I am looking for good diophantine approximations for a specific class of irrational numbers.

Let $e^{2 \pi i \theta}$ be a complex algebraic number. I would like a result to the effect that $\theta$ can be approximated well; more specifically, for any constant $k$, I would like for the inequality

$|n \theta -m| < \frac{1}{k n}$

to have infinitely many integer solutions in $n$ and $m$.

What I know is that Hurwitz's theorem guarantees a value of $k$ of at least $\sqrt{5}$, and that Khinchin's theorem asserts that, for any given $k$, the inequality $|n \alpha -m| < \frac{1}{k n}$ will have infinitely many solutions for $\textit{almost}$ $\textit{all}\ $ irrational numbers $\alpha$.

Are there any other relevant results I can use here? And is it plausible to conjecture that irrational numbers of the form $\theta$ are somehow mysteriously guaranteed to have good approximations (i.e., with any value of $k$) as given above?


1 Answer 1


Your inequality $\left| \theta - \dfrac{m}{n}\right| < \dfrac{1}{kn^2}$ has infinitely many integer solutions iff the continued fraction of $\theta$ has unbounded elements. In particular, this would not be true if $\theta$ was algebraic of degree $2$ (i.e. a root of a quadratic polynomial over the integers). However, by the Gelfond-Schneider theorem, any irrational $\theta$ such that $e^{2\pi i \theta} = (-1)^{2\theta}$ is algebraic must be transcendental. But there are also uncountably many transcendental numbers whose continued fractions have bounded elements.

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    $\begingroup$ Or, in plain English, "We have no idea and nobody else does either". Unfortunately, this is not an unusual answer to a random naive question in number theory :(. $\endgroup$
    – fedja
    Jun 2, 2012 at 1:53
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    $\begingroup$ No. Baker's theorem does give information about linear forms in logs of algebraic numbers and, in particular, to this kind of number. Probably not as strong as the OP wants, but highly non-trivial. Actually I am not sure what the OP wants. Does he want his numbers to be non-generic? Why would anyone expect that? $\endgroup$ Jun 2, 2012 at 2:57
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    $\begingroup$ @Robert: Thanks a lot, that does seem useful. Let me make sure I fully understand. I we write $k(\theta)$ to denote the sup of the constants $k$ such that the approximation inequality has infinitely many (n,m)-solutions, I believe that what you are saying is that $k(\theta)$ is infinite iff $\theta$ has unbounded elements in its continued fraction representation. Is that correct? And if so, supposing that the elements in the continued fraction are bounded by $B(\theta)$, say, is there a relationship between $k(\theta)$ and $B(theta)$? Any reference you could point me to? $\endgroup$ Jun 2, 2012 at 3:28
  • $\begingroup$ @Felipe, re. what does OP want? I would ideally like someone to say, "Yes, for that class of transcendental numbers, for any value of $k$ there are always infinitely many $(n,m)$-solutions to the inequality", or "No, sorry, here's a counterexample that has $k$ bounded by 17 (say)". Preferably with pointers to the literature... :) $\endgroup$ Jun 2, 2012 at 3:42
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    $\begingroup$ If $a_k$ are the elements and $p_k/q_k$ the convergents of the simple continued fraction of $\theta$, then $$\dfrac{1}{q_k^2 (a_{k+1}+2)} < \left| \theta - \dfrac{p_k}{q_k}\right| \le \dfrac{1}{q_k^2 a_{k+1}}$$ Moreover, every irreducible rational fraction $p/q$ with $|\theta - p/q| < 1/(2q^2)$ is a convergent. $\endgroup$ Jun 3, 2012 at 6:12

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