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.