# Completeness Axiom

#### Bleys

I'm having a little touble understanding application of the completeness axiom to certain subsets of real numbers. In a problem in a book (Fundamentals of Real Analysis by Haggarty), it asks you to show that the set S={a + b*sqrt(2) : a,b are rational} is not complete. As a hint, it tells you to show that sqrt(3) is not in the set. That's fine. What I don't understand is how it follows, from this last fact, that the set is not complete.
I know the Completeness axiom applies to bounded sets, so how can it be applied here. I tried following a similar approach to proving that Q is not complete, but I'm not sure if it's correct. It went something like this:
Take a subset of S such that a + b*sqrt(2) < sqrt(3), a,b rational. Then obviously the supremum (which is sqrt(3)) is not part of S and so it's not complete. Is this valid? It seems really trivial to me...
If anyone could enlighten me, that would be great.

#### Dragonfall

He means "complete" in the sense that every Cauchy sequence has a limit. So all you need to do is find a Cauchy sequence that converges to something outside S, say $$\sqrt{3}$$.

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#### Bleys

Well, completeness hasn't been defined using sequences yet in this part of the book, so I'd rather not use methods that aren't required. The only things given are the axioms for a complete ordered field and some properties of sets (Archimedean Postulate and density of Q).

#### Bleys

is it not enough to show that a subset of the set {a + b*sqrt(2) < sqrt(3) : a,b rational} does not have a least upper bound within the set?
So for example if you let a=0, the subset is {b*sqrt(2) < sqrt(3) : b rational} which is equivalent to {b < sqrt(3/2) : rational}, which does not have a rational least upper bound. Or is that not valid?

#### Dragonfall

That's enough, and valid.

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