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Math Help - Mathematical Induction

  1. #1
    Junior Member Cthul's Avatar
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    Mathematical Induction


    My friend needs help and requires this work to be done by today. I hope it won't trouble anyone to elaborate how to do so.
    EDIT: Where it says 2^{2}, it's supposed to be 2^{N} I apologies.
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  2. #2
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    The induction hypothesis is
    k < 2^k

    You assume that you have already proved this for k. Now you use this to prove k+1 < 2^{k+1}

    So k+1 < 2^{k} + 1 just adds 1 to both sides of the induction hypothesis.
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  3. #3
    Junior Member Cthul's Avatar
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    Quote Originally Posted by friend
    then they change 2^{k+1} into 2^{k}+2^{k} because 1<2^k
    and set that to 2\times2^k=2^{k+1}
    He still doesn't get it.
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  4. #4
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    \displaystyle<br />
k<2^k \ \mbox{multiple by 2}\Rightarrow \ 2k<2*2^k\Rightarrow k+k<2^{k+1}

    Since \displaystyle k\geq n\geq 1, we can take the 2nd k and set it equal to 1 without loss of generality.

    \displaystyle k+1<2^{k+1}
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  5. #5
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    Quote Originally Posted by Cthul View Post

    My friend needs help and requires this work to be done by today. I hope it won't trouble anyone to elaborate how to do so.
    EDIT: Where it says 2^{2}, it's supposed to be 2^{N} I apologies.

    P(k)

    k<2^k


    P(k+1)

    Replace k with k+1

    k+1<2^{k+1}\;\;\;?


    Proof

    k<2^k\Rightarrow\ 2(k)<2\left(2^k\right)

    2k<2^{k+1}

    2k=k+k\Rightarrow\ (k+k)<2^{k+1}

    Since it is proven for the base case k=0 or k=1,

    then (k+1)<(k+k) for k>1

    Therefore, if k<2^k

    then (k+1)<2^{k+1} since (k+1)<(k+k) for k>1
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