# optimality of 2 in a continued fraction theorem

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.

-

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.
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! – KCd Jul 26 '12 at 20:40
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. – Gerry Myerson Jul 26 '12 at 23:51