# 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 ${a}_{n}\ne {b}_{m}$ 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 $N\ge 1$ 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 $N$.

$$ |

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 $$ 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 |
---|---|

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 |