Proving that the Archimedean axiom is true

  • #1

Homework Statement


Show that the Archimedean axiom O5 follows from the Least Upper Bound Property O6, together with the other axioms for the reals.

Homework 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.

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 [itex]\geq[/itex] na, or b/a [itex]\geq[/itex] 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.
 
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Answers and Replies

  • #2
22,129
3,297
That [itex]S=\sup(\mathbb{N})[/itex] means that there is a natural number n that is close to S. But then n+1>S...

Try to formalize this.
 
  • #3
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
22,129
3,297
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?

Yes, that is correct. How would you choose n though?
 
  • #5
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
22,129
3,297
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?

Yes, but you need to state why such an n exists. You probably know it, but I want to make sure.
 
  • #7
Oh. Um, it exists because S is the supremum of the set?
 
  • #8
22,129
3,297
Oh. Um, it exists because S is the supremum of the set?

Yes, do you understand why?
 
  • #9
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).
 

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