Let R be the reals. Let S ⊆ R have n ≥ 2 elements. Let A_{S} = { x ∈ R **:** x = (s + t)/2 for some s, t ∈ S with s ≠ t}. What is the smallest possible |A_{S}|?

**Solution**

Answer: 2n - 3.

*Easy.*

{1, 2, ... , n} gives 2n - 3 averages, namely 1 1/2, 2, 2 1/2, ... , n - 1/2, so it remains to show that we cannot do better.

Suppose a_{1} < a_{2} < ... < a_{n+1}. Let S be the set {a_{1}, a_{2}, ... , a_{n}} and S' the set S ∪ {a_{n+1}}. Then x = (a_{n} + a_{n+1})/2 > a_{n}, so x ∉ A_{S}. Similarly, y = (a_{n+1} + a_{n-1})/2 > (a_{n} + a_{n-1})/2, which is the largest element of A_{S}, so y ∉ A_{S}. Hence |A_{S'}| ≥ |A_{S}| + 2.

© John Scholes

jscholes@kalva.demon.co.uk

24 Dec 1998