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Math Help - sets and relatively prime integers

  1. #1
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    sets and relatively prime integers

    If S is any set of n + 1 integers selected from 1, 2, 3,\cdot\cdot\cdot,2n + 1,

    prove that S contains two relatively prime integers.
    Prove that the result does not hold if S contains only n integers.

    I think that the pigeonhole principle might be a way to prove this, but I don't know to determine what are the pigeons and what are the pigeonholes.

    Thanks for the help!
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  2. #2
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    Quote Originally Posted by keityo View Post
    If S is any set of n + 1 integers selected from 1, 2, 3,\cdot\cdot\cdot,2n + 1,
    prove that S contains two relatively prime integers.
    Prove that the result does not hold if S contains only n integers.
    You need to notice that if \{a,b\}\subseteq \{1,2,\cdots,2n+1\} then \gcd(a,b)\le n-1.
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  3. #3
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    Why is then true?

    I just had an idea: if S has only n integers, then the set {2,4,6,\cdot\cdot\cdot,2n} will have n integers, and all will be not relatively prime, which proves the second statement.

    If S has n+1 integers, then one of the integers must be odd because there can only be n even numbers in the set {1,2,...,2n + 1} Thus, 2 and that odd number will be relatively prime, which proves the second statement.

    Is this proof ok?
    Thanks again.
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  4. #4
    MHF Contributor Bruno J.'s Avatar
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    Use the pigeon-hole principle! Amongst the following n sets, there must be one containing two of your numbers :

    \{1,2\}, \{3,4\},\hdots \{2n-1, 2n, 2n+1\}

    Now just remark that any two integers which lie in the same part of this partition are relatively prime.
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  5. #5
    MHF Contributor chiph588@'s Avatar
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    For your second part:  \{2,4,6,8,...,2n\} .
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