proof of Beatty's theorem

We define $a_n:=np$ and $b_n:=nq$ . Since $p$ and $q$ are irrational, so are $a_n$ and $b_n$ .

It is also the case that $a_n\neq b_m$ for all $m$ and $n$ , for if $np=mq$ then $q=1+\frac{n}{m}$ would be rational.

The theorem is equivalent with the statement that for each integer $N\geq 1$ exactly $1$ element of $\{a_n\}\cup\{b_n\}$ lies in $(N,N+1)$ .

Choose $N$ integer. Let $s(N)$ be the number of elements of $\{a_n\}\cup\{b_n\}$ less than $N$ .

$$a_n<N\Leftrightarrow np<N\Leftrightarrow n<\frac{N}{p}$$

So there are $\lfloor{\frac{N}{p}}\rfloor$ elements of $\{a_n\}$ less than $N$ and likewise $\lfloor{\frac{N}{q}}\rfloor$ elements of $\{b_n\}$ .

By definition,

and summing these inequalities gives $N-2<s(N)<N$ which gives that $s(N)=N-1$ since $s(N)$ is integer.

The number of elements of $\{a_n\}\cup\{b_n\}$ lying in $(N,N+1)$ is then $s(N+1)-s(N)=1$ .