I was doing this question and the was curious if my method is valid or if I missed anything out. The question: say we have a subset $X \subseteq \{1, \dots, 2n\}$ where $|X| = n + 1$, show that there are 2 elements of $X$ that one is a multiple of the other.
If we prove that there will be an element that is double another element we will satisfy the proof. We have $n$ 'holes' where each whole is a pair of numbers where one is double the other. As we have $n + 1$ 'pigeons' that are the elements of $X$ by pigeon hole principle, there will always be a pair that one is double the other and so a multiple.
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1Does this answer your question? Prove that there are two numbers among them such that one divides the other. Similiar idea has been used in the answer to this one. The way you wrote this seems vague and confusing. – bb_823 Apr 20 '24 at 17:29
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Your proof doesn’t quite work because the holes may not be disjoint from one another. Try your proof with $n=3$ and you’ll see what goes wrong — which “hole” contains $5$? – Robert Shore Apr 20 '24 at 17:31