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Basic Proof

  1. Feb 22, 2008 #1
    For all integers n; n^2+n+1 is odd.

    Prove the statement directly from definition of the terms & do not use any previously known facts.

    My main problem is if n^2+n+1=n(n+1)+1 then can I say n is even & n+1 is odd or would that imply that I'm using previously known facts. In the text there is a section on parity, but I'd hate to have to have a proof inside of a proof. Then even worse, would I then have to go through & prove what even & odd are? Not really sure what "directly from the definition of the terms" implies. Any ideas? Its obvious the professor would know best what she meant but there is no way to contact her atm.
  2. jcsd
  3. Feb 22, 2008 #2
    Well what they DON'T want you to do is say "well if n is odd then n^2 is odd, odd number plus an even number(the odd number n and then +1 is even)is always an odd number

    if n is even then n^2 is even, even + odd = odd again(n+1 would be odd in the case of even n) That's what they DON'T want to see

    You knew before hand as a fact that even * even = even, odd * odd = odd, and even+odd=odd. You would have to prove THOSE if you wanted to use 'em. And you could!
  4. Feb 22, 2008 #3
    Obviously n is either even or odd right? In each case what is the parity of n(n+1), and hence what is the parity of n^2+n+1?
  5. Feb 22, 2008 #4
    What I did was break it down by cases. But first I said consecutive integers must be even & odd or odd & even, by parity. Then Case 1 was n is odd & hence n+1 is even. Case two n is even & n+1 two is odd (these were drawn out showing the algebraic steps). Then I showed how an odd times an even always equaled an even, 2(integer). But I just got a big x on my paper that said refer to directions, not proven directly from definition of terms & the word parity was underlined.

    So does that mean I would have to prove consecutive integers are even/odd since I set it up to n(n+1)+1? Sorry for the confusion, as I'm a bit confused as well.
  6. Feb 22, 2008 #5


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    Or, to be very precise:

    n is either even or odd.

    case 1: If n is even, n= 2m for some integer m. Then n2+ n+ 1= (2m)2+(2m)+ 1= 4m2+ 2m+ 1= 2(2m2+ m)+ 1, an odd number.

    case 2: If n is odd, n= 2m+1 for some integer m. Then n2+ n+ 1= (2m+1)2+ (2m+1)+ 1= 4m2+ 2m+ 1+ 2m+ 1+ 1= 4m2+ 4m+ 3= 2(2m2+ 2m+ 1)+ 1, again an odd number.
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