Beatty's Theorem/Proof 1

Proof
We have been given that $r > 1$.

Let $\dfrac 1 r + \dfrac 1 s = 1$.

Then:
 * $s = \dfrac r {r - 1}$

It is to be shown that every positive integer lies in exactly one of the two Beatty sequences $\mathcal B_r$ and $\mathcal B_s$.

Consider the ordinal positions occupied by all the fractions $\dfrac j r$ and $\dfrac k s$ when they are jointly listed in nondecreasing order for positive integers $j$ and $k$.

that $\dfrac j r = \dfrac k s$ for some $j, k \in \Z_{>0}$.

Then:
 * $\dfrac r s = \dfrac j k$

which is rational.

But also:
 * $\dfrac r s = r \left({1 - \dfrac 1 r}\right) = r - 1$

which is not rational.

Therefore, no two of the numbers occupy the same position.

Consider some $\dfrac j r$.

There are $j$ numbers $\dfrac i r \le \dfrac j r$.

There are also $\left\lfloor{\dfrac {j s} r}\right\rfloor$ numbers $\dfrac k s \le \dfrac j r$.

So the position of $\dfrac j r$ in the list is $j + \left\lfloor{\dfrac {j s} r}\right\rfloor$.

The equation $\dfrac 1 r + \dfrac 1 s = 1$ implies:


 * $j + \left\lfloor{\dfrac {j s} r}\right\rfloor = j + \left\lfloor{j \left({s - 1}\right)}\right\rfloor = \left\lfloor{j s}\right\rfloor$

Likewise, the position of $\dfrac k s$ in the list is $\left\lfloor{k r}\right\rfloor$.

It is concluded that every positive integer corresponding to every position in the list is of the form $\left\lfloor{n r}\right\rfloor$ or of the form $\left\lfloor{nr}\right\rfloor$, but not both.

The converse statement is also true: if $p$ and $q$ are two real numbers such that every positive integer occurs precisely once in the above list, then $p$ and $q$ are irrational and the sum of their reciprocals is $1$.