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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".

Pavel
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  • If you square first you get all the $6^{th}$ roots of $1$ including $1$. – Mark Bennet Oct 18 '14 at 17:45
  • See http://math.stackexchange.com/questions/956541/what-is-8-frac23 – 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, http://www.wolframalpha.com/input/?i=%28-1%29^{\frac26} and http://www.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

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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.

Carl Mummert
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Pavel
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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.

Rory Daulton
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  • 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 09 '15 at 01:54