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Some help on mathematical induction

  1. May 14, 2008 #1
    help guys i am really stumped on this question.
    prove that

    if "n" is an integer , then n^2-n+2 is even
     
  2. jcsd
  3. May 14, 2008 #2

    disregardthat

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    Suppose n is odd. What can you say about each member of n^2-n+2 ?
    Suppose n is even. What can you say about each member of n^2-n+2?
     
  4. May 14, 2008 #3
    That is one way of thinking about the problem. If you want to do this via induction, however, what is the first step you need to take? Also, are you trying to prove that fact for n is any integer or just a positive integer?
     
  5. May 15, 2008 #4
    to prove by principle of mathemetical induction
    step 1:
    put n=1
    1^2-1+2=1^3=1
    which is false
     
  6. May 15, 2008 #5
    I think what is meant by n^2-n+2 is [tex]n^2 - n +2[/tex], not [tex]n^{2-n+2}[/tex] as you seem to use.
     
  7. May 15, 2008 #6

    CRGreathouse

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    A proof by induction would take the following form:

    1. For n = 1, [itex]n^2-n+2[/itex] is even.
    (subproof)
    2. Suppose [itex]n^2-n+2[/itex] is even. Then [itex](n+1)^2-(n+1)+2[/itex] is even.
    (subproof)
    3. By induction (base case 1, inductive step 2), [itex]n^2-n+2[/itex] is even for all natural numbers n.

    If you can fill in the subproofs you're done.

    (If you really want all integers, of course, you'll have to prove them too -- maybe by induction on [itex](-n)^2-(-n)+2.[/itex])
     
  8. May 15, 2008 #7
    :)

    hmm...My suggestion :

    N^2 - N + 2
    = N(N-1) + 2

    For any integer N, if N is even , (N-1) is odd therefore N(N-1) is divisible by 2 and is an even #, so , N(N-1) + 2 is even

    For any integer N, if N is odd , (N-1) is even therefore N(N-1) is divisible by 2 and is an even #, so , N(N-1) + 2 is even

    Is this logical?
     
  9. May 15, 2008 #8
    For n=1, n^2 - n +2 = 2 is even.

    Now, suppose that for some integer k, that [tex]k^2 -k + 2[/tex] is even. Then for the next integer k+1

    [tex](k+1)^2 -(k+1) +2 = k^2 + 2k +1 - k -1 +2 = k^2 + k +2[/tex]

    Now

    [tex] k^2 + k + 2 = (k^2 -k + 2) + 2k[/tex]

    On the RHS, the stuff in bracket is what we started with, we assumed it was even. 2k is even for all integers k. An even number plus an even number is an even number. Therefore for the next integer, it will also be even.

    Therefore for all integers n, [tex]n^2 -n +2[/tex] is even
     
  10. May 23, 2008 #9
    Quick question about induction. Could you prove it true for n = 1. Then prove it for n = n + 1. Then prove it for n = n - 1. Would that be enough to prove it for all integers?
     
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