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#1 h Apr 29, 2025, 4:59 AM • 1 Y

Y by Euler_Gauss

Let $x_1,\ldots ,x_{2n-1}$ be distinct irrational numbers. Show that there exists $y_1,\ldots ,y_n\in\{x_1,\ldots ,x_{2n-1}\},$ such that $\forall S\subset [n]\neq\varnothing,$
$\sum_{s\in S}x_s\notin\mathbb Q.$

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yofro

#2 h Apr 29, 2025, 6:49 PM

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I think the problem statement should require $\sum_{s\in S}\mathbf{y_s} \notin \mathbb{Q}$ , rather than $x_s$ .

The idea is to characterize which $x_i$ possess a new additive structure. We build $x_1, x_2, \dots, x_{2n-1}$ in that order, and if $x_j$ does not belong in some new additive "area" of $\overline{\mathbb{Q}}$ , then we can simply write it as what it is: some $\mathbb{Q}$ -linear combination of $\{x_1, x_2, \dots, x_{j-1}\}$ . The most natural way to represent this is via a linear mapping $\sigma: \{x_1, x_2, \dots, x_{2n-1}\}\to \mathbb{Q}^{2n-1}$ . We have that $\sigma(x_1)=(1,0,0,\dots,0)^T$ , and that $\sigma(x_i)$ is the $i$ th basis element of $\mathbb{R}^{2n-1}$ in the standard basis, unless it is a linear combination of the previous $x_j$ , in which case we recover $\sigma(x_i)$ from the linearity of $\sigma$ . We artificially constructed $\sigma$ such that it possesses the following property: with $I\subset [2n-1]$ ,
$\sum_{i\in I}x_i\in \mathbb{Q}\iff \sum_{i\in I}\sigma(x_i)=(0,0,0,\dots,0)^T.$ Since all the $\sigma(x_i)$ are distinct the problem is as follows:

Quote:

We have $2n-1$ distinct points in $\mathbb{Q}^{2n-1}$ . Show that some $n$ of these are such that no subset of them has the origin as their centroid.

But this problem is easy: consider a plane $P$ through the origin which doesn't intersect any of the points. Then by pigeonhole, we may choose $n$ points on one side of $P$ , say "above" $P$ . Then the centroid of any subset of these points is also above $P$ , and this is satisfactory.

This post has been edited 1 time. Last edited by yofro, Apr 29, 2025, 7:05 PM
Reason: added clarity

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