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Don't know if I got this right. Prove n^2>n+1

  1. Apr 23, 2008 #1
    1. The problem statement, all variables and given/known data
    The principal of mathematical induction can be extended as follows. A list P(m),P(m+1)... of propositions is true provided 1)P(m) is true, 2) P(n+1) is true whenever P(n) is true and n>(or =) m

    I have to use the above to prove that n^2>n+1 for n>(or equal to) 2


    2. Relevant equations
    n^2>n+1 for n>(or equal to) 2




    3. The attempt at a solution

    so I said m=n+1

    Then since I assume that the original statement implies that I hold m constant and increase n by 1

    Inductive step (n+1)^2>(n+1)
    => n+1>1 True b/c {1,2,3,...n}=N
     
  2. jcsd
  3. Apr 23, 2008 #2

    rock.freak667

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    Assume true for n=k

    [itex]k^2>k+1 for k \geq 2[/itex]

    +(2k+1)
    [itex]k^2+2k+1>k+1+2k+1 [/itex]

    and [itex]k^2+2k+1=(k+1)^2[/itex] which is what you need on the left side. Deal with the right side now.
     
  4. Apr 23, 2008 #3
    Right side: (k+1+2k+1)=(2k+k+2)>(k+2)

    Right?
     
  5. Apr 23, 2008 #4

    rock.freak667

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    2k+k+2>k+2 => 3k>k which is true so that 2k+k+2>k+2 is true

    and now you have

    (k+1)^2>2k+k+2>k+2
     
  6. Apr 23, 2008 #5
    Why do you have to show that many steps?
     
  7. Apr 23, 2008 #6

    rock.freak667

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    Because that is to show how P(k) => P(k+1) instead of putting n=k+1 in the formula and showing it is true.
     
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