TITLE: Is being rational decidable?
QUESTION [21 upvotes]: Given a real number uniquely defined by a finite system of equations and inequalities with rational coefficients involving the standard elementary functions only. Is it decidable whether this number is rational?
Edit: The system of equations and inequalities may have $n$ real-valued variables, and the real number to be defined is $\xi$. By decomposition using intermediate variables, we may suppose that each inequality is of the form $x_i\,\omega\, c$ where $\omega\in\{<,>,\le,\ge\}$ and $c$ is an integer, and each equation is of the form $x_i=\xi$, or $x_i=c$ where $c$ is an integer, or $x_i=x_j\circ x_k$ with $\circ\in\{+,-,*,/\}$, or $x_i=\phi(x_k)$ where $\phi\in E$ with a specified set $E$ of elementary functions, such as $E=\{\exp,\log\}$. The answer to the question might depend on the choice of $E$.
I assume that $\xi$ is uniquely determined (hence defined) by this system of equations and inequalities. One may even assume that one has a proof for this, should this help.
The algorithm may be defined according to any of various traditional models, specifying a sensible set of primitive operations (including the rational operations [e.g., on bits or on reals] and branching on an expression). I am more interested in the applicable techniques than in an answer for a particular model.
Note that the Wikipedia article on the constant problem 
claims (by reference to 
D.H. Bailey, Mathematics of Computation 50 (1988), 275–281)
a result from which decidability would follow, but this claim is spurious.

REPLY [4 votes]: I found a paper by Babai, Just, and Meyer auf der Heide that answers a closely related question. 
They prove among others (see the discussion on p.101) that there are no anlytic computation trees (ACTs) that accept the complement of the rationals in the reals, and hence would prove the irrationality of a number.