Problem 1517

1. Define the Vector Space:
Let $L$ be the vector space over the scalar field $\mathbb{Q}$ generated by $S \cup \{1\}$. Since $S$ consists of $2n-1$ different irrational numbers, and adding $1$ (a rational number) to the set, the dimension of $L$ is at least $2$.

2. Dual Space and Hyperplane:
The dual space of $L$, which consists of all linear functionals on $L$, also has a dimension at least $2$. Therefore, we can find a vector $v \in L$ that is not collinear with $1$. This means $v$ is not a scalar multiple of $1$.

3. **Hyperplane $H$:**
Consider the hyperplane $H$ in $L$ that is orthogonal to $v$. By definition, $H$ is the set of all vectors in $L$ that are orthogonal to $v$.

4. Half-Spaces:
The hyperplane $H$ divides $L$ into two half-spaces. Since $S$ consists of $2n-1$ points, at least one of these half-spaces must contain at least $n$ points from $S$. Let these points be $x_1, x_2, \ldots, x_n$.

5. Projections and Irrationality:
The projections of $x_i$ (for $i = 1, 2, \ldots, n$) on $v$ have the same sign and none of them equals zero. This implies that for any non-negative rational numbers $a_1, a_2, \ldots, a_n$ with $a_1 + a_2 + \cdots + a_n > 0$, the linear combination $a_1 x_1 + a_2 x_2 + \cdots + a_n x_n$ cannot be rational.

To see why, assume for contradiction that $a_1 x_1 + a_2 x_2 + \cdots + a_n x_n$ is rational. Since $x_i$ are irrational and their projections on $v$ have the same sign, the linear combination would also be irrational unless all $a_i$ are zero, which contradicts the condition $a_1 + a_2 + \cdots + a_n > 0$.

Therefore, we have shown that there exist $n$ different elements $x_1, x_2, \ldots, x_n \in S$ such that for all non-negative rational numbers $a_1, a_2, \ldots, a_n$ with $a_1 + a_2 + \cdots + a_n > 0$, the sum $a_1 x_1 + a_2 x_2 + \cdots + a_n x_n$ is irrational.

$\blacksquare$