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.

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

 $a_{n}

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

Title proof of Beatty’s theorem ProofOfBeattysTheorem 2013-03-22 13:18:58 2013-03-22 13:18:58 lieven (1075) lieven (1075) 8 lieven (1075) Proof msc 11B83