What we learn here:

### 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

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

• Geometrically the absolute value of a number $x$ is its distance from $0$ regardless of the direction.
• 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 $3-\sqrt{3}\approx1.27$], we have $|2|=2,\ |0|=0,\ |3-\sqrt{3}|=3-\sqrt{3}$ but $-2$ and $\sqrt{3}-3$ are negative, so we have$|-2|=-(-2)=2,\quad|\sqrt{3}-3|=-(\sqrt{3}-3)=3-\sqrt{3}.$

### 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:

1. $|a|\geq0$
2. $|a|=0$ if and only if $a=0$
3. $||a||=|a|$ (the absolute value of the absolute value of $a$ is the absolute value of $a$ because $|a|\ge0$)
4. $|-a|=|a|$.
5. $|ab|=|a|\ |b|$.
6. $\left|\dfrac{a}{b}\right|=\dfrac{|a|}{|b|}$    (provided $b\neq0$)
7. If $r>0$
 $|x|\le r$ is equivalent to $-r\le x\le r$ (i) $|x|\ge r$ is equivalent to $-x\le r$  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)

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

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

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

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

$\boxed{|a|-|b|\leq|a+b|\leq|a|+|b|.}$

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
$|a|-|b|\leq|a-b|\cancel{+|b|}\cancel{-|b|}$

It follows from (4) that
9.  $|a-b|=|b-a|$

### 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 $\delta>0$. Then

\begin{align*}
I & =\{x|\ |x-a|<\delta\}\\
& =\{x|\ -\delta<x-a<\delta\}\\
& =\{x|\ a-\delta<x<a+\delta\}
\end{align*}

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

Such a set is called a neighborhood (or more precisely the $\delta$-neighborhood) of $a$ and $\delta$ is called the radius of the neighborhood. The $\delta$-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

$0<|x-a|<\delta$

or

$J=\{x|\ 0<|x-a|<\delta\}$

Here we have two inequalities

$0<|x-a|\quad\text{and}\quad|x-a|<\delta$

Recall that the absolute value is always nonnegative (that is $0\le|t-a|$ 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{align*}
That is $J$ is the $\delta$-neighborhood of $a$ with the midpoint $a$ removed. The set $J$ is called the deleted $\boldsymbol{\delta}$neighborhood or punctured neighborhood of $a$. The deleted $\delta$-neighborhood of $a$ is shown in the following figure. The deleted δ-neighborhood of a is the δ-neighborhood of $a$ minus the midpoint $a$.