Beatty's Theorem/Proof 2

Collisions
there exist integers $j > 0, k, m$ such that:
 * $j = \left\lfloor {k \cdot r} \right\rfloor = \left\lfloor {m \cdot s} \right\rfloor$

This is equivalent to the inequalities:
 * $j \le k \cdot r < j + 1$

and:
 * $j \le m \cdot s < j + 1$

As $r$ and $s$ are irrational, equality cannot happen.

So:


 * $j < k \cdot r < j + 1$

and:
 * $j < m \cdot s < j + 1$

which leads to:
 * $\dfrac j r < k < \dfrac {j + 1} r$

and:
 * $\dfrac j s < m < \dfrac {j + 1} s$

Adding these together and using the by hypothesis:


 * $j < k + m < j + 1$

Thus there is an integer strictly between two adjacent integers.

This is impossible.

Thus the supposition must be false.

Anti-collisions
that there exist integers $j > 0, k, m$ such that:


 * $k \cdot r < j$

and:
 * $j + 1 \le \left({k + 1}\right) \cdot r$

and:
 * $m \cdot s < j$

and:
 * $j + 1 \le \left({m + 1}\right) \cdot s$

Since $j + 1 \ne 0$, and $r$ and $s$ are irrational, equality cannot happen.

So:


 * $k \cdot r < j$

and:
 * $j + 1 < \left({k + 1}\right) \cdot r$

and:
 * $m \cdot s < j$

and:
 * $j + 1 < \left({m + 1}\right) \cdot s$

Then:
 * $k < \dfrac j r$:

and:
 * $\dfrac {j + 1} r < k + 1$

and:
 * $m < \dfrac j s$

and:
 * $\dfrac {j + 1} s < m + 1$

Adding corresponding inequalities:
 * $k + m < j$

and:
 * $j + 1 < k + m + 2$


 * $k + m < j < k + m + 1$

which is also impossible.

Thus the supposition is false.