unREAListic! Topic: Algebra. Level: AIME/Olympiad.

Problem: (WOOT Test 5 - #3) Find all non-empty sets $ A $ of real numbers satisfying the following property -

For all real numbers $ x,y $, if $ x+y \in A $ then $ xy \in A $.

Solution: We claim that $ A $ must be the set of all real numbers.

We will prove this by showing that
(1) $ 0 \in A $
(2) $ -n \in A $ for all positive reals $ n $
(3) $ m \in A $ for all positive reals $ m $

(1) Since $ A $ is non-empty, we know there exists at least one element $ a \in A $. Then take $ x = 0, y = a $, to get $ x+y = a \in A \Rightarrow xy = 0 \in A $.

(2) Take $ x = \sqrt{n}, y = -\sqrt{n} $, and we have, from (1), $ x+y = 0 \in A \Rightarrow xy = -n \in A $.

(3) Take $ x = -m, y = -1 $, and we have, from (2), $ x+y = -m-1 \in A \Rightarrow xy = m \in A $.

Hence all reals must be in $ A $. QED.

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Comment: During the actual test, I had a slightly more convoluted solution, but the thought process was very much the same. I found that that given any element $ a $, I could take $ x,y $ such that one of them was very negative and the other very positive to get all the negative reals and obviously zero. From there, it's not a big step to see that all positive reals are in $ A $ as well.