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Thread: Have I got this right?

  1. #1
    Sep 2008

    Have I got this right?

    Taking my first tottering steps with induction here. Of course, the math lecture tomorrow will handle this, but I like to keep a little bit ahead so I hope it's okay to post a thread here and ask if I've understood the fundamentals!
    Now, I've spent the last hour reading the chapter and done this exercise, so basically, if someone could just point out any errors I'd be grateful:

    "Prove by induction that the sum of all the n first uneven integers equals n^2."

    The sum of the first n integers = 1 + 3 + 5 +...+(2n-1) --> n(2n-1)

    Assume that n(2n-1) = n^2 for all n >= 1

    I: For n = 1 we have 1*(2*1-1) = 1 and 1^2 = 1
    The assumption is true for n = 1

    II: Assume that the statement is true for n = x
    If so, then
    (x+1)(2*(x+1)-1) = (x+1)^2

    Proof: (x+1)(2*(x+1)-1) = (x+1)^2 QED

    III: According to the induction axiom, it follows from I and II that
    n(2n-1) = n^2 is true for all n >= 1

    A bit heavy on the formalia perhaps, but hey - freshman calculating.
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  2. #2
    Junior Member
    Sep 2008
    I believe the BS you've fallen for / trying to propagate is to prove
    n^2 = n(2n-1) for all n because

    (a) 1^2 = 1(2*1-1) --> OK
    (b) if x^2 = x(2x-1) then (x+1)^2 = (x+1)(2x+1)

    LHS = (x+1)^2 = x^2 + 2x + 1 = x(2x-1) + 2x + 1 = 2x^2 + x + 1
    Not sure how you equate it to RHS

    (x+1)(2*(x+1)-1) = (x+1)^2
    Proof: (x+1)(2*(x+1)-1) = (x+1)^2 QED
    is restating the same line. Not a proof.
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