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Liouville's Theorem states that for any algebraic $\alpha \in \mathbb{R}$ of degree $n$, there exists a positive constant $c:=c(\alpha)$ such that $$\left\lvert\alpha-\frac{p}{q}\right\rvert>\frac{c}{q^n}$$ for any $p \in \mathbb{Z}$ and $q \in \mathbb{N}.$

One can find an effective lower bound for $c(\alpha).$ In the special case that $\alpha$ is a quadratic irrational, Exercise 27 in the following set of notes of Jorn Steuding

http://www.math.uni-bremen.de/~bos/dioph.pdf

yields $$c(\alpha) \gg \frac{1}{(1+|\alpha|)H(\alpha)} .$$ Here if $m_{\alpha}(x):=x^2+bx+c \in \mathbf{Q}[x]$ is the minimal polynomial of $\alpha$, the height $H(\alpha)$ is defined as the maximum of $|b|$ and $|c|.$ My question is whether one can find a better lower bound for $c(\alpha)$ when $\alpha$ is a quadratic irrational or if this is best possible.

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  • $\begingroup$ I would start with the ultra-classic Hardy - Wright. I vaguely remember some related discussion in the final notes of the relevant chapters, with references $\endgroup$
    – user24527
    Commented Jun 27, 2012 at 19:48

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Let's assume $\alpha > 0$. If $\alpha$ is a quadratic irrational, its simple continued fraction $a_0 + \dfrac{1}{a_1 + \frac{1}{a_2+\ldots}}$ is eventually periodic. Every $p/q$ (in lowest terms) with $\left|\alpha - \dfrac{p}{q}\right| < \dfrac{1}{2q^2}$ is a convergent of $\alpha$, and for the $n$'th convergent $$ \dfrac{1}{q_n^2 (a_{n+1}+2)} < \left| \alpha - \dfrac{p_n}{q_n} \right| \le \dfrac{1}{q_n^2 a_{n+1}}$$ Thus $\dfrac{1}{a_M+2} \le c(\alpha) \le \dfrac{1}{a_M}$ where $a_M$ is the largest element in the continued fraction of $\alpha$.

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  • $\begingroup$ Thank you Robert. Your answer proves that an equivalent formulation (up to absolute constants) of my question is: Is there a good upper bound about the maximum of the elements in the continued fraction of a quadratic irrational ? $\endgroup$
    – Dr. Pi
    Commented Jun 27, 2012 at 23:23
  • $\begingroup$ In the case where the quadratic irrational is simply $\sqrt d$, the largest element in the continued fraction is twice the integer part of $\sqrt d$. $\endgroup$
    – Gerry Myerson
    Commented Jun 27, 2012 at 23:50
  • $\begingroup$ Indeed Lagrange showed that the largest element for $(P+\sqrt D)/Q$ is less that $2\sqrt D$. $\endgroup$
    – Gerry Myerson
    Commented Jun 28, 2012 at 0:03
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The constant $c=1/\sqrt5$ (with $n=2$) works for any $\alpha$. If $\alpha$ is not, roughly speaking, the golden ratio, then $c$ can be improved to $1/2\sqrt2$, etc. If one removes a certain infinite sequence of quadratic irrationals, one can take $c=1/3$, but this is the best you can do in a general setting.

A nice exposition can be found in [Cassels, An Introduction to Diophantine Approximation]. You may also start with a Wikipedia article.

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  • $\begingroup$ Dear Nikita, I am afraid that Liouville's is about how bad the approximation of an irrational by a rational can be. Unfortunately you refer to Hurwitz's result, as well as Markoff's spectrum, both of which mention how good this approximation can be. That is the opposite of what I am asking. I am really sorry for this. $\endgroup$
    – Dr. Pi
    Commented Jun 27, 2012 at 23:17
  • $\begingroup$ OK, noted. Still, if you look at all the quadratic irrationals in Markoff's spectrum, these constants are exact. $\endgroup$
    – Nikita Sidorov
    Commented Jun 28, 2012 at 0:19
  • $\begingroup$ Indeed you are right. However, the reason for my question was that I had a specific application in mind, where no information regarding $\alpha$ exists, apart of course from the fact that $\alpha$ is a quadratic irrational. $\endgroup$
    – Dr. Pi
    Commented Jun 28, 2012 at 0:47

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