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.x1x2x3 ... , y = 0.y1y2y3 ... , z = 0.z1z2z3 ... , where every xi, yi, zi is 0 or 1. Then after n operations (assuming a number > 1/2 at all stages) the three numbers are simply x = 0.xn+1xn+2xn+3 ... , y = 0.yn+1yn+2yn+3 ... , z = 0.zn+1zn+2zn+3 ... . So we have to choose the xi, yi, zi so that (1) for each i, exactly one of xi, yi, zi 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: xi = 1 for i = 1, 4, 9, 16, ... ; yi = 1 for i = 2, 5, 10, 17, ...; zi = 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