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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.

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ Nice counterexample, thanks! $\endgroup$ – agb Feb 8 '15 at 18:58
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    $\begingroup$ 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

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