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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.

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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!

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I don't know much analysis (had to look up supremum, trichotomy of reals, least upper bound) but I followed roughly. How does the first part of the second paragraph relate to the second? I can see how the least upper bound of $\{x^q | q\in\mathbb{Q},\ q<r\}$ would be arbitrarily close to (equal in the limit?) $x^r$, but how does this relate to the supremum, $s$, of $S_a$? Thanks for your help! –  Bob George Jan 24 '13 at 4:46
    
So the supremum $s$ of $S_a$ is in fact the $n^{th}$ root of $a.$ After this, you can define $x^{p/q}$ as $\left(x^{1/q}\right)^p$ for rationals. And then, you can extend to reals by continuity. The supremum of the last set is unrelated to $S_a.$ And yes, you are correct that it is equal in limit (that follow from the definition of supremum). –  lyj Jan 24 '13 at 6:06
    
Awesome! Thanks so much for your time and help; I really appreciate it. Now I can sleep at night...and there might be analysis in my future. –  Bob George Jan 24 '13 at 6:17
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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.

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