How to define an irrational to the power an irrational How to define an irrational number to the power an irrational number ?. At max I can understand the definition of an irrational to the power rational, but how to define an irrational number to the power of irrational. 
 A: Assume that $a$ is positive. One way to define $a^x$ is to take a sequence $r_0,r_1,r_2,\dots$ of rationals with limit $x$, and show that the sequence $a^{r_0},a^{r_1}, a^{r_2},\dots$ has a limit independent of the chosen sequence. Then we can define $a^x$ to be that limit. This is technically somewhat unpleasant, and verifying that the function so-defined has the right properties takes some work.
We also need to have done prior work in defining $a^r$ for rational $r$, and verifying that the right properties hold.  
An alternative is to define the (natural) logarithm function via an integral, verify its basic properties, and define the exponential function as the inverse of the logarithm function. Then we define $a^x$ by $a^x=\exp(x\ln a)$.
Alternately, we define the exponential function $\exp(x)$ using the power series $1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$, or as the solution of a certain initial value problem. Then we define the natural logarithm function as the inverse of the exponential function, and use the same definition of $a^x$ as in the preceding paragraph. 
A: I would write this as a comment, but I apparently don't have enough points:
We can use the fact that a continuous function  f defined on a dense subset  has (under certain conditions;  f must actually be uniformly-continuous) a unique extension to the whole space. In this case, if you can define $a^x$ on the dense subset $\mathbb Q^2$ of 
$\mathbb R^2$ , then you can extend the definition uniquely to the whole of $\mathbb R^2$ using sequential continuity.
A: Following Rudin's Principles of Mathematical Analysis, it's easy.
First, show there's an $n$th root by considering the l.u.b. of numbers whose $n$th powers are no more the target number.
Then, show that the $n$th root is unique, by noticing that $n$th power is one-to-one function.
Define, temporarily,
$$
x^{\frac{k}{l}} = (x^\frac{1}{l})^k
$$
Then, notice that for any $x$, and integer exponents, $x^{ab} = (x^a)^b = (x^b)^a$, because this kind of exponentiation is defined by multiplication.
Therefore if $r$ is rational, with two different representations $r = \frac{p}{q} = \frac{m}{n}$, then
\begin{eqnarray}
(x^r)^{qn}
   &=& (x^\frac{p}{q})^{qn}
   = (x^\frac{1}{q})^{qpn}
   = \Big[(x^\frac{1}{q})^{q}\Big]^{pn}
   = x^{pn}
\end{eqnarray}
But also
\begin{eqnarray}
(x^r)^{qn}
   &=& (x^\frac{m}{n})^{nq}
   = (x^\frac{1}{n})^{nmq}
   = \Big[(x^\frac{1}{n})^{n}\Big]^{mq}
   = x^{mq}
\end{eqnarray}
Now, $np = mq$, since the fractions are equal, so both ways of expressing $r$ as a fraction give the same value for of $x^r$ (again because raising to integer exponents is one-to-one).
Since there's a well-defined rational exponent, all you have to do is define:
$$
a^x = \sup \{a^r : r \in \mathbb{Q}, r \le x \}
$$
