S is a set of real numbers which is closed under multiplication. S = A ∪ B, and A ∩ B = ∅. If a, b, c ∈ A, then abc ∈ A. Similarly, if a, b, c ∈ B, then abc ∈ B. Show that at least one of the A, B is closed under multiplication.
Solution
Trivial.
Suppose the result is false, then we can find a, b, c, d such that a, b ∈ A, but ab ∈ B, and c, d ∈ B, but cd ∈ A. But now a.b.cd is a product of three elements of A and hence in A, whereas ab.c.d is a product of three elements of B and hence in B. Contradiction.
© John Scholes
jscholes@kalva.demon.co.uk
12 Dec 1998