What does $x^\pi$ mean? I was just wondering, what does $x^\pi$ or for that matter, $x$ raised to any irrational number mean? For example, I want to represent $x^2$ then that would mean $x * x$ or if I want to do $x^\frac{2}{3}$ then that would mean, I first square it then cube root it but what does that mean for irrantional numbers such as $\pi$?
Thanks!
 A: If $0< x$ we have:
$$x^\pi=\sup\{x^r\mid r<\pi, r\text { is a rational number}  \}$$
A: From a very abstract viewpoint, the position $$x^\pi = e^{\pi \log x}$$ is a circular definition: after all $e$ is as much a real number as $x$. There are two ways to break this cirle:


*

*$e^{\pi \log x}$ means $\exp (\pi \log x)$, provided you have defined the exponential function in an indipendent way;

*we study the theory real numbers, which includes the construction of $b^x$ for $b>0$ and $x \in \mathbb{R}$. This is not trivial at all, as can be seen in W. Rudin's Principle of mathematical analysis.

A: It's better to think of it as rather
$$x^{\pi} = e^{\pi \log{x}}$$
A: Ultimately, it's nothing near so nice as the rational case. If we're given an irrational number $\alpha$, then $$x^\alpha:=\underset{r\to\alpha}{\lim_{r\in\Bbb Q}}x^r,$$ so long as this limit is defined (for example, it doesn't work when $x$ is negative, and it doesn't work when $x=0$ and $\alpha$ is negative).
Alternately, it is at times more beneficial to define it as $$x^\alpha:=e^{\alpha\ln x}.$$ These are equivalent definitions for $x>0$. We can even do a bit better and define $$x^\alpha:=\lim_{t\to x^+}e^{\alpha\ln t}.$$ That definition works precisely when the original definition does--namely, whenever $x>0$, and whenever $x=0$ and $\alpha>0$. Of course, both of the latter two definitions do require some independent definition of $e^w$, such as $$e^w:=\lim_{n\to\infty}\left(1+\frac wn\right)^n.$$
A: Ron already gave the canonical answer: $x^\alpha = e^{\alpha \log{x}}$ by definition, and then you may refer to the very nice properties of $e^x$. If you prefer, think at two approximating series of rational numbers $\{a_n\}$ and $\{b_n\}$ converging from above and from below to $\pi$, compute $\{x^{a_n}\}$ and $\{x^{b_n}\}$ and see that they actually converge to a single value.
At least as far as I am concerned, there is not a lot of difference asking what $x^\alpha$ is and asking what an irrational number $\alpha$ is.
