Contents

1 Introduction

2 Definition

3 Simplify expressions involving absolute values

4 Remove the absolute value symbols when

the variable is inside the bars

5 Alternative definition

6 Elementary properties

7 Properties of absolute value under basic operations

8 Properties of absolute value with equations and Inequalities.

Proof of the rules

1 Introduction

2 Definition

3 Simplify expressions involving absolute values

4 Remove the absolute value symbols when

the variable is inside the bars

5 Alternative definition

6 Elementary properties

7 Properties of absolute value under basic operations

8 Properties of absolute value with equations and Inequalities.

Proof of the rules

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.

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

Place your mouse over the expression to see the answer.

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

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
$$|3-\pi|=-(3-\pi) $$

You take into account that they behave as grouping symbols. Thus, we agree perform operations inside absolute value symbols.

Do all operations inside the absolute value bars.

Perform all calculations inside bars using the rule for the order of operations.

1º Grouping symbols; 2º Exponents and radicals; 3º Multiplications and divisions;

4º Additions and subtractions

$$\begin{array}{rcl} 2|3-3\cdot 5| +6 &=& 2|3-15| +6\\ &=& 2|-12| +6\\ \end{array}$$

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

Press the button to see how to perform the calculation of the above example in any software.

The following example illustrates the role of the variable to remove the bars

a)

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.

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

Answer.

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

$x<-3$ 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 $$ |2x+6|=2x+6$$

Determine the values of the variable, $x$, for which the expression inside the bars is negative.

We solve the inequality: $$ 3 {\;< \;} x $$ We have, $(3-x)$ is negative iff $x>3$.

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 $x\leq 3$

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
$$|x|=\sqrt{x^2}$$

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.

From property 3 applied to $a-b$ we have

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.

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

Let $c>0$ and let $x$ be real number.

$$|x| = c \quad \Leftrightarrow \quad x =- c \quad o \quad x = c $$

Let $c>0$ and let $x$ and $y$ be real numbers.

$$|x| =|y| \quad \Leftrightarrow \quad x =y \quad o \quad x =-y $$

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

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

$(\Rightarrow )$ 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 $-c\leq x$.

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 $|x| \leq c$

♦ $x \geq 0$

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

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

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

Let $x$ and $y$ be real numbers.

$$|x| < |y| \quad \Leftrightarrow \quad x ^2 < y^2 $$

$(\Rightarrow )$ 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