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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.

goblin GONE
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  • Your last statement is incorrect. $i^2 = 1$ – Don Larynx Sep 21 '13 at 04:26
  • @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 GONE Sep 21 '13 at 04:40
  • @Bitrex, my comment above is addressed to you also. – goblin GONE Sep 21 '13 at 04:41
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    Without $(-1)^{2/3}$ there would be no discrete Fourier transform and our calculators would be much slower – Cocopuffs Sep 21 '13 at 05:02
  • @Cocopuffs, care to elaborate? – goblin GONE Sep 21 '13 at 05:04
  • I would avoid defining fractional powers of negative numbers in all courses that don't use complex numbers even though there is a natural continuous branch, when the denominator is odd. Too many questions like this emerge. Do they appear? I don't immediately recall a use for functions like $x^{1/5}$ for negative $x$. The complex (multibranched) functions are all over the place though. Any smartphone is doing arithmetic with all the 2048 complex values of $1^{1/2048}$ by bucketloads :-) – Jyrki Lahtonen Sep 21 '13 at 05:04
  • Well, Cocopuffs beat me to it! – Jyrki Lahtonen Sep 21 '13 at 05:05
  • @user18921 I'm referring to the Schönhage-Strassen algorithm. It's not as useful for pocket calculator applications but it surely comes up in applied math. Very roughly speaking, instead of performing "long multiplication" on a number in binary representation, a trick with roots of one makes it enough to compute the digitwise products alone – Cocopuffs Sep 21 '13 at 05:33
  • @user18921 Here's an example of a practical use for roots of negative unity in an electrical engineering context: http://en.wikipedia.org/wiki/Butterworth_filter#Transfer_function – MattyZ Sep 21 '13 at 05:41
  • @Bitrex, I'm not sure I see the relevance. In the link you posted, they use $(-1)^{1/n}$ as an abuse of notation for ${z \mid z^n = -1}.$ To formalize this, we might assert that $w^{(1/n)}$ is notation for the (multi?)set of solutions $z$ to the equation $z^{n} = w$. However, this new function we've just defined does not generalize our usual definitions for real exponentiation. For example $1^{1/2} = 1$, but $1^{(1/2)} = {-1,1}$. – goblin GONE Sep 21 '13 at 06:53
  • @JyrkiLahtonen, yes I'm aware that the multivalued functions are of great importance. However, they do not generalize the usual definitions, see my comment above. – goblin GONE Sep 21 '13 at 06:54
  • @Cocopuffs, nope sorry I was wrong. I cannot find anywhere on that page where a negative number is raised to a non-integral power. – goblin GONE Sep 21 '13 at 06:59
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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 07:29
  • @JyrkiLahtonen, okay thanks for your input. You're a lecturer by the sounds of things? – goblin GONE Sep 21 '13 at 07:32
  • @JyrkiLahtonen, ah I see. Your profile says you're a tenured lecturer. Glad to have you on the website! – goblin GONE Sep 21 '13 at 07:46
  • Yes, I'm a lecturer. Also guilty of bad proofreading. What I meant above is that extending the definitio of the function $f(x)=x^{1/3}$ from non-negative reals to all reals is IMVHO rather pointless. – Jyrki Lahtonen Sep 21 '13 at 09:59
  • I think while it is perfectly reasonable to ask about the existence of and properties of the square of an element whose cube is blah, the 'analytic' contexts far outnumber the algebraic ones as evidenced by all the use of irrational powers and absolute value -- considering $\mathbb{R}$ and exponents as varying continuously, restricting to positive arguments only happens to be a restriction at all for integer exponents and odd roots, so one is not missing out on much. BTW @JyrkiLahtonen, in light of confessing to bad proofreading I couldn't resist mentioning 'definitio'. I believe you! – Vandermonde Oct 19 '15 at 03:44

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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 $\exp(z) = \exp(3z)^{1/3}$.