Quote from: benmachine on December 11, 2007, 01:26:39 PM

-1^0.5 is i, as you said, because x^(p/q) is the qth root of x ^ p and the square root of -1 is imaginary.
However, -1^(2/3) is the cube root of -1 squared, which IS real: -1.
see http://www.google.co.uk/search?q=(-1)^0.2
It becomes clear with a little investigation that all -1^(p/q) where p and q are coprime and q is odd are real, and equal 1 where p is even and -1 when p is odd. However, this doesn't really resolve the situation since the graph is still discontinuous and now seems to have an infinite number of distinct real results between x = 0 and 1, despite having an equally infinite number of x values where clearly no real y exists. I still have no idea what the graph looks like.

How is exponentiation of a negative number defined for non-rational exponents? I don't know of any such definition. There are much more non-rational numbers than there are rational numbers (the former is not countably infinity, the latter is). So unless you have a definition of this function, it is mostly undefined on the reals.

As posted before, on the rationals this function is very discontinuous: the set of values for which its image is 1 is dense and so is the set of values for which the image is -1. How the graph looks depends on your drawing style I guess. If you only draw a line if all intermediate reals are on the line, you should draw nothing. If having a dense set of points suffices (although it still has "more" holes than points), you can draw two lines as pointed out before. Maybe you better draw two gray lines

If you just consider it as a function on integers, it nicely alternates between 1 and -1. That is probably how this function is used most often.