what does $|x-2| < 1$ mean?

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

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I added the tag intuition to the question because it seems like you're looking for an intuitive way to understand the formula. –  Git Gud Jul 12 '13 at 16:48
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. –  JB King Jul 12 '13 at 16:59
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. –  Kaz Jul 12 '13 at 19:01

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 $a$.

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thank you, and |x-2| = |-x-2| ? or |x+2| ? or |-x+2| ? –  themhz Jul 12 '13 at 16:51
@themhz Add that to the question. –  Git Gud Jul 12 '13 at 16:51
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? –  themhz Jul 12 '13 at 17:01
"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. –  celtschk Jul 12 '13 at 17:04
|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. –  JB King Jul 12 '13 at 20:08

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 $$1\lt x\lt 3.$$

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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?

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that's very intuitive thank you –  themhz Jul 12 '13 at 17:02
You're welcome. –  Sami Ben Romdhane Jul 12 '13 at 18:23

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 $1<x<3$.

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I hope you are feeling better, Babak! ;-) –  amWhy Jul 13 '13 at 0:35

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 $x=0$)

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