This is a challenge problem. It has been approved by Mr Fantastic.
This problem is from some putnam comptition. I got the problem out of a book that I have. It's fun.
let be a set that is closed under multiplication; that is, for any 2 elements of A, their product is also in A.
now, let , be disjoint. Also, suppose that the union of these 2 sets is A its self. Finaly, suppose that both sets T and Q have the following property: for any 3 elements that comes from one of those sets, the product is in the set that the 3 came from. The 'three elements that come from one of those sets need not be distinct.
Prove that at least one of the subsets is closed under multiplication.
Moderator approved challenge question.
I like this one :D
Suppose the contrary: that there exist such that and such that .
Then, since and , plus the fact that is closed under multiplication,
it implies that and .
Therefore, from the conditions on and regarding multiplication,
it follows that and , a contradiction to .