I am very confused about the following:
whenever I put in into wolfram alpha the answer it gives me is "indeterminate", is it not possible to simplify fractional exponents or something? if the exponent is 2/6 can i simplify it to 1/3 and get -1 as the answer? Provided I'm working with real numbers.
What I put in wolfram alpha was "(-1)^(1/3) real numbers" to which it answered "-1", then when I tried "(-1)^(2/6) real numbers" it answered "indeterminate".
On
The usual definition (used in high schools) of $x^{a/b}$ is $\sqrt[b]{x^a}$. Therefore, your expression $(-1)^{2/6}$ could mean either:
$$(-1)^{2/6}=\sqrt[6]{(-1)^2}=\sqrt[6]{1}=1$$ or $$(-1)^{2/6}=(-1)^{1/3}=\sqrt[3]{(-1)^1}=\sqrt[3]{-1}=-1$$
That seems pretty indeterminate to me!
You could complain that it is obvious that the fraction $2/6$ should be simplified before applying the definition of a rational power, but it apparently is not obvious to Wolfram Alpha. Wolfram does not know you or your particular definition, it could be argued that Wolfram took the correct approach.
I figured it out, apparently wolfram first takes the 6th root of -1 and since (-1)^(1/6) is indeterminate since we're dealing with real numbers and can't take an even root of a negative number, it says "indeterminate".
Also, Rory's first solution is absolutely wrong (unless he corrects if when you see this), because we can't take a root of a real negative number raised to a power greater than 1, if what he wrote were true I can then prove that -1 = 1 like so: -1 = (-1)^(1/3) = (-1)^(2/6) = (1)^(1/6) = 1.