Let S be the set of all collections of 3 (not necessarily distinct) positive irrational numbers with sum 1. If A = {x, y, z} ∈ S and x > 1/2, define A' = {2x - 1, 2y, 2z}. Does repeated application of this operation necessarily give a collection with all elements < 1/2?

**Solution**

Answer: no.

Write the three numbers in binary: x = 0.x_{1}x_{2}x_{3} ... , y = 0.y_{1}y_{2}y_{3} ... , z = 0.z_{1}z_{2}z_{3} ... , where every x_{i}, y_{i}, z_{i} is 0 or 1. Then after n operations (assuming a number > 1/2 at all stages) the three numbers are simply x = 0.x_{n+1}x_{n+2}x_{n+3} ... , y = 0.y_{n+1}y_{n+2}y_{n+3} ... , z = 0.z_{n+1}z_{n+2}z_{n+3} ... . So we have to choose the x_{i}, y_{i}, z_{i} so that (1) for each i, exactly one of x_{i}, y_{i}, z_{i} is 1, (2) x, y, z are irrational. To achieve (2) we just have to ensure that there is no periodicity. So, for example, we could take: x_{i} = 1 for i = 1, 4, 9, 16, ... ; y_{i} = 1 for i = 2, 5, 10, 17, ...; z_{i} = 1 if i is not a square or a square plus 1.

[If the triples are not required to be irrational, we have the even simpler solution: 1/7, 2/7, 4/7.]

© John Scholes

jscholes@kalva.demon.co.uk

30 Nov 1999