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Thread: Fibonacci number Induction Proof

  1. #1
    Newbie Pi R Squared's Avatar
    Oct 2009

    Question Fibonacci number Induction Proof

    Prove every third Fibonacci number is even.-- The Fibonacci numbers are the sequence 1,1,2,3,5,8,13,..., given by $\displaystyle f_n$ where $\displaystyle f$(1)=1, and $\displaystyle f$(2)=1, and $\displaystyle f_(n)=f(n-1)+(n-2)($ for all $\displaystyle n$.

    This is the problem...

    This is what I have...
    Base Case:
    f(3k-2) if k=1 then f(3(1)-2)=f(1)=1 odd #
    f(3k-1) f(3(1)-1)=f(2)=1 odd #
    f(3k) f(3(1) =f(3)=2 even #

    so f(1)+f(2)=f(3)
    an odd+odd=even

    Inductive step:

    Suppose it is true for k+1
    f(3(k+1)) =f(3K+3)

    I do not know where to go from here...
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  2. #2
    MHF Contributor
    Oct 2009
    First let me note that it's very good that you made your induction statement (let's call it $\displaystyle P(k)$) consist of three parts:

    $\displaystyle P(k)$ is ($\displaystyle f(3k-2)$ is odd) and ($\displaystyle f(3k-1)$ is odd) and ($\displaystyle f(3k)$ is even)

    The original assertion to prove (let's call it $\displaystyle Q(k)$) is just:

    $\displaystyle Q(k)$ is ($\displaystyle f(3k)$ is even)

    However, this is not enough to prove the induction step, which is the following statement:

    for all $\displaystyle k \ge 1$, $\displaystyle Q(k)$ implies $\displaystyle Q(k+1)$

    Knowing $\displaystyle Q(k)$ just is not enough to be able to prove $\displaystyle Q(k+1)$. Therefore, we strengthen the induction statement from $\displaystyle Q(k)$ to $\displaystyle P(k)$. Now, in proving that $\displaystyle P(k)$ implies $\displaystyle P(k+1)$, we have more to prove (three substatements of $\displaystyle P(k+1)$ instead of just one of $\displaystyle Q(k+1)$), but we also have a stronger assumption $\displaystyle P(k)$. This technique -- strengthening the induction statement in order for the induction step to go through -- is very common in mathematics.

    End of a long digression. Concerning your proof, why do you say "Suppose it is true for k+1"? Inductive step is usually "suppose it is true for k; we need to prove it for k+1", i.e., one has to prove $\displaystyle P(k)$ implies $\displaystyle P(k+1)$ for all $\displaystyle k\ge 1$. So write carefully what you know, i.e., $\displaystyle P(k)$, and what you need to prove, i.e., $\displaystyle P(k+1)$, and the proof should be pretty straightforward.
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