# Absolute Value and Distance

If a is a real number, the absolute value of a is

The absolute value of a number cannot be negative. For example, let a = -4. Then, -4 < 0, because you have

| a | = | -4 | = -(-4) = 4.

Remember that the symbol -a does not necessarily mean that is -a negative.

Operations with Absolute Value

Let a and b be real numbers and let n be a positive integer.

Properties of Inequalities and Absolute Value

Let a and b be real numbers and let k be a positive real number.

1. -| a | ≤ a ≤ | a |

2. | a | ≤ k if and only if -k ≤ a ≤ k.

3. k ≤ | a | if and only if k ≤ a or a ≤ -k.

4. Triangle Inequality: | a + b | ≤ | a | + | b |

Properties 2 and 3 are also true if ≤ is replaced by <.

Example 1

Solving an Absolute Value Inequality

Solve | x - 3| ≤ 2.

Solution

Using the second property of inequalities and absolute value, you can rewrite the original inequality as a double inequality.

 -2 ≤ x - 3 ≤ 2 Write as double inequality. -2 + 3 ≤ x - 3 + 3 ≤ 2 + 3 Add 3. 1 ≤ x ≤ 5 Simplify.

The solution set is [1, 5], as shown in the figure below.

Example 2

A Two-Interval Solution Set

Solve 3 < | x + 2 |

Solution

Using the third property of inequalities and absolute value, you can rewrite the original inequality as two linear inequalities.

 3 < x + 2 or x + 2 < -3 1 < x or x < -5

The solution set is the union of the disjoint intervals (-∞, -5) and (1,) as shown in the figure below.

Examples 1 and 2 illustrate the general results shown in the figure below.

Note that if d > 0, the solution set for the inequality | x - a | ≤ d is a single interval, whereas the solution set for the inequality | x - a | ≥ d is the union of two disjoint intervals.

The distance between two points a and b on the real line is given by d = | a - b | = | b - a |

The directed distance from a to b is b - a and the directed distance from b to a is as shown in the figure below.

Example 3

Distance on the Real Line

a. The distance between -3 and 4 is or

| 4 - (-3)| = | 7 | = 7 or | -3 - 4 | = | -7 | = 7.

(See the figure below.)

b. The directed distance from -3 to 4 is 4 - (-3) = 7.

c. The directed distance from 4 to -3 is -3 - 4 = -7.

The midpoint of an interval with endpoints a and b is the average value of a and b. That is,

To show that this is the midpoint, you need only show that (a + b)/2 is equidistant from a and b.