Lately I was reading a bit about continued fractions and came up with a question that I couldn't find an answer for.

Here are the definitions I will use in the question:

Let $x \in \mathbb{R}$. A fraction $\frac{p}{q}$ (assume $q > 0$) is said to be a rational best approximation of $x$ if $$\left| x - \frac{p}{q}\right| \leq \left|x - \frac{p'}{q'}\right|$$ for all $p', q' \in \mathbb{Z}, 1 \leq q' \leq q$.

Then $\frac{p}{q}$ is called a good approximation of $x$ if $$ \left|x - \frac{p}{q}\right| < \frac{1}{q^2}. $$ Now I know that every convergent of the continued fraction for $x$ is both a best approximation and a good approximation.

On the other hand: Not every best approximation for $x$ is given through a convergent of its continued fraction (take e.g. $13/4$, which is not a convergent of $\pi$ but a rational best approximation).

My question is: Is every good approximation given through a convergent? By checking a few examples of best rational approximations which are no convergents I got the feeling that this could be true, but I did not find a definite answer on this.

  • $\begingroup$ agb: Just put a comment on my answer but forgot to put your handle on it, so you likely don't get a notice... $\endgroup$ – coffeemath Feb 8 '15 at 19:39

The first few convergents for $\sqrt{3}$ are $1,2,5/3,7/4,19/11.$ In particular none have denominator $7,$ but $\sqrt{3}-12/7 \approx 0.0177$ while $1/49 \approx 0.0204.$ So in this example we have a good approximation which is not a convergent.

Added: If one defines a "very good" approximation of an irrational $x$ as one for which $|x-p/q|<1/(2q^2),$ then it is known that any very good approximation to an irrational must be one of the convergents to it. The above example of the good approximation $12/7$ is not close by a margin of $1/2\cdot 49$ to $\sqrt{3},$ as would be expected by this known result.

  • $\begingroup$ Nice counterexample, thanks! $\endgroup$ – agb Feb 8 '15 at 18:58
  • 1
    $\begingroup$ I think the topic of Farey series, at least for approximating numbers in $(0,1),$ is relevant here. Once one takes off the integer part, $\sqrt{3}-1 \approx .73205,$ while $12/7-1=5/7\approx .7142.$ Adjacent terms in the Farey series of order $n$ are within about $1/n^2$ of each other, and it seems related to your question except the requirement to be a convergent restricts the denominators, which have to get large for convergents. See wiki page en.wikipedia.org/wiki/Farey_sequence $\endgroup$ – coffeemath Feb 8 '15 at 19:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.