**Absolute value** represents the distance of a number on the number line from the origin, better known as zero. The absolute value of a number is never negative. It ignores in which direction from zero the number lies, it only matters how far it is. This distance is also called a module of a number.

For example, if look at the points: $\ A (4)$, $\ B(2)$, $C(-2)$ marked on the number line we can observe their distance from the origin.

We can see that the distance between points $A$ and $O$ is $4$, so the absolute value of the number $4$ equals $4$. The distance both from the points $B$ and $C$ to $O$ is $2$, so their aboslute value equals $2$. Notice how the absolute value of positive numbers is exactly that number, and the absolute value of negative numbers is the positive version of that number.

The same applies to all real numbers, rational, irrational, integers and natural numbers.

What about the zero? The absolute value of $0$ is $0$ since it is at that point from which we measure the distance in question. This is why we don’t say that the absolute value of a number is positive: Zero is neither negative nor positive.

### Properties of absolute value

Let take $x$, a real number. The absolute value of $x$ is denoted by $| x |$ and it is defined as:

Whether $x$ is positive or negative, its absolute value will always be positive.

$\mid x \mid = 0$ if and only if $ x= 0$

$\mid x \mid \ge 0$ for every $ x \in \mathbb{R}$

An absolute value also has some properties:

- $\mid a \cdot b \mid = \mid a \mid \cdot \mid b \mid$ for every $ a, b \in \mathbb{R}$
- $\mid \frac{a}{b} \mid = \frac{\mid a \mid}{\mid b \mid}$, $ b \not= 0$ for every $ a \in \mathbb{R}$, $ b \in \mathbb{R} \setminus \{ 0 \}$
- $\mid a + b \mid \le \mid a \mid + \mid b \mid$, for every $ a, b \in \mathbb{R}$

Are those equalities true?

- The product of two positive numbers and the product of two negative numbers will always be positive, and if one is negative and the other positive the product is negative. If we first multiply two numbers and then take their absolute value it will always be non-negative because of the way the absolute value is defined. If we first take the absolute value of two numbers and than multiply them, the product will always be non-negative. Since the absolute value only changes the sign the output in both cases is the same.
- Similarly, a quotient is positive if both the divisor and dividend are either postive or negative, so it doesn’t matter if you divide two positive numbers, or if you remove the minus sign from the result.
- Why is there an inequality? Well, the logic we used for multiplication and division cannot be applied here, because it’s not the same if you add two positive numbers, and if you have subtraction and then you make it positive. For example:

**Example 1**. Calculate the value of the given expression if:

$\ a = 2$, $\ b = -5$ ,

$\ |(|a|-|b|)| = ?$.

First, we’ll include the values of $a$ and $b$ into the task:

$\ |(|2|-|-5|)| = ?$

When you have more than one absolute value, you treat them just like parenthesis; first solve inside the absolute value and then outside.

$\ |(|2|-|-5|)| = |2-5| = |-3| = 3$

**Example 2.** Calculate the value of given expression if:

$ a = -\frac{-1}{2}$ , $b = \frac{4}{3}$. $\frac{2|a – 3b|}{|b|} = ?$

$\frac{2 \mid -\frac{1}{2} – 4 \mid}{\mid \frac{4}{3} \mid} $

$=\frac{2 \mid -\frac{9}{2} \mid}{\frac{4}{3}}$

$ = \frac{9}{\frac{4}{3}} $

$= \frac{27}{4}$

## Graph of absolute value

**Example 1**.

How would you draw the graph of $\ f(x)= |x|$?

Let’s take a look at what happens with the number from $-3$ to $3$.

The graph is symmetric with respect to the $*y$* –*axis.*

In the next example we will graphically review the difference between $\|a+b|$ and $\ |a| + |b|$.

**Example 2**.

We’ll compare two functions: $\ f(x) = |x – 2|$ and $\ g(x) = |x| – 2$.

Parts of two different functions overlap. Why? The constant inside the absolute value moves the whole graph to the right if that constant is negative and to the left if it is positive. But a constant outside of the absolute value, such as the $-2$ in the function $g(x)$, the graph will move upward if that value is positive, and downward if it’s negative. The $x$ coordinate of the vertex (the point in which the two lines connect, the tip of the graph) is exactly the value by which the graph is shifted left or right, and the $y$ coordinate is the value by which the graph is shifted up or down.

There are only two more options for a simple graph with absolute value, if the variable is multiplied by a constant greater than or lesser than zero.

**Example 3**.

Let’s compare two functions: : $\ f(x) = 2 |x|$, $ g(x) = \frac{1}{2}|x|$ , $\ z(x) = -2 |x|$.

From here, you can conclude that if you have smaller coefficients multiplying an absolute value, its graph will become more widespread, and if you have larger coefficients, it will become narrower. And if you multiply the function with a negative coefficient, the graph will move to the negative side of $y$ – axis.

## Simple absolute equations and inequalities

__Example 1.__

Solve: $\|x| + 2 = 4$.

These equations are solved in a similar way as are the ones without the absolute values. That means, by moving the unknowns, whether they are bound with a module or not, to the left side, and the constants to the right. Remember to change the sign of the constant due to the side change and this is what we get:

$\ |x| = 2$. However, since the unknowns are bound by the absolute value this equation has two solutions: $2$ and $-2$. That is because $\ |2| = |-2| = 2$.

__Example 2.__

$\mid -0.2x \mid + \mid -\frac{3}{4} \mid = 2$ -> $\mid -\frac{1}{5}x \mid + \frac{3}{4} = 2$

$\mid -\frac{1}{5}x \mid = \frac{5}{4}$ here we can use property number 1 $ (\mid a * b \mid = \mid a \mid * \mid b \mid)$

$\mid -\frac{1}{5} \mid \mid x \mid = \frac{5}{4}$

$\frac{1}{5} \mid x \mid = \frac{5}{4} \setminus 5$

$\mid x \mid = \frac{25}{4} \rightarrow x_1 = \frac{25}{4}, x_2 = -\frac{25}{4}$

__Example 3.__

$\ |-2x – 2| = 2$

First we have to know for what values of $x$ will the value of the whole module become negative, and for what positive. That is done by “searching zeroes” (equating the expression with $0$, and then solving that equation) from the module:

Choose some random number from the interval $\ <-\infty, -1 >$ and then see the value of the module in that case.

For example we’ll take $\ -2: -2(-2) – 2 = 2$ which is greater than zero. And that means that for the interval $<-∞, -1 >$ everything inside module will remain the same, and

1. Case : $\ x < -1$ -> everything inside the module remains the same: $\ – 2x – 2 = 2$ -> $\ x = – 2$

2. Case : $\ x > -1$ -> everything inside the module changes sign $\ 2x + 2 = 2$ -> $\ x = 0$

To check our solution, we can simply replace the x with an appropriate value:

For -2: $\ |-2(-2)-2| = 2$ , $\ |2| = 2$

For 0: $\ |0-2| = 2$ , $\ |2| = 2$

Also, you should always check if your x is from the interval you are looking for. If it’s not, the $x$ is not included in the set of solutions. $-2 < -1, 0 > -1$. Since $-2$ and $0$ are parts of our interval, they are included in our set of solutions.

**Example 4.**

$|\frac{x-1}{x-2}|=\frac{1}{2}$

Solution:

We use properties: $|a|=|b| \Leftrightarrow$ $a=b$ or $a=-b$

$\frac{|x-1|}{|x-2|}=\frac{1}{2}$ $/\cdot 2|x-2|, x \neq 2$

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

$|2x-2|=|x-2|$

Now, we solve this like before:

1. $2x-2=x-2$

$2x-2=x-2 /+(-x+2)$

$2x-2-x+2=x-2-x+2$

$x=0$

2. $2x-2=-x+2$

$2x-2=-x+2 /+(x-2)$

$2x-2+x-2=-x+2+x-2$

$x=\frac{4}{3}$

__Example 5.__

$\ |x| > 2$

Now, variable $x$ can again be either positive or negative. So we’ll break it down to cases:

Case 1: for $\ x > 0$ -> $\ x > 2$

Case 2: for $\ x > 0$ -> $\ – x > 2$ -> $\ x < – 2$

And the final solution is the union of those two intervals: $\ <-\infty, -2 > U < 2, \infty >$.

The solution represented graphically:

__Example 6.__

$\ |x + 3| > 4$ (the same as the equations)

$\ x + 3 = 0 \Rightarrow \ x = – 3$

For $x = 0$ we have $\ 0 + 3 = 3 > 0$.

1. Case: $\ x > 3 \Rightarrow$ everything inside module remains the same $ \Rightarrow \ x + 3 > 4$ , $\ x > 1$

2. Case: $\ x < 3 \Rightarrow$ everything inside module changes sign $\Rightarrow \ – x – 3 > 4$ , $\ – x > 7$ , $\ x > – 7$

Here we have to look out for the conditions:

In the first case, we are looking only at numbers greater than three. Since we got a solution $x > 1$, we have to look at their intersection, which is $\ x > 3$

In the second case, we are only looking at numbers lesser than $3$, and we got the solution $x > -7$. Once again, we are looking for the intersection. If you’re having trouble imagining their intersection, you can always help yourself by drawing it on the number line:

So our solution for the second case is $\ <-7, 3 >$

And the solution to the whole task is: $\ <-7, 3 > $.

## Absolute value worksheets

**Solve and graph functions - Integers** (1.2 MiB, 757 hits)

**Solve and graph functions - Decimals** (1.2 MiB, 635 hits)

**Solve and graph functions - Fractions** (1.3 MiB, 625 hits)

**Equations - Integers** (509.2 KiB, 683 hits)

**Equations - Decimals** (542.2 KiB, 621 hits)

**Equations - Fractions** (784.8 KiB, 589 hits)