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 $a_{n}\eq b_{m}$ 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,

\\begin{array}[]{ccccc}\\frac{N}{p}-1&<&\\lfloor{\\frac{N}{p}}\\rfloor&<&\\frac{N}{p%}\\\\\\frac{N}{q}-1&<&\\lfloor{\\frac{N}{q}}\\rfloor&<&\\frac{N}{q}\\\\\\end{array}

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 $s(N+1)-s(N)=1$ .