Let f be a function on 
to given by:
The function just-defined is called the absolute value function. For every x ∈ the image of x under f is denoted by |x|. Thus,
The graph of the absolute value function is shown below.
Illustration 1
Compute |-4|.
Answer
Since -4 < 0, by the definition of the absolute value function, |x| = -x holds. Hence |-4| = -(-4) = 4.
Compute |¾|.
Answer
Since ¾ ≥ 0, by the definition of the absolute value function, |x| = x holds.. Thus, |¾| = ¾.
Compute |0|.
Answer
Since 0 ≥ 0, by the definition of the absolute value function, |x| = x holds.. So, |0| = 0.
Compute |-√5|.
Answer
Since -√5 < 0, by the definition of the absolute value function, |x| = -x holds. Hence |-√5| = -(-√5) = √5.
From Illustration 1, we may conjecture that the absolute value of a real number cannot be negative. Indeed, it is one of the properties of the absolute value function. Look at Theorem 1.
Theorem 1
If x is a real number, then:
a) |x| ≥ 0
b) |x| = 0 if and only if x = 0
Part a) of the theorem states that the absolute value of a real number cannot be negative. In addition, it can be inferred from part b) that i) |0| = 0 and ii) the only real number whose absolute value is zero is zero. As an example of how ii) is applied, suppose that |p + 3| = 0 is given. How to find p? Since the only real number whose absolute value is zero is zero, it must hold that p + 3 = 0, there is no other possibility. Therefore, p = -3.
Triangle Inequality
Let a and b be real numbers. Then |a + b| ≤ |a| + |b|.
Illustration 2
Let a = -3 and b = 5.
Note that a + b = -3 + 5 = 2. It follows that |a + b| = |2| = 2.
On the other hand, |a| + |b| = |-3| + |5| = 3 + 5 = 8.
We have verified that |-3 + 5| ≤ |-3| + |5|.
What is the relationship between triangle inequality and a triangle?
If a, b, and c are the lengths of the sides of a triangle, then the following apply: a < b + c, b < a + c, and c < a + b. In other words, for a triangle is possible to be constructed the length of any side must be less than the sum of the length of the other two sides. We cannot construct a triangle (on a plane) whose sides are 5 cm, 7 cm, and 15 cm in length because 15 cm > 5 cm + 7 cm.
Theorem 2
Let a, b ∈ . Then,
a) |-a| = |a|
b) |a – b| = |b – a|
Sample Problem 1
Let a, b, and c be real numbers.
Given |a – b| < ε/2 and |c – b| < ε/2, prove that |a – c| < ε.
Answer
Note that |a – c| = |(a – b) + (b – c)|.
By triangle inequality, |(a – b) + (b – c)| ≤ |a – b| + |b – c| and furthermore |a – c| ≤ |a – b| + |b – c|
By part b of Theorem 2, |b – c| = |c – b|. This results in:
|a – c| ≤ |a – b| + |c – b| ……………………………………………………………………………………………………………………………………………… (*)
From (*), |a – b| < ε/2, and |c – b| < ε/2., it follows that:
|a – c| ≤ |a – b| + |c – b| < ε/2 + ε/2 = ε
So, |a – c| < ε.
Theorem 3
Let a, x ∈ . Then,
a) |x| < a ⇔ -a < x < a ; a > 0
b) |x| ≤ a ⇔ -a ≤ x ≤ a ; a ≥ 0
c) |x| > a ⇔ x < -a or x > a ; a > 0
d) |x| ≥ a ⇔ x ≤ -a or x ≥ a ; a ≥ 0
Here is an example of how to apply Theorem 3 in solving inequalities involving absolute values.
Sample Problem 2
Find the solution set of |2x – 5| < 7, x ∈ .
Answer
By Theorem 3 part a, |2x – 5| < 7 ⇔ -7 < 2x – 5 < 7.
Add 5 to each part of -7 < 2x – 5 < 7.
-7 + 5 < 2x – 5 + 5 < 7 + 5
-2 < 2x < 12
Divide each part of the inequality -2 < 2x < 12 by 2. Accordingly, we have -1 < x < 6.
Thus, the solution set is 
.
It can also be easily proved that if a, b ∈ then:
and
; b ≠ 0
