Suppose $y$ is a quadratic irrational. I want to show that there exists $M>0$ such that for all $p, q$, $q>0$,

$$\left| \frac{p}{q} - y \right| > \frac{M}{q^2}.$$

I'd like a reminder on how to do this. Say $y=[a_0, a_1, ...]$ is the continued fraction expansion with $p_n, q_n$ defined as usual. I tried using the facts

  • The continued fraction of $y$ is periodic, hence bounded

  • For $q < q_n$, $\left|\frac{p}{q} - y \right| > \left|\frac{p_n}{q_n} - y \right|$ (i.e. the continued fraction's approximant is the best)

  • $\left|\frac{p_n}{q_n} - y \right| \leq \frac{1}{q_n q_{n+1}}$

Then if the $a_i$s are bounded, say $\max_i a_i = a$, then we can write $\frac{1}{q_n q_{n+1}} \leq \frac{1}{aq_n^2 + q_{n}q_{n-1}} + \frac{1}{aq_n^2}$. Set $M = \frac{1}{a}$ and we are done.

But some of the inequalities are backwards in showing this result - particularly in the third fact, disallowing us chaining them together. I've forgotten how this is done, can someone please show where I've gone wrong and correct the proof?

  • $\begingroup$ You don't need the apparatus of continued fractions to show a lower bound, in the case of quadratic irrationals. I wrote an answer, a bit long: what matters is there is a binary quadratic form with a "minimum" integer value; then that the same form factors when the irrational roots are thrown in. I should check Cusick and Flahive... $\endgroup$
    – Will Jagy
    May 15 at 18:06
  • 1
    $\begingroup$ This is a particular case of the Liouville's Lemma. $\endgroup$
    – rtybase
    May 15 at 18:44
  • $\begingroup$ Good; Khinchin's little book, theorem 27 on page 45. archive.org/details/khinchin-continued-fractions/page/44/mode/… $\endgroup$
    – Will Jagy
    May 15 at 19:13

1 Answer 1


The thing that quadratic irrational gives you is this: your irrational, call it $r,$ is one of the roots of $Ar^2 + Br +C,$ where $A,B,C$ are integers, next $ \delta^2 = B^2 - 4AC >0,$ while this quantity is not a perfect square, that is $\delta$ is itself irrational. Call the roots $r,s.$ name $r = \frac{-B + \delta}{2A}$ while $s = \frac{-B - \delta}{2A}.$
Note $|r-s| = \frac{\delta}{|A|}.$

We make an integer quadratic form $$ A x^2 + B xy + C y^2.$$ As $\delta$ is irrational, for $x,y$ not both zero, $$ |A x^2 + B xy + C y^2| \; \geq 1 \; \; .$$

At the same time, the form factors using $\delta.$ $$ |A||x-ry | |x-sy | \geq 1 $$ Divide through by $y^2$ $$ |A| |\frac{x}{y} - r| |\frac{x}{y} - s| \geq \frac{1}{y^2} $$

Finally, let us demand that the convergent $x/y$ be close enough to $r,$ in this form: require $|\frac{x}{y} - r| < \frac{\delta}{|A|}.$ We note $ \frac{x}{y} - s = (\frac{x}{y} - r) + (r-s)$ Thus $ |\frac{x}{y} - s| < \frac{2\delta}{|A|}.$ or $\frac{2\delta}{|A|} > |\frac{x}{y} - s| $ From $ |A| |\frac{x}{y} - r| |\frac{x}{y} - s| \geq \frac{1}{y^2} $ we reach $$ |A| |\frac{x}{y} - r| \frac{2\delta}{|A|} > |A| |\frac{x}{y} - r| |\frac{x}{y} - s| \geq \frac{1}{y^2} $$ or $$ 2 \delta |\frac{x}{y} - r| >\frac{1}{y^2}$$ so that $$ |\frac{x}{y} - r| >\frac{1}{2 \delta y^2}$$

Reminder: we demanded that the rational approximation be closer to $r$ than the (fixed) distance between the roots, called $r,s.$ We are demanding $|\frac{x}{y} - r| < \frac{\delta}{|A|}.$ Now that I think of it, a poor approximation proves itself: if $|\frac{x}{y} - r| \geq \frac{\delta}{|A|}$ then $|\frac{x}{y} - r| \geq \frac{\delta}{|A|y^2}$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .