# Why does $|x_1| = |x_2| \implies x_1 = \pm x_2$

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?

• try $x_1=|x_2|$ and see what you get – JonMark Perry Jul 29 '15 at 5:04
• you ought to add $x\in \mathbb{R}$ aswell – JonMark Perry Jul 29 '15 at 5:06
• 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. – colormegone Jul 29 '15 at 5:07
• well, you are confused, it's the other way around, the right side is untouched, $\mp x_1 = x_2$ – Mirko Jul 29 '15 at 5:08
• Or stand on the other side of your screen... – colormegone Jul 29 '15 at 5:08

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.

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 $x_1=\pm x_2$.

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 $x_2 = \pm(-x_1)$.

• 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 $x_1 = \pm x_2$. – N. F. Taussig Jul 30 '15 at 15:44

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 $y=\pm \mp x=\pm x$.

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

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 $x=\pm y$.