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I'm reading an English translation of Khinchin's Continued Fractions and I may have found an error in Theorem 1, page 4.

Khinchin observes that if we simplify a finite continued fraction $[a_0; a_1, ... a_n]$ as $p/q$, and the continued fraction $[a_1; a_2, ... a_n]$ as $p'/q'$, then because of $p/q = a_0 + \frac 1 {p'/q'}$, we can choose values of $p$ and $q$ such that:

$$p = a_0 p' + q', \quad q= p' \tag {*}$$

And uses this to recursively define a standard representation of a finite continued fraction in the form $p/q$.

Khinchin defines $[a_0; a_1, ... a_k]$ as the $k$-th convergent of a continued fraction $[a_0; a_1, ... a_n]$ where $n\geq k$.

Theorem 1 states, where $p_k/q_k$ is the standard representation of the $k$-th convergent:

$$p_k = a_k p_{k-1} + p_{k-2}$$ $$q_k = a_k q_{k-1} + q_{k-2}$$

Now Khinchin writes $p'_r/q'_r$ for the $r$-th convergent. This is where I first got suspicious... isn't he already writing $p_r/q_r$ for the that? Then he says:

On the basis of the formulas in $(*)$, $$p_n = q_0p'_{n-1} + q'_{n-1}$$ $$q_n=p'_{n-1}$$

What? That isn't what $(*)$ says at all! That recurrence relates $[a_0; a_1, ... a_n]$ to $[a_1; a_2, ... a_n]$, here Khinchin is trying to relate $[a_0; a_1, ... a_n]$ to $[a_0; a_1, .. a_{n-1}]$. It seems to me that Khinchin or the translator is using the word "convergent" wrongly. The recurrence relations forming the basis of Theorem 1 seem to be true when "convergent" is interpreted using the definition that I gave above, but I only verified that for $k=2$.

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2 Answers 2

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The problem turned out to be a very simple reading error on my part.

In fact Khinchin's $p'_r/q'_r$ is not the $r^{\rm th}$ convergent of the original continued fraction

$$[a_0; a_1, ... a_n],$$

but of the continued fraction:

$$[a_1; a_2, ..., a_n],$$

in which case his claim is obvious.

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in the middle of page 3, not numbered, he says

$$ [a_0; a_1, \ldots, a_n] = [a_0; r_1] = a_0 + \frac{1}{r_1}. $$ Here, $$ r_1 = [a_1; a_2, \ldots, a_n] $$

which is consistent with your (*), this being formula (6) on page 4.

Meanwhile, I have gotten good use out of these for years, I recommend you get accustomed to this slang for writing the convergents to some $\sqrt N.$ Notice how the first two convergents are the (legal) $0/1$ and the illegal $1/0.$ Here, for convergent $p/q,$ the integer directly below is $p^2 - 13 q^2.$

$$ \sqrt {13} $$

$$ \begin{array}{cccccccccccccccccccccccccccccc} & & 3 & & 1 & & 1 & & 1 & & 1 & & 6 & & 1 & & 1 & & 1 & & 1 & & 6 & \\ \frac{0}{1} & \frac{1}{0} & & \frac{3}{1} & & \frac{4}{1} & & \frac{7}{2} & & \frac{11}{3} & & \frac{18}{5} & & \frac{119}{33} & & \frac{137}{38} & & \frac{256}{71} & & \frac{393}{109} & & \frac{649}{180} & & \frac{4287}{1189} \\ \\ & 1 & & -4 & & 3 & & -3 & & 4 & & -1 & & 4 & & -3 & & 3 & & -4 & & 1 & & -4 \end{array} $$

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  • $\begingroup$ Yes, but the situation to which formula (6) is applied in the proof of Theorem 1 doesn't seem to be of that form. $\endgroup$
    – Jack M
    May 12, 2015 at 21:01
  • $\begingroup$ @JackM, the important thing is Theorem 1. I recommend you practice with some square roots of positive integers, as i have edited into my answer, then see if you can go back and justify (6). All that is needed to find the "digits" above, the 3;1,1,1,1,6,1,1,1,1,6 is a calculator that takes a positive real number, takes its fractional part, takes the reciprocal of that, and reports that as the new real number. The "digits," the $a_n,$ are the integer parts of those. $\endgroup$
    – Will Jagy
    May 12, 2015 at 21:09
  • $\begingroup$ @JackM, meanwhile, I think Khinchin is correct in every particular, but I can live with it if (6) turns out to be wrong. I know how to do these by hand and how to prove theorems with them. $\endgroup$
    – Will Jagy
    May 12, 2015 at 21:12
  • $\begingroup$ @WillJagy, what is the bottom line in your computation? When I compute convergents, I write down everything except for that part, and I can't tell what you're doing there. To be clear, I'm talking about the line: 1, -4, 3, -3, ... $\endgroup$ Jul 1, 2015 at 2:40
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    $\begingroup$ Ah, I see now where you explained that. Thank you. It's the number-field norm of the algebraic integer $p+q\sqrt{13}$. $\endgroup$ Jul 1, 2015 at 3:04

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