Prove $a_n = \lfloor ni \rfloor$ for some irrational $i$ has no pattern

Question

I have a sequence $a_n = \lfloor ni \rfloor$ for an irrational $i$ and I want to prove that there is no 'pattern' to these terms.

In particular, I have $1 < i <2$ and I want to prove that $\forall a,b \in \mathbb{N}, \exists k \in \mathbb{N}$ such that $a +bk \neq a_n$ for any $n$.
This makes sense to me intuitively because $a_n - a_{n-1}$ will be 1 or 2 'randomly', so the sequence must eventually 'skip over' one of the terms of $a+bk$.

My approach has been to assume that it holds, then show that $i$ must be rational for a contradiction. However, the floor is messing up the simple divisibility techniques I would use.

Answer 1

This is a case of the equidistribution theorem. If the sequence repeats, subdivide $[0,1]$ finely enough and some pieces will get nothing.