Why is $|x|$ defined as $\sqrt{x^2}$ instead of $(\sqrt{x})^2$? I can't seem to understand this even though it might be utterly simple for some people. For me, saying $|x|=\sqrt{x^2}$ is a bit weird since $\sqrt{x^2}$ doesn't force positivity as there are always two possible square roots of a number, 
 $\sqrt{x}=+\sqrt x, -\sqrt{x}$. Then why is it still used everywhere? Is there another way of interpreting this?
I think $|x|=(\sqrt x)^2$ does the job much better.
 A: No, $\sqrt{x^2}$ does, in fact, force positivity, because $\sqrt{a}$, where $a$ is a positive number, is defined to be the positive square root.  Yes, $a$ has two square roots, but the positive one is $\sqrt{a}$, and the negative one is $-\sqrt{a}$.  
If $x$ is negative, we normally don't write $\sqrt{x}$, because things just got pushed into the complex plane, and roots get crazy their.
A: One needs to distinguish two concepts.

*

*A square root of $a$ is any number whose square is $a$. This definition is valid in any number system, not necessarily restricted to non-negative real numbers.
Also, this definition never tells that such number is unique, and in fact, there are exactly two square roots for a given non-zero number either in $\mathbb{R}$ (real numbers) or in $\mathbb{C}$ (complex numbers).


*The square root of $a$ for $a \geq 0$ is defined as the unique non-negative square root of $a$. This square root is denoted by $\sqrt{a}$.
Assuming OP has knowledge on complex numbers, the principal square root of a complex number $z$ can be defined as the square root $w$ of $z$ with $\arg(w) \in (-\pi/2, \pi/2]$. This $w$ is denoted as $\sqrt{z}$. It is easy to check that this definition of $\sqrt{z}$ extends the previous one when $z \in [0, \infty)$.
If $x$ is a square root of $a$, then by definition we necessarily have $x^2 = a$. In particular, $(\sqrt{a})^2 = a$ always holds by definition.
Finally, for each $x \in \mathbb{R}$,

*

*Square roots of $x^2$ are $x$ and $-x$.


*$\sqrt{x^2}$ is the unique non-negative square root of $x^2$.
So if $x \geq 0$, then $x$ is non-negative and $\sqrt{x^2} = x$. If $x \leq 0$, then $-x$ is non-negative and $\sqrt{x^2} = -x$. Combining both cases, we obtain $\sqrt{x^2} = |x|$.
A: You can see that $\sqrt{x^2}$ is defined for all $x\in\Bbb R$, while $(\sqrt{x})^2$ is defined only for $x\ge 0$. And that's a huge difference.
Furthermore, $|x|$ is not defined as $(\sqrt x)^2$. Instead, $|x|$ is defined as $x$ if $x\ge 0$ and as $-x$ if $x<0$.
A: First of all, $\sqrt{x^2}$ is necessarily positive.  There is an implicit + sign.
Secondly, the notation is meant to represent the magnitude of a given vector.  Sure, your definition works in the case of a scalar.  But for vector magnitudes(with 2 or more components), your definition breaks down(i. e. $|\binom{x}{y}| = \sqrt{x^2+y^2} \ne (\sqrt{x+y})^2$).
