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:
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 repeatyou 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 $\,2\,$:
$$
\cssId{s35}{\overbrace{\strut |x-1|}^{\text{want numbers $x$ whose distance from $1\ \ldots \ $ }}}
\cssId{s36}{\overbrace{\strut \ \ < 2\ \ }^{\text{is less than two}}}
$$
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 $\,1\,$:
$$
\begin{gather}
\cssId{s57}{|x+3| \ge 1}\cr
\cssId{s58}{|x - (-3)| \ge 1}\cr\cr
\cssId{s59}{\overbrace{\strut |x-(-3)|}^{\text{want numbers $x$ whose distance from $-3\ \ldots \ $ }}}
\cssId{s60}{\overbrace{\strut \ \ \ge 1\ \ }^{\text{is greater than or equal to $\,1$}}}
\end{gather}
$$
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 inequalityit 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 resulting solution set is often called a ‘punctured neighborhood of $\,2\,$’:
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: $\,2-\delta < x < 2+\delta \ \ \text{ and }\ \ x\ne 2\,$
For fun, zip up to WolframAlpha and type in any of these:
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