proof of Beatty’s theorem

We define an:=np and bn:=nq. Since p and q are irrational, so are an and bn.

It is also the case that anbm for all m and n, for if np=mq then q=1+nm would be rational.

The theorem is equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath with the statement that for each integer N1 exactly 1 element of {an}{bn} lies in (N,N+1).

Choose N integer. Let s(N) be the number of elements of {an}{bn} less than N.


So there are Np elements of {an} less than N and likewise Nq elements of {bn}.

By definition,


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

The number of elements of {an}{bn} lying in (N,N+1) is then s(N+1)-s(N)=1.

Title proof of Beatty’s theorem
Canonical name ProofOfBeattysTheorem
Date of creation 2013-03-22 13:18:58
Last modified on 2013-03-22 13:18:58
Owner lieven (1075)
Last modified by lieven (1075)
Numerical id 8
Author lieven (1075)
Entry type Proof
Classification msc 11B83