# 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

Staff Emeritus
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

Staff Emeritus
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

Staff Emeritus
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

Staff Emeritus
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).