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Unsure whether question already answered - there are over 1000 questions on this forum with the "continued-fractions" string.

I request a proof (or hints) that an infinite continued fraction can not converge to a rational number. Using http://www.math.uchicago.edu/~may/VIGRE/VIGRE2008/REUPapers/Yang.pdf as a reference, I followed pages 1-7 (through theorem 1.19), but can not determine how to use the referenced theorems 1.16 or 1.19 to create the desired proof.

Note: I am aware that all rational numbers have a (unique) finite continued fraction, as discussed at https://en.wikipedia.org/wiki/Euclidean_algorithm and https://en.wikipedia.org/wiki/Continued_fraction. However, I don't think that the Euclidean Algorithm eliminates the possibility of an infinite continued fraction converging to a rational number.

user2661923
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  • The convergents of the continued fraction for $\alpha \in \mathbb{R}^+$ are the best rational approximations of the number, in the sense that any better approximation has a larger denominator when written as a fraction. You can't have a better approximation than equality. –  Mar 14 '18 at 00:15
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    Also, this isn't twitter. You can use full words in your title..... –  Mar 14 '18 at 00:19
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    It's important that it is a simple continued fraction. An infinite general continued fraction can definitely converge to a rational. – law-of-fives Mar 14 '18 at 00:21
  • to law-of-fives thanks for the specification. to user296602 although your comment seems intuitively reasonable, I don't understand how to turn this into a formal proof. Also, I looked at the bottom of https://crypto.stanford.edu/pbc/notes/contfrac/converge.html, but that reasoning compared (p/q), a convergent of the finite continued fraction with (p_n/q_n), a convergent in the (mythical) infinite convergent fraction. This seems invalid to me. – user2661923 Mar 14 '18 at 00:44

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Next day, better mood. Khinchin does, indeed, do a better job. Theorem 11 on page 12 says, with the $a_0; a_1, a_2,...$ (positive) integers, all convergents $p/q$ are irreducible, that is $\gcd(p,q)=1.$ ASSUME the limiting value of an infinite (simple) continued fraction is rational $m/n$ with $\gcd(m,n) = 1.$

Theorem 9 on page 9 says $$ \left| \frac{m}{n} - \frac{p}{q} \right| < \frac{1}{q^2} \; \; , $$ since the apparatus has $q_{k+1} > q_k.$

Your continued fraction is infinite, which means we can take a convergent far enough out so that $$ q > n \; \; . $$ Then $$ \left| \frac{m}{n} - \frac{p}{q} \right| = \left| \frac{mq-np}{nq} \right| < \frac{1}{q^2} \; \; . $$ In turn, this says $$ |mq-np| < \frac{nq}{q^2} = \frac{n}{q} < 1. $$ As $|mq-np|$ is a non-negative integer, we find $$ mq=np. $$ As $\gcd(p,q) = 1,$ it follows that $$ q | n \; \; . $$ However, this contradicts $q > n,$ then contradicts the assumption that the limit of the infinite continued fraction is a rational number.

Will Jagy
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  • wow. i'm going to flag, requesting that you be given more points. i was frustrated because i had looked at 6 different cont frac sources, the REU pdf seemed the best of the 6, and I didn't want to start over with Khinchin. Through thm 11 of the Khinchin pdf, it seems great. – user2661923 Mar 14 '18 at 22:57