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?

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    $\begingroup$ 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 $k$.

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  • $\begingroup$ 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 $\left|\frac pq-x\right|\lt\frac1q$. 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 exponentially-sized 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

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