irrational root of a number number

Question

Let $x>0$, and let $\alpha$ be an irrational number. Can we make sense of $x^{\alpha}$ ? What about the case $x<0$ ?

Answer 1

Assuming everything here is real, there are several ways to make sense of $x^\alpha$.

The one most people learn in school is (possibly a simplification of) the following: Let $a_1, a_2, a_3, \ldots$ be a sequence of rational numbers converging to $\alpha$. Then we define $$ x^\alpha = \lim_{n\to\infty}x^{a_n} $$ One should, of course, make sure that any sequence converging to $\alpha$ gives the same limit.

The one I think is most often used as a definition by mathematicians is simply $$ x^\alpha = \exp(\alpha\cdot\ln x) $$ where both $\exp$ and $\ln$ may be defined several different ways, for instance $$ \exp(a) = 1 + a + \frac{a^2}2 + \frac{a^3}6 + \cdots + \frac{a^n}{n!} + \cdots\\ \ln x = \int_1^x\frac1tdt $$ There are many definitions to choose from here, though.

Then we have this one, which is not so common, but somewhat of a favourite of mine, if one knows a little group theory: Consider $(\Bbb R, +)$, the real numbers with the operation of addition and $(\Bbb R^+, \cdot)$, the positive real numbers with multiplication. As groups, these are isomorphic, and there are many different homomorphisms between them, but each one may be uniquely determined by where it sends $1\in (\Bbb R, +)$ (and we probably need to assume that it's "nice" for some suitable notion of niceness, like being continuous, or monotonic).

One of these homomorphisms $f: (\Bbb R, +)\to (\Bbb R^+, \cdot)$ has $f(1) = x$. We then define $$ x^\alpha = f(\alpha) $$