IMC 1994 · Day 1 · P3

Let $I$ be the set of irrational numbers, $\mathbb{Q}$ – the set of rational numbers, $\mathbb{Q}^+ = \mathbb{Q} \cap [0,\infty)$. We work by induction. For $n = 1$ the statement is trivial. Let it be true for $n - 1$. We start to prove it for $n$. From the induction argument there are $n-1$ different elements $x_1, x_2, \dots, x_{n-1} \in S$ such that $$\tag{1} a_1 x_1 + a_2 x_2 + \cdots + a_{n-1} x_{n-1} \in I$$ for all $a_1, a_2, \dots, a_n \in \mathbb{Q}^+$ with $a_1 + a_2 + \cdots + a_{n-1} > 0$.

Denote the other elements of $S$ by $x_n, x_{n+1}, \dots, x_{2n-1}$. Assume the statement is not true for $n$. Then for $k = 0,1,\dots,n-1$ there are $r_k \in \mathbb{Q}$ such that $$\tag{2} \sum_{i=1}^{n-1} b_{ik} x_i + c_k x_{n+k} = r_k \quad \text{for some } b_{ik}, c_k \in \mathbb{Q}^+,\ \sum_{i=1}^{n-1} b_{ik} + c_k > 0.$$ Also $$\tag{3} \sum_{k=0}^{n-1} d_k x_{n+k} = R \quad \text{for some } d_k \in \mathbb{Q}^+,\ \sum_{k=0}^{n-1} d_k > 0,\ R \in \mathbb{Q}.$$ If in (2) $c_k = 0$ then (2) contradicts (1). Thus $c_k \ne 0$ and without loss of generality one may take $c_k = 1$. In (2) also $\sum\limits_{i=1}^{n-1} b_{ik} > 0$ in view of $x_{n+k} \in I$. Replacing (2) in (3) we get $$\sum_{k=0}^{n-1} d_k \left( -\sum_{i=1}^{n-1} b_{ik} x_i + r_k \right) = R \quad \text{or} \quad \sum_{i=1}^{n-1} \left( \sum_{k=0}^{n-1} d_k b_{ik} \right) x_i \in \mathbb{Q},$$ which contradicts (1) because of the conditions on $b$'s and $d$'s.