Absolute Value and Distance

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.

Definition 1.

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 & \text{if }x>0\\ 0 & \text{if }x=0\\ -x & \text{if }x<0 \end{cases}

Geometrically the absolute value of a number $x$ is its distance from 0 regardless of the direction.

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} .

In computer languages and mathematical packages, the absolute value of $x$ is often denoted by abs(x).
We have by definition: $- | x | \leq x \leq | x |$ 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 above definition, it follows that for every real numbers $a$ and $b$ we have:

  |
  a
  |
  ≥
  0

  |
  a
  |
  =
  0
 if and only if 
  a
  =
  0

  |
  |
  a
  |
  |
  =
  |
  a
  |
 (because 
  |
  a
  |
  ≥
  0
)

  |
  −
  a
  |
  =
  |
  a
  |
.

  |
  a
  b
  |
  =
  |
  a
  |
   
  |
  b
  |
.

  
    |
    
      
        a
        b
      
    
    |
  
  =
  
    
      
        |
        a
        |
      
      
        |
        b
        |
      
    
  
 (provided 
  b
  ≠
  0
)
If 
  r
  >
  0

|x|\le r\qquad\text{is equivalent to}\qquad-r\le x\le r  \tag{i}r\le|x|\qquad\text{is equivalent to}\qquad x\le-r\quad\text{or }\quad r\le x  \tag{ii}|x|=r\qquad\text{is equivalent to}\qquad x=r\quad\text{or}\quad x=-r  \tag{iii}

Geometric 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 .

Number line showing that |x| ≤ r means x lies between −r and r, with the interval [−r, r] highlighted and |x| marked as the distance from 0 to x. — Geometric interpretation of |x| ≤ r: the value x lies within distance r from the origin, i.e., −r ≤ x ≤ r.

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

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

If $a$ and $b$ are either both positive, both negative, or at least one of them is zero, then $| a + b | = | a | + | b |$ . Otherwise, when $a$ and $b$ have opposite signs, $| a + b | < | a | + | b |$ .

Proof of the triangle inequality

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

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

This is what we were trying to prove.

◻

Example 2.

Show that

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

Solution

This result follows from the triangle inequality and the fact that

| - b | = | b |

(Property 4). Using the triangle inequality, we have:

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

By subtracting

| b |

from both sides of the inequality, we obtain:

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

Combining the above inequality with the standard triangle inequality

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

, we conclude that:

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

Example 3.

Prove that for all real number $a$ and $b$ , we have

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

Solution

This follows directly from the triangle inequality and the fact that

| - b | = | b |

. Applying the triangle inequality to

a + (- b)

, we get:

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

Thus,

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

holds for all real numbers

a

and

b

Example 4.

Prove that for all real numbers $a$ and $b$ ,

| a | - | b | \leq | a - b | .

Solution

We start by adding and subtracting

b

and then applying the triangle inequality. Specifically:

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

Subtracting

| b |

from both sides gives:

| a | - | b | \leq | a - b | .

Thus, the inequality

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

holds for all real numbers

a

and

b

It follows from (4) that

$| a - b | = | b - a |$

Distance Between Points on the Real Line

Look at the following figure. The distance between $- 2$ and $10$ is $12$ units. The distance can be calculated as the difference $10 - (- 2)$ , which involves subtracting the smaller value from the larger one. However, since the absolute value of $10 - (- 2)$ equals $| 12 | = 12$ , and similarly, $| - 2 - 10 | = | - 12 | = 12$ , we can use the absolute value function to find the distance between two points without worrying about which number is larger or smaller.

Illustration for Absolute Value and Distance — **Figure 1**

Definition 2.

If $P$ and $Q$ are two points located on a number line with coordinates $a$ and $b$ , the distance between $P$ and $Q$ , denoted by $d (P, Q)$ , is calculated as:

d (P, Q) = | b - a |

Since $| b - a |$ is the same as $| a - b |$ , it follows that the distance from $P$ to $Q$ is equal to the distance from $Q$ to $P$ :

$d (P, Q) = d (Q, P) .$

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

$| x - a | < r$

is equivalent to

$- r < x - a < r$

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 5.

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 \begin{aligned} I & =\{x|\ |x-2|<0.6\} & =\{x|\ -0.6 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|<\delta\}\\ & =\{x|\ -\delta

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 <math xmlns= — **Figure 3** 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|\quad0<|x-a|<\delta\} & =\{x|\ |x-a|<\delta\quad\text{and}\quad 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 neighborhood or punctured neighborhood of $a$ . The deleted $δ$ -neighborhood of $a$ is shown in the following figure.

The deleted <math xmlns= — **Figure 4** The deleted $δ$ -neighrbood of $a$ is the $δ$ -neighborhood of $a$ minus the midpoint $a$ .

Absolute Value

Distance Between Points on the Real Line

Neighborhoods

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