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](https://en.wikipedia.org/wiki/Roth%27s_theorem#Generalizations).


**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](https://mathoverflow.net/questions/383353/distribution-of-the-first-occurrence-of-a-maximum-record-run-of-zeros-in-the-d), 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.



Algebraic and rational parts of a real number
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](https://en.wikipedia.org/wiki/Continued_fraction#Some_useful_theorems)): 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](https://en.wikipedia.org/wiki/Diophantine_approximation#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 $N-2$.


