# Talk:Ordered field

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Hello, I don't quite understand two things: How does a*a >= 0 follow from the two axioms? Why can you follow from this, that the complex numbers can't be ordered? After all, there should be a total order in the first place.--80.139.83.37 20:08, 22 March 2006 (UTC)

Off-hand, I think a proof might go something like this:

First, I want a corollary that says that ${\displaystyle -1^{2}=1}$. By definition, 1 is the multiplicative identity and -1 is the additive inverse of 1. So 1 + (-1) = 0 and for all x, 1x = x. Also, by definition, -x = (-1)x which is the additive inverse of x (ie x + (-x) = 0). Therefore let x be given and consider ${\displaystyle y=-1^{2}x}$. ${\displaystyle y=-1^{2}x=(-1)(-1)x=(-1)(-x)}$, which is the additive inverse of -x, meaning the y+(-x) = 0. But by definition, the additive inverse of x is -x and vice versa, so therefore ${\displaystyle y=x=(-1^{2})x}$ and thus ${\displaystyle -1^{2}}$ is the multiplicative identity for all x, which is 1. Therefore ${\displaystyle -1^{2}=1}$

From that it follows that if multiplication is cummatative (see below) and ab = ba that for all x, ${\displaystyle (-x)^{2}=(-x)(-x)=(-1)(x)(-1)(x)=(-1)(-1)(x)(x)=(-1^{2})(x^{2})=1(x^{2})=x^{2}}$. So ${\displaystyle -x^{2}=x^{2}}$.

Thus there are two cases:

1. If 0 ≤ x, then by the second axiom ${\displaystyle 0<=xx=x^{2}}$.
2. If x ≤ 0, then it follows that x ≤ 0 ≤ -x. So ${\displaystyle 0<=(-x)^{2}=x^{2}}$

Therefore for all x, ${\displaystyle 0<=x^{2}}$

Note that it's very possible I made a mistake or faulty assumption above, so if you spot a problem with my proof above please feel free to correct it. Dugwiki 18:16, 13 November 2006 (UTC)

Please: Don't write ${\displaystyle -1^{2}}$ when you mean ${\displaystyle (-1)^{2}}$. In standard notation, those mean two different things. The first is −1; the second is +1. (Also, please don't write ${\displaystyle <=\,}$ when you mean ${\displaystyle \leq \,}$. Michael Hardy 20:28, 13 November 2006 (UTC)

## Dubious

Any idea what the axiom of continuity is? — Arthur Rubin | (talk) 19:48, 22 August 2007 (UTC)

You don't know what Dedekind's axiom of continuity is? Think of any condition that an ordered field has to satisfy to be the real field. That should be equivalent to the axiom of continuity.218.133.184.93 23:39, 22 August 2007 (UTC)

Circular argument.
Dedekind's axiom of continuity is that which an ordered field has to satisfy to be the real field.
An ordered field is the reals if and only if satisfies the axiom of continuity.
If you want to include it, you need to create an article for it. A Google search for "axiom of continuity" seems to show multiple meanings, some most of which seem to be equivalent to being Archimedian. That's not adequate. — Arthur Rubin | (talk) 00:14, 23 August 2007 (UTC)

As I said, think of a condition for an ordered field to be real field. You know that axiom of Archimedes is not enough.218.133.184.93 01:19, 23 August 2007 (UTC)

Sources define the axiom of continuity as being the Archimedean property. Choose another term for the "axiom" (wrong, in context). — Arthur Rubin | (talk) 01:21, 23 August 2007 (UTC)

I have a question to you, Arthur. What is the necessary and sufficient condition for the ordered field to be the real field?218.133.184.93 01:26, 24 August 2007 (UTC)

It's not up to me to find one. If you can specify such a condition, and can find a source that it is such a condition, go ahead, and add it to the article. Until then, out it goes. — Arthur Rubin | (talk) 17:35, 24 August 2007 (UTC)

If you don't know it, then you'd better learn it. It's so elementary math.218.133.184.93 02:02, 25 August 2007 (UTC)

There probably is an elementary (actually, pseudo-elementary) description of the reals. However, it's not called the axiom of continuity. I've created the proper redirect for the axiom of continuity. By the way, removing the {{dubious}} term is considered vandalism. Do you want to be blocked again? — Arthur Rubin | (talk) 04:01, 25 August 2007 (UTC)
Since WAREL is having trouble expressing himself, here are three, of different levels of sophistication:

## RCF

The following conditions are equivalent when R is an ordered field.

1. R is a real closed field.
2. R is a maximal ordered field.
3. R is a maximal real field.

That doesn't seem right, although the latter two conditions aren't defined. Any idea what was intended? — Arthur Rubin | (talk) 00:36, 23 August 2007 (UTC)

• A real closed field cannot be algebraically extended and remain orderable. This is the meaning of "closed"; in this context, "real" = "analogous to a subfield of the reals" = "orderable". Please don't copy this in; it can be made more precise. Septentrionalis PMAnderson 23:24, 25 August 2007 (UTC)
I marked the equivalence as dubious, and WAREL deleted it. I would have been perfectly happy if it had remained while discussion was occurring. — Arthur Rubin | (talk) 14:16, 26 August 2007 (UTC)
I meant "don't copy in what I said"; I have no problem with WAREL's three-fold exquivalence, although I'm not sure that the "maximal" terms are common enough to be worth it. Septentrionalis PMAnderson 17:38, 26 August 2007 (UTC)

## Counter examples

I think it would be interesting to list some of examples that *aren't* ordered fields, this might make it clearer why even needing a definition of ordered field is useful. Oblivioid (talk) 22:38, 1 December 2008 (UTC)

There are already a number of non-examples in the Which fields can be ordered? section. --Zundark (talk) 22:47, 1 December 2008 (UTC)

## the subset F

What is meant by "the subset F" in Def 2?--agr (talk) 13:12, 21 April 2011 (UTC)

## Ramanujan–Nagell equation

Is there a relation between the fact that Q2 has a square root of -7 and the Ramanujan–Nagell equation? GeoffreyT2000 (talk) 02:17, 19 May 2015 (UTC)