TITLE: What does $2^x$ really mean when $x$ is not an integer?
QUESTION [209 upvotes]: We all know that $2^5$ means $2\times 2\times 2\times 2\times 2 = 32$, but what does $2^\pi$ mean? How is it possible to calculate that without using a calculator? I am really curious about this, so please let me know what you think.

REPLY [6 votes]: We assume that $\mathbb{R}$ is constructed using equivalence classes of rational Cauchy-sequences and $\alpha^\frac{1}{n}$ is defined as the unique non-negative solution of $x^n = \alpha$, where $\alpha$ is non-negative. For rational $a=\frac{p}{q}$, $p,q \in \mathbb{Z}$, $2^a$ is defined by $(2^\frac{1}{q})^p$. We first show that $2^a$ from $\mathbb{Q}$ to $\mathbb{R}$ is a strictly increasing function. Take $a,b\in\mathbb{Q}$, $a<b$. Then $a = \frac{p_1}{q_1}$, $q_1 > 0$ and $b = \frac{p_2}{q_2}$, $q_2 > 0$, where $p_i,q_i\in \mathbb{Z}$. The following lines are equivalent
\begin{eqnarray}
2^a & < & 2^b \\
2^\frac{p_1}{q_1} & < & 2^\frac{p_2}{q_2} \\
(2^\frac{1}{q_1 q_2})^{p_1 q_2} & < & (2^\frac{1}{q_1 q_2})^{p_2 q_1} \\
p_1 q_2 & < & p_2 q_1 \\
\frac{p_1}{q_1} & < & \frac{p_2}{q_2} \\
a & < & b \ .
\end{eqnarray}
The fourth line is the definition of the ordering relation $<$ in $\mathbb{Q}$. Hence $2^a$ is strictly increasing. Take now any Cauchy-sequence $\{a_n\}_{n = 0}^\infty \subset \mathbb{Q}$. Because $a_n$ is a Cauchy-sequence, it is bounded and there is an integer $m$ s.t. $a_{n_k} < m$. Assume that $2^\frac{1}{j}$ does not converge to $1$ as $j \rightarrow \infty$. The sequence is decreasing and bounded below. It has a limit. Note that the limit cannot be reached because the sequence is strictly decreasing. Let the limit be $\beta > 1$. Then we have $2^\frac{1}{j} > \beta$ for every $j \in \mathbb{Z}^+$. We can now estimate
\begin{eqnarray}
2 = x_j^j > \beta^j \rightarrow \infty \ ,
\end{eqnarray}
that is a contradiction. Hence
\begin{equation}
\lim_{j \rightarrow \infty} 2^\frac{1}{j} = 1 \ .
\end{equation}
We next estimate the sequence $a_n$. Choose $\epsilon > 0$. Choose $j \in \mathbb{Z}^+$ s.t. $ 2^m (2^\frac{1}{j} - 1) < \epsilon$. Choose $N \in \mathbb{N}$ s.t. $k,l>N$ implies $|a_k-a_l| < \frac{1}{j}$. We estimate
\begin{eqnarray}
|2^{a_k}-2^{a_l}| & = & 0 < \epsilon \ \ , \ a_k = a_l \ , \\
|2^{a_k}-2^{a_l}| & = & 2^{a_l} (2^{a_k-a_l} - 1) <  2^m (2^\frac{1}{j}-1) < \epsilon \ \ , \ a_k > a_l \ , \\
|2^{a_k}-2^{a_l}| & = & 2^{a_l}-2^{a_k} = 2^{a_k} (2^{a_l-a_k} - 1)  <  2^m (2^\frac{1}{j}-1) < \epsilon \ \ , \ a_k < a_l \ . 
\end{eqnarray}
Hence $2^{a_k}$ is a Cauchy-sequence and it converges to a limit. Continuity of $2^a$ can be shown as follows. Assume that $a_n$ converges to a limit $a\in\mathbb{Q}$. Then replace $a_l$ by $a$ in estimates and obtain the definition of convergence.
We define
\begin{equation}
2^x = \lim_{n \rightarrow \infty} 2^{a_n} \ .
\end{equation}