We have $2n+1$ irrational numbers, then exists $n+1$ of them such that every subset of this set with $n+1$ elements has the sum an irrational number.

Question

Show that if we are given a set $S$ containing $2n+1$ irrational numbers, there exists a subset $T\subset S$ containing $n+1$ elements, such that every non-empty subset of $T$ sums to an irrational number.

I tried to consider an equivalence relation on real numbers: two elements are equivalent iff the difference of them is irrational. Then $n+1$ equivalence classes each have a representative that is "positive" or "negative" ... but I don't know hot to continue.

Any idea? Thank you!

Answer 1

I assume that the statement is as follows. My version of the problem is stated a little bit awkwardly because I want to include the cases where the given irrationals are not necessarily distinct. My proof is based on the Axiom of Choice.

Problem. Given are $2n+1$ irrational numbers $t_0,t_1,t_2,\ldots,t_{2n}$. Prove that there exist $n+1$ of them, say, $$t_{i_0},t_{i_1},t_{i_2},\ldots,t_{i_n}\,,$$ with indices $i_0,i_1,i_2,\ldots,i_n$ satisfying $0\leq i_0<i_1<i_2<\ldots<i_n\leq 2n$, such that any sum of the form $$t_{i_{j_0}}+t_{i_{j_1}}+t_{i_{j_2}}+\ldots+t_{i_{j_k}}\,,$$for some $k\in\{0,1,2,\ldots,n\}$ and indices $j_0,j_1,j_2,\ldots,j_k$ with $0\leq j_1<j_2<\ldots<j_k\leq n$, is irrational.

Pick a basis $\{1\}\cup\mathcal{B}$ of $\mathbb{R}$ over $\mathbb{Q}$. Well order $\mathcal{B}$ with an order $\triangleleft$. For each $b\in\{1\}\cup\mathcal{B}$, let $\pi_b:\mathbb{R}\to\mathbb{Q}$ be the projection sending $x=\sum\limits_{a\in\{1\}\cup\mathcal{B}}r_aa$ to $r_b$, where $r_a\in\mathbb{Q}$ for $a\in\{1\}\cup\mathcal{B}$ with finitely many nonzero terms. Then, there exists a lexicographic ordering $\prec$ on the quotient space $\mathbb{R}/\mathbb{Q}$ as follows. For $x,y\in\mathbb{R}/\mathbb{Q}$, we say $x\prec y$ if there exists $b\in \mathcal{B}$ such that $\pi_b(y-x)>0$ and for every $a\in\mathcal{B}$ such that $a\triangleleft b$, $\pi_a(y-x)= 0$. Show that $\prec$ is a total order on $\mathbb{R}/\mathbb{Q}$, and it is compatible with addition, that is, if $x,y,z,w\in\mathbb{R}/\mathbb{Q}$ satisfy $x\preceq y$ and $z\preceq w$, then $$x+z\preceq y+w\,.$$

We say that $x\in\mathbb{R}/\mathbb{Q}$ is pozitiv if $0\prec x$, and $x$ is negaziv if $x\prec 0$. Since the $2n+1$ numbers are irrational, there images under the quotient map $\mathbb{R}\to\mathbb{R}/\mathbb{Q}$ are nonvanishing. Therefore, at least $n+1$ are pozitiv, or at least $n+1$ of them are negaziv. Then, take $n+1$ pozitiv elements, or $n+1$ negaziv elements to complete the proof.

Answer 2

We may solve this through a suitable labelling technique: we associate to each element of $\{a_1,\ldots,a_{2n+1}\}$ a vector in $\mathbb{R}^{2n+1}$. We start by associating to $a_1$ the vector $(1,0,0,0,\ldots)$, then we process $a_2,a_3,\ldots$ according to the following algorithm:

• If all the subsets of $\{a_1,\ldots,a_n\}$ which include $a_n$ have irrational sums, we associate to $a_n$ the vector $v_n$ with a $1$ in the $n$-th position, zeroes anywhere else (encoding the idea that $a_n$ gives something new with respect to the previous additive structure);
• Otherwise we consider the smallest (according to the lexicographic order) subset of $\{a_1,\ldots,a_n\}$, including $a_n$, with a rational sum: assuming it is $\{a_{k_1},a_{k_2},\ldots,a_{k_h},a_n\}$, we associate to $a_n$ the vector $-v_{k_1}-v_{k_2}-\ldots-v_{k_h}$ (encoding the idea that "$a_n$ does not bring something new with respect to the previous additive structure")

If you manage to prove that the sum of the coordinates of $v_n$ is never zero, you have that at least $n+1$ vectors among $v_1,\ldots,v_{2n+1}$ belong to the same half-space $\sum x_k>0$ or $\sum x_k<0$ and they are associated to a $(n+1)$-subset of $\{a_1,\ldots,a_{2n+1}\}$ with the wanted property.