# Simplifying fractional exponents

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".

• If you square first you get all the $6^{th}$ roots of $1$ including $1$. – Mark Bennet Oct 18 '14 at 17:45
• – lab bhattacharjee Oct 18 '14 at 17:48
• @labbhattacharjee ok, so, then why does wolfram say "indeterminate" for (-1)^(2/6) and "-1" for (-1)^(1/3)? Keep in mind that I'm talking about real numbers! – Pavel Oct 18 '14 at 17:51
• @paulpaul1076, wolframalpha.com/input/?i=%28-1%29^{\frac26} and wolframalpha.com/input/?i=%28-1%29^{\frac13} – lab bhattacharjee Oct 18 '14 at 17:57
• @labbhattacharjee once again, I am talking about real numbers! add "real numbers" to what you wrote and you get "indeterminate" for "2/6" exponent and "-1" for "1/3" exponent, also I edited my question – Pavel Oct 18 '14 at 18:00

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.

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.

• Then why does it not answer "1" instead of "indeterminate"? – Pavel Oct 18 '14 at 18:35
• Because there are two obvious ways of calculating the answer, and those give different answers. If you asked me which answer were correct, I would choose $-1$. So $1$ is not at all the obvious correct answer. I made a slight edit to my answer that might help. – Rory Daulton Oct 18 '14 at 18:37
• Yeah, I got it and posted my own answer, thanks. – Pavel Oct 18 '14 at 18:40
• Also it's indeterminate not because it can take 2 approaches but because the answer to (-1)^(1/6) is unknown, as i said in my post – Pavel Oct 18 '14 at 18:45
• I have actually talked about this with someone, and he said that expression x^(a/b) = b roots from x^a is only valid when x is not negative, so your first solution is incorrect – Pavel Jan 9 '15 at 1:54