a, b, c are positive integers such that b > 2a, c > 2b. Show that there is a real k such that the fractional parts of ka, kb, kc all exceed 1/3 and do not exceed 2/3.
Solution
Let In denote the interval ( (n + 1/3)/a, (n + 2/3)/a ]. Similarly, let Jn denote the interval ( (n + 1/3)/b, (n + 2/3)/b ], and Kn denote the interval ( (n + 1/3)/c, (n + 2/3)/c ]. We wish to show that there is some real value k which belongs to intervals of all three types, for k ∈ In is equivalent to (n + 1/3) < ak ≤ (n + 2/3) and hence to the fractional part of ak being > 1/3 and ≤ 2/3. Similarly for the other intervals.
The intervals Kn are spaced a distance 2/(3c) < 1/(3b) apart. The intervals Jn all have width 1/(3b), so every Jn must intersect some Km. We claim that we can find a Jn which is contained in some Is. It follows that a common point of Jn and Km also belongs to Is and we are done. It remains to prove the claim.
Jn ⊆ Is is equivalent to (s + 1/3)/a ≤ (n + 1/3)/b and (n + 2/3)/b ≤ (s + 2/3)/a, or (3s+1)/(3n+1) ≤ a/b ≤ (3s+2)/(3n+2). Now a/b lies in the interval (0, 1/2). We show that for any real number h in (0, 1/2) we can find s, n such that (3s+1)/(3n+1) ≤ h ≤ (3s+2)/(3n+2).
For if h ≤ 1/4, then we can take the smallest positive integer N such that 1/(3.2N + 1) < h. Then 2/(3.2N + 2) = 1/(3.2N-1 + 1) ≥ h. [Note that if N = 1, this is still true since 1/3 > 1/4 ≥ h.]. Hence h belongs to the interval [1/(3.2N + 1), 2/(3.2N + 2) ) and we can put s = 0, n = 2N.
For N = 0, 1, 2, 3, ... we have (3.2N - 2)/(3.2N+1- 2) = 1/4, 2/5, 5/11, ... . The sequence obviously tends to 1/2. So for any h satsifying 1/4 < h < 1/2 take the smallest non-negative integer N such that (3.2N - 2)/(3.2N+1 - 2) ≤ h. So h < (3.2N+1 - 2)/(3.2N+2 - 2) = (3.2N - 1)/(3.2N+1 - 1) = (3s+2)/(3n+2), with s = (2N - 1), n = (2N+1 - 1). Also (3.2N - 2)/(3.2N+1 - 2) = (3s+1)/(3n+1), so (3s+1)/(3n+1) ≤ h < (3s+2)/(3n+2), as required.
Note that we do not require a, b, c to be integral.
© John Scholes
jscholes@kalva.demon.co.uk
10 Oct 2002