Homework Help: Set of Powers Unbound from Above

1. Sep 1, 2008

e(ho0n3

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. Sep 1, 2008

morphism

If an < k for all n, then what happens when you take logs of both sides?

3. Sep 1, 2008

e(ho0n3

I'm not allowed to use logarithms yet unfortunately, but I like the argument.

4. Sep 1, 2008

morphism

Look at successive differences between the an's. What can you say?

5. Sep 1, 2008

e(ho0n3

The differences should be increasing. I don't understand how this helps.

6. Sep 3, 2008

e(ho0n3

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?

7. Sep 4, 2008

HallsofIvy

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?

8. Sep 4, 2008

e(ho0n3

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?

9. Sep 5, 2008

morphism

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).

10. Sep 5, 2008

e(ho0n3

Ah. Unfortunately I haven't dealt with Cauchy sequences yet so I can't use any of its results.