<s>[INST] Can fractional/decimal radicals/roots exist? For questions like "What is the 1/2th root of x would the answer be $x^2$?
My logic is that since $$
\sqrt[\cfrac{1}{2}]{x}=x^{1/{(\cfrac{1}{2}})}
$$
Which simplifies to $x^2$.
So as a general rule it could be $$
\sqrt[\cfrac{1}{a}]{x}=x^{1/{(\cfrac{1}{a}})}
=x^a
$$
And with a different denominator $$\sqrt[\cfrac{b}{a}]{x}=x^{1/{(\cfrac{b}{a}})}
=x^{\cfrac{a}{b}}$$
This corresponds to how decimal/fractional exponents denote radicals (their inverse) while fractional radicals are easier shown with exponents.
Example : (2/3rd root of 4)
$$\sqrt[\cfrac{2}{3}]{4}=4^{1/{(\cfrac{2}{3}})}
=4^{\cfrac{3}{2}}= 8$$
Example (22/7th root of π) :
$$\sqrt[\cfrac{22}{7}]{π}=π^{1/{(\cfrac{22}{7}})}
=π^{\cfrac{7}{22}}
\approx 1.439$$
Example (1/2th root of 1/4) :
$$\sqrt[\cfrac{1}{2}]{\cfrac{1}{4}}=\cfrac{1}{4}^{1/(\cfrac{2}{1})}
=\cfrac{1}{4}^{(\cfrac{2}{1})}
=\cfrac{1}{4}^{2}
=\cfrac{1}{16}
$$
 [/INST] The definition is:
$$\sqrt[a]{x} = x^{\frac{1}{a}}.$$
Hence, if $a = \frac{1}{b}$, then:
$$\sqrt[\frac{1}{b}]{x} = x^{b}.$$
As a consequence:
$$\sqrt[\frac{1}{2}]{x} = x^2.$$

Notice that $a = \frac{1}{b}$ implies that, if for example $b = \frac{22}{91}$, then $a = \frac{91}{22}$. If for example $b = \sqrt{3}$, then $a = \frac{1}{\sqrt{3}}$. And so on.
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