Originally Posted by

**patrick** Is there a proof of the Archimedean Property that doesn't involve the words "least upper bound" or "completeness axiom"?

Naturally, there should be since we use the Archimedean Property in the proof that R complete implies R has the least upper bound property.

If we assume the least upper bound property, then it is easy to prove the Archimedean Property. However, it must be the case that by simply assuming that R is complete we can deduce the Archimedean Property. Otherwise, the usual proof that complete implies least upper bound would be circular.

Any ideas on how to do this?

Well, let's define R to be the set of equivalence classes of Cauchy sequences of rational numbers. With a little thought, we see that given this definition, we can approximate any real number by a sequence of rational numbers. If x < y are positive rational numbers, then by the previous comment we can find

0 < (p/q) < x < y < (a/b)

for some positive integers p,q,a,b. Then, set n = aq to obtain n(p/q) > a/b and therefore

nx > n(p/q) > (a/b) > y ,

as desired. This proof seems like it works and properly avoids mention of the least upper bound property. Any objections to this line of reasoning? :)