If $|x| = 1 $ why this implies that $ x = \pm 1 $ What is the rationale behind saying that $|x|=1$ implies $x=\pm 1$?
 A: 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.$\text{ Hence, } x = \pm 1$.
A: 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 $x$).
A: 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.
A: If thinks in this as $f(x)= |x|$, you need to see which are the preimages of $1$.
Graphicaly you could to get this:

Blue : $f(x)=|x|$
Red : $g(x)=1$
Thus, $f(x)=1$ only on the intersection points of these two functions, is say on $x=\pm 1$
A: $|1| = 1$ and $|-1| = 1$, and there are no other real numbers $x$ such that $|x|=1$.
