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So I stumbled upon this weird result when experimenting with fractional exponents. Suppose you have some negative, real number, for example -8. We know $\sqrt{-8}$ is not a real number.But $$\sqrt{-8} = {(-8)}^{\frac1{2}}$$ and $$\frac1{2} = \frac2{4}$$ so $$\sqrt{-8} = {(-8)}^\frac2{4} = ((-8)^2)^{1/4} = 64^{1/4}$$ which is obviously real!
I can't seem to find out what the error of the proof is, any thoughts are appreciated!

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    $\begingroup$ it's not the case that $(-8)^{2/4} = ((-8)^2)^{1/4}$. these exponentiation rules do not necessarily hold for complex numbers, mostly because raising something to the $\frac{1}{4}$ power is not well defined. For example, you can't say $1^{1/2} = 1$, since also $(-1)^2 = 1$. $\endgroup$ – mathworker21 Nov 23 '18 at 12:43
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You're using the power rule $a^{bc} = (a^b)^c$, but this rule does not hold in general unless $a$ is a positive real (and $b, c$ are both real too), or $b, c$ are both integers.

The trouble is that extending exponentiation to complex numbers most naturally produces a multiple-valued function -- and if you use such a definition, you can generally choose values for each of the exponentiations that make $(a^b)^c$ come out the same as $a^{bc}$ -- but multiple-valued functions don't really work with equational reasoning, so something still has to give. The most you could conclude is that there is some multiple-valued function whose value for some input can be both $\sqrt8$ and $i\sqrt 8$, but that does not make those two outputs equal.

If you want a single-valued exponentiation operation for complex numbers, it will necessarily contain more-or-less arbitrary discontinuities and branch choices that makes $a^{bc}= (a^b)^c$ fail for some values.

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A square root is specifically a quantity whose square is equal to the argument of the square root. While it is true that $x^2=-8$ implies $x^4=+64$, the converse does not hold. While the latter equation has real solutions, those real solutions are extraneous to the defining equation for any square root of $-8$.

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  • $\begingroup$ I agree with this point, but this does not answer my question completely. I am using square roots and n-roots only as exponents and not as solutions to equations. $\endgroup$ – Magnus E-F Nov 23 '18 at 12:58
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You tell it yourself, $\sqrt{-8}$ is not a real number. So you have to wonder what is the legitimacy of raising that beast to a power: what's the meaning of

$$\left(\sqrt{-8}\right)^2 ?$$

Mathematicians have found answers to this question, but it turns out that

$$\left(\sqrt{-8}\right)^{2}=\sqrt{(-8)^2}$$ cannot hold !

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