Is $\le$ a binary operation? I am a little confused with $\le$. Is it a binary operation?
I know the definition of binary operation that is any function from $A×A \rightarrow A$ and $A$ non empty.
Kindly help me to understand the concept.
 A: A binary operation on a set $A$ is a function $A \times A \to A$.  You need to be able to plug any elements of $A$ into the function and get an element of $A$ back.  Addition on the naturals is a binary operation.  You can take any two naturals, add them, and get a natural as the result.  
A relation on $A$ is a set of ordered pairs that is a subset of $A \times A$.  It just tells you which pairs are related in this way.  $\le$ is a relation on the naturals.  For some pairs of naturals $(x,y), x \le y$.  For some, not.  Those pairs where $x \le y$ are members of the relation.  Another relation on the naturals is $x|y$, where $x$ is related to $y$ if $x$ divides $y$.
A: A binary operation is, as you said, a function from $A \times A \to A$. An example of this might be $+ : \mathbb{N} \times \mathbb{N} \to \mathbb{N}$. Binary operations are functions - they take in two inputs, and give you an output.
What you are considering is a binary relation, which is a subset $R \subseteq \mathbb{N} \times \mathbb{N}$. Given two elements $m,n \in \mathbb{N}$, we can ask if $(m,n) \in R$. Thus, relations give us a way of answering true/false questions. In your case, $\leq$ takes in two naturals (or objects in $A$), and says "yes" or "no".
So, in $\mathbb{N}$, we might write $\leq \subseteq \mathbb{N} \times \mathbb{N}$ where $(m,n) \in \leq$ if and only if $m$ is actually less than or equal to $n$.

I hope this helps ^_^
A: No.
It is a binary relation, since it specifies a property of $a$ in relation to $b$ when $a\le b$. This is often written as $\le\subseteq A\times A$, where $A$ is the domain of the relation.
