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