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 TwoInterval 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.
