# Unique representation of reals by (infinite) application of the (+,-,/,*,^) operations to elements of $\mathbb Q$?

There are many ways to express $\pi$ by infinite application of some simple operation (+,-,/,*,^) .

Is there a method that represents all real numbers uniquely? By method, I mean a restriction to certain operations applied in a certain way, such that all real numbers are expressible, but they cannot be expressed in two different ways.

Examples of representations of $\pi$:

$\pi= {3 + \cfrac{1^2}{6 + \cfrac{3^2}{6 + \cfrac{5^2}{6 + \cfrac{7^2}{6 + \ddots\,}}}}}$

$\pi = \cfrac{4}{1 + \cfrac{1^2}{3 + \cfrac{2^2}{5 + \cfrac{3^2}{7 + \cfrac{4^2}{9 + \ddots}}}}}$

What I would like the most is a natural way to restrict the class of infinite products to one that is unique for every real number. The generalized prime factorization representation $x = \prod_{i \in \mathbb N} p_i^{f(i)}$ where $p_i$ is the $i$th prime number, and $f : \mathbb N \rightarrow \mathbb Z$ is a function is an example of an infinite product representation. I.e.: What should $f$ satisfy, if it should be possible to represent all real numbers uniquely? Here, it would be nice if the representation coincided with the following property: The only $f_a$ describing the elements $a \in \mathbb Q$ corresponds to the exponents of the primefactors in the numerator and denominator in the representation of $x = p/q$.

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You can use the continued fraction representation. You can use the decimal expansion, if you avoid infinitely repeating $9$'s. You can use a harmonic sum: for $x \gt 0$ add up all the terms in $\sum_n \frac 1n$ that keep it less than $x$, for negative $x$ subtract them. None of these are products, but the last two are sums, so you can represent $\ln x$ and then you can take the exponential.
For any $x>0$ there should be infinitely many $A\subseteq \Bbb N$ such that $x=\sum_{n\in A}1/n$. However we should be able to order the subsets of $\Bbb N$ in such a way that we can choose the minimal such $A$ uniquely, e.g. $$A\le B \iff \min \big(A\setminus (A\cap B)\big)\le \min \big(B\setminus (A\cap B)\big).$$ (This minimal $A$ for which $\sum=x$ will also be the quickest converging to it.) – anon Oct 19 '12 at 2:36
@anon: yes, but I was thinking of the greedy algorithm. Take $\frac 1n$ if it fits. Then it is unique. Maybe that wasn't clear in my answer. – Ross Millikan Oct 19 '12 at 2:40