continuity extension of exponential $f(x)= a^x$ Consider the exponential function $f(x) = a^x$, where $f: \mathbb{Q} \to \mathbb{R}$. My problem is to show that it has unique extension and how am I going to define this one? Also, I used a convergent sequence of rational numbers to irrational numbers. But how can I be sure that the corresponding function values converge uniquely? 
 A: http://en.wikipedia.org/wiki/Cauchy-continuous_function
So you just have to show that for any Cauchy sequence (x1, x2, ...) that (f(x1), f(x2), ...) is Cauchy in R. 
A: Assume the $n$-th root function of a positive (nonnegative) real is known (and you can build it by monotonicity of the integer $n$-th power and completeness of reals). You can reduce the problem to the "fundamental" limit $$a^q\to1\text{ when }q\to0,$$ by using the power laws $a$ (consider first $a>1$ and if $a^{1/n}\geq1+\rho$ for some $\rho$ and all $n$, then for the binomial theorem $a=(a^{1/n})^n=(1+\rho)^n\geq 1+n\rho$ against Archimedean property of rationals).  Then use the so-called semigroup property $$a^{q+\eta}=a^qa^\eta\to a^q1=a^q$$ for fixed $a\in\mathbb R$, $q\in \mathbb Q$ and variable $\eta\in\mathbb Q$ as $\eta\to0$.  This proves that the exponential with base $a>1$ is continuous on $\mathbb Q$
a dense metric subspace of $\mathbb R$ and therefore $q\mapsto a^q$ can be uniquely extended by continuity to $\mathbb R$.  This general result can be proved (a bit tedious) by using Cauchy sequences of rationals "converging" to irrational numbers as sequences in $\mathbb R$.
