Let k be the positive root of the equation x2 = 1991 x + 1. For positive integers m, n define m # n = mn + [km] [kn]. Show that the operation # is associative.
Solution
Put K = 1991, so k(k - K) = 1. Evidently k = K/2 + √( (K/2)2 + 1), so 0 < k - K < 1. By definition, 0 ≤ ka - [ka] < 1, and 0 ≤ kb - [kb] < 1 for any integers a, b. So multiplying, 0 ≤ (k - K) (ka - [ka]) (kb - [kb]) < 1. Expanding, and using k(k - K) = 1, we get 0 ≤ k(a # b) - (a [kb] + b [ka] + K [ka] [kb] ) < 1, so [ k(a # b) ] = a [kb] + b [ka] + K [ka] [kb].
Hence (a # b) # c = (a # b)c + [ k(a # b) ] [kc] = abc + [ka] [kb] c + a [kb] [kc] + [ka] b [kc] + K [ka] [kb] [kc]. But this expression is symmetrical in a, b, c, so (a # b) # c = (b # c) # a. But (b # c) # a is obviously the same as a # (b # c), so # is associative.
© John Scholes
jscholes@kalva.demon.co.uk
2 Jan 2003
Last corrected/updated 2Jan 03