# If $y$ is a quadratic irrational, then there exists $M > 0$ such that $\forall p, q$, $q > 0$, $\left| \frac{p}{q} - y \right| > \frac{M}{q^2}$.

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?

• 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... May 15 at 18:06
• This is a particular case of the Liouville's Lemma. May 15 at 18:44
• Good; Khinchin's little book, theorem 27 on page 45. archive.org/details/khinchin-continued-fractions/page/44/mode/… May 15 at 19:13

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 $$|\frac{x}{y} - r| \geq \frac{\delta}{|A|y^2}$$