## Wednesday, January 18, 2006

### 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.

#### 1 comment:

1. That's an amazing solution!