Something that many early students, including myself, take for granted is that $$x^\frac{3}{2}=\sqrt{x^3}=(\sqrt{x})^3$$
but is this true? Is exponentiation "commutative" and does a fractional exponent mean the same thing as a root?
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For positive $x$, yes. I advise avoiding the notation $x^a$ if $x$ is not positive and $a$ is not an integer though. – Angina Seng Dec 19 '19 at 06:04
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In the reals it is true that $(a^b)^c=a^{bc}=(a^c)^b$. This works for all $a \gt 0$ and any real $b,c$, so fractions are included.
It is not true in the complex numbers because there are multiple roots of a number and you need to be careful to pick the correct one. We avoid the root sign in the complex field for that reason.
Ross Millikan
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No, this is not true in general.
You could create pointless results like this. Take for example $\sqrt{-2}$ which is not definied in $\mathbb{R}$.
But we have that $\sqrt{(-2)^2}=\sqrt{4}=2$, while $(\sqrt{-2})^2$ does not make sense.
Cornman
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It makes sense, but the result is -2 rather than 2. There is an intermediate imaginary term. – Cristobol Polychronopolis Dec 19 '19 at 14:08