TITLE: Does $X^{2/2} = X^{1/1}$?
QUESTION [6 upvotes]: I'm having a bit of a hard time wrapping my head around how the following that I have just learned:
$\sqrt{X^2} = |X|$, and I totally understand why.
But, when expressed as an exponent, doesn't this really just mean the following:
$X^{2/2} = |X|$, if this is the case, and I simplify the rational exponent, I would get:
$X^{1/1}$ or $X^1$, which does not equal $|X|$.
Also, if I apply the following rule of a radical function:
$\sqrt[n]{P^Q} = (\sqrt[n]P)^Q$ where n is the index of the root and $Q$ is the power of the radicand, then this should mean that:
$\sqrt{X^2} = (\sqrt{X})^2$, but the $(\sqrt{X})^2$ does not equal $|X|$ and has a domain where $X > 0$, while the $\sqrt{X^2}$ has a domain equal to all real values for $X$.
Does this mean that when $X$ is raised to an even-numbered power and is the radicand in a radical expression, that one should not simplify the rational exponent or one should not rewrite the radical expression such that the power of the radicand $X$ now lies outside of the root function?
Any replies will be greatly appreciated.

REPLY [3 votes]: This is due to the fact that there is a slight difference between $\sqrt{x}$ and $x^{1/2}$
I recommend looking up the term "Principle Root", with a basic introduction here.
In essence, for positive numbers there are always two answers to $x^{1/2}$, namely $\pm \sqrt{x}$.
From that, note that the $\sqrt{x}$ function always gives the positive root for a positive argument, and is thus a true function... it outputs one root for each argument $x$. However, the function $x^{1/2}$ outputs two values for each argument $x$, and is thus NOT a true function. Nuances like these are what is messing with your argument!