Difference between $y = x^{1/2}$ & $y = x^{2/4}$ The other day something occurred to me when graphing $y = x^{1/2}$. 
I understand that this is equivalent to $y = \sqrt{x}$ & this can't have negative values for $x$. But is it not also equivalent to $y = x^{2/4}$
which in turn is $y = \sqrt[4]{x^2}$ which would allow negative values for $x$?
I know the easy answer here is to say you should simplify $\frac24$ first but is there a deeper mathematical explanation for what looks to me to be a bit of a paradox?
 A: I think this is an interesting question. Technically, anywhere that those expressions ($x^{1/2}$, $x^{2/4}$, $\sqrt[2]{x}$, and $\sqrt[4]{x^2}$) are defined and make sense, they'll have the same value. So there is no real issue in confusing these while making calculations.
However, I think that if I were asked to write down the maximal domain of definition of $f(x) = \sqrt[2]{x}$, I would say $x \geq 0$, wheras for $g(x) = \sqrt[4]{x^2}$, I would say that it's defined on all of the real line, and the graph of $g(x)$ looks like the graph of $f(x)$ along with its reflection over the $y$-axis.
A: $x^{2/2} = (x^2)^{1/2} = {\sqrt {x^2}} = |x|.$
$x^{2/4} = (x^{2/2})^{1/2} = \sqrt {|x|}$
So by doubling the values you're essentially taking the absolute value of x.
A: I probably like Zwim's comment & link the most. 
In summary, Zwim's link states that you run in to issues when manipulating the indice of a function in the form $f(x) = x^{p/q}$. So you need conditions to prevent any issues. The 2 key conditions that relate to this question are that the function is only defined for x < 0 if the greatest common denominator $gcd(p,q)=1$ & q is odd. 
