Let $\alpha$ be an irrational number, $n\geq 1$ and


$ X\_n=\lbrace (x,y) \in {\mathbb Z}^2 | |y| \leq n, \ x+y\alpha >0 \rbrace$


Now let $(x\_n,y\_n)$ minimize the quantity $x+y\alpha$ on $X\_n$. This
pair is unique because $\alpha$ is irrational.
 Now the problem is to deduce all the "location" of $\alpha$ from
the three integers $n,x\_n$ and $y\_n$ only.


The two simplest cases are :


$x\_n=0,y\_n=1$. This is equivalent to $0 < \alpha < \frac{1}{n+1}$.


$x\_n=1,y\_n=-2$. This is equivalent to $\frac{1}{2}-\frac{1}{2m} < \alpha
< \frac{1}{2}$ (where $m$ is the smallest odd integer $>n$).


More generally, it seems that for any triple $(n,x,y)$, the set $Y(n,x,y)$
of all irrationals $\alpha$ yielding $x\_n=x,y\_n=y$, is either empty
or an interval $[A(n,x,y),B(n,x,y)] \setminus {\mathbb Q}$. Is that true,
and are there recursive formulas to compute $A(n,x,y)$ and $B(n,x,y)$ ?


This problem is certainly related to continued fractions and best approximations
in the usual sense, but I don't see how to make the connection effective.



Best rational approximation in a special sense
The dots $\{ x+ y \alpha \colon |y|\leq n \}$ move continuously with $\alpha$, so we should consider when $x\_n+y\_n \alpha$ can *stop* being the smallest positive dot. First, note that $y\_n = 0$ is impossible. We need $x\_n+y\_n \alpha > 0$, so that $\alpha > - x\_n/y\_n$ if $y > 0$ and $\alpha < -x\_n /y\_n$ if $y\_n < 0$. At the other extreme, $x\_n+y\_n\alpha$ can lose its crown if two of the dots are equal $x\_n+y\_n \alpha = x' + y' \alpha$, i.e., if $\alpha$ is fraction with denominator at most $n-1$. Note that the minimizer really does change when we get equality since the slopes are different.


So, for $y\_n > 0$, you get $A(n,x\_n,y\_n) = -x\_n / y\_n$ and $B(n,x\_n,y\_n)$ is the smallest fraction that is both larger than $-x\_n/y\_n$ and has a denominator less than $n$. If $y\_n < 0$, then $B(n,x\_n,y\_n)= -x\_n/y\_n$ and $A(n,x\_n,y\_n)$ is the largest fraction that is both smaller than $-x\_n/y\_n$ and has denominator less than $n$.


Example: say $n=102, x\_n=52, y\_n = 37$. This happens if and only if $\alpha$ is in the interval $(-52/37, a/b)$, where $a/b$ is only slightly larger than -52/37 and $0 < b < 102$. To find $a,b$, we first translate -52/37 so that it lies in (0,1): it translates to 22/37. Now from the theory of Farey fractions, we know that the fraction after 22/37 (call it $c/d$) in $F\_{101}$ satisfies $22d-37c=\pm 1$, and also $37+d > 101$. Solving this modulo 101 (cleverly chosen to be prime, so that the solutions are unique) gives $c/d=41/69$ or $c/d=82/69$ (which is clearly wrong, since $c/d<1$). Translating back gives $a/b=41/69-2=-97/69$. So the final conclusion is that $\alpha$ is in $(-52/37,-97/69)$.


