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Let $x$ denote an arbitrary real number. Then $x^n$ makes sense for arbitrary $n \in \mathbb{N},$ via the obvious recursive definition. We can extend this definition by asserting that if $x$ is furthermore non-zero, then $x^n$ additionally make sense for $n \in \mathbb{Z} \setminus \mathbb{N}$, and is defined by iterative multiplication of $1/x$. We can go one step further by asserting that if $x$ is furthermore positive, then $x^y$ additionally makes sense for arbitrary $y \in \mathbb{R} \setminus \mathbb{Z}$ and is defined by $x^y = e^{y \log x}$ in this case.

We then have all the usual exponential laws that we know and love, as well as closure:

  1. if $n \in \mathbb{N}$ and $x$ is real, then $x^n$ is real.
  2. if $n \in \mathbb{Z}$ and $x$ is real non-zero, then $x^n$ is real-nonzero.
  3. if $y \in \mathbb{R}$ and $x$ is positive, then $x^y$ is positive.

However, in high school we go one step further, learning that $(-1)^{1/3}$ is well-defined, and equals the unique real $x \in \mathbb{R}$ such that $x^3 = -1$. So too is $(-1)^{2/3},$ which equals $((-1)^2)^{1/3}$, or equivalently $((-1)^{1/3})^2.$ However, $(-1)^{3/2}$ is ill-defined because we cannot take the square root of a negative number.

Is this fourth and final step we learn in high school actually useful? The theory seems much more complicated than the theory of steps 1, 2 and 3, because for example $$-1 = (-1)^{2/2} \neq ((-1)^{1/2})^{2} = \mathrm{undef}.$$

So what I want to know is, do expressions like $(-1)^{2/3}$ show up naturally in pure or applied math?

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Your last statement is incorrect. $i^2 = 1$ –  Don Larynx Sep 21 '13 at 4:26
    
@Jossie -1..... –  Bitrex Sep 21 '13 at 4:28
    
@Josie, if we go to the complex numbers, we have to make a choice of branch cut, and no matter how we make it, familiar properties of exponentiation fail. So either way, my point still stands. Things get messier, and my question, are there situations where we actually have to deal with this messiness? –  goblin Sep 21 '13 at 4:40
    
@Bitrex, my comment above is addressed to you also. –  goblin Sep 21 '13 at 4:41
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Yeah, I was also trying to say that IMVHO extending the definition of $x^{1/3}$ is mostly pointless. As it happens I am reviewing the power concept and related rules in my freshman calculus course just now. And I won't be mentioning this possibility. Exactly for the reasons that you brought up - some of the familiar rules will break down. –  Jyrki Lahtonen Sep 21 '13 at 7:29

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