α and β are positive irrational numbers satisfying 1/α + 1/β = 1. Let an = [n α] and bn = [n β], for n = 1, 2, 3, ... . Show that the sequences an and bn are disjoint and that every positive integer belongs to one or the other.
β positive implies α > 1, so [α (n+1)] > [α n] and the sequence does not contain any integers twice. Similarly for [βn].
If k appears in both sequences, then for some m, n we have: k < n α < k + 1, k < m β < k + 1. Hence k/α < n < (k+1)/α, k/β < m < (k+1)/β. Adding gives k < m+n < k + 1. Contradiction. So an integer appears in at most one sequence.
Suppose k does not appear in the sequence [n α]. Then for some n we have n α < k, and k + 1 < (n + 1) α. The first inequality implies that n < k/α = k - k/β. Hence (k - n) β > k. The second inequality implies that n + 1 > (k + 1)/α = k + 1 - (k + 1)/β. Hence (k - n) β < k + 1. So [(k - n) β ] = k.
20th Putnam 1959
© John Scholes
15 Feb 2002