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Math Help - Induction n^2 > 2n for n>2

  1. #1
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    Induction n^2 > 2n for n>2

    Prove n^2 > 2n for n>2 by induction

    Check n = 3
    9 > 6
    Say it is true for n
    now check for n+1

    This is where I am worried that what I'm doing isn't enough proof

    (n+1)^2 > 2(n+1)

    n^2+2n+1 > 2n +2

    by cancelling

    n^2 > 1

    we know that n > 2 so n^2 is always going to be greater than 1.

    Does this suffice?
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  2. #2
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    Hi

    The idea behind what you use is a non-inductive proof, which is:

    Let n>2 be an integer, n^2>2n \Leftrightarrow n>2 (the equivalence is true because n positive integer). Since we assumed n>2, we have what we want.


    In a usual proof by induction, you assume your induction hypothesis (case n) and try to show the statement for n+1, using the case n.

    So for your proof, writing (n+1)^2=n^2+2n+1 is a good thing. What you have to do now is to obtain n^2+2n+1>2(n+1), using the induction hypothesis:  n^2>2n
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  3. #3
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    Ok how bout this

    Step 1 as above

    Step 2 (I need to prove (n+1)^2 > 2(n+1)


    (n+1)^2=n^2 +2n +1
    >2n + 2n +1 (Using induction)
    = 4n +1
    = 2(n+1) +2n -1 (2n-1) is always positive for n>2
    >2(n+1)

    Should i be using k instead?
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  4. #4
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    What you've done is correct, no need to choose another letter.
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