# Way to prove inequality without Archimedean?

1. Sep 4, 2008

### quasar_4

Hi everyone,

How might someone construct a proof that for a, b $$\in$$ R such that a < b, there exists a real number c such that a < c < b, without using the Archimedean principle?

I know if we can assume the Archimedean principle that we can easily prove this. But is it possible to follow logic that just uses order axioms to prove it?

Just as a reminder, the Archimedean principle says that if I have reals a, b such that a > 0 and b> 0, then there exists some positive integer n such that an > b.

Thanks.

2. Sep 5, 2008

### morphism

Did you really mean for c to be a real number, as opposed to say a rational number? Because this is trivial, e.g. take c=(a+b)/2.

3. Sep 5, 2008

### quasar_4

oops, that's right. We want to worry about the rational case, not real...

I've been frustrated - the "text" we're using in our class hasn't introduced the Archimedean principle or the well-ordering principle with the completeness axiom. I guess you really don't need them for it, but it sure help to make proofs on the denseness of the rationals. We do have the fact that the set A containing a is bounded by B, so there must exist a sup(A). But I'm not sure how to get past that point, unless I put some kind of archimedean lemma into my proof.

4. Sep 5, 2008

### HallsofIvy

Staff Emeritus
I believe there exist non-Archimedean ordered fields in which it is not true that, given a< b, there exist c such that a< c< b. If that is correct, then, no, any proof must use the Archimedean property.