### Introduction

The concept of absolute value or modulus of a number can think as the distance between a point with coordinate $a$ and the origin on a number line.

Remember the distance is always positive or 0.

The distance between a point and the origen ignores the sign of the coordinate.
Algebraically, the distance between $-2$, negative, and zero on the number line is the additive inverse $-(-2)$. If $a=-2$, this distance can be written as $-a$. In general, for any point with coordinate $a$ negative, the distance from the origin is equal to $-a$. However, for any point with positive coordinate, $a$, the distance is the same number. The absolute value of a number $a$ behaves in a similar way.

### Defining the absolute value of a number

The absolute value of a real number $x$, denoted by $|x|$, is itself when $x$ is positive or zero and it is $-x$ when $x$ is negative. The minus sign reverses its sign.
This we write as follows

Definition $$|x|=\left\{\begin{matrix} x ,& \text{if }x\geq 0\\ -x, & \text{if }x< 0 \end{matrix}\right.$$

$|x|$ is read as the absolute value of .$x$

Examples Rewrite the expression without the absolute value symbols.

In some numbers is not so obvious its sign. You have to take care when you remove the absolute value symbols.

Example Rewrite the expression without absolute value bars. $$|3-\pi|$$ Solution
Determine the sign of $3-\pi$
We approximated the number $\pi$ by $3,14$ $$3-\pi \approx 3-3,14=-0,14$$ The number is negative.

Use the definition of absolute value.
The number $3 - \pi$ is negative, the absolute value of ·$3 - \pi$ is the opposite $$|3-\pi|=-(3-\pi)$$

How simplify expressions containing absolute values?
You take into account that they behave as grouping symbols. Thus, we agree perform operations inside absolute value symbols.

Example Simplify $$2|3-3\cdot 5| +6$$ Solution
Click to see the development of the step.
Do all operations inside the absolute value bars.
Step 1
Perform all calculations inside bars using the rule for the order of operations.
1º Grouping symbols; 2º Exponents and radicals; 3º Multiplications and divisions;
$$\begin{array}{rcl} 2|3-3\cdot 5| +6 &=& 2|3-15| +6\\ &=& 2|-12| +6\\ \end{array}$$
Use the definition of absolute value in order to remove the bars.
Step 2 $$\begin{array}{rcl} 2|3-3\cdot 5| +6 &=& 2|-12| +6\\ &=& 2\cdot 12 +6\\ \end{array}$$
Determine the value of the resulting expression.
Step 3 $$\begin{array}{rcl} 2|3-3\cdot 5| +6 &=& 24 +6\\ &=& 30\\ \end{array}$$

Many calculators have the Abs key to evaluate absolute values, too computer application programs have the mathematical functions: Abs( ).

Press the button to see how to perform the calculation of the above example in any software.
Example           2*abs(3-3*5)+6

How to remove the absolute value symbols when the variable is inside the bars?
The following example illustrates the role of the variable to remove the bars

Example
a)
Determine the values of the variable for which $|2x+6|=-(2x+6)$.
Solution
You want to determine the values of the variable in which the absolute value changes the sign to the expression $2x+6$. This occurs if and only if the expression between the bars is negative.
Write the condition " expression between the bars is negative" as an inequality.
$$2x+6<0$$

Solve the inequality.
$$\begin{array}{rl} 2x+6&<&0\\ 2x&<&-6\\ x&<&-3 \end{array}$$

If $x <-3$ then $$|2x+6|=-(2x+6)$$

b) Find all values of $x$ that make $|2x+6|=2x+6$?

$x<-3$ if and only if $2x+6$ is negative.
Hence,
$x\geq -3$ iff $2x+6$ is positive or 0.

Consequently, if $x\geq -3$ then $$|2x+6|=2x+6$$

Exercise Write $|3-x|$ in an equivalent form that does no contain absolute value bars.
Solution .
Determine the values of the variable, $x$, for which the expression inside the bars is negative.
Step 1 $(3-x)$ es negative iff $3-x<0$.
We solve the inequality: $$3 {\;< \;} x$$ We have, $(3-x)$ is negative iff $x>3$.
Point out all values of the variable for which the expression inside the absolute value symbols is positive or zero.
Step 2
In the complement of above set: $x\leq 3$. This is,

$3-x\geq 0$ ( $3-x$ is positive or 0) if and only if $x\leq 3$
Write $|3-x|$ as the definition.
Step 3 $$|3-x|=\left\{\begin{matrix} 3-x ,& \text{if }x\leq 3\\ -(3-x), & \text{if }x >3 \end{matrix}\right.$$

Since $x^2$ is a positive number and $\sqrt{* }$ denote the positive square root , we have the following algebraic expression to define the absolute value
Alternative definition $$|x|=\sqrt{x^2}$$

Example
Hover over the expression to verify that the example is agreed with the definitions

This definition is useful in order to prove some properties of the absolute value.

### Properties of absolute value

#### Immediate properties

First, we enunciate some elementary properties of the absolute value. They follow from the definition

Properties   Let $a$ be a real number.
1  $|a| =0 \; \Leftrightarrow \; a=0$
2  $|a| \geq 0$
3  $|-a|= |a|$

From property 3 applied to $a-b$ we have
Property   Let $a$ and $b$ be real numbers.
4  $|b-a|= |a-b|$

#### Properties of the absolute value under elementary operations

Proposition   Let $a$ and $b$ be real numbers.
5  $|a\cdot b| =|a| |b|$
6  $|\frac ab| =\frac{|a|}{ |b|}$, $\quad b\neq 0$
7  $|a^n| =|a|^n, \quad$
for any nonnegative integer $n$

Proof 5
We wiil use the definition with root and properties of exponents and radicals.
$$|ab|=\sqrt{(ab)^2}=\sqrt{a^2}\sqrt{b^2}=|a||b|$$

The proof 6 is similar to 5.

Proof 7   The proposition is evident for $n=0$. For positive integer n, we will use the definition of positive Integer exponents and property of the absolute value of a product $$|a^n|=|a\cdot a \cdots a|=|a|\cdot |a| \cdots |a|= |a|^n$$

Inequality triangle   Let $a$ and $b$ be real numbers. $$|a+b|\leq |a|+|b|$$

#### Properties useful to solve absolute value equations and inequalities

These rules allow to transform the problem into solving equations or inequalities without absolute value.

Theorem
Let $c>0$ and let $x$ be real number.
$$|x| = c \quad \Leftrightarrow \quad x =- c \quad o \quad x = c$$

Theorem
Let $c>0$ and let $x$ and $y$ be real numbers.
$$|x| =|y| \quad \Leftrightarrow \quad x =y \quad o \quad x =-y$$
If two numbers are equals in absolute value then they are equals or opposites of each other.

Theorem
Let $c$ be a positive real number and let $x$ be real number.

$$\begin{array}{ll} a)& |x| \leq c \quad \Leftrightarrow \quad -c \leq x \leq c \\ b)& |x| < c \quad \Leftrightarrow \quad -c < x < c \end{array}$$
Proof $a$
$(\Rightarrow )$ Suppose that $|x|\leq c$.We must prove thate $-c\leq x$ and $x\leq c$

We have $x\leq |x|$, then, by transitivity $x\leq c$

To see the other inequality observe that $-x\leq |x|$, hence $x\geq -|x|$ and we have $-|x|\geq -c$. Again by transitivity we get $x\geq -c$, this is $-c\leq x$.
$(\Leftarrow)$ Asumimos que $-c \leq x \leq c$. We will consider the two possibilities.

♦ $x \geq 0$
We will use that $x \leq c$ Since $|x|=x$ we have that $|x| \leq c$

♦ $x < 0$
In this case we have $|x|=-x$. We use that $-c \leq x$, this is $c \geq -x$, to obtain that $c \geq |x|$, and can be written as $|x| \leq c$

The next theorem follows from the above theorem
Theorem
Let $c$ be a positive real number and let $x$ be real number.

$$\begin{array}{ll} a) & |x| \geq c \quad \Leftrightarrow \quad x\leq -c \quad o \quad x \geq c \\ b) & |x| > c \quad \Leftrightarrow \quad x > -c \quad o \quad x> c \end{array}$$

Theorem
Let $x$ and $y$ be real numbers.
$$|x| < |y| \quad \Leftrightarrow \quad x ^2 < y^2$$
Proof
$(\Rightarrow )$ Since $|x|< |y|$ we have $$|x|\cdot |x|< |y| \cdot |x| \quad y \quad |x|\cdot |y| < |y| \cdot |y|$$ By transitivity $$|x|\cdot |x| < |y| \cdot |y|$$ This is $$|x|^2 < |y|^2$$ $$|x^2|< |y^2|$$ Since $x^2$ is positive or zero, we have $$x^2 < y^2$$
$(\Leftarrow)$ \begin{array}{lcl} x^2< y^2 & \Rightarrow & |x^2|< |y^2| \\ &\Rightarrow & |x|^2< |y|^2 \\ &\Rightarrow & |x|^2- |y|^2< 0 \\ &\Rightarrow & (|x|- |y|)(|x|+ |y|) < 0 \\ &\Rightarrow & (|x|- |y|) < 0 \\ &\Rightarrow & |x|< |y| \end{array}

Definition and properties of the absolute value