TITLE: Premise of Minkowski's theorem: "Irrational" Line
QUESTION [5 upvotes]: If we take a straight line $L$ in $\Bbb{R}^2$ of the form $y=ax$ where $a\in\Bbb{R}\setminus{\Bbb{Q}}.$
Then we see that (except at the origin $(0,0)$) $$L\cap \Bbb{Z}^2=\emptyset.$$
But now if the line becomes a strip of width $r>0$ denoted $S$ how can we be sure that $$S\cap \Bbb{Z}^2\ne\emptyset?$$
I know about Minkowski's theorem but in this particular case is there a high school argument? Perhaps by writing the continued fraction of $a?$

REPLY [2 votes]: Let $m\in\mathbb{N}$ be such that $1/m<r$.  Given $m+1$ real numbers, there must be two of them whose fractional parts are within an interval of length $1/m$, by pigeonhole.  In particular, there must be two different integers $n$ and $n'$ such that the fractional parts of $an$ and $an'$ differ by less than $1/m$.  Then $a(n-n')$ will be within $1/m$ of an integer $k$.  The point $(n-n',k)\in\mathbb{Z}^2$ is then in your strip $S$, since it is within $r$ of $(n-n',a(n-n'))$.