When I do a search on Wolfram|Alpha, I get that $x≥0$, except I thought that this isn't exactly the case. Because it is defined for all real numbers greater than zero, but there are infinitely many negative numbers for which the function is also defined. So why can't we extend the domain to numbers less than zero? It isn't continuous, I understand, but I suppose this is linked to the domain of the Dirichlet function?
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Well, it's clearly devined for negative integers... For non-integers, however, you get into multivalue problems. For instance, is $(-1/2)^{-1/2}$ equal to $0.707i$, or $-0.707i$? – Arthur Jan 03 '17 at 10:06
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@Arthur how about some values with real solutions that are non-integers? Does the domain contain these values? – rb612 Jan 03 '17 at 10:07
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Well, $(-2.6)^{-13/5}$ has five different possible values in the complex plane. Which one do you choose? Well, here there is one that is actually a negative number because $5$ is odd, but try $-2.75$ instead, for instance. – Arthur Jan 03 '17 at 10:09
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@Arthur, my apologies, that was a mistake. I mean to extend this to any non-integer negative value in which the has a real output. – rb612 Jan 03 '17 at 10:10
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Well, could give a real value to any rational number that has an odd denominator, but then you get into philosophical issues with $\frac12 = \frac24$. – Arthur Jan 03 '17 at 10:11