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Saudi Arabian Mathematical Competitions

Algebra Difficulty 5.4 AIME, harder Prove it Saudi Arabia

Let SS be a given set of real numbers such that:
i) 1∈S1 \in S,
ii) for any a,b∈Sa, b \in S (not necessarily different), then a−b∈Sa-b \in S,
iii) for a∈Sa \in S, a≠0a \neq 0 then 1a∈S\frac{1}{a} \in S.
Prove that for any a,b∈Sa, b \in S then ab∈Sa b \in S.

Solution

If a=0∈Sa=0 \in S or a=1∈Sa=1 \in S then for any b∈Sb \in S we have ab∈Sa b \in S which is obvious. So we can suppose that a,b∉{0;1}a, b \notin \{0 ; 1\}.
From 1∈S1 \in S, a∈Sa \in S we have 1−a∈S→(1−a)−1=−a∈S1-a \in S \rightarrow (1-a)-1 = -a \in S, then
b−(−a)=a+b∈S. b-(-a) = a+b \in S.
We also have a−1∈Sa-1 \in S so 1a,1a−1∈S\frac{1}{a}, \frac{1}{a-1} \in S then
1a−1−1a=1a(a−1)∈S→a(a−1)∈S. \frac{1}{a-1} - \frac{1}{a} = \frac{1}{a(a-1)} \in S \rightarrow a(a-1) \in S.
Since a∈Sa \in S, a(a−1)∈Sa(a-1) \in S then a2=a(a−1)+a∈Sa^{2} = a(a-1) + a \in S. Similarly, we also have b2∈Sb^{2} \in S. Thus a,b∈S→a+b∈S→(a+b)2∈Sa, b \in S \rightarrow a+b \in S \rightarrow (a+b)^{2} \in S.
Finally, (a+b)2−a2=2ab+b2∈S(a+b)^{2} - a^{2} = 2 a b + b^{2} \in S and (2ab+b2)−b2=2ab∈S\left(2 a b + b^{2}\right) - b^{2} = 2 a b \in S which implies that
12ab+12ab=1ab∈S→ab∈S. \frac{1}{2 a b} + \frac{1}{2 a b} = \frac{1}{a b} \in S \rightarrow a b \in S.

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