Homework Help: Proving that the Archimedean axiom is true

1. Sep 13, 2011

major_maths

1. The problem statement, all variables and given/known data
Show that the Archimedean axiom O5 follows from the Least Upper Bound Property O6, together with the other axioms for the reals.

2. Relevant equations
O5 = [if a,b > 0, then there is a positive integer n such that b<a+a+a+...+a (n summands)] or [if a,b > 0, then b < na or b/a < n]

O6 = if A is any nonempty subset of R that is bounded above, then there is a least upper bound for A.

3. The attempt at a solution
My teacher told us to do this as a proof by contradiction so that's the format I'll be doing.

Suppose the Archimedean axiom is false towards a proof by contradiction. Therefore, there exists some a,b > 0 such that b $\geq$ na, or b/a $\geq$ n.
Then the set, say N, is bounded above by b/a and so sup(N) exists. Write sup(N) = S.

And then I can't figure out how to finish this proof.

Last edited: Sep 13, 2011
2. Sep 13, 2011

micromass

That $S=\sup(\mathbb{N})$ means that there is a natural number n that is close to S. But then n+1>S...

Try to formalize this.

3. Sep 13, 2011

major_maths

If n+1>S, then there exists a natural number not bounded above by S.
This is a contradiction as the set N is the set of whole positive integers and adding 1 would not exclude any n previously in the set N.

Is that right?

4. Sep 13, 2011

micromass

Yes, that is correct. How would you choose n though?

5. Sep 13, 2011

major_maths

I would choose n to be close to S, or S-1<n<S.

So for any n bounded above by S but greater than S-1, if n+1>S, then there exists a natural number not bounded above by S.

And so on, and so on. Is that specific enough?

6. Sep 13, 2011

micromass

Yes, but you need to state why such an n exists. You probably know it, but I want to make sure.

7. Sep 13, 2011

major_maths

Oh. Um, it exists because S is the supremum of the set?

8. Sep 13, 2011

micromass

Yes, do you understand why?

9. Sep 13, 2011

major_maths

S is the supremum of the set because the set is bounded above by b/a, which is what sup(N[/]) is defined as at the beginning of the proof (b/a ≥ n).