A question on subgroup-restricted irrationality measures The irrationality measure μ(x) of a positive irrational number x is defined to be the supremum of the exponents e such that |x - p/q| < 1/q^e has an infinite number of solutions p/q. By the Dirichlet approximation theorem, the irrationality measure of any irrational number is at least 2, and by what Yann Bugeaud calls "an easy covering argument" (http://www-irma.u-strasbg.fr/~bugeaud/travaux/IrratTM1.pdf) it is measure 2 for almost all x.
If S is a subgroup of the group of positive rational numbers under multiplication which is dense in the positive reals, we can similarly define an S-restricted irrationality measure by confining p/q to S. If S is finitely generated, and in particular if S is generated by the first n primes, can we find the S-restricted irrationality measure ξ such that almost all x have measure ξ?
 A: I think the irrationality exponent is 0 for a typical number. Here's a rough argument.
Suppose you've decided to use k primes. Imagine you've chosen a $\theta$. Now you're looking
for coprime $p$ and $q$, both of size approximately $e^N$, such that $p/q\approx\theta$.
Both $p$ and $q$ are supposed to be products involving only the first $k$ primes. 
How many possible $p$'s are there? If we look at $p_1^{\alpha_1}p_2^{\alpha_2}\ldots
p_k^{\alpha_k}$, we require $\alpha_1\log p_1+\ldots+\alpha_k\log p_k\approx n$. 
There are something like $n^k$ solutions of the right size. So if you're looking at $p/q$, there are at most $n^{2k}$ possibilities (actually quite a bit less than this because of cancellation, but don't worry about this for now). 
This means that you should expect them to be very roughly spread $n^{-2k}$ apart. 
So you're getting $n^{-2k}$ approximation at a "cost" of having a denominator $e^n$. 
Since $n^{-2k}$ is much bigger than any fixed power of $e^{-n}$, I think the irrationality measure should be 0 almost everywhere, independently of $k$.
