45th Putnam 1984

Let S_n be a set with n elements. Can we find a binary operation * on S which satisfies (1) right cancellation: a*c = b*c implies a = b (for all a, b, c), and (2) non-associativity: a * (b * c) ≠ (a * b) * c for all a, b, c? Note that we are not just requiring that * is not associative, but that it is never associative.

Solution

Answer: yes.

Let the elements be a₁, ... , a_n. Define a_i*b = a_i+1 for all b (taking a_n+1 to be a₁). Then we certainly have right cancellation, because if a_i*c = a_j*c, then a_i+1 = a_j+1, so i = j. We also have non-associativity, because a_i*(b*c) = a_i+1, but (a_i*b)*c = a_i+1*c = a_i+2.

45th Putnam 1984

© John Scholes
jscholes@kalva.demon.co.uk
7 Jan 2001