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Well, this question may seem silly at first, but I'll make my point clear. Suppose $n \in \Bbb N$ and suppose $a \in \Bbb R$ is any number. Then the definition of $a^n$ is clear for any $a$ we choose. Indeed we define:

$$a^n = \prod_{k=1}^na$$

And even if $a$ is negative this has a meaning. Then we extend the definition for $n \in \Bbb Z$ and for $n \in \Bbb Q$. When we are to define to $n \in\Bbb R$ we define it as:

$$a^x=e^{x\ln a}$$

That's fine, but $\ln $ is a function defined on $\Bbb R^+$ so that if we try to compute $(-5)^\pi$ we'll get into trouble because this would be:

$$(-5)^{\pi}=e^{\pi \ln(-5)}$$

But $\ln (-5)$ is undefined. In that case, the function that $f : A \subset \Bbb R^2 \to \Bbb R$ given by $f(a,x) = a^x$ would be undefined if $a < 0$, so that $A = \Bbb R^+ \times \Bbb R$. What I thought was: we can extend this function when $a$ is negative and $x$ is rational. In that case we would set it to the old definition of exponentiation, since we would have a real raised to a rational.

So, for negative base and irrational exponent the exponential remains undefined?

Thanks very much in advance!

Gold
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  • . . . and I believe some programming languages will give you an error message if you try to raise an irrational number to a power. – Michael Hardy Jun 09 '13 at 03:10
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    @MichaelHardy: all the ones I know accept positive numbers to powers (using $a^b=e^{b \log a}$) and negative numbers to integral powers (using multiply), but not negative numbers to non-integral powers. – Ross Millikan Jun 09 '13 at 03:33
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    I don't see how you are extending the definition to $n \in \mathbb Q$ if $a < 0$, unless you already want to pass to the complex numbers here. Take for instance $a = -1$ and $n = 1/2$. – treble Jun 09 '13 at 03:37

2 Answers2

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It depends what number system you are working in. In the reals, there is no answer. In the complex numbers, there are many answers. You can take the principal branch of the logarithm function, just saying that the imaginary part of the result will be in the range $(-\pi,\pi]$, in which case it is well defined. But if $e^z=w$, it is also true that $e^{z+2i\pi k}=w$ for $k \in \Bbb Z$ so you could say $\log w=z + 2i \pi k$

Ross Millikan
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If $a < 0$ and $a \ne -1$, then it is possible to define $f(x) = a^x$, provided we restrict our domain to $a \in \{m / n : m$ is an integer and $n$ an odd positive integer$\}$. In this case, $a^{m/n} = \sqrt[n]{a^m}$.

For example, $(-8)^{-2/3} = 1/\sqrt[3]{(-8)^2} = 1/4$.

Tobia
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user264158
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