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 $x \ge y$, then the distance between them is just the difference, $\,x - y\,$.
For example, the distance between $\,-3\,$ 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:
That is, to find the distance between any two real numbers:
- subtract them in any order
- take the absolute value of the result
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:
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 $\,2\,$:
$$
\cssId{s35}{\overbrace{\strut |x-1|}^{\text{want numbers $x$ whose distance from $1\ \ldots \ $ }}}
\cssId{s36}{\overbrace{\strut \ \ < 2\ \ }^{\text{is less than two}}}
$$
- mark the number $\,1\,$ on a number line; we want all numbers that are less than $\,2\,$ units away
- starting at $\,1\,$, go $\,2\,$ units to the right; mark $\,1 + 2 = 3\,$ on the number line
- starting at $\,1\,$, go $\,2\,$ units to the left; mark $\,1 - 2 = -1\,$ on the number line
- since we want the distance from $\,1\,$ to be strictly less than $\,2\,$, put hollow dots at the endpoints
- shade everything between $\,-1\,$ and $\,3\,$
interval notation for solution set:
$\,(-1,3)$
sentence form of solution:
$\,-1 < x < 3\,$
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 $\,1\,$:
$$
\begin{gather}
\cssId{s57}{|x+3| \ge 1}\cr
\cssId{s58}{|x - (-3)| \ge 1}\cr\cr
\cssId{s59}{\overbrace{\strut |x-(-3)|}^{\text{want numbers $x$ whose distance from $-3\ \ldots \ $ }}}
\cssId{s60}{\overbrace{\strut \ \ \ge 1\ \ }^{\text{is greater than or equal to $\,1$}}}
\end{gather}
$$
- mark the number $\,-3\,$ on a number line; we want all numbers that are $\,1\,$ or more units away
- starting at $\,-3\,$, go $\,1\,$ unit to the right; mark $\,-3 + 1 = -2\,$ on the number line
- starting at $\,-3\,$, go $\,1\,$ unit to the left; mark $\,-3 - 1 = -4\,$ on the number line
- since we want the distance from $\,-3\,$ to be greater than or equal to $\,1\,$, put solid dots at the endpoints
- shade everything to the right of $\,-2\,$ and to the left of $\,-4\,$
interval notation for solution set:
$\,(-\infty,-4] \cup [-2,\infty)$
sentence form of solution:
$\,x \le -4 \ \ \text{ or }\ \ x \ge -2\,$
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:
$$
\cssId{s86}{0 < |x - 2| < \delta}\qquad \qquad \qquad \qquad \qquad
\cssId{s87}{\text{ is equivalent to }}\qquad \qquad \qquad \qquad \qquad
\cssId{s88}{0 < |x - 2|\ \ \text{ and }\ \ |x - 2| < \delta}
$$
Recall that the only time an ‘and’ sentence is true is when both subsentences are true.
So, here's what you should think:
$$
\cssId{s91}{\overbrace{\strut |x - 2|}^{\text{we want numbers $x$ whose distance from $2\ \ldots$ } }}
\cssId{s92}{\overbrace{\strut \ \ > 0\ \ }^{\text{is greater than $0$} }}
\ \ \ \
\cssId{s93}{\overbrace{\strut \text{ and }}^{\text{ and }}}\ \ \
\cssId{s94}{\overbrace{\strut |x - 2|}^{\text{ whose distance from $2\ \ldots$ } }}
\cssId{s95}{\overbrace{\strut < \delta}^{\text{is less than delta}}}
$$
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 top number line (labeled ‘$\,0 < |x - 2|\,$’)
has everything shaded except $\,2\,$;
these are the numbers whose distance from $\,2\,$ is strictly greater than zero -
the middle number line (labeled ‘$\,|x - 2| < \delta\,$’)
shades all numbers that are less than $\,\delta\,$ units from $\,2\,$:
-- starting at $\,2\,$, go $\,\delta\,$ to the right; mark $\,2+\delta\,$ on the number line
-- starting at $\,2\,$, go $\,\delta\,$ to the left; mark $\,2 - \delta\,$ on the number line
-- shade everything between $\,2-\delta\,$ and $\,2+\delta\,$ -
the bottom number line (labeled ‘$\,0 < |x-2| < \delta\,$’)
shows the intersection (the overlap) of the shaded sets on the top and middle,
which is the desired solution set
The resulting solution set is often called a ‘punctured neighborhood of $\,2\,$’:
- it is a neighborhood of $\,2\,$ (an open set that contains $\,2\,$) ...
- ... except that the $\,2\,$ is missing—it is punctured at $\,2\,$
Now that you've seen all the details, you should never have to do this much work again!
You should merely think:
-
the ‘$\,|x-2| < \delta\,$’ part gives all the numbers between $\,2-\delta\,$ and $\,2+\delta\,$
(mark the $\,2\,$, go $\,\delta\,$ to the left and $\,\delta\,$ to the right, shade everything in between) - the ‘$\,0 < |x-2|\,$’ part takes away the $\,2\,$ (put a hollow dot at $\,2\,$)
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:
-
|x - 1| < 2
(look carefully—the vertical bar used for absolute value should be somewhere on your keyboard) - |x + 3| >= 1
- 0 < |x - 2| < 0.3
by practicing the exercise at the bottom of this page.
When you're done practicing, move on to:
Circles