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. |

I’m not sure if the Putnam was easier back in the day like the AMC/AIME, but this A6 was ridiculously simple. The problem popped up in Larson’s book 1.9.5 with some hints, which I’ll outline the solution with.

First, note that 1 is in our set. This can be seen several ways, with the official solution being much more elegant (only one of -1 or 1 can be in the set, and we also need to maintain multiplicative closedness). With this, we can prove that all positive integers are in the set.

Finally, assume that there exists a negative rational number in the set. Then simply add it to itself however many times is in the denominator, and it’ll be a negative integer; contradiction.