Is $\mathbb{R}$ the only ordered field? 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.
 A: 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.
A: Any Archimedean ordered field is (isomorphic to) a subfield of $\mathbb R$, though.
A: 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 $\mathbb{R}$.
A: 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.
