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Proving a properties of fibonacci numbers

  1. May 10, 2013 #1
    Let p[n] be the following property of Fibonacci numbers:
    p[n]: f[itex]_{n+1}[/itex] * f[itex]_{n-1}[/itex] - (f[itex]_{n}[/itex])[itex]^{2}[/itex] = (-1)[itex]^{n}[/itex]. Prove p[n] by induction.

    This is the proof I wrote. i used regular induction. Is weak induction sufficient to prove it or do I need to prove this by strong induction?


    BASE STEP: p[1]: f[itex]_{2}[/itex] * f[itex]_{0}[/itex] - (f[itex]_{1}[/itex])[itex]^{2}[/itex] = (1 * 0) - 1 = -1

    and (-1)[itex]^{1}[/itex] = -1


    assume P[k]: f[itex]_{k+1}[/itex] * f[itex]_{k-1}[/itex] - (f[itex]_{k}[/itex])[itex]^{2}[/itex] = (-1)[itex]^{k}[/itex]

    show p[k+1]: f[itex]_{k+2}[/itex] * f[itex]_{k}[/itex] - (f[itex]_{k+1}[/itex])[itex]^{2}[/itex] = (-1)[itex]^{k+1}[/itex]

    We know f[itex]_{k+2}[/itex] = f[itex]_{k+1}[/itex] + f[itex]_{k}[/itex]

    p[K+1]: f[itex]_{k+2}[/itex] * f[itex]_{k}[/itex] - (f[itex]_{k+1}[/itex])[itex]^{2}[/itex] = (-1)[itex]^{k+1}[/itex]

    substitute f[itex]_{k+2}[/itex] = f[itex]_{k+1}[/itex] + f[itex]_{k}[/itex] into p[k+1]

    p[k+1]: (f[itex]_{k+1}[/itex] + f[itex]_{k}[/itex]) * f[itex]_{k}[/itex] - (f[itex]_{k+1}[/itex])[itex]^{2}[/itex] = (-1)[itex]^{k+1}[/itex]

    = f[itex]_{k}[/itex] * f[itex]_{k+1}[/itex] + (f[itex]_{k}[/itex])[itex]^{2}[/itex] - (f[itex]_{k+1}[/itex])[itex]^{2}[/itex]

    We know p[k]: f[itex]_{k+1}[/itex] * f[itex]_{k-1}[/itex] - (f[itex]_{k}[/itex])[itex]^{2}[/itex] = (-1)[itex]^{k}[/itex]

    We can rewrite p[k] as: (f[itex]_{k}[/itex])[itex]^{2}[/itex] = f[itex]_{k+1}[/itex] * f[itex]_{k-1}[/itex] - (-1)[itex]^{k}[/itex]

    We substitue (f[itex]_{k}[/itex])[itex]^{2}[/itex] from p[k] into the (f[itex]_{k}[/itex])[itex]^{2}[/itex] in p[k+1]

    = f[itex]_{k+1}[/itex] + f[itex]_{k}[/itex] + f[itex]_{k+1}[/itex] * f[itex]_{k-1}[/itex] - (-1)[itex]^{k}[/itex] - f[itex]_{k+1}[/itex][itex]^{2}[/itex])

    = f[itex]_{k+1}[/itex] (f[itex]_{k+1}[/itex] + f[itex]_{k}[/itex]) (-1)[itex]^{k}[/itex] - f[itex]_{k+1}[/itex][itex]^{2}[/itex])

    = (f[itex]_{k+1}[/itex])[itex]^{2}[/itex]) - (-1)[itex]^{k}[/itex] - (f[itex]_{k+1}[/itex])[itex]^{2}[/itex])

    = - ((-1)[itex]^{k}[/itex])

    = (-1)[itex]^{k+1}[/itex]

    Does this proof strong enough or would I have to do this by strong induction? If by strong induction, how do I got about proving it?
  2. jcsd
  3. May 10, 2013 #2


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    Staff: Mentor

    Every valid proof with weak induction is a valid proof with strong induction, too (and this is the first time I saw specific words for them). Your proof uses valid steps only, therefore it is valid. There is no need to do more.
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