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.
Best Answer
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
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 $\mathbb{R}$.