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 $=$?
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Sign up to join this communityI 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 $=$?
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.
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)
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.)