I have to prove this fact (found in an article without proof). Let $\alpha \in \mathbb{R}$ be an irrational number. Let $\alpha = [a_0;a_1,a_2,\ldots]$ be the continued fraction expansion. We call $\frac{p_n}{q_n}$ the simplified fraction of the $n$-th convergent, or in other words, $$\frac{p_n}{q_n} = [a_0;a_1,\ldots,a_n].$$ Furthermore we put $\theta_n = q_n\alpha - p_n$. It can be easily proved that $\text{sgn}(\theta_{n}) = (-1)^n$ and that $\left|\theta_n\right| > \left|\theta_{n+1}\right|$.

The theorem that I want to prove states that for every $\beta \in \left]-\alpha,-\alpha +1\right[$ exists and it is unique a sequence of integers $(d_n)_{n\in \mathbb{N}}$ with $$0 \leq d_0 < a_0 \quad \text{and} \quad 0 \leq d_i \leq a_{i+1} \quad \text{ if } i > 0 \quad (1)$$ $$ d_i = a_{i+1} \quad \implies \quad d_{i-1} = 0 \quad \quad \quad \text{if } i > 0 \quad (2)$$ $$ d_{2i} \neq a_{i+1} \quad \text{for infinitely many } i $$ and with the propriety that $$\beta = \sum_{i=0}^{+\infty}\left(d_i\theta_i\right)$$

Note: maybe could be useful to know the proof of Zeckendorf's theorem, which can be found here http://en.wikipedia.org/wiki/Zeckendorf%27s_theorem, since this is a sort of generalization of it.

  • $\begingroup$ For the sake of completeness, you could give bibliographic details on the article in which you found this. $\endgroup$ Sep 27 '14 at 23:55
  • $\begingroup$ Sure! J.-M. Deshouullers A. B ́ıro ́ and V. T. So ́s. Good approximation and cha- racterization of subgroups of R/Z. Studia Scientiarum Mathematicarum Hungarica, 38:105–110, 2001. Pag. 107. In the article there is also a reference about this to V. T. Sós, On the theory of Diophantine approximations. II. Inhomogeneous problems, Acta Math. Acad. Sci. Hungar. 9 (1958), 229-241. $\endgroup$
    – Nisba
    Sep 28 '14 at 10:53

This doesn't seem quite correct. For one thing the final condition does not make sense as written. It should probably be " It is not the case that $d_{2i}=a_{2i+1}$ (so $d_{2i-1}=0$) for all large enough $i$ " (equivalently "$d_{2i} \lt a_{2i+1}$ infinitely often") . One would also want the similar condition for the situation $d_{2i-1}=a_{2i}.$

Perhaps $d_0$ can be any integer, positive or negative. Otherwise there is certainly an upper bound (perhap$(a_0+1)\theta_1$) on the $\beta$ which can be expressed.

HOWEVER, something along these lines should be true. You could likely discover then prove it by (imagining that) you are looking at an example such as $\sqrt{3}=[3;3,3,3,\cdots]$, considering the sums with $d_i=0$ for $i \gt N$ and where they are: The following would seem to need to be true along with the obvious extensions. Only the indicated inequalities actually need to be justified. $$0 \lt \theta_0 \lt 2\theta_0 \lt 3\theta_0$$ $$(d_0-1)\theta_0 \lt^? d_0\theta_0+3\theta_1 \lt d_0\theta_0+2\theta_1 \lt d_0\theta_0+\theta_1 \lt d_0\theta_0 $$

$$d_1\theta_1 \lt d_1\theta_1+\theta_2 \lt d_1\theta_1+2\theta_2\lt d_1\theta_1+3\theta_2 \lt^? (d_1-1)\theta_1$$

Describe what happens (for general $\alpha$), prove it using the properties of continued fractions, figure out what breaks down with $d_i=a_{i+1}$ and $d_{i-1} \ne 0$ and with having the (correct) final conditions fail.

Later A alternating in sign sequence $|\theta_0| \gt |\theta_2| \gt \cdots$ from a continued fraction has many properties but perhaps only some of them are needed for the result you wish (whatever it is). It might be helpful to just analyze what has to be true for the result to hold.


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