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Is exponentiation by rational numbers defined only for simple fractions?

$(-32)^{\frac{2}{10}}=\sqrt[10]{(-32)^2}=\sqrt[10]{1024}=\pm2$ (and $8$ other complex roots)

$(-32)^{\frac{1}{5}}=\sqrt[5]{(-32)^1}=\sqrt[5]{-32}=-2$ (and $4$ other complex roots)

How do you settle this conflict?

barak manos
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  • $x^{\frac{2}{10}} = \sqrt[10]{x^2}$ isn't correct. You could write $x^{\frac{2}{10}} = \left(\sqrt[10]{x}\right)^2$. But basically, $x^y = \exp (y\log x)$, and for rational $y = \frac{p}{q}$, you get $q$ possible values if the representation $y = \frac{p}{q}$ is reduced. – Daniel Fischer Nov 20 '14 at 11:13
  • @DanielFischer: Thanks, but can you please explain what is not correct about it? – barak manos Nov 20 '14 at 11:15
  • $\sqrt\cdot$ is by definition the positive square root. See http://en.wikipedia.org/wiki/Square_root. The case $\root{10}\of\cdot$ is similar. – Martín-Blas Pérez Pinilla Nov 20 '14 at 11:15
  • @Martín-BlasPérezPinilla: What about $\sqrt[5]{-32}$ then? – barak manos Nov 20 '14 at 11:16
  • For $n\in\Bbb N$, $x\to x^{2n+1}$ is bijective but $x\to x^{2n}$ isn't. You can define a unique $2n+1$-root for any real number. – Martín-Blas Pérez Pinilla Nov 20 '14 at 11:18
  • @Martín-BlasPérezPinilla In complex analysis, $\sqrt[n]{,\cdot,}$ means any branch of the $n$-th root. One should specify which is used, unless the choice is irrelevant. – Daniel Fischer Nov 20 '14 at 11:19
  • @barakmanos The identity $x^{ab} = (x^a)^b$ doesn't generally hold. Using it requires that one specifies branches of the powers for which the identity holds. – Daniel Fischer Nov 20 '14 at 11:22
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    @DanielFischer, true. But in this context no complex numbers are required. The "high-school-ism" $\sqrt\cdots=\pm\cdots$ is the source of many evils. – Martín-Blas Pérez Pinilla Nov 20 '14 at 11:24
  • Imagine putting bread into two similar toasters. You set the toasters to equivalent settings but for some reason one of the toast is burnt while the other is cooked to perfection. This is simply because the way the toasters handled bread was different. – Nick Nov 20 '14 at 11:59
  • @Nick: $\frac{2}{10}=\frac{1}{5}$. If to take your toasters analogy, then the breads are burnt at the same rate. – barak manos Nov 20 '14 at 12:21
  • Yes, but the output is still different. Why? The machine involved here is not as direct or simple as a toaster. Basically to exponentiate something with $\frac{2}{10}$, you pass the the thing over to the machine that does $\sqrt[10]{\star}$ and then to the machine that does $(\star)^2$. This need not get you the same resultant that the $\sqrt[5]{\star}$ machine gets you. – Nick Nov 20 '14 at 12:36

2 Answers2

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Of the several different but related definitions for exponentiation, the one that accepts a rational exponent should be phrased to exclude negative bases, in order to avoid exactly this problem.

In fact, the only one of the usually encountered definitions that give meaning to a negative integer to a non-integral power is the one for complex numbers, which comes with its own problems of being multi-valued.

One could choose to define $a^{1/n}$ for arbitrary real $a$ and odd $n$ as a root extraction -- but again this doesn't scale well to other rational exponents, and in general it is not really thought to be worth the trouble to maintain this special case when one can just write $\sqrt[n]{a}$ instead.

Alternatively, one would worm around the problem by defining $a^{p/q}=\sqrt[q]{a^p}$ if and only if $p/q$ is in lowest terms, but I'm not aware of any use of this somewhat tortuous interpretation except as a way to gain the warm fuzzies of having defined exponentiation with a slightly larger domain than the next guy uses.

Community
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hmakholm left over Monica
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2

Please read this. In general, $c^a=e^{a \log c}$ when $c\in\mathbb{C}$ and $a$ is a fraction will produce a set, so when you say $c_1^{a_1}=c_2^{a_2}$, you are comparing two sets. In particular,$$(-32)^\frac{1}{5}=2 (-1)^\frac{1}{5}=2 e^{i(\pi+2k\pi)\frac{1}{5}}=\{2e^{i\frac{\pi}{5}+n\alpha}|n=0,1,2,3,4,\alpha=\frac{2\pi}{5}\}$$

$$(-32)^\frac{2}{10}=2 (-1)^\frac{2}{10}=2 e^{i(\pi+2k\pi)\frac{2}{10}}=\{2e^{i\frac{\pi}{5}+n\alpha}|n=0,1,2,3,4,\alpha=\frac{2\pi}{5}\}$$

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John
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  • Thanks. My problem is with the fact that $\frac{2}{10}=\frac{1}{5}$, and I can't see how your answer addresses this problem. – barak manos Nov 20 '14 at 14:31
  • What I mean is that if $x<0$, to compute $x^{ab}$, you can't compute $x^a=y$ then compute $y^b$. Instead, you need to compute it using the definition, $e^{ab\log x}$ – John Nov 20 '14 at 14:48
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    If your definition of $a^b$ requires knowing the "numerator" and "demonimator" of $b$, then $a^b$ will not be well-defined... – GEdgar Nov 20 '14 at 14:49
  • @GEdgar Sorry, I don't quite understand your meaning. May you give me an example? Thanks. – John Nov 20 '14 at 14:52
  • The example is in the question! – GEdgar Nov 20 '14 at 15:30