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I'm giving some lectures on continued fractions to high school and college students, and I discussed the standard theorem that, for a real number $\alpha$ and integers $p$ and $q$ with $q \not= 0$, if $|\alpha-p/q| < 1/(2q^2)$ then $p/q$ is a convergent in the continued fraction expansion of $\alpha$. Someone in the audience asked if 2 is optimal: is there a positive number $c < 2$ such that, for every $\alpha$ (well, of course the case of real interest is irrational $\alpha$), when $|\alpha - p/q| < 1/(cq^2)$ it is guaranteed that $p/q$ is a convergent to the continued fraction expansion of $\alpha$?

Please note this is not answered by the theorem of Hurwitz, which says that an irrational $\alpha$ has $|\alpha - p_k/q_k| < 1/(\sqrt{5}q_k^2)$ for infinitely many convergents $p_k/q_k$, and that $\sqrt{5}$ is optimal: all $\alpha$ whose cont. frac. expansion ends with an infinite string of repeating 1's fail to satisfy such a property if $\sqrt{5}$ is replaced by any larger number. For the question the student at my lecture is asking, an optimal parameter is at most 2, not at least 2.

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2 Answers 2

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2 is optimal. Let $\alpha=[a,2,\beta]$, where $\beta$ is a (large) irrational, and let $p/q=[a,1]=(a+1)/1$. Then $p/q$ is not a convergent to $\alpha$, and $${p\over q}-\alpha={1\over2-{1\over \beta+1}}$$ which is ${1\over(2-\epsilon)q^2}$ since $q=1$.

More complicated examples can be constructed. I think this works, though I haven't done all the calculations. Let $\alpha=[a_0,a_1,\dots,a_n,m,2,\beta]$ with $m$ and $\beta$ large, let $p/q=[a_0,a_1,\dots,a_n,m,1]$. Then again $p/q$ is not a convergent to $\alpha$, while $$\left|{p\over q}-\alpha\right|={1\over(2-\epsilon)q^2}$$ for $m$ and $\beta$ sufficiently large.

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  • $\begingroup$ Gerry: I agree with the first answer, although (perhaps you felt the same way) it seemed kind of cheap to get away with an integer for $p/q$. In your second example I didn't want to check the general case, but I checked $\alpha = [1,m,2,\beta]$. Then $p/q = (m+2)/(m+1)$, with $q = m+1$ and I find that $\alpha - p/q = A/(2q^2)$, where $A = 2(m+1)(\beta+1)/((2m+1)\beta+m) = (1 + 1/\beta + 1/m + 1/(2m\beta))/(1 + 1/(2\beta) + 1/(2m))$, which indeed is nearly 1 (from above) when the integer $m$ and irrational $\beta$ are both sufficiently large. Thus $A = 1+\delta$ for small $\delta$. Thanks! $\endgroup$
    – KCd
    Jul 26, 2012 at 20:40
  • $\begingroup$ Thanks for asking the question. I suspect one can strengthen the result by finding a single $\alpha$ such that there is an infinite sequence of $p_n/q_n$, none a convergent, with $\alpha-(p_n/q_n)=(1+\delta_n)/(2q^2)$, $\delta_n$ going to zero. I'm thinking $\alpha=[1,b_1,2,b_2,2,b_3,2,\dots]$ with $b_1\lt b_2\lt b_3\dots$. Perhaps the $b_i$ have to increase rapidly. $\endgroup$ Jul 26, 2012 at 23:51
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Roth's theorem, for which he won the 1958 Fields Medal, implies that for all algebraic numbers the exponent limit of 2 is optimal.

Theorem (Thue-Siegel-Roth). Suppose $x$ and $\alpha$ are real numbers, $\alpha >2$. If there are infinitely many reduced fractions $p,q$ satisfying $$ |x − \frac{p}{q}| < \frac{1}{q^{\alpha}}$$ then $x$ is transcendental.

As I understand it, this theorem was key in demonstrating the existence of many more (classes of) transcendental numbers (which were first proposed by Leibniz in the 1600s), and thus ending the work of approximating with rationals, that was begun by Louisville in 1844, and marking a major milestone in identifying classes of transcendental numbers that was begun by Euler in 1748.

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  • $\begingroup$ The existence of transcendental numbers was known in the 19th century, first for $e$ by Hermite and later in great abundance by cardinality reasoning by Cantor. Did you mean something else by your last sentence? $\endgroup$
    – KCd
    Jul 12, 2018 at 10:22
  • $\begingroup$ I see. The focus of my question in this post was the optimality of $2$ as a coefficient rather than as an exponent. $\endgroup$
    – KCd
    Jul 12, 2018 at 12:44
  • $\begingroup$ @ KCd. My bad. I mis-read the question. Sorry! $\endgroup$ Jul 12, 2018 at 12:53
  • $\begingroup$ Okay. I edited the title of the question to make it clearer. $\endgroup$
    – KCd
    Jul 12, 2018 at 16:41

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