1.5 Absolute Value

What we learn here:

Definition of the absolute value
Properties of the absolute value
Distance between numbers
Neighborhoods and deleted neighborhoods

Definition of the Absolute Value

It is frequently desirable to measure how large a quantity is, regardless of its sign. In such cases, we use merely the absolute value of the quantity.

The absolute value (or modulus) of a real number $x$ , is a nonnegative real number denoted by $| x |$ , and defined as follows

$| x | = {\begin{cases} x & if x > 0 \\ 0 & if x = 0 \\ - x & if x < 0 \end{cases}$

Geometrically the absolute value of a number $x$ is its distance from $0$ regardless of the direction.
In computer languages and mathematical packages, the absolute value of $x$ is often denoted by abs(x).

Example 1

Find

| 2 |, | - 2 |, | 0 |, | 3 - \sqrt{3} |, | \sqrt{3} - 3 |

Solution

Because

2, 0, 3 - \sqrt{3}

are nonnegative [

\sqrt{3} \approx 1.73

, so

3 - \sqrt{3} \approx 1.27

], we have

| 2 | = 2, | 0 | = 0, | 3 - \sqrt{3} | = 3 - \sqrt{3}

but

- 2

and

\sqrt{3} - 3

are negative, so we have

| - 2 | = - (- 2) = 2, | \sqrt{3} - 3 | = - (\sqrt{3} - 3) = 3 - \sqrt{3} .

Properties of the Absolute Value

By definition, we have
$- | x | \leq x \leq | x |$
because if $x > 0$ , then $| x | = x$ and we have the sign of equality on the right and the sign of inequality on the left (a positive number is larger than a negative one). If $x < 0$ , then $| x | = - x$ , and we have the sign of equality on the left and the sign of inequality on the right. [Note that $a \leq b$ means $a < b$ or $a = b$ ]

From the definition of the absolute value, it follows that for every real numbers $a$ and $b$ we have:

$| a | \geq 0$
$| a | = 0$ if and only if $a = 0$
$| | a | | = | a |$ (the absolute value of the absolute value of $a$ is the absolute value of $a$ because $| a | \geq 0$ )
$| - a | = | a |$ .
$| a b | = | a | | b |$ .
$| \frac{a}{b} | = \frac{| a |}{| b |}$ (provided $b \neq 0$ )
If $r > 0$

$\| x \| \leq r$	is equivalent to	$- r \leq x \leq r$	(i)
$\| x \| \geq r$	is equivalent to	$- x \leq r$ or $r \geq x$	(ii)
$\| x \| = r$	is equivalent to	$x = r$ or $x = - r$	(iii)

Geometrical interpretation

Let

r > 0

and

| x | \leq r .

The above relationship implies that

x

is nearer to 0 than

r

, and you can see from the following figure that this is possible if and only if

x

lies between

- r

and

r

- r \leq x \leq r .

You can show the significance of the other two statements geometrically.

8. $| a + b | \leq | a | + | b |$ (known as triangle inequality)

more about triangular inequality

In particular, if $a$ and $b$ have the same signs (both are positive or negative) or at least one of them is zero, then $| a + b | = | a | + | b |$ ; otherwise $| a + b | < | a | + | b |$ . The proof of the triangular inequality is as follows: we know

$- | a | \leq a \leq | a |$

and

$- | b | \leq b \leq | b | .$

Adding two inequalities we obtain

$- (| a | + | b |) \leq a + b \leq | a | + | b | .$

Let $c = | a | + | b |$ and $x = a + b$ . Because $- c \leq x \leq c$ , it follows from (1) that $| x | \leq c$ or

$| x | = | a + b | \leq | a | + | b | .$

This is what we were trying to prove.

If follows from the triangular property and (4) that

$| a | = | (a + b) - b | \leq | a + b | + | - b | = | a + b | + | b |$

$\Rightarrow | a | - | b | \leq | a + b | .$

Therefore, always we have

$| a | - | b | \leq | a + b | \leq | a | + | b | .$

Example 2

Prove that for every real number $a$ and $b$

$| a - b | \leq | a | + | b |$

Solution

Using the triangular inequality and the fact that

| - b | = | b |

| a - b | = | a + (- b) | \leq | a | + | - b | = | a | + | b | .

Example 3

Prove that for every real number $a$ and $b$

$| a | - | b | \leq | a - b |$

Solution

We add and subtract

b

and then use the triangular inequality

| a | = | a - b + b | \leq | a - b | + | b |

Now if we subtract

| b |

from both sides, we get

| a | - | b | \leq | a - b | + | b | - | b |

It follows from (4) that

| a - b | = | b - a |

Distance Between Numbers

By distance between two numbers, we often mean the value of the larger less the smaller. Thus the distance between 4 and 10 is 6, and the distance between 10 and 4 is also 6 .

Let $a$ and $b$ be two real numbers. Then, the distance between the points $a$ and $b$ on the number line, denoted by $d (a, b)$ , is

$d (a, b) = | a - b | .$

Because $| a - b | = | b - a |$ , the distance from $a$ to $b$ is the same as the distance from $b$ to $a$ $d (a, b) = d (b, a)$

It follows from Equation (ii) that for any real number $a$ and any positive number $r$ ,

$| x - a | < r$

is equivalent to

$- r < x - a < r$

or if we add $a$ to each side:

$a - r < x < a + r$

This means that the distance of $x$ from $a$ is less than $r$ if and only if $x$ is between $a - r$ and $a + r$ .

Example 4

Find the set

I

, if it consists of all points whose distance from the point 2 is less than 0.6.

Solution

As discussed above

This set is graphed in the following figure.

Neighborhoods

Let $I$ be the set of all points whose distance from a fixed point $a$ is less than a number $δ > 0$ . Then

$\begin{aligned} I & = {x | | x - a | < δ} \\ = {x | - δ < x - a < δ} \\ = {x | a - δ < x < a + δ} \end{aligned}$

[For the second equation, we used the fact that $| t | < δ$ is equivalent to $- δ < t < δ$ and for the third equation, we added $a$ to each side]

Such a set is called a neighborhood (or more precisely the $δ$ -neighborhood) of $a$ and $δ$ is called the radius of the neighborhood. The $δ$ -neighborhood of $a$ is shown in the following figure.

The δ-neighborhood of a is the set of all points whose distance from a is less than δ >0 . The radius of this neighborhood is δ.

Now let’s consider the set of all points such that

$0 < | x - a | < δ$

$J = {x | 0 < | x - a | < δ}$

Here we have two inequalities

$0 < | x - a | and | x - a | < δ$

Recall that the absolute value is always nonnegative (that is $0 \leq | t - a |$ for all $t$ ) , so

$0 < | x - a |$

means

$| x - a | \neq 0$

or equivalently

$x \neq a$

[Recall that $| t | = 0$ if and only if $t = 0$ ]. Therefore,

$\begin{aligned} J & = {x | 0 < | x - a | < δ} \\ = {x | | x - a | < δ and x \neq a} \end{aligned}$

That is $J$ is the $δ$ -neighborhood of $a$ with the midpoint $a$ removed. The set $J$ is called the deleted $δ$ –neighborhood or punctured neighborhood of $a$ . The deleted $δ$ -neighborhood of $a$ is shown in the following figure.