ABSOLUTE VALUE AS DISTANCE BETWEEN TWO NUMBERS

To find the distance between any two real numbers,
take the greater number (the one lying farthest to the right on a number line)
and subtract the lesser number (the one lying farthest to the left).

That is, if [beautiful math coming... please be patient] $x \ge y$, then the distance between them is just the difference, $\,x - y\,$.

For example, the distance between [beautiful math coming... please be patient] $\,-3\,$ and $\,5\,$ is:   $\,5 - (-3) = 8\,$

But what if you goof up the order of subtraction?
In this case, you get a number with the right size (distance from zero), but the wrong sign:   $\,-3 - 5 = -8\,$
To get from $\,-8\,$ to the answer you really want, just take the absolute value!

Thus, absolute value provides a convenient way to talk about the distance between any two real numbers:

DISTANCE BETWEEN TWO REAL NUMBERS absolute value of the difference
Let $\,x\,$ and $\,y\,$ be real numbers. Then: [beautiful math coming... please be patient] $$ |x-y| = \text{the distance between } \,x\, \text{ and } \,y $$

That is, to find the distance between any two real numbers:

Notice that you don't need to know which number is greater (if the numbers are different) to find the distance between them.
And, of course, if two numbers are equal, then the distance between them is zero.

Notice that $\,x - y\,$ and $\,y - x\,$ are opposites, since $\,x - y = -(y - x)\,$.
Opposites have the same distance from zero, hence have the same absolute value.
(Remember: absolute value gives distance from zero!)
Thus, $\,|x - y\,| = |y - x|\,$. So we repeat—you can subtract in whatever order you want!

By recognizing an absolute value expression as representing a distance between two numbers,
many simple absolute value sentences can be solved ‘by inspection’—just look at them, think, and write down the solution set!
Here are some examples:

EXAMPLE:   solving a ‘less than’ absolute value sentence by inspection

Solve:   $|x - 1| < 2$

Solution:
View $\,|x - 1|\,$ as the distance between $\,x\,$ and $\,1\,$.
Here, we want all numbers $\,x\,$ whose distance from $\,1\,$ is less than $\,2\,$: $$ \overbrace{\strut |x-1|}^{\text{want numbers $x$ whose distance from $1\ \ldots \ $    }} \overbrace{\strut \ \ < 2\ \ }^{\text{is less than two}} $$

interval notation for solution set:   $\,(-1,3)$
sentence form of solution:   $\,-1 < x < 3\,$

EXAMPLE:   solving a ‘greater than’ absolute value sentence by inspection

Solve:   $|x + 3| \ge 1$

Solution:
First, rewrite $\,|x+3|\,$ so there is a difference inside the absolute value:   $\,|x + 3\,| = |x - (-3)|\,$
Thus, $\,|x+3|\,$ represents the distance between $\,x\,$ and $\,-3\,$.
Here, we want all numbers $\,x\,$ whose distance from $\,-3\,$ is greater than or equal to $\,1\,$: $$ \begin{gather} |x+3| \ge 1\cr |x - (-3)| \ge 1\cr\cr \overbrace{\strut |x-(-3)|}^{\text{want numbers $x$ whose distance from $-3\ \ldots \ $    }} \overbrace{\strut \ \ \ge 1\ \ }^{\text{is greater than or equal to $\,1$}} \end{gather} $$

interval notation for solution set:   $\,(-\infty,-4] \cup [-2,\infty)$
sentence form of solution:   $\,x \le -4 \ \ \text{ or }\ \ x \ge -2\,$

EXAMPLE:   solving a compound absolute value sentence by inspection

This final example is important in Calculus.
When you study the precise definition of a limit (which is the central concept in Calculus),
then you'll see this type of absolute value sentence.

Solve:   $\,0 < |x - 2| < \delta\,$

Solution:
The symbol   ‘$\,\delta\,$’   is the lowercase Greek letter delta.
It is used in this context to denote a small positive number.

The sentence   ‘$\,0 < |x - 2| < \delta\,$’   is a compound inequality—it uses more than one inequality symbol.
This type of compound inequality is a shorthand for a mathematical ‘and’ sentence: $$ 0 < |x - 2| < \delta\qquad \qquad \qquad \qquad \qquad \text{ is equivalent to }\qquad \qquad \qquad \qquad \qquad 0 < |x - 2|\ \ \text{ and }\ \ |x - 2| < \delta $$ Recall that the only time an ‘and’ sentence is true is when both subsentences are true.
So, here's what you should think: $$ \overbrace{\strut |x - 2|}^{\text{we want numbers $x$ whose distance from $2\ \ldots$    } } \overbrace{\strut \ \ > 0\ \ }^{\text{is greater than $0$} } \ \ \ \ \overbrace{\strut \text{ and }}^{\text{  and  }}\ \ \ \overbrace{\strut |x - 2|}^{\text{    whose distance from $2\ \ldots$    } } \overbrace{\strut < \delta}^{\text{is less than delta}} $$

That is, the solutions to ‘$\,0 < |x - 2| < \delta\,$’ are all numbers that are within $\,\delta\,$ units of $\,2\,$, but not equal to $\,2\,$.

More precisely, study the number lines below:

The resulting solution set is often called a ‘punctured neighborhood of $\,2\,$’:


Now that you've seen all the details, you should never have to do this much work again!
You should merely think: Easy!

interval notation for solution set:   $\,(2-\delta,2) \cup (2,2+\delta)$

sentence form of solution:   $\,2-\delta < x < 2 \ \ \text{ or }\ \ 2 < x < 2+\delta\,$

equivalent sentence form of solution:   $\,2-\delta < x < 2+\delta \ \ \text{ and }\ \ x\ne 2\,$

Solving Absolute Value Sentences at WolframAlpha

For fun, zip up to WolframAlpha and type in any of these:

Pretty amazing, isn't it?

Master the ideas from this section
by practicing the exercise at the bottom of this page.

When you're done practicing, move on to:
Circles
On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.
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