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Thread: proof by induction 5^n-1 is divisible by 4

  1. #1
    Member Jskid's Avatar
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    proof by induction 5^n-1 is divisible by 4

    Use mathematical induction to prove that the assertion is true for $\displaystyle n\ge 1$. $\displaystyle 5^n-1$ is divisible by $\displaystyle 4$.

    basis $\displaystyle n_{o}=1$ then $\displaystyle 5^1-1=4$ is divisible by $\displaystyle 4$.

    IH let $\displaystyle 1\le l<k$ and assume $\displaystyle 5^l-1$ is divisible by $\displaystyle 4$.

    Induction Step
    $\displaystyle 5^{k}-1=(5)5^{k-1}-1$
    Since $\displaystyle 5^{k-1}-1$ is divisble by $\displaystyle 4$ (by the induction hypothesis) then $\displaystyle (5)5^{k-1}-1$ is also divisible by $\displaystyle 4$

    Is this right?
    What is wrong with the latex
    Last edited by CaptainBlack; Nov 28th 2011 at 06:38 PM. Reason: fix LaTeX
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  2. #2
    Master Of Puppets
    pickslides's Avatar
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    Re: proof by induction 5^n-1 is divisible by 4

    The induction step should have n = k+1
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  3. #3
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    Re: proof by induction 5^n-1 is divisible by 4

    IH let $\displaystyle 1\le l<k$ and assume $\displaystyle 5^l-1$ is divisible by $\displaystyle 4$.
    If this is a proof by induction you need to go a step at a time, not prove it for all numbers at once.

    I'd say this. "Assume $\displaystyle 4 |(5^k-1)$. We want to show $\displaystyle 4|( 5^k^+^1-1)$."

    Since $\displaystyle 5^{k-1}-1$ is divisble by $\displaystyle 4$ (by the induction hypothesis) then $\displaystyle (5)5^{k-1}-1$ is also divisible by $\displaystyle 4$
    Not quite, no. You have the formula wrong, and you need to show your work; how do you know 4 divides $\displaystyle 5^k^+^1-1$ for certain?
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