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What is the rationale behind saying that $|x|=1$ implies $x=\pm 1$?

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Because $+1$ and $-1$ are the only integers whose distance from $0$ is $1$? Note $|1|=1=|-1|$, so $x$ can take either value. –  yunone Dec 9 '10 at 18:59
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please don't use the title as an integral part of your message. Make the body of your message self-contained, even if that means you have to write the same thing twice, once in the title and once in the body. –  Arturo Magidin Dec 9 '10 at 19:18
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@yunone: Even better, they are the only real numbers. –  Sean Tilson Dec 10 '10 at 1:07
    
@Sean, ah yes of course, very true. –  yunone Dec 10 '10 at 7:56

5 Answers 5

As in other's answers, the absolute value of $x$ can be defined as the distance between $x$ and 0. When $x$ is a real number, we can think of $x$ as being on a number line, and there are two locations on a number line that are 1 unit away from 0: 1 unit in the positive direction (called 1) and 1 unit in the negative direction (called "-1").

If $x$ can be a complex number, we can think of $x$ as being a point in the plane, and there are a whole lot more locations that are a distance of 1 from 0, such as $\frac{3}{5}+\frac{4}{5}i$ (1 and -1 are also still 1 unit away from 0, so they are still possible values of $x$).

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+1,for the complex part. –  Quixotic Dec 9 '10 at 19:38
    
the fact that you assume x is real is integral to the argument, thanks for making this point! –  Sean Tilson Dec 10 '10 at 1:09
up vote 6 down vote accepted

Absolute value of a number is its distance from zero on the number line. The absolute value of a number n is denoted by, |n|.

So here if $|x| = 1$ implies, in the number line the distance from zero on the number is 1.$\text{ Hence, } x = \pm 1$.

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You should think of the absolute value as being the distance a point is away from 0. So with $|x| = 1$, we want all of the points that are one unit away from 0. We see that 1 and -1 are the only values for x that satisfies this. Therefore, with $|x| = 1$ we have that $x \in\{-1,1\}$.

I hope this helps.

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Well, to be precise, you should conclude that $x \in \{ -1,1\}$, not that $x$ equals that set. $x$ is a number, not a set. –  Fredrik Meyer Dec 9 '10 at 21:08
    
@Fredrik -- You are correct, thank you for catching that. –  Tyler Clark Dec 14 '10 at 22:46

If thinks in this as $f(x)= |x|$, you need to see which are the preimages of $1$.

Graphicaly you could to get this: alt text

Blue : $f(x)=|x|$
Red : $g(x)=1$

Thus, $f(x)=1$ only on the intersection points of these two functions, is say on $x=\pm 1$

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$|1| = 1$ and $|-1| = 1$, and there are no other real numbers $x$ such that $|x|=1$.

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