Let $\alpha$ be a positive real number. Does it make sense to define the closest rational to $\alpha$ as the number $R(\alpha)=\frac{p_1}{p_2}$ such that $p_1,p_2$ are positive co-prime integers minimizing $p_2 \cdot |p_2\alpha - p_1|$? Clearly, there are going to be some irrational numbers for which this makes no sense, for instance $\alpha=\frac{1+\sqrt{5}}{2}$. My guess is that such numbers are rare: there are countably infinitely many of them, or their set has Lebesgue measure 0. Do we have $R(\pi) = \frac{355}{113}$? I looked at the fist 12 convergents of $\pi$, and $\frac{355}{113}$ achieves the minimum.

This is related to approximations of irrational numbers, continued fractions, and the irrationality measure of a number. More generally, we could define the closest algebraic number of degree $d$, as the number $R_d(\alpha)$ defined as follows.

$$P^*= \arg\min_P\Big(H(P)\Big)^{\mu(d)}\cdot |P(\alpha)|,\\ R_d(\alpha) = \Big(H(P^*)\Big)^{\mu(d)}\cdot |P^*(\alpha)| $$

where $P$ is any polynomial of degree $d$ with integer coefficients, with highest and lowest coefficients not equal to $0$, and $H(P)$ is the height of $P$, that is, its highest coefficient in absolute value. To make this work for most $\alpha$, how should we choose $\mu(d)$? Does $\mu(d)=d$ work? It seems to work if $d=1$.

As of now, as far as I know, all results involving $d>1$ are conjectures. Related material includes the Wirsing conjecture. See also the "Generalizations" section in the Wikipedia article on Roth's theorem, here.

Update on Feb 8, 2021: It is possible that the best approximation, if $d$ is an even integer, may be a complex number. Also, see my new question here, about approximations by dyadic fractions. The plan is to look at approximations using the first $n$ digits of $\alpha\in [0, 1]$ in base $b$, where (say) $b=\sqrt{2}$, focusing on values of $n$ where a long run of zeros start, leading to approximations of transcendental numbers by quadratic irrationals. This is briefly discussed in the comments in my new question.

  • $\begingroup$ It would be interesting to see which real numbers $\alpha$ have $R(\alpha)=R(\pi)$, assuming $R(\pi)$ exists. $\endgroup$ Feb 4, 2021 at 20:11
  • $\begingroup$ Since every countable set has measure 0, saying "either X is countable, or X has measure 0" is kind of redundant (and either-or reads to me, as a non-native speaker anyway, as a dichotomy, so exactly one of the options is true). $\endgroup$
    – Asaf Karagila
    Feb 5, 2021 at 11:13
  • $\begingroup$ @Asaf: unless I am mistaken, a set can have measure 0 yet be non countable. For instance, consider the set of all real numbers and their representation in base 2. For each real, add a 0 in the binary digits in positions 1, 3, 5, and so on. The transformed numbers are in bijection with real numbers, but none of them is a normal number. So it is an (uncountable) subset of non-normal numbers. And non-normal numbers have measure 0, so that set also has measure 0. $\endgroup$ Feb 5, 2021 at 16:43
  • $\begingroup$ Yes, but you're not reading my [previous] comment. $\endgroup$
    – Asaf Karagila
    Feb 5, 2021 at 16:43
  • $\begingroup$ I'll try to fix my wording, but true, I am not a native speaker. $\endgroup$ Feb 5, 2021 at 16:45

1 Answer 1


Let $\alpha$ be an irrational. We shall consider its continued fraction $[a_0;a_1,a_2,\dots]$. Recall some basic results about convergents of continued fractions (see e.g. here): letting $p_n,q_n$ be the sequence of numerators and denominators of convergents, for any $n>1$ we have $q_{n+1}=a_{n+1}q_n+q_{n-1}>a_{n+1}q_n$ and $$\left|\alpha-\frac{p_n}{q_n}\right|<\frac{1}{q_nq_{n+1}}<\frac{1}{a_{n+1}q_n^2},$$ hence $q_n|q_n\alpha-p_n|<\frac{1}{a_{n+1}}$. Therefore if $a_{n+1}$ are unbounded, then $q_n|q_n\alpha-p_n|$ does not attain a minimum. Therefore $R(\alpha)$ does not exist for those $\alpha$.

Therefore $R(\alpha)$ can only exist for badly approximable numbers. It is known that those numbers form a set of measure zero, and most natural constants besides quadratic irrationalities, including $\pi$ and all higher degree algebraic irrationals, are conjectured to not lie in it. Therefore $R(\pi)$ probably doesn't exist.

On the other hand, if $\alpha$ is badly approximable, then this still doesn't necessarily mean $R(\alpha)$ necessarily exists, as you note with $\alpha=\frac{1+\sqrt{5}}{2}$. In fact I believe it won't exist for any quadratic irrational. However, using the bound $$\left|\alpha-\frac{p_n}{q_n}\right|>\frac{1}{q_n(q_{n+1}+q_n)}>\frac{1}{(a_{n+1}+2)q_n^2},$$ we at the very least get that for those numbers the quantity $q_n|q_n\alpha-p_n|$ is bounded away from zero (note that $q|q\alpha-p|$ can only be smaller than $1/2$ if $p/q$ is a convergent, so we don't lose much from looking at just looking at convergents).

Last remark I have is that there are uncountably many $\alpha$ for which $R(\alpha)$ exists. Indeed, from the above considerations it follows easily that this is the case if for some $N$ we have that the continued fraction of $\alpha$ contains a partial denominator $N$, but from some point on all denominators are at most $N-2$.

  • 4
    $\begingroup$ The famously accurate approximation $\pi\approx\frac{355}{113}$ comes from the 5th partial denominator of $\pi$, equal to 292, being exceptionally large. According to OEIS the next largest one occurs at position 308 and is equal to 436. So you would need to do a lot of computation to actually realize $R(\pi)\neq \frac{355}{113}$. $\endgroup$
    – Wojowu
    Feb 4, 2021 at 20:57
  • $\begingroup$ Thank you for the great insights. I was wondering if there might be a function $f(q)$ such that minimizing $f(q) |q\alpha -p|$ works for most $\alpha$. If $f(q)=q$ does not work, I would think $f(q)=q^3$ would work, but then the best approximation might be the first convergent, which is useless. It has to be something like $f(q)=q^\nu$ with $\nu>1$ as small as possible. $\endgroup$ Feb 4, 2021 at 23:52
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    $\begingroup$ Almost all numbers have irrationality measure $2$, and so as long as $f(q)$ grows at least as fast as $q^\nu$ for any $\nu>1$ (indeed, weaker conditions suffice), then $R(\alpha)$ will be defined for almost all $\alpha$. I believe that for any such $f$ and any $n$, the set of $\alpha$ with $R(\alpha)$ equal to the $n$-th convergent will have positive measure. $\endgroup$
    – Wojowu
    Feb 5, 2021 at 10:32

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