Is "=" an Operator? I know that $+$, $-$, $\times$, and $/$ are all operators. But is $=$ an operator?
For example, in the equation:

$5 \times 5 = 25$

I know $\times$ is an operator, but is $=$?
 A: In mathematical jargon, an operator is usually a function that takes some members of a set $S$ (most often, but not always two members of $S$) and yields another member of the same set $S$.  $+, -,$ and $ \times$ are examples.  
In contrast, the $=$ sign is not a function, and “$4=5$” cannot be evaluated to yield another number.  Instead, “$4=5$” is an assertion that $4$ and $5$ are equal.  The $=$ symbol denotes the relation of equality, and for each $a$ and $b$ one has either that $a=b$ is true or that it is false.  
In computer programming languages, there is a more unified approach. There is a special set of “boolean values”, which are considered true and false, and which are values, in the same way that numbers are values.  In this view,  $=$ (and $<, >,$ and the rest) is considered to be a function whose result is either a true or a false value.
(Some languages even dispense with the special true and false values, and define = to be a function which yields the number $1$ if its arguments are equal and the number $0$ otherwise.)  In a computer programming language, one can typically write something like 
(5 < 4) = (3 > 7)

which the computer will consider to yield a true value.  In the mathematical view, an expression like $(5<4)= (3>7)$ is at least puzzling, and probably meaningless, because mathematics does not usually construe relation symbols as representing functions.
A: The '$=$' sign can be an treated as an operator if Iverson brackets convention is used. For example, 
$$\sum_{i=0}^{n}[i=2] = 1$$ or
$$\sum_{i=0}^{n}[gcd(i,n)=2]$$
Using this notation, each time the condition inside the brackets is true, $1$ is returned.
(See this article: http://en.wikipedia.org/wiki/Iverson_bracket)
A: Yes, it is an operator!
Usually - and that is the only way if encountered so far - is that equality is defined as relation within a set.
(There is as well equality w.r.t. set theory but I guess that is a somewhat different story.)
So first extend they're all relations; unary, binary, ternary, finitary and so on.
Now, in order to call it an operator or more generally a map it must be cosurjective and coinjective:
$$\forall a\in A\exists a_0\in A:\quad a=a_0$$
$$\forall a\in A:\quad a=a'\land a=a''\implies a'=a''$$
(That is every element will be mapped somewhere and at most to one point.)
But that is the case for the equality relation. So it is a map!
(In fact, it is nothing but the identity map.)
