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Thread: Proof by induction question

  1. #1
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    Proof by induction question

    Hello guys

    I'm having a problem with this real simple proof by induction example.

    Proof is that $5^n>n^5$ for all integers $n>6$


    1. anchor step $P(6)$: $5^6=15625=7776=6^5$

    This is pretty obvious, ok so far


    2. inductive step
    Assume $P(n)$
    $5^{n+1}=5\cdot5^n>5n^5$
    Ok here is the problem. we get $5^{n+1}=5\cdot5^n$ from sub-in $n+1$ into $5^n$ and law of exponentials
    But how do we get the other side $5n^5$ ???
    $5n^5$ is then used in the last part of the proof:


    Since $n\geq6$ and $P(n)$ is true

    $(n+1)^5=n^5(\frac{n+1}{n})=n^5(1+\frac{1}{n}) \leq n^5(1+\frac{1}{6})^5=n^5(2.16...)<5n^5$

    Together we have $5^{n+1}>(n+1)^5$
    So we've shown that $P(n)\Rightarrow P(n+1)$
    Thanks guys for the help
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  2. #2
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    Re: Proof by induction question

    Quote Originally Posted by shakra View Post
    Hello guys

    I'm having a problem with this real simple proof by induction example.

    Proof is that $5^n>n^5$ for all integers $n>6$




    This is pretty obvious, ok so far



    Ok here is the problem. we get $5^{n+1}=5\cdot5^n$ from sub-in $n+1$ into $5^n$ and law of exponentials
    But how do we get the other side $5n^5$ ???

    Same comment as the other proof you posted. You are assuming that 5^n>n^5 so under that assumption 5*5^n>5*n^5

    $5n^5$ is then used in the last part of the proof:



    Thanks guys for the help
    see comment in red
    Thanks from topsquark
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  3. #3
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    Re: Proof by induction question

    Since when is 15 625 = 7776???
    Thanks from topsquark
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  4. #4
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    Re: Proof by induction question

    I guess that was meant to be 15625 > 7776.
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