Complete Fields property

[Bachelier]

This is not a homework problem. I came across this in an analysis book:

In a complete ordered field (not specifically R) a member that is not zero is positive ⇔ this member is a square.

WHY?

How can we prove it?

Is this similar to Hilbert's 17th problem?

[Fredrik]

All complete ordered fields are isomorphic, so you can call any one of them "the" field of real numbers. What you're saying is that if x≠0, then x>0 and there exists a y such that y2=x. The first part of this is clearly not true for x=-1, so I assume that you meant to say that if x>0, then there exists a y such that y2=x, i.e. that every positive real number has a square root.

I don't immediately see how to prove this, and I have to go out for a while. But I might try to prove it later if no one else has posted the proof or a link to it by then.

[Bachelier]

that every positive real number has a square root.

I guess we can rephrase the question like that.

But it is a bi-condition. It says: a non-zero element in a complete field is positive ⇔ it is a square.

[mathwonk]

if complete means lub complete then all are isomorphic but not if it only means cauchy complete. then you have to say complete archimedean ordered field. (of course lub complete implies archimedean)

[Fredrik]

It says: a non-zero element in a complete field is positive ⇔ it is a square.
If this is to be interpreted as "x is non-zero and positive, if and only if x is a square", then the word "non-zero" adds nothing, since "positive" in this context should mean >0 which implies ≠0.

if complete means lub complete then all are isomorphic but not if it only means cauchy complete. then you have to say complete archimedean ordered field. (of course lub complete implies archimedean)
Thanks for explaining that. I've been wondering why some people throw the word "archimedean" in there.

[Bachelier]

If this is to be interpreted as "x is non-zero and positive, if and only if x is a square", then the word "non-zero" adds nothing, since "positive" in this context should mean >0 which implies ≠0.



True. I guess it helps in limiting the cases in the converse proof.

I still don't see where to start going from x is positive then there exist a y s.t. x= y2

[Fredrik]

I'm sure this is proved in lots of books, but I decided to leave my Rudin on my bookshelf and just think about it. It's clear that the axioms for ordered fields aren't enough, because ℚ is an ordered field that has an element that doesn't have a square root. So we have to use that supremum thingy.

Can't we just define y=sup{z|z2<x} when x>1, y=inf {z|z2>x} when x<1, and y=1 when x=1? (I haven't thought this through to the end).

Edit: The full proof is on page 10 of Rudin's "Principles of mathematical analysis", but you can probably find it any analysis book. He doesn't treat x>1 and x<1 separately. I'm not sure why I did. After defining y=sup{z|z>0,z2<x}, he proves that y2=x by showing that y2<x and y2>x both lead to contradictions. (D'oh, I didn't include the condition z>0 in my definition).

[Bachelier]

I'm sure this is proved in lots of books, but I decided to leave my Rudin on my bookshelf and just think about it. It's clear that the axioms for ordered fields aren't enough, because ℚ is an ordered field that has an element that doesn't have a square root. So we have to use that supremum thingy.

Can't we just define y=sup{z|z2<x} when x>1, y=inf {z|z2>x} when x<1, and y=1 when x=1? (I haven't thought this through to the end).

Edit: The full proof is on page 10 of Rudin's "Principles of mathematical analysis", but you can probably find it any analysis book. He doesn't treat x>1 and x<1 separately. I'm not sure why I did. After defining y=sup{z|z>0,z2<x}, he proves that y2=x by showing that y2<x and y2>x both lead to contradictions. (D'oh, I didn't include the condition z>0 in my definition).

Makes sense. Thanks for making that connection. I haven't digested well that part of the proof. time for a review.
I will type something later here to have a complete proof.

[Bachelier]

if complete means lub complete then all are isomorphic but not if it only means cauchy complete. then you have to say complete archimedean ordered field. (of course lub complete implies archimedean)

What would be the difference between Every Cauchy seq converges completeness and Least Upper Bound completeness. Is it because one is topological and the other isn't?

thx.