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://wwwirma.ustrasbg.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 Srestricted 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 Srestricted irrationality measure ξ such that almost all x have measure ξ?

2$\begingroup$ Did you see my answer? Does this seem right? I was a little surprised that I wrote this over a week ago and there's no response of any kind. $\endgroup$ – Anthony Quas Dec 4 '13 at 22:53
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 $k$.

$\begingroup$ This doesn't make sense to me  it seems like the measure has to be at least $1$, since for every $q$ there's some $p$ with $\left\frac pqx\right\lt\frac1q$. Certainly this holds for e.g. the dyadics. $\endgroup$ – Steven Stadnicki Apr 25 '14 at 21:28

$\begingroup$ (Wait, silly mistake on my part  the dyadics aren't a subgroup under multiplication, only addition, as e.g. the inverse of $\frac32$ isn't in this group. The group of powers of 2, of course, isn't dense in $\mathbb{Q}^+$.) $\endgroup$ – Steven Stadnicki Apr 26 '14 at 1:45

$\begingroup$ @Steven Stadnicki: Is my argument convincing to you? It seemed to be ignored by the OP. $\endgroup$ – Anthony Quas Apr 26 '14 at 6:59

$\begingroup$ For the most part; the one concern I see is in the spacing, since there's no guarantee of uniformity across the range. Still, it looks like the core concept  there are only polynomially many numbers in an exponentiallysized range, so you can't expect any sort of density of the sort needed here  is a solid one. $\endgroup$ – Steven Stadnicki Apr 26 '14 at 7:17