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One can motivate Roth's theorem as follows. On $[0,1]$ consider the function $f$ that takes $x$ to the cardinality of the set

{ $p/q : |x - p/q | < 1/q^{2 + \epsilon}$ } .

Now one can see that $\int f dx <\infty$ by overestimating with a sum over q. This means that $f$ has finite values except on a set $E$ of measure 0. A random countable set would miss E, so one can hope that the irrational real algebraic numbers in $[0,1]$ will too.

But say now that we are only interested in denominators of the form $q=10^j$. The same heuristic strategy produces a conjecture much stronger than what Roth's Theorem offers. To wit, defining again a new function $f$ by the cardinalities of {$p/q, q = 10^j : |x - p/q | < \frac{1}{q \cdot (\ln q)^{1 + \epsilon}}$ } leads again to a function with a finite integral, and so the the prediction that algebraic numbers will have only finitely many approximations this good by numbers of the form $p/10^j$.

With so much room between Roth and and this prediction, my question is whether any result in the literature improves upon Roth in the direction of the prediction? More generally, are there known Roth variants where, at the expense of restricting the denominators $q$, one still has finiteness for a weaker standard of approximation?

David Feldman
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    is your question (at the end) just about irrational real algebraic numbers? – mathworker21 Dec 06 '23 at 21:02
  • Roth's theorem is famously ineffective, so attempting to improve the denominator from $q^{2 + \varepsilon}$ to something like $q^2 (\log q)^\kappa$ for some $\kappa > 0$ would be extremely difficult. – Stanley Yao Xiao Dec 06 '23 at 21:14
  • @mathworker21 Yes. – David Feldman Dec 06 '23 at 23:05
  • @StanleyYaoXiao I think you want $\kappa > 1$ and then that's Lang's conjectures, but my question is in a different direction, because I'm not asking about all rational approximators. – David Feldman Dec 06 '23 at 23:05
  • Such a bound is basically the same as asking that the decimal expansion of $x$ not have any long strings of zeroes too early, and is similar to the question of whether $x$ is normal https://en.wikipedia.org/wiki/Normal_number . Almost nothing is known about proving concrete numbers are normal, so I would suspect nothing like this is known. See https://mathoverflow.net/questions/23547 for similar discussion. – David E Speyer Dec 07 '23 at 02:04
  • @DavidESpeyer Well that was indeed what I was thinking about. But only "similar" right? My conjecture could be false without scuttling normality, right? I mean that the infinitely many "long strings of zeroes" could arise so rarely that they would not spoil the statistics. So the two questions are mutually independent for all we know. – David Feldman Dec 07 '23 at 06:49
  • Agreed, I don't see a reduction in other direction. – David E Speyer Dec 07 '23 at 12:34

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