# Do expressions like $(-1)^{2/3}$ show up naturally in pure or applied math?

Background. Let $x$ denote an arbitrary real number. Then $x^n$ can be defined for each $n \in \mathbb{N}$ as follows:

$$x^n = \underbrace{x \times \cdots \times x}_n$$

If $x$ is furthermore non-zero, then we can extend the above definition by declaring that $x^k$ makes sense for each $k \in \mathbb{Z},$ by defining:

$$x^{-n} = \underbrace{(1/x) \times \cdots \times (1/x)}_n$$

for all positive integers $n$.

If $x$ is furthermore positive, we can go one step further and declare that $x^y$ makes sense for each $y \in \mathbb{R},$ by defining that

$$x^y = e^{y \log x}$$

for all non-integral real numbers $y$.

The result. The outcome of all this is rather pretty. We observe that all the usual exponential laws hold, and we get some nice closure results, too. In particular:

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.

The fourth step. However, in high school we go one step further. For example, we learn 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 $((-1)^{1/3})^2.$

Unfortunately, once we make this fourth step, the theory suddenly becomes more complicated, because the usual exponential laws don't all hold. For example, $$-1 = (-1)^{2/2} = (-1)^{(1/2) \cdot 2} \neq ((-1)^{1/2})^{2} = \mathrm{undef}.$$

Question. Is this fourth and final step we learn in high school actually useful? More precisely, do expressions like $(-1)^{2/3}$ show up naturally in pure or applied math? If so, where?

If so, I'd like to see specific examples.

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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
Without $(-1)^{2/3}$ there would be no discrete Fourier transform and our calculators would be much slower – Cocopuffs Sep 21 '13 at 5:02
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

Yes; in settings where a number $x$ has a unique $n$-th root, it can very well be useful to define $x^{1/n}$ to be that number.

The usual exponential laws do still hold if you don't try to take them too far; for example, if $x$ and $x^m$ both have unique $n$-th roots, then $(x^m)^{1/n} = (x^{1/n})^m$. In fact, we can say more: if $x$ has a unique $n$-th root, then so does $x^m$. And as a partial converse, if $x^m$ has a unique $n$-th root and $\gcd(m,n) = 1$, then $x$ has a unique $n$-th root as well.

I posit that in most purely real contexts where you would allow $x^3$ to be defined when $x$ is a negative real number, then you should also allow $x^{1/3}$ to be defined for negative real numbers.

I have recently made use of this, in the setting of $p$-adic analysis. Specifically, I was interested in the structure of $\mathbf{Z}_3[\sqrt{3}]$.

The power series for $\exp(z)$ is only defined for those elements such that $z \equiv 0 \pmod 3$; however, for this ring, it is convenient to extend the definition to all $z \equiv 0 \pmod{\sqrt{3}}$ by using $\exp(z) = \exp(3z)^{1/3}$.

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+1, interesting. – goblin Oct 20 '15 at 2:18