I am trying to prove that at least 2 members of an n-length subset of {1,2,...2n-1} will be relatively prime. The way I am thinking about it is that if 1 is included, then 1 is relatively prime with any other member. If 1 is not included, then there are only 2n-2 choices. Because of spacing, two of these must be consecutive and since any two consecutive numbers are relatively prime, this proves the original theorem. I will prove that any two consecutive numbers are relatively prime with Euclidean algorithm, but I don't know how to prove that there must be consecutive numbers even though I know its true.
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Are you familiar with the pigeonhole principle? – saulspatz Feb 26 '18 at 06:04
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The result follows from considering the consecutive pairs $$\{2, 3\}, \{4, 5\}, \ldots, \{2n-4, 2n-3\}, \{2n-2, 2n-1\}.$$ Clearly, there are $n-1$ such sets, so by the pigeonhole principle, choosing $n$ numbers from the set $\{2, 3, \ldots, 2n-1\}$ must mean choosing both numbers from one of these pairs.
Think of the maximum number of numbers you can choose from the above pairs such that no pair has both of its numbers selected. This means choosing one number from each pair, but there are only $n-1$ such pairs. If we choose one more number, it will mean it must be taken among one of the pairs already chosen.
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I think you should have started with the pair ${1,2}$ and ended with ${2n-3,2n-2}$, since we're allowed to choose $1$, but not $2n-1$. – Arthur Feb 26 '18 at 06:11
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The title and the content of the original post have different last-elements. – Air Conditioner Feb 26 '18 at 06:12