A Disjointness Criterion for Beatty’s Sequences

A DISJOINTNESS CRITERION FOR BEATTY’S SEQUENCES 1. Introduction Let α be an irrational number with α > 1. We denote S(α) by S(α) = {bnαc|n ∈ N}. In 1926, Sam Beatty [1] proved that if α, β are positive irrational numbers, then the disjoint union of S(α) and S(β) is N if and only if α1 + β1 = 1. It is easy to see that if α, β are positive irrational numbers and l k + =1 α β for some positive integers k, l, then S(α) ∩ S(β) = φ. In this note, we shall show that the converse of this statement is also valid. In addition, we deduce that S(α) ∩ S(β) is either empty or infinite. Theorem. For irrational numbers α, β > 1, S(α) ∩ S(β) = φ if and only if k l + =1 α β for some positive integers k, l. 2. Lemmas We denote B(x, y, r) by an open ball in R2 centered at (x, y) with radius r. For a real number x, (x) denotes the fractional part of x. For real numbers α and β, we define S(α, β) by
S(α, β) = { (nα), (nβ) |n ∈ N}. Lemma 1.
Let m, n,
r be positive integers. Then, bnαc = bmβc = r − 1 if and only if αr ≤ α1 and βr ≤ β1 . Proof. We observe that bnαc = bmβc = r − 1 is equivalent to any of the following statements: nα < r ≤ nα + 1, mβ < r ≤ mβ + 1; r 1 r 1 n< ≤n+ , m< ≤m+ ; α α β β r r 1 1 0< ≤ , 0< ≤ . α α β β Thus, Lemma 1 follows. 1 2 Lemma 2 (Kronecker’s Theorem). If 1, α, β are linearly independent over Q, then the set S(α, β) is dense in [0, 1]2 . Proof. See [2], p382.  Lemma 3. Let α > 1, β > 1 be irrational numbers satisfying αk + βl = m with
k, l, m relatively prime integers and l > 0. Then S α1 , β1 is dense in [0, 1]2 ∩ {(x, y)|kx + ly ∈ Z}. Proof. Let (x, y) ∈ [0, 1]2 be such that kx + ly = z for some z ∈ Z and 0 < ε < d/2, where d is the distance between two lines, kx + ly = 0, kx + ly = 1. Using the pigeon hole principle, we get          n1 n2 n1 n2 − , − v = (v1 , v2 ) = α α β β with positive integers n1 , n2 (n1 < n2 ), and |v| < ε < d2 . Since (k, l, m) = 1, for any z ∈ Z, there is a triple (n, z1 , z2 ) with n ∈ N, z1 ∈ Z and z2 ∈ Z such that mn + kz1 + lz2 = z From |v| < d, we have kv1 + lv2 = 0. Then, we get  n  n + z1 + N v 1 + l + z2 + N v2 = mn + kz1 + lz2 = z k α β for any N ∈ N. Hence, we can find a positive integer N and integers u1 , u2 such that n  n + z1 + N v1 , + z2 + N v2 ∈ B(u1 + x, u2 + y, ε) α β Thus, Lemma 3 follows.  3. Proof of the Theorem Let α > 1, β > 1 be irrational numbers satisfying S(α) ∩ S(β) = φ. 1, α1 , β1 are either linearly independent over Q or linearly dependent over Q. The latter case we multiply a nonzero integer to get αk + βl = m, with k, l, m relatively prime integers, and l > 0. Since k 6= 0, we divide the latter into two cases k < 0, and k > 0. Case 1. 1, 1 1 α, β are linearly independent over Q.
By Lemma 2, S α1 , β1 is dense in [0, 1]2 . Then we have 1 1  1i  1i S , ∩ 0, × 0, α β α β is an infinite set. This implies r α ≤ 1 r 1 , ≤ α β β for infinitely many positive integers r. Using this and Lemma 1, we have S(α)∩S(β) is an infinite set. Case 2. k α + l β = m, with k, l, m relatively prime integers, l > 0 and k < 0. A DISJOINTNESS CRITERION FOR BEATTY’S SEQUENCES 3 The set  1i  1i {(x, y)|kx + ly ∈ Z} ∩ 0, × 0, α β contains a line segment of kx + ly = 0. By Lemma 3, the set 1 1  1i  1i S ∩ 0, × 0, , α β α β is an infinite set. This implies S(α) ∩ S(β) is an infinite set as in Case 1. Case 3. k α + l β = m with k, l, m relatively prime integers, l > 0 and k > 0. Since S(α) ∩ S(β) = φ, we have 1 1  1i  1i (1) S ∩ 0, × 0, =φ , α β α β by Lemma 1. It follows that  1i  1i {(x, y)|kx + ly ∈ Z} ∩ 0, × 0, α β does not contain any line segment, otherwise it contradicts (1) by Lemma 3. This implies 1 k1 1 ≤− + β lα l which is equivalent to m ≤ 1. Thus, we obtain m = 1. By Cases 1–3, we complete the proof of the Theorem. References 1. Sam, Beatty, Problem 3173, Amer. Math. Monthly, 1926, p159. 2. G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, 4th edition, Oxford At The Clarendon Press, 1960.