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Functions question, must at least one be even?

  1. Nov 4, 2006 #1
    Hello everyone.

    I'm having some troubles seeing how I would apply the even case for this problem.

    The problem is:
    If n + 1 integers are chosen from the set {1,2,3,....,2n} where n is a postive integer, must at least one of them be even? why?

    Well the book did the exact same problem but instead of even they said must at least one of them be odd? why?
    Well there explanation was the following:
    Yes. There are n even integers in the set {1,2,3,....,2n}, namely 2(=2x1), 4(=2x2), 6(=2x3),...,2n(2xn). So the maximum number of even integers that can be chosen is n. Thus if n + 1 integers are chosen, at least one of them must be odd.

    If I apply the same concept to my problem, don;t' they answer my question in explaning that at least one of them must be odd?
    Yes. There are n even integers in the set {1,2,3,....,2n}, namely 2(=2x1), 4(=2x2), 6(=2x3),...,2n(2xn). So the maximum number of even integers that can be chosen is n. Thus if n+1 integers are chosen, at least one must be even.

    Or do I have to work in reverse and say:

    yes. There are 2n+1 odd integers in the set {1,2,3...,2n},
    namely 1(=2(0)+1),
    3(=2(1)+1),
    5(=2(2)+1),.....
    but when it comes to 2n, i can't express that as odd can i?
    or do i just say, 2n+1?
    So the maximum number of odd integers that can be chosen is 2n+1. Thus if n+1 integers are chosen, at least one must be even.
     
  2. jcsd
  3. Nov 4, 2006 #2

    0rthodontist

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    There are n odd integers in the numbers 1...2n. It is almost exactly the same argument your book gave.
     
  4. Nov 4, 2006 #3

    matt grime

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    No, there aren't. There are n even integers in there, and n odd integers. There aren't even 2n+1 integers in the range 1 to 2n.
     
  5. Nov 5, 2006 #4
    Thanks for the responce, i wrote:

    yes. There are n odd integers in the set {1,2,3,...,2n},1(=1x1), 3=(3x1), 5(=5x1),...

    But i'm stuck here, i can't say 2n(=2xn) Because thats not an odd number is it, the odd number would be 2n+1 I thought...or 2n-1, but if that is correct then I can go on saying:
    so the maximum of odd integers that can be chosen is n. Thus if n + 1 integers are chosen, at least one of them must be even.

    If its in the range of 1 to 2n, would the larger odd number be 2(n-1)?
     
  6. Nov 5, 2006 #5

    0rthodontist

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    Is 2(n-1) odd?
     
  7. Nov 5, 2006 #6
    alright i got it i think:

    yes. There are n odd integers in the set {1,2,3,...2n}, 1(=2(1-1)), 3(=2(2-1)),....,2n-1(=2(n-1)). It follows that if at least n + 1 integers are chosen, one is sure to be even.
     
  8. Nov 5, 2006 #7

    0rthodontist

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    Does 1 = 2(1-1)?
     
  9. Nov 5, 2006 #8
    oops i got my ( )'s mixed up i ment
    1 (=(2x1)-1), 3(=(2x2)-1), ..., 2n-1(= (2n)-1), now is it okay?
     
  10. Nov 6, 2006 #9

    0rthodontist

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    Yeah, looks fine.
     
  11. Nov 6, 2006 #10
    wee thanks!
     
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