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I was doing a 'prove this is not surjective' practice problem and the step leading from my hypothesis, as listed, to the conclusion was not defined. I don't recall being exposed to a situation where both sides of an equation had absolute values applied, and I had to solve for one.

Basically what I'm asking is: why does removing the absolute value notation from both sides of the equation lead to the left side being untouched and the other side being plus/minus?

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  • $\begingroup$ try $x_1=|x_2|$ and see what you get $\endgroup$ – JonMark Perry Jul 29 '15 at 5:04
  • $\begingroup$ you ought to add $x\in \mathbb{R}$ aswell $\endgroup$ – JonMark Perry Jul 29 '15 at 5:06
  • $\begingroup$ It's not that the left side of the equation is "left untouched"; it's that the two possibilities are $ \ x_1 \ = \ \pm x_2 \ $ and $ \ x_1 \ = \ \mp x_2 \ $ , which are redundant statements. $\endgroup$ – colormegone Jul 29 '15 at 5:07
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    $\begingroup$ well, you are confused, it's the other way around, the right side is untouched, $\mp x_1 = x_2$ $\endgroup$ – Mirko Jul 29 '15 at 5:08
  • $\begingroup$ Or stand on the other side of your screen... $\endgroup$ – colormegone Jul 29 '15 at 5:08
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If $x$ is a real number, then you can think of $|x|$ as simply being the distance from $x$ to $0$ on the number line. Thus if two real numbers have the same absolute value, then there are only two possibilities: either they are the same, or they are negatives of each other.

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To give you comprehensive and correct, but confusing answer, here is why.

There are four cases to consider.

Case 1. $x_1\ge0$ and $x_2\ge0$. Then the equality becomes $x_1=x_2$ .

Case 2. $x_1\ge0$ and $x_2<0$. Then the equality becomes $x_1=-x_2$ .

Case 3. $x_1<0$ and $x_2\ge0$. Then the equality becomes $-x_1=x_2$ which is equivalent to $x_1=-x_2$ .

Case 4. $x_1<0$ and $x_2<0$. Then the equality becomes $-x_1=-x_2$ which is equivalent to $x_1=x_2$ .

So instead of four cases, we end up with just two: Either $x_1=x_2$ or $x_1=-x_2$. This is abbreviated, by way of convention, as $x_1=\pm x_2$.

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There's a good amount of symmetry at work here.

The most intuitive argument comes from realizing that $\lvert x \rvert$ can be interpreted as the distance between a real number, $x$, and $0$. In this case, there are only two ways that two numbers can have the same distance from $0$: If they are in fact the same number, or negatives of one another (QED).

A messier algebraic option would be to attempt solving $\lvert x_1 \rvert = \lvert x_2 \rvert$ using the definition of the absolute value. Since

$$\lvert x_1 \rvert = \begin{cases}x_1, & x_1 \ge 0 \\ -x_1, & x_1 < 0\end{cases},$$

we have two equations to solve (assuming solutions exist): $x_1 = \lvert x_2 \rvert$, or $-x_1 = \lvert x_2 \rvert$. The former has solutions $x_2 = \pm x_1$, the latter has solutions $x_2 = \pm(-x_1)$.

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    $\begingroup$ Another algebraic option is to take $|x| = \sqrt{x^2}$, so $|x_1| = |x_2| \Rightarrow x_1^2 = x^2 \Rightarrow x_1^2 - x^2 = 0$. Factoring as a difference of squares yields $(x_1 - x_2)(x_1 + x_2) = 0$, from which we conclude $x_1 = \pm x_2$. $\endgroup$ – N. F. Taussig Jul 30 '15 at 15:44
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Surely you know that if you're solving for $x$ in $|x|=y$, you get that $x=\pm y$. You can check this by graphing the function $y=|x|$.

So if you have $|x|=|y|$ and you want to solve for $x$, then you know that $x=\pm |y|$, or $\mp x=|y|$. Then you can apply this again, solving for $y$ now, that $y=\pm \mp x=\pm x$.

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Symbol mode '$| \ \ \ |$' represents a positive value irrespective of the sign of a real number inside it

Eg: $|x|=\text{magnitude(positive value) of 'x'}$ while $x$ may be positive or negative real number as well $$|x|=2\implies x=\pm 2\iff |2|=2\ \text{or}\ |-2|=2$$

Hence, in general $|x_1|=|x_2|\implies x_1=\pm x_2\ \text{or}\ x_2=\pm x_1$ $\color{blue}{\forall\ x_1, x_2\in R}$

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Let $x\in\mathbb{R}$. Define the $||$ function as $||:\mathbb{R}\to\mathbb{R^+}\cup\{0\}$ as:

$$ |x|= \begin{cases} x& x\ge0 \\ -x&x\le0 \end{cases} $$

so $|3|=|-3|=3$.

Then if $|x|=|y|, x=\pm y$.

We can write any number of possibilites, example $\mp x=\pm y$, but it is convention to use $x=\pm y$.

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