Prove that if you pick 10 numbers from 1 to 1000, then there exists a pair where the larger of the pair is at most twice the other.
I'm imagining this has to deal with either a proof by contrapositive or contradiction. In the case of contradiction I would assume that there exists a set of 10 numbers from 1 to 1000 then for all pairs the larger of the pair is greater than twice the smaller. But I cant seem to find a flaw with that. I also tried creating a set of 10 where the numbers were evenly spaced out as far as possible (100, 200, 300) but it holds.
Any help would be great, thanks.
2007-10-18
21:06:48
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6 answers
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asked by
David
2
in
Mathematics