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Natural numbers proof help

  1. Feb 5, 2009 #1
    If n is a natural number then n2+n+3 is odd.

    This is what I have and wanted to know if I was doing it right or not:

    Let n be a member of the natural numbers. If n is even, then n=2k, k member of natural numbers, and n2+n+3
    =(2k)2+2k+3
    =4k2+2k+3
    = 2(2k2+k+1)+1, where (2k2+k+1) is a member of the natural numers. This means that when n is even, n2+n+3 is odd.

    If n is odd, then n=2j+1 where j is a member of the natural numbers
    and n2+n+3
    =(2j+1)2+(2j+1)+3
    =4j2+4j+1+2j+4
    =4j2+6j+5
    =2(2j2+3j+2)+1, where (2j2+3j+2) is a memeber of the natural numbers. This means that when n is odd, n2+n+3.

    Is this ok? Thanks for the help!
     
  2. jcsd
  3. Feb 5, 2009 #2
    It looks ok, but it's overly complicated. A simpler argument is that n^2+n=n(n+1) is always even (because clearly one of n, n+1 is), therefore, n^2 + n + 3 is always odd.
     
  4. Feb 5, 2009 #3

    HallsofIvy

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    I would say that what you give is a perfectly good proof and because it is yours it is the one you should submit. Of course, you should then be aware of Preno's simpler proof.
     
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