|$$1-7<0 \Rightarrow$$|1-7|=7-1$

### Find absolute value

ExpressionAnalysisCalculation
$|a-b|$
if $a>b$
$a-b\geqslant 0 \Rightarrow$$|a-b| = a-b |m-n| if m < n m-n<0 \Rightarrow$$|m-n|={\color{red}-}(m-n)$
${\color{red}{-}}(m-n) = n-m$
$|\alpha-\beta|$
if $\alpha>\beta$
$\alpha-\beta\geqslant 0 \Rightarrow$$|\alpha-\beta|=\alpha-\beta |x-y| if y>x x-y<0 \Rightarrow$$|x-y|=y-x$

$a,\,\, b,\,\, m,\,\, n,\,\, \alpha,\,\, \beta,\,\, x,\,\, y \in \mathbf{R}$

### Find absolute value

ExpressionAnalysisCalculation
$|x-2|$
if $x>2$
$x-2\geqslant 0 \Rightarrow$$|x-2| = x-2 |5-y| if 5 > y 5-y<0 \Rightarrow$$|5-y|={\color{red}-}(5-y)$
${\color{red}{-}}(5-y) = y-5$
$|\alpha-8|$
if $8 < \alpha$
$\alpha-8\geqslant 0 \Rightarrow$$|\alpha-8|=\alpha-8 |z-1| if 1 > z z-1<0 \Rightarrow$$|z-1|=1-z$

$x,\,\, y,\,\, \alpha,\,\, z \in \mathbf{R}$

## Geometric interpretation

For any real number the absolute value means
how far a number is from zero

$|{\color{blue}{-4}}|={\color{green}{4}}\qquad$ $\qquad|{\color{red}{4}}|={\color{green}{4}}$

### Modulus as a value of distance

Geometric interpretationAnalytical expression
$|{\color{red}2}|={\color{green}2}$
$|{\color{blue}{-3}}|={\color{green}3}$
$|{\color{blue}{-2}}|={\color{green}2}$

### Modulus of Subtraction

For any two real numbers
absolute value of their subtraction
means how far the numbers are from each other

$|{\color{blue}{a}} - {\color{red}{b}} |={\color{green}{c}}$

### Modulus of Subtraction

Geometric interpretationAnalytical expression
$|{\color{red}1}-{\color{red}3}|={\color{green}2}$
$|{\color{blue}{-2}}-{\color{red}2}|={\color{green}4}$
$|{\color{blue}{-3}}-({\color{blue}{-1}})|={\color{green}2}$

## Absolute value Equations (A)

$$\bbox[4pt]{\Large{|x-a| = b}}$$

Using the definition of the modulus

If $\,\,x-a\geqslant 0\,\,$ then $\,\,x-a = b$

If $\,\,x-a < 0\,\,$ then $\,\,{\color{red}{-1}}\cdot(x-a)=b$

$$\bbox[2pt]{|x-a|= b \Leftrightarrow \left[ ({\color{blue}{\text{or}}}) \begin{array}{l} \left\lbrace ({\color{blue}{\text{and}}}) \begin{array}{l} x-a\geqslant 0\\ x-a = b \end{array} \right. \\ \left\lbrace ({\color{blue}{\text{and}}}) \begin{array}{l} x-a< 0\\ x-a = -b \end{array} \right. \end{array} \right.}$$

### $|x-2|=5$

 $\left[\begin{array}{l} \left\lbrace \begin{array}{l} x-2\geqslant 0\\ x-2 = 5 \end{array}\right. \\ \left\lbrace \begin{array}{l} x- 2 < 0\\ -(x- 2) = 5 \end{array}\right. \end{array}\right. ;$ $\left[ \begin{array}{l} \left\lbrace \begin{array}{l} x\geqslant 2\\ x = 7 \end{array} \right. \\ \left\lbrace \begin{array}{l} x < 2\\ x = -3 \end{array} \right. \end{array}\right. ;$ $\begin{array}{l} \begin{array}{l} 7\geqslant 2\\ x_{1} = 7 \end{array} \\ \begin{array}{l} -3 < 2\\ x_{2} = -3 \end{array} \end{array}$

$x\in \left\lbrace -3; 7\right\rbrace$

$|{\color{blue}{-3}} -2| = |-5| =5;\qquad |{\color{red}{7}} -2| = |5| =5$

### $|2x-1|=3$

 $\left[\begin{array}{l} \left\lbrace \begin{array}{l} 2x-1\geqslant 0\\ 2x-1 = 3 \end{array}\right. \\ \left\lbrace \begin{array}{l} 2x- 1 < 0\\ -2x+1 = 3 \end{array}\right. \end{array}\right. ;$ $\left[ \begin{array}{l} \left\lbrace \begin{array}{l} x\geqslant 0.5\\ x = 2 \end{array} \right. \\ \left\lbrace \begin{array}{l} x < 0.5\\ x = -1 \end{array} \right. \end{array}\right. ;$ $\begin{array}{l} \begin{array}{l} 2\geqslant 0.5\\ x_{1} = 2 \end{array} \\ \begin{array}{l} -1 < 0.5\\ x_{2} = -1 \end{array} \end{array}$

$x\in \left\lbrace -1; 2\right\rbrace$

$|2\cdot ({\color{blue}{-1}}) -1| = |-3| =3;\qquad |2\cdot{\color{red}{2}} -1| = |3| =3$

### $|2x-4|=1+x$

 $\left[\begin{array}{l} \left\lbrace \begin{array}{l} 2x-4\geqslant 0\\ 2x-4 = 1+x \end{array}\right. \\ \left\lbrace \begin{array}{l} 2x- 4 < 0\\ -2x+4 = 1+x \end{array}\right. \end{array}\right. ;$ $\left[ \begin{array}{l} \left\lbrace \begin{array}{l} x\geqslant 2\\ x = 5 \end{array} \right. \\ \left\lbrace \begin{array}{l} x < 2\\ x = 1 \end{array} \right. \end{array}\right. ;$ $\begin{array}{l} \begin{array}{l} 5\geqslant 2\\ x_{1} = 5 \end{array} \\ \begin{array}{l} 1 < 2\\ x_{2} = 1 \end{array} \end{array}$

$x\in \left\lbrace 1; 5\right\rbrace$

$|2\cdot {\color{blue}{1}} -4| = |-2| = 1+{\color{blue}{1}};$

$|2\cdot{\color{red}{5}} -4| = |6| =1+{\color{red}{5}}$

### $|x-2|=4+x$

 $\left[\begin{array}{l} \left\lbrace \begin{array}{l} x-2\geqslant 0\\ x-2 = 4+x \end{array}\right. \\ \left\lbrace \begin{array}{l} x- 2 < 0\\ -x+2 = 4+x \end{array}\right. \end{array}\right. ;$ $\left[ \begin{array}{l} \left\lbrace \begin{array}{l} x\geqslant 2\\ {\color{red}{0=6}} \end{array} \right. \\ \left\lbrace \begin{array}{l} x < 2\\ x = -1 \end{array} \right. \end{array}\right. ;$ $\begin{array}{l} \begin{array}{l} \\ \varnothing \end{array} \\ \begin{array}{l} -1 < 2\\ x_{1} = -1 \end{array} \end{array}$

$x\in \left\lbrace -1\right\rbrace$

$|{\color{blue}{-1}} -2| = 4+({\color{blue}{-1}})$

### $|2x-1|=x-2$

 $\left[\begin{array}{l} \left\lbrace \begin{array}{l} 2x-1\geqslant 0\\ 2x-1=x-2 \end{array}\right. \\ \left\lbrace \begin{array}{l} 2x-1 < 0\\ -2x+1=x-2 \end{array}\right. \end{array}\right. ;$ $\left[ \begin{array}{l} \left\lbrace \begin{array}{l} x\geqslant 0.5\\ x=-1 \end{array} \right. \\ \left\lbrace \begin{array}{l} x < 0.5\\ x = 1 \end{array} \right. \end{array}\right. ;$ $\begin{array}{l} \begin{array}{l} {\color{red}{-1 < 0.5}}\\ \varnothing \end{array} \\ \begin{array}{l} {\color{red}{1 > 0.5}}\\ \varnothing \end{array} \end{array}$

$\varnothing$

### $|x+3|=x^2+x-6$

 $\left[\begin{array}{l} \left\lbrace \begin{array}{l} x+3\geqslant 0\\ x+3=x^2+x-6 \end{array}\right. \\ \left\lbrace \begin{array}{l} x+3 < 0\\ -x-3=x^2+x-6 \end{array}\right. \end{array}\right. ;$ $\left[ \begin{array}{l} \left\lbrace \begin{array}{l} x\geqslant -3\\ x^2-9=0 \end{array} \right. \\ \left\lbrace \begin{array}{l} x < -3\\ x^2+2x-3=0 \end{array} \right. \end{array}\right. ;$ $\left[ \begin{array}{l} \left\lbrace \begin{array}{l} x\geqslant -3\\ (x-3)(x+3)=0 \end{array} \right. \\ \left\lbrace \begin{array}{l} x < -3\\ (x-1)(x+3)=0 \end{array} \right. \end{array}\right. ;$ $\begin{array}{l} \begin{array}{rl} -3&\geqslant -3;\,\, x_{1}=-3 \\ 3&\geqslant -3;\,\, x_{2}=3 \end{array} \\ \begin{array}{rl} 1&{\color{red}{>}} -3;\\ -3&{\color{red}{=}} -3; \end{array} \quad \varnothing \end{array}$

$x\in \left\lbrace -3; 3\right\rbrace$

### $|x-3|=4x-3-x^2$

 $\left[\begin{array}{l} \left\lbrace \begin{array}{l} x-3\geqslant 0\\ x-3=4x-3-x^2 \end{array}\right. \\ \left\lbrace \begin{array}{l} x-3 < 0\\ 3-x=4x-3-x^2 \end{array}\right. \end{array}\right. ;$ $\left[ \begin{array}{l} \left\lbrace \begin{array}{l} x\geqslant 3\\ x^2-3x=0 \end{array} \right. \\ \left\lbrace \begin{array}{l} x < 3\\ x^2-5x-6=0 \end{array} \right. \end{array}\right. ;$ $\left[ \begin{array}{l} \left\lbrace \begin{array}{l} x\geqslant 3\\ x(x-3)=0 \end{array} \right. \\ \left\lbrace \begin{array}{l} x < 3\\ (x-3)(x-2)=0 \end{array} \right. \end{array}\right. ;$ $\begin{array}{l} \begin{array}{ll} \,\,\,\,0 {\color{red}{<}} 3;& \\ 3 \geqslant 3;&\,\, x_{1}=3 \end{array} \\ \begin{array}{rl} 3{\color{red}{=}}3;& \\ 2 < 3;&\,\, x_{2}=2 \end{array} \quad \end{array}$

$x\in \left\lbrace 2; 3\right\rbrace$

## Absolute value Equations (G)

$$\bbox[4pt]{\Large{|{\color{purple}{x-a}}| = {\color{green}{b}}}}$$

Using geometric interpretation

${\color{green}{b}}\geqslant 0$ is the value of the distance

$$\bbox[2pt]{|x-a|= b \Leftrightarrow \left[ \begin{array}{l} {\color{purple}{x-a}} = {\color{red}{b}}\\ {\color{purple}{x-a}} = {\color{blue}{-b}} \end{array} \right.}$$

### $|{\color{purple}{x-2}}|={\color{green}{5}}$

 $\left[\begin{array}{l} {\color{purple}{x-2}} = {\color{red}{5}} \\ {\color{purple}{x-2}} = {\color{blue}{-5}} \end{array}\right. ;$ $\left[\begin{array}{l} x = {\color{red}{7}} \\ x = {\color{blue}{-3}} \end{array}\right. ;$

$x\in \left\lbrace -3; 7\right\rbrace$

$|{\color{blue}{-3}} -2| = |-5| ={\color{green}{5}};\qquad |{\color{red}{7}} -2| = |5| ={\color{green}{5}}$

### $|{\color{purple}{2x-1}}|={\color{green}{3}}$

 $\left[\begin{array}{l} {\color{purple}{2x-1}} = {\color{red}{3}} \\ {\color{purple}{2x-1}} = {\color{blue}{-3}} \end{array}\right. ;$ $\left[\begin{array}{l} 2x = 4 \\ 2x = -2 \end{array}\right. ;$ $\left[\begin{array}{l} x = {\color{red}{2}} \\ x = {\color{blue}{-1}} \end{array}\right. ;$

$x\in \left\lbrace -1; 2\right\rbrace$

$|2\cdot ({\color{blue}{-1}}) -1| = |-3| ={\color{green}{3}};\qquad |2\cdot{\color{red}{2}} -1| = |3| ={\color{green}{3}}$

### $|2x-4|=1+x$

${\color{green}{1+x\geqslant 0}}$ as the Value of the Distance

 $\left\lbrace\begin{array}{l} {\color{green}{1+x\geqslant 0}}\qquad \text{!!!} \\ \left[\begin{array}{l} 2x-4= {\color{red}{+(1+x)}}\\ 2x-4 = {\color{blue}{-(1+x)}} \end{array}\right. \end{array}\right. ;$ $\left\lbrace\begin{array}{l} x\geqslant -1 \\ \left[\begin{array}{l} x=5\\ x = 1 \end{array}\right. \end{array}\right. ;$ $\begin{array}{l} x\geqslant -1 \\ \begin{array}{l} x_{1}=5 \\ x_{2} = 1 \end{array} \end{array}$

$x\in \left\lbrace 1; 5\right\rbrace$

### $|x-2|=4+x$

${\color{green}{4+x\geqslant 0}}$ as the Value of the Distance

 $\left\lbrace\begin{array}{l} {\color{green}{4+x\geqslant 0}}\qquad \text{!!!} \\ \left[\begin{array}{l} x-2= {\color{red}{+(4+x)}}\\ x-2 = {\color{blue}{-(4+x)}} \end{array}\right. \end{array}\right. ;$ $\left\lbrace\begin{array}{l} x\geqslant -4 \\ \left[\begin{array}{l} \varnothing \\ x = -1 \end{array}\right. \end{array}\right. ;$ $\begin{array}{l} x\geqslant -4 \\ \begin{array}{l} \ldots \\ x_{1} = -1 \end{array} \end{array}$

$x\in \left\lbrace -1\right\rbrace$

### $|2x-1|=x-2$

${\color{green}{x-2\geqslant 0}}$ as the Value of the Distance

 $\left\lbrace\begin{array}{l} {\color{green}{x-2\geqslant 0}} \\ \left[\begin{array}{l} 2x-1= {\color{red}{x-2}}\\ 2x-1 = {\color{blue}{2-x}} \end{array}\right. \end{array}\right. ;$ $\left\lbrace\begin{array}{l} x\geqslant 2 \\ \left[\begin{array}{l} x=-1 \\ x = 1 \end{array}\right. \end{array}\right. ;$ $\begin{array}{l} x\geqslant 2 \\ \begin{array}{l} {\color{red}{-1 < 2}} \\\,\,\,\,\, {\color{red}{1 < 2}} \end{array} \end{array}$

$\varnothing$

### $|x+3|=x^2+x-6$

$x^2+x-6\geqslant 0$ as the Value of the Distance

 $\left\lbrace\begin{array}{l} x^2+x-6\geqslant 0 \\ \left[\begin{array}{l} x+3= {\color{red}{x^2+x-6}}\\ x+3 = {\color{blue}{6-x-x^2}} \end{array}\right. \end{array}\right. ;$ $\left\lbrace\begin{array}{l} (x-2)(x+3)\geqslant 0 \\ \left[\begin{array}{l} (x-3)(x+3)=0 \\ (x-1)(x+3)=0 \end{array}\right. \end{array}\right.$

$x\in \left\lbrace -3; 3\right\rbrace$

### $|x-3|=4x-3-x^2$

$4x-3-x^2\geqslant 0$ as the Value of the Distance

 $\left\lbrace\begin{array}{l} 4x-3-x^2\geqslant 0 \\ \left[\begin{array}{l} x-3= {\color{red}{4x-3-x^2}}\\ x-3 = {\color{blue}{x^2-4x+3}} \end{array}\right. \end{array}\right. ;$ $\left\lbrace\begin{array}{l} (x-1)(x-3)\leqslant 0 \\ \left[\begin{array}{l} x(x-3)=0 \\ (x-2)(x-3)=0 \end{array}\right. \end{array}\right.$

$x\in \left\lbrace 2; 3\right\rbrace$