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Proof Related to the Binomial Theorem

  1. Nov 30, 2011 #1
    1. The problem statement, all variables and given/known data
    Use the above to prove that given a rational number a > 1 and A any other rational number, there exists b ε N such that ab > A.

    2. Relevant equations
    The above refers to the proving, by use of both induction and binomial theorem, that (1+a)n ≥ 1+na.

    Binomial Theorem: (i=0 to n)Ʃ(n choose i)ai

    3. The attempt at a solution

    So I tried using the binomial theorem to get the value aN.
    I get that aN must be greater than (i=0 to n-1)Ʃ(n choose i)ai
    So how do I choose an N so that this holds?
    Could you just let N > (i=0 to n-1)Ʃ(n choose i)ai?
     
  2. jcsd
  3. Nov 30, 2011 #2

    Office_Shredder

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    Let's keep it simple and use this result:
    (1+a)n ≥ 1+na

    The a in your problem and the a in the result they tell you to use are not going to be the same number. What should the relationship between them be?
     
  4. Nov 30, 2011 #3

    Thanks for the reply!

    So, would the a in my problem be equivalent to (1+a), and then you would choose b = 1?
    Then you would get (1+a)b ≥ 1+ba,
    And using the expansion formula, they are equivalent when b = 1, so as a result b ≥ 1?
     
  5. Nov 30, 2011 #4

    Office_Shredder

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    No, there's a difference between A and a.


    The objective now is to show that given any a>0, A>0, there exists some b such that (1+a)b>A. Use the fact that (1+a)b[/sub]>1+ab at this point
     
  6. Nov 30, 2011 #5
    Sorry for the slowness...
    It looks awfully similar to a Cauchy sequence proof.

    But I do realize (?) that I am trying to solve for b in terms of the given/fixed values a and A.

    When I look at the "hint" that was given, I keep seeing that (i=2 to b)Ʃ(b choose i)ai ≥ 0

    Does this train of thought lead me anywhere?
     
    Last edited: Nov 30, 2011
  7. Nov 30, 2011 #6
    Oh!
    Would it work if we just choose b > (A-1)/a?

    If it is I feel bad for missing something so simple. Looking too hard perhaps.
     
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