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Prove that the product of 2 consecutive natural numbers is even

  1. Dec 3, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove that the product of two consecutive natural numbers is even.

    2. The attempt at a solution
    Hi, I'm just starting to work with proofs by induction, I'm just wondering if this is a valid technique, and/or if I am being too verbose in my proof, thanks!

    Let [itex]P(n)=n(n+1)[/itex] is even
    Let [itex]f(n)=n(n+1)[/itex]

    Let [itex]n=1[/itex]
    Since [itex]1(1+1)=2[/itex], and 2 is even, it follows that [itex]P(1)[/itex] is true.
    Suppose [itex]P(k)[/itex] is true, that is, suppose [itex]k(k+1)[/itex] is even.
    Then,
    [itex]f(k+1)-f(k)=(k+1)((k+1)+1)-k(k+1)[/itex]
    [itex]f(k+1)-f(k)=(k+1)(k+2)-k(k+1)[/itex]
    [itex]f(k+1)-f(k)=(k+1)(k+2-k)[/itex]
    [itex]f(k+1)-f(k)=2(k+1)[/itex]
    [itex]f(k+1)=2(k+1)+f(k)[/itex]

    Since by our induction hypothesis, [itex]f(k)[/itex] is even, it follows that the sum, [itex]f(k+1)[/itex], of [itex]2(k+1)[/itex] and [itex]f(k)[/itex] is also even. Since [itex]P(1)[/itex] is true, and [itex]P(k+1)[/itex] is true if [itex]P(k)[/itex] is true, by induction, [itex]P(n)[/itex] is true for all natural numbers [itex]n[/itex].
     
    Last edited: Dec 3, 2012
  2. jcsd
  3. Dec 3, 2012 #2

    lurflurf

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    That is right. You might not need so many algebra steps. The telescoping sum is just a fancy way to see that

    [tex]n(n+1)=2 \sum_{k=1}^n k[/tex]
     
  4. Dec 3, 2012 #3

    Mentallic

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    If you're going to be explaining P(k) is even by assumption, why not explicitly make it so? Let P(k) = k(k+1) = 2n where n is some natural number, so then you can plug that in later to obtain

    P(k+1) = 2(k+1+n)

    And then say, since n and k are assumed to be natural numbers, P(k+1) is even... etc.
     
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