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Let $S = \{1,2,3,4,5,6,7,8,9\}$. Prove that every subset of $S$ with $6$ or more elements must contain two numbers whose difference is equal to $5$.

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What have you tried? –  Jonas Meyer Jan 11 '12 at 3:42
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@Q123 I suggest you look up the pigeonhole principle. –  Alex Becker Jan 11 '12 at 3:47
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welcome to MathSE. I see that you are relatively new here. So I wanted to let you know a few things about MathSE. We like to know what you've tried on a problem. These sort of pleasantries usually result in more and better answers. Finally, I should add that posting questions in the imperative (i.e. Compute all such, Prove that...) is considered rude by some of the members, so it would be nice of you to change that wording. Thank you –  Arturo Magidin Jan 11 '12 at 3:48
    
@AlexBecker Thank you. –  Q123 Jan 11 '12 at 16:00

1 Answer 1

Well, your set can be broken into 4 subsets of 2 (with one element left over) such that the difference of the 2 elements is 5. Can you take it from there?

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Well, that was stupid of me. Fixed. –  Mike Jan 11 '12 at 4:35

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