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Let $x_1,x_2,\ldots,x_{2n-1}\in\mathbb{R}\setminus\mathbb{Q}$. Claim: there exists a nonempty subset $S\subset\{1,2,\ldots,2n-1\}$ such that $\lvert S \rvert \geq n$ and for any $S'\subset S$ with $S'\neq\varnothing$, we have $\sum_{i\in S'}x_i\in\mathbb{R}\setminus\mathbb{Q}$.

This is an exercise from a linear algebra (not number theory).

Lavendula
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  • Shouldn't that be $\forall S^{\prime} \neq \varnothing \subset S$? :-) (otherwise you are referring to every possible empty set) – Dominique Oct 21 '24 at 13:49
  • It doesn't matter how difficult it is to prove that the sum of two or more particular numbers is irrational. You don't have particular numbers to work with, just arbitrary numbers. – David K Oct 21 '24 at 13:58
  • The syntax error has been corrected. The example of transcendental numbers is where my wondering about its provable rises. – Lavendula Oct 21 '24 at 14:07
  • I suggest: try it for small $n$. – lulu Oct 21 '24 at 14:07
  • Do you want it for any set of $x$'s or for some set of $x$'s? For some is true. Almost any set will work but an easy one to prove is square roots of various primes. There are sets of $x$'s that will fail because you have constructed them carefully. Let $n$ of them be $k+\sqrt 2$ for various integers $k$ and the rest $k-\sqrt 2$ for example – Ross Millikan Oct 21 '24 at 14:16
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    @RossMillikan in the example you constructed, taking $S$ to be the $n$ numbers of the first type would satisfy the given requirement, so I don’t see how it would fail. – Pranay Oct 21 '24 at 14:20
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    Please edit to include the source of the problem. If we just want some example, then take any transcendental, $\pi$ for instance, and look at ${\pi, \pi^2, \pi^3, \cdots, \pi^{2n-1}}$. No non-empty sum of these can be rational , as such a relation would imply that $\pi$ was algebraic. – lulu Oct 21 '24 at 15:27
  • @lulu I am more likely to interpret the first sentence "Let x_1,x_2,\cdots,x_{2n-1}\in\mathbb{R}\setminus\mathbb{Q}" as "For all x_1,x_2,\cdots,x_{2n-1}\in\mathbb{R}\setminus\mathbb{Q}". But I cannot change the problem because it is exactly what our teaching assistant gave us as our homework.(Maybe he ignore the ambiguity?) – Lavendula Oct 21 '24 at 15:45
  • @Lavendula If this is your homework problem, please provide more thoughts of what you've been trying. – Calvin Lin Oct 21 '24 at 15:52
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    This question is similar to: 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.. If you believe it’s different, please [edit] the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. – Calvin Lin Oct 21 '24 at 15:53
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    I think it can also be solved by some linear algebra: Assume that it is not true. Then there is some matrix $A\in\mathbb{Q}^{N\times (2n-1)}$ with $N \geq 2n-1$ and with entries $0,1$ and with $\ker A = 0$ and $Ax = b \in \mathbb{Q}^{N}$ (where $x = (x_1,\dots,x_{2n-1})$). Some submatrix $\tilde A$ is invertible and therefore $x = \tilde A^{-1}\tilde b \in \mathbb{Q}^{2n-1}$, a contradiction. – psl2Z Oct 21 '24 at 16:01
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    When a problem starts with, "Let $x_1,x_2,\cdots,x_{2n-1}\in\mathbb{R}\setminus\mathbb{Q}$," I would interpret this as presenting us with a set of numbers that we did not choose and that we know nothing else about. Hence the only statements you can prove are statements that are true for all such sets. If the exercise were one where you are meant to choose numbers to make the statement true, it would presumably have been worded differently. – David K Oct 21 '24 at 16:54
  • The claim in your linear-algebra exercise is true (as proved in the linked answers) for all $x_1,\ldots,x_{2n-1}\in\mathbb{R}\setminus\mathbb{Q}$. For that very reason, the claim is of absolutely no use in determining whether a particular sum of irrational numbers is rational. The modified question seems completely pointless. Look elsewhere to determine the rationality of any specific sum. – David K Oct 22 '24 at 18:15
  • Does your "next question" have anything to do with your "initial question"? You have somehow edited most of the words in the question text while making the entire thing even more obscure and unintelligible. – David K Oct 31 '24 at 04:35
  • @DavidK Has nothing to do with initial. That's just two different questions. What may be common is about solving $e+\pi$. The "obscure and unintelligible" is about separate abstract algebra from number theory. – Lavendula Oct 31 '24 at 04:44
  • "Has nothing to do with initial." That makes more sense now. Since it is really a different question, it should be asked as a different question, not as a part of this one. (Although I still think it is unlikely to get the kind of answer you want.) – David K Oct 31 '24 at 04:55

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