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A field, in mathematics, means a set $A$, which is an abelian group under an operation "$\ast$", $(A,\ast)$, which is further a commutative ring with an additional operation, $+$, defined on it (A,*,+), which when provides assurance that for every element $x\in A$, there exists an element $y$ such that $x+y= e$, where $e$ is the identity element for the operation $+$. Now I read the definition of an ordered field as follow

A field $F$, is said to be an ordered field if there exists two disjoint subsets $P$ and $-P$ (where $-P= \{-x\mid x\in P\,\,\,\}$ ) such that the union of $P$, $\{0\}$ and $-P$ is $F$, and an element $b$ of $F$ is said to be greater than element $a$ if $b-a$ belongs to $P$, less than if it $b-a$ belongs to $-P$ and equal if $b-a=0$.

Now given this definition with mentioning of "0", and "-", operation I am convinced to ask that "Do ordered fields always contain $\mathbb{R}$ or I can say is $\mathbb{R}$ the only ordered field? It will be better if anyone attaches a reference to his answer.

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$\mathbb{Q}$ is an ordered field, and so is every subfield of $\mathbb{R}$. –  Olivier Bégassat Sep 19 '11 at 18:55
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The rational numbers, $\mathbb{Q}$, are an ordered field that does not contain $\mathbb{R}$, and are not equal to $\mathbb{R}$. –  Zev Chonoles Sep 19 '11 at 18:56
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$\mathbb{Q}$ is an ordered field. So is the function field $\mathbb{Q}(x)$ under a variety of different orderings, such as the one where $\mathbb{x}$ is a positive infinite number. –  Hurkyl Sep 19 '11 at 18:56
    
I see we all wrote comments at the same time. –  jspecter Sep 19 '11 at 18:58
    
Also, I would guess that if $F$ is an ordered field, then $F(X)$ is too defining $P$ to be the set of fractions such that the denominator has leading coefficient $=1$ and the numerator has positive leading coefficient. –  Olivier Bégassat Sep 19 '11 at 18:59
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No, there are many other ordered fields. See the Wikipedia page on ordered fields; some examples of ordered fields that are not $\mathbb{R}$ are

  • The rationals, $\mathbb{Q}$
  • The real algebraic numbers, $\overline{\mathbb{Q}}\cap\mathbb{R}$
  • The rational functions over $\mathbb{Q}$, i.e. $\mathbb{Q}(x)$
  • The rational functions over $\mathbb{R}$, i.e. $\mathbb{R}(x)$

Note that $\mathbb{Q}\not\supseteq\mathbb{R}$, and $\mathbb{R}(x)\not\subseteq\mathbb{R}$, so ordered fields need not contain, or be contained in, $\mathbb{R}$. Furthermore, as Hurkyl points out below, $\mathbb{Q}(x)$ neither contains nor is contained in $\mathbb{R}$.

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@Zevchonoles- Dear Zev, Thanks a lot for your effort. But actually I intend to investigate can there be ordered fields (if not then fields) which are sets containing no real numbers. All the examples that you have provided contain real numbers. So could you provide any such example? –  Primeczar Sep 19 '11 at 19:06
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It might be worth mentioning that while there are many real fields. Any real closed field (a real field without any nontrivial algebraic real extensions) is elementarily equivalent to $\mathbb{R}.$ So if a first order statement in the language of ordered fields is true in any real closed field it is true in the real numbers and vise-versa. So in the case you're working with a real closed field and a statement in first order logic, you may as well assume you are working in $\mathbb{R}.$ –  jspecter Sep 19 '11 at 19:10
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@Primeczar: That depends what you mean by "contain real numbers". For example, if I take some set $S$ that is in bijection with $\mathbb{R}$, but such that $S\cap \mathbb{R}=\varnothing$, I can make $S$ into a field that is isomorphic to $\mathbb{R}$ by "copying" the field structure of $\mathbb{R}$ and putting it onto $S$, and so $S$ will be an ordered field, but it will not contain any real numbers. On the other hand, any ordered field must be of characteristic $0$, and therefore contain (an isomorphic copy of) the rational numbers $\mathbb{Q}$, which "are" real numbers. –  Zev Chonoles Sep 19 '11 at 19:10
    
@Zev: It might be worth mentioning $\mathbb{Q}(x)$ in your examples too, to provide one that is $\subseteq$-incomparable with $\mathbb{R}$. –  Hurkyl Sep 19 '11 at 19:15
    
@Hurkyl: Excellent point, I have edited my answer to include that. –  Zev Chonoles Sep 19 '11 at 19:18
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Any ordered field is characteristic 0 and contains an isomorphic copy of the rational numbers. Any field automorphism fixes this copy of the rationals. As a result, little is lost by saying that any ordered field contains the rational numbers, therefore contains "some" real numbers. So this:

"But actually I intend to investigate can there be ordered fields (if not then fields) which are sets containing no real numbers. All the examples that you have provided contain real numbers. So could you provide any such example?"

is not going to work, they all contain infinitely many "real numbers," although, as noted, perhaps not "all" real numbers.

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Any Archimedean ordered field is (isomorphic to) a subfield of $\mathbb R$, though.

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The standard spelling is Archimedean, ending with -ean, not with -ian, just like Euclidean. There's an obvious reason for that if you consider the etymology. –  Michael Hardy Sep 19 '11 at 23:53
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The axioms that you mention that define an ordered field are formulated in first order logic. Any set of first order axioms that have an infinite model actually has models of any infinite size. This is the upward version of the Loewenheim-Skolem theorem. So there are arbitrarily large ordered fields.

The Archimedean axiom cannot be formulated in first order logic. This axiom keeps the fields small, as mentioned above: Every Archimedean ordered field is isomorphic to a subfield of the reals.

On the other hand, if you add some completeness axiom (every bounded subset has a least upper bound), the field cannot be too small. Every complete Archimedean ordered field is isomorphic to the reals. The completeness axiom also cannot be formalized in first order logic.

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"Every" is an interesting word, though. If "every subset" means ones that can be defined by the theory -- the "real algebraic subsets", then you can formalize completeness and you get real closed fields. On the other hand if "every subset" means every set-theoretic subset of a model, then you get the real numbers of your set theory as you point out. –  Hurkyl Sep 19 '11 at 20:04
    
Of course I meant every nonempty subset, but that was clear, I suppose. And you are right, if every means real algebraic, then the theory even has countable models. –  Stefan Geschke Sep 19 '11 at 20:14
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