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Jul 11, 2022 at 1:47 vote accept Dominic van der Zypen
Jul 10, 2022 at 21:38 answer added user485571 timeline score: 3
Jul 10, 2022 at 7:33 comment added Gerry Myerson For fixed $r$ and $c$, $c>2$, those approximations don't necessarily come from convergents to a continued fraction, but (as noted earlier) I suspect the $n$ that work are still few and far between.
Jul 10, 2022 at 7:29 comment added Dominic van der Zypen That's very interesting, thanks @GerryMyerson! - Does your argument extend to $r$ irrational and $|r - (m/n)| < c(n^{-2})$ for $c > 2$?
Jul 10, 2022 at 1:09 comment added Gerry Myerson Note that if $r$ is irrational, and $|r-(m/n)|<(2n^2)^{-1}$, then necessarily $m/n$ is a convergent of the continued fraction for $r$. For a given $r$, such $n$ form a very sparse set. I suspect that for any fixed $r$ and $c$ the $n$ for which $|r-(m/n)|<cn^{-2}$ will also be sparse.
Jul 10, 2022 at 1:00 history edited Dan Turetsky
This question does not relate to computability theory, so I have removed that tag.
Jul 9, 2022 at 20:40 history asked Dominic van der Zypen CC BY-SA 4.0