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Set of Powers Unbound from Above

  1. Sep 1, 2008 #1
    The problem statement, all variables and given/known data
    Let a > 1 and let S = {a, a2, a3, ...}. Prove that S is not bounded from above.

    Relevant theorems
    The least upper bound property of the reals and these consequences:
    (1) For any real x there is an integer n satisfying n > x .
    (2) For any positive real x there is a positive integer n satisfying 1/n < x.
    (3) For any real x there is an integer n satisfying n ≤ x < n + 1.
    (4) For any real x and positive integer N, there is an integer n satisfying n/N ≤ x < (n+1)/N.
    (5) For any real x and e, there is a rational r satisfying |x - r| < e.

    The attempt at a solution
    I was thinking of a proof by contradiction by I can't think of anything contradictory. Any tips?
     
  2. jcsd
  3. Sep 1, 2008 #2

    morphism

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    If an < k for all n, then what happens when you take logs of both sides?
     
  4. Sep 1, 2008 #3
    I'm not allowed to use logarithms yet unfortunately, but I like the argument.
     
  5. Sep 1, 2008 #4

    morphism

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    Look at successive differences between the an's. What can you say?
     
  6. Sep 1, 2008 #5
    The differences should be increasing. I don't understand how this helps.
     
  7. Sep 3, 2008 #6
    The following just dawned on me: Suppose S is bounded above. By the LUB property, S has a least upper bound s. Therefore, for any n, an ≤ s but also an+1 = a(an) ≤ s. This latter inequality implies that an ≤ s/a < s. We thus have that s/a is an upper bound smaller than s, which is a contradiction. Ergo, S can't be bounded above.

    Is this correct?
     
  8. Sep 4, 2008 #7

    HallsofIvy

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    No, there is a logical error in your proof. The fact that s is the least upper bound for the entire sequence does not imply that there is not some smaller number larger than an for a specific n.

    However, you can say that there must exist some specific value N such that s- aN< 1. Now what can you say about aN+1?
     
  9. Sep 4, 2008 #8
    But I didn't state that...I think. Please tell me exactly where the logical error is:
    1. Let n be a positive integer.
    2. We have that an ≤ s.
    3. We also have that an+1 ≤ s.
    4. Thus, we have that an ≤ s/a.
    5. Since n is arbitrary, an ≤ s/a for all n.
    6. Ergo, s/a < s is a lower bound of S. Contradiction.
    How? Which of the consequences I that I wrote in the first post are you using to deduce this?
     
  10. Sep 5, 2008 #9

    morphism

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    I don't see anything wrong with that.

    By the way, my previous post was hinting towards a proof of the fact that {a^n} isn't Cauchy, hence not convergent, hence cannot be bounded (because it's monotone).
     
  11. Sep 5, 2008 #10
    Ah. Unfortunately I haven't dealt with Cauchy sequences yet so I can't use any of its results.
     
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