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Math Help - Mathematical Induction Proof

  1. #1
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    Mathematical Induction Proof

    Prove that for all the natural numbers n that 2^n > n

    Base Case is easy
    Then the inductive step you have 2^k > k as the inductive hypothesis
    show that p[k+1] holds
    2^(k+1) > k+1
    on the left side 2^(k+1) = 2^k * 2
    but idk what else to do
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  2. #2
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    Quote Originally Posted by amiv4 View Post
    Prove that for all the natural numbers n that 2^n > n

    Base Case is easy
    Then the inductive step you have 2^k > k as the inductive hypothesis
    show that p[k+1] holds
    2^(k+1) > k+1
    on the left side 2^(k+1) = 2^k * 2
    but idk what else to do
    Hi amiv4,

    2^{k+1}>k+1 ?

    (2)2^k>k+1 ?

    2^k+2^k>k+1 ?

    If 2^k>k then 2^k+2^k>k+1

    since 2^k>1
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