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Math Help - [SOLVED] Archimedian Property proof

  1. #1
    Member ilikedmath's Avatar
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    Question [SOLVED] Archimedian Property proof

    Statement to prove:
    If x > 0, show there exists n in N (the set of all natural numbers) such that 1/(2^n) < x.

    My work on the proof so far:
    Let x > 0. By the Archimedian Property, we know if ε > 0, there exists an n in N such that 1/n < ε.
    Take x = ε . So there exists an n in N such that 1/n < x.

    That is as far as I've gotten. I am stuck as to how I can algebraically manipulate the inequality to get the 2 in there somehow and to get the final form of 1/(2^n) < x.
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  2. #2
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    For k such that \frac{1}{k} < x, choose n such that 2^n > k. Then \frac{1}{2^n} < \frac{1}{k} < x.
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    let n>1, so 2^n>n therefore 1/(2^n)<1/n<x.
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  4. #4
    Member ilikedmath's Avatar
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    Clarification

    Quote Originally Posted by icemanfan View Post
    For k such that \frac{1}{k} < x, choose n such that 2^n > k. Then \frac{1}{2^n} < \frac{1}{k} < x.
    So I can arbitrarily choose an n such that 2^n > k?
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  5. #5
    Member ilikedmath's Avatar
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    Clarification

    Quote Originally Posted by chiph588@ View Post
    let n>1, so 2^n>n therefore 1/(2^n)<1/n<x.
    So I can just state that n > 1? And how do I know 2^n > n? That looks familiar. I think we proved something like that before with math induction.
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  6. #6
    MHF Contributor chiph588@'s Avatar
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    actually, n >= 0 (Typo)

    2^n > n is proved by induction

    base case: 2^0 > 1

    inductive case: suppose 2^n > n
    then 2*2^n > 2n
    2^(n+1) > 2n > n+1

    Therefore 2^n > n
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  7. #7
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    Quote Originally Posted by ilikedmath View Post
    So I can arbitrarily choose an n such that 2^n > k?
    The choice is not entirely arbitrary, but since 2^n has no upper bound, for any given k, there must be some value of n for which 2^n > k.
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  8. #8
    Member ilikedmath's Avatar
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    I think I got it

    I think I got it [I have a proof within a proof]:

    My proof: Let x > 0. By induction 2^n > n
    (Proof: Base case: for n = 1, 2^1 > 1. Check. Inductive step: Assume true for n = k, k a natural number. That is 2^k > k is true. To show this true for k + 1, we need to show 2^(k+1) > k +1 which implies 2*2^k > k +1. By assumption that 2^k > k, then 2*2^k > 2k. But for all k ≥ 1, 2k ≥ k + 1, so
    2^(k+1) > 2k ≥ k+1. Therefore 2^(k+1) > k + 1, and 2^n > n. QED)
    So since 2^n > n, then 1/2^n < 1/n. By the Archimedian Property, 1/n < x. Therefore
    1/2^n < 1/n < x which implies 1/2^n < x. QED.
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