What is the meaning of exponentiation to real powers for which series do not exist?

Question

If we have $a\in\mathbb{R}$ and $b\in\mathbb{R}$, what do we mean to say when we write $b^a$? I understand if $a\in\mathbb{Z}$, we're expressing $$ b^a = \prod_0^a b $$ However, if $a$ is of the form $1/n,\ n\in\mathbb{N}$ what do we mean by $b^{1/n}$? Do we mean to express this as $\sqrt[n]{b}$ and then use the limit of a sum of a series (if such a series can even be guaranteed to exist)? Furthermore, what if $a\in\mathbb{R} \backslash \mathbb{Q}$? I know often a real can be expressed as a series of rationals, but can we guarantee this series of rationals exists? If so, we could we could simply write: $$ b^a = b^{\sum_{i=0}^\infty A_i}= \prod_{i=0}^\infty b^{A_i} \quad A_i\in\mathbb{Q}$$ Of course this wouldn't resolve the issue of $a = 1/n,\ n\in\mathbb{N}$. This seems elementary, but I'm stuck. Thanks for your help!

By the way, I'm a physics undergrad so I'm fairly mathematically able but have little pure math background (yet). This is curiosity, not HW.

Answer 1

This is a question that is answered in most analysis courses. You first define $x^{n}$ for natural numbers $n,$ then extend this to all integers by the natural condition $x^{m+n} = x^m x^n$ for integers $m, n.$ What you have to do next is prove that $n^{th}$ roots exist. How do we do this? Well, assuming you know a little bit of analysis...

Suppose that $a \in \mathbb{R}^{+}$ (positive reals) and consider the set $S_{a} = \{x \in \mathbb{R}\mid x \ge 0,\, x^{n} \le a\}.$ From the least upper bound property, there is a supremum of $S_{a},$ say $s.$ It's an exercise to show that $s^{n} = a.$ Here's the idea: use the trichotomy of reals and show that if $s>a,$ it is not the least upper bound. Similarly, show that if $s < a,$ then it is not even an upper bound. Following this, you can define $x^{r}$ for $r\in \mathbb{R}\backslash \mathbb{Q}$ as the least upper bound of $\{x^q \mid q \in \mathbb{Q}, q < r\},$ or the limit of a sequence, but they're equivalent.

So in fact, your question was far from "elementary" in the sense that most students will never even get to this point - it's a wonderful thing to be curious!

Answer 2

If $b > 0$ and $a \in {\mathbf R}$, pick a sequence of rational numbers $a_1, a_2, a_3, \dots$ such that $a_n \rightarrow a$ as $n \rightarrow \infty$. Then we define $b^a$ to be $\lim_{n \rightarrow \infty} b^{a_n}$. It turns out this limit is the same no matter what sequence of rationals $a_n$ you pick that converges to $a$. In this way, rational powers of $b$ are extended to real powers "by continuity", that is, by using a definition that we expect should work if the function $f(x) = b^x$ made sense as a continuous function.