Given a strictly positive integer $A$, let $D(A)$ denote the set of all real quadratic algebraic numbers with a continued fraction having almost all coefficients $\leq A$.

Consider the field $Q_A$ generated by all elements of $D(A)$. One has $Q_1=\mathbb Q[\sqrt{5}]\subset Q_2\subset Q_3,\dots$.

The inclusion $Q_1\subset Q_2$ is strict since $Q_2$ contains for example $\sqrt{2}=[1;2,2,2,\dots]$ and $\sqrt{3}=[1;1,2,1,2,\dots]$.

Are there other strict inclusions? Are there cases of equality? (I ignore for example if $Q_2$ is a proper subfield of the field $\mathbb Q[\sqrt{\mathbb N}]$ generated by all real quadratic number-fields.)

A related question: Given a real quadratic algebraic number $\alpha$ with continued fraction expansion $\alpha=[a_0;a_1,a_2,\dots]$ consider the mean value $\mu(\alpha)=\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{j=1}^na_j$. Since $[a_0;a_1,\dots]$ is ultimately periodic, this mean value is a well-defined rational number $\geq 1$ if $\alpha$ is irrational.

Are there examples of quadratic number-fields $\mathbb Q[\sqrt{N}]$ such that $\inf_{\alpha\in \mathbb Q[\sqrt{N}]\setminus\mathbb Q}\mu(\alpha)>1$?

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    $\begingroup$ Two questions about the question : (1) I guess "almost all" = "all but finitely many" ? (2) are you sure $Q_A$ will be a number field (ie: finite extension of $\mathbb Q$)? $\endgroup$ May 17, 2012 at 10:20
  • $\begingroup$ Almost all means indeed all but finitely many (it neglects thus the non-periodic part). Thank you for your observation concerning $Q_A$. It is a field but probably not finitely generated. (I have corrected this mistake). $\endgroup$ May 17, 2012 at 10:28
  • $\begingroup$ In response to your final question: Are there examples of quadratic number-fields $\mathbb Q[\sqrt{N}]$ other than $N=5$ such that $\inf_{\alpha\in \mathbb Q[\sqrt{N}]\setminus\mathbb Q}\mu(\alpha)=1$? I don't know, but it is not clear to me that there are. $\endgroup$ May 17, 2012 at 20:52

2 Answers 2


In this paper of McMullen, he asks on p. 842 whether all real quadratic fields contain infinitely many continued fractions with coefficients bounded by 2? In this case, one would have $Q_d=Q_2=\mathbb{Q}(\sqrt{\mathbb{N}})$ for all $d\geq 2$. See also McMullen's slides from his talk at MSRI in February.


Nothing is known for sure but it seems unlikely that there are any other inclusions. For $d \ge 2$ there are irrational non-quadratic numbers in $Q_d$ (of course they are algebraic). It seems likely that most, or probably all, such algebraic numbers have unbounded partial quotients which are distributed according to the Gauss–Kuzmin distribution: they are $1$ about $41 \%$ of the time and $k$ with probability $$p(k)=-\log_2 \left(1-\frac{1}{(1+k)^2} \right).$$ This is the case with probability $1$ for a uniformly distributed real number. However, there is no proof that there is even one algebraic number with unbounded partial quotients.

I have no idea about your last question. It might be more natural to ask about $\lim_{n\rightarrow\infty}\left(\prod_{j=1}^na_j\right)^{\frac{1}{n}}$.

  • $\begingroup$ Concerning the first question: I do not care for numbers with unbounded coefficients. The question is essentially: Is there a smallest integer $A$ such that every real quadratic number field can is generated by a number having a continued fraction involving only coefficients $\leq A$. Replacing the arithmetic by the geometric sum in the last question changes (essentially) nothing since the question is for quadratic reals which have eventually periodic continued fraction expansions. $\endgroup$ May 18, 2012 at 8:21

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