23rd Putnam 1962

Problem A6

X is a subset of the rationals which is closed under addition and multiplication. 0 ∉ X. For any rational x ≠ 0, just one of x, -x ∈ X. Show that X is the set of all positive rationals.



Either x or -x belongs to X. X is closed under multiplication, so the square x2 = (-x)2 belongs to X. In particular, 1 belongs to X. Hence by repeated addition all positive integers must belong to X. Suppose positive rational m/n does not belong to X. Then -m/n does, and hence by repeated addition -m. So m does not belong to X. Contradiction. So X contains all positive rationals. But if x is in X, -x is not, so X does not contain any negative rationals and hence X is just the set of all positive rationals.



23rd Putnam 1962

© John Scholes
5 Feb 2002