Discrete rational rotations on the two dimensional torus

Question

It is well known (Kronecker's Theorem) that "irrational rotations" are dense on $[0,1)$. That is, the set $$ \{ x+nr\mod 1 : n \in \mathbb{N} \} $$ is dense on $[0,1)$, provided that $r$ is irrational. This theorem is relatively easy to prove.

On the two dimensional torus $\mathbb{T}=[0,1)\times[0,1)$ (with opposite edges identified), the following result is true. The set $$ \{ (x+nr \mod 1,x+nr' \mod 1) \in \mathbb{T} : n \in \mathbb{N} \} $$ is dense in $\mathbb{T}$ if and only if $\{r, r', 1\}$ are rationally independent (i.e., if there exist integers $a$ and $b$ such that $ar+br'$ is an integer, then $a=b=0$). I have seen a very complicated proof of this. Is there an "easy" proof? That is, something that one could assign for reading to an undergraduate (say a senior)?

Answer 1

I have seen IMHO quite accessible proofs of this fact in number theory books. When I was an advanced high school kid (I had found my calling), I saw a proof of this with hints in Joe Roberts' lovely book, typeset in calligraphic font, Elementary Number Theory - A Problem Oriented Approach. IIRC I managed to follow the proof given there, but this was among the more taxing problems. As an undergraduate I had the pleasure of giving a talk about this at a seminar going through Apostol's Modular Functions and Automorphic Forms in Number Theory. It is in one of the late chapters, and I recall enjoying that chapter and the exercises therein immensely.

I don't know if this is helpful to you. This stuff is certainly not too demanding for an undergraduate in that it doesn't rely on any deep theory. But I wouldn't assign this to someone who hasn't shown a real interest in thinking things through for him/herself. You know your clients better.

Answer 2

I proved it by first proving $1, r, r'$ must be rationally independent by pairs (consider the circles $x=0$ and $y=0$, and you first have to prove the continuous version of the theorem, which only requires $r$ and $r'$ to be rationally independent (without the $1$)). Then you only need to consider the case $r' = ar + b$ for rational $a, b$. In this case you can show there is a finite number of lines of the type $y = qx + c$, with rational $q$ and $c$, which include all points in the orbit.

If you do it for specific values, for example: $r = \pi, r' = \frac{\pi}4 + \frac35$, you'll get the idea.