10
$\begingroup$

I am studying some inequality properties of absolute values and I bumped into some expressions like $|x-2| < 1$ that I just can't get the meaning of them.

Lets say I have this expression

$$ |x|<1.$$

This means that $x$ must be somewhere less than $1$ or greater than $-1$ which means that

$$-1 < x < 1.$$

So basically $|x|<1$ and $-1 < x < 1$ are the same thing.

$$|x|<1 \iff -1 < x < 1 \iff\text{"Somewhere less that $1$ or greater than $-1$" or between $-1$ and $1$}$$

Now lets say I have

$$ |x-2| < 1.$$

This means that the result of the expression $|x-2|$ must be less than $1$ or greater than $-1$? What does that also mean for $x$? Is it that $x$ must be a value that when we subtract $2$ the result has to stay withing the bound of $-1$ or $1$ or less than zero? If $x =5$ the statement fails because $3 <1$ is false. So it has to determine a boundary of $x$'s that satisfy this equation right?

if $|x| = |-x|$

what can this mean for

$|x-2| = |-x-2|$ or $|x+2|$ or $|-x+2|$ ?

Thank you

$\endgroup$
3
  • 2
    $\begingroup$ I added the tag intuition to the question because it seems like you're looking for an intuitive way to understand the formula. $\endgroup$
    – Git Gud
    Jul 12 '13 at 16:48
  • 2
    $\begingroup$ Correction to: "Somewhere less that 1 or greater than -1" or between -1 and 1. The second part is correct. The part in quotes should have AND as the conjunction. Otherwise, one could use 4 which is "greater than -1" yet isn't less than 1. $\endgroup$
    – JB King
    Jul 12 '13 at 16:59
  • 3
    $\begingroup$ It means that $x$ is less than 1 away from 2 in either direction on the number line, or any direction on the complex plane. $\endgroup$
    – Kaz
    Jul 12 '13 at 19:01
20
$\begingroup$

The geometric interpretation, in $\Bbb R$, for $|x-a|<b$ is '$x$ is at a distance smaller than $b$ from $a$'.

In your particular example, $|x-2|<1$, it means that $x$ is at a distance of at most $1$ from $2$ and it (the distance) never reaches $1$.


To interpret $|x-2|=|-x-2|$, I find useful to first note that $|-x-2|=|x-(-2)|$ (why?). The equality $|x-2|=|x-(-2)|$ says that $x$ is at equal distance between $2$ and $-2$.

More generally, $|x-a|=|x-b|$ says that $x$ is at the same distance between $a$ and $b$.


To summarize, read $|x-a|$ as the distance between $x$ and $a$.

$\endgroup$
10
  • $\begingroup$ thank you, and |x-2| = |-x-2| ? or |x+2| ? or |-x+2| ? $\endgroup$
    – themhz
    Jul 12 '13 at 16:51
  • $\begingroup$ @themhz Add that to the question. $\endgroup$
    – Git Gud
    Jul 12 '13 at 16:51
  • $\begingroup$ so |x-2| => say x=3 then 3-2 = 1 or x=-3 then -3-(-2)=-3+2 =-1 so |-1|=|1|=1 so |x-2| = |-x+2| =|2-x| right? $\endgroup$
    – themhz
    Jul 12 '13 at 17:01
  • $\begingroup$ "it means that x is at a distance of at most 1 from 2 and it never reaches 2." Of course it reaches 2. It won't reach 1 or 3, though. $\endgroup$
    – celtschk
    Jul 12 '13 at 17:04
  • 1
    $\begingroup$ |x-2| is not the same as |-x-2| as the latter could be expressed as |x+2| which is quite different if you consider evaluating this for a few different values of x. $\endgroup$
    – JB King
    Jul 12 '13 at 20:08
11
$\begingroup$

Hint

Denote $x-2$ by $y$ then $$|x-2|<1\iff |y|<1$$ and you find exactly your first inequality. Can you take it from here?

$\endgroup$
3
  • $\begingroup$ that's very intuitive thank you $\endgroup$
    – themhz
    Jul 12 '13 at 17:02
  • $\begingroup$ You're welcome. $\endgroup$
    – user63181
    Jul 12 '13 at 18:23
  • $\begingroup$ Excellent answer! $\endgroup$
    – amWhy
    May 19 '14 at 11:30
11
$\begingroup$

The way to think of $$|x-y|$$ is as the distance from $x$ to $y$. For example, is $|x-y| = |y-x|$? Yes, it is, because the distance from $x$ to $y$ is the same as the distance from $y$ to $x$.

Is $$|x-y| + |y-z| = |x-z|?$$ This says that the distance from $x$ to $y$, plus the distance from $y$ to $z$, is equal to the distance from $x$ to $z$. That would mean that $y$ was on the direct path from $x$ to $z$. So we would expect it to be false if $y$ was not on this direct path; say if $x = 2, z=4,$ but $y = 17$. And indeed $|2-17| + |17-4| \ne |2-4|$, so the equation above is not always true. But we might guess from this understanding that $$|x-y| + |y-z| \ge |x-z|,$$ with equality occurring just if $y$ is between $x$ and $z$. And in fact this is always true.

With this idea, what does $$|x|$$ mean? It should be the same as $$|x-0|,$$ which is the distance from $x$ to 0. And that is correct.

Now what does $$|x-2| < 1$$ mean? It means that the distance from $x$ to 2 is less than 1. So another way to write this is $$1\lt x\lt 3.$$

$\endgroup$
7
$\begingroup$

We know that: $$|x-2|= \left\{ \begin{array}{ll} -x+2 & \quad x < 2 \\ x-2 & \quad x \ge 2 \end{array} \right.$$ Now if we have to do $|x-2|<1$ so: $$x\ge2\to x-2<1\to x<3\\\ x<2\to 2-x<1\to x>1$$ This means that , overall, we have $1<x<3$.

$\endgroup$
1
  • $\begingroup$ I hope you are feeling better, Babak! ;-) $\endgroup$
    – amWhy
    Jul 13 '13 at 0:35
4
$\begingroup$

Just as you replace $|x|<1$ with $-1<x<1$, you can replace $|x-2|<1$ with $-1<x-2<1$. Adding $2$ on all three parts leaves the order unchanged, so $-1+2<x-2+2<1+2$, i.e. $1<x<3$.

To see what signs are correct in $|x-2|=|\pm x\pm 2|$, note that the absolute value does not change when we replace its argument with its negative. The argument here is $x-2$, the negative thereof is $-(x-2)$ and that can be simplified to $-x+2$. Thus $|x-2|=|-x+2|$. (Of course in rare cases it may also be true that $|x-2|=|x+2|$, namely precisely when $x=0$)

$\endgroup$
1
$\begingroup$

Draw a graph of $$f(x) = |x|$$ Then ask yourself, how does one obtain $$g(x) = |x-2|$$ from the graph of $f(x)$. Once you figure out how the graph looks, the question that you asked is simply asking to find all the possible $x$ values such that $g(x) < 1$. You can draw a horizontal line $y=1$ on the graph of $g(x)$ and observe the $x$ values that satisfy.

$\endgroup$
1
1
$\begingroup$

$|x-2| < 1\iff-1<x-2<1\iff2-1<2+x-2<2+1\iff1<x<3$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.