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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 $\mathbb{R}$.2011-09-19
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    The rational numbers, $\mathbb{Q}$, are an ordered field that does not contain $\mathbb{R}$, and are not equal to $\mathbb{R}$.2011-09-19
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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.2011-09-19
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    I see we all wrote comments at the same time.2011-09-19
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    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.2011-09-19
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    There are ordered fields which extend $\mathbb R$ in the sense of allowing infinitesimals. You can have such field extending $\mathbb Q$ without including $\mathbb R$ as well.2011-09-19
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    @Primeczar: Your definition of field is incorrect; the field is not an abelian group under $\ast$ (multiplication), because $0$ has no inverse. If $F$ is a field, it *is* true that $F-\{0\}$ is an abelian group under multiplication, though.2011-09-19
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    There are a number of equivalent properties you can add to the condition of being an ordered field to uniquely characterize $\mathbb R$; one of these is the least upper bound property, and another is Archimedean + Cauchy sequences converge.2011-09-19

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