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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 $=$?

Artem
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warspyking
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    = is a relation: http://en.wikipedia.org/wiki/Finitary_relation – Qiaochu Yuan Nov 06 '14 at 19:01
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    = can be an operator, in some computer languages, that returns "True" or "False". In mathematics, however, it is a relation. – vadim123 Nov 06 '14 at 19:07
  • If you want to think of $=$ as an operator here, you have to then think of $55=25$ as an expression (the result of applying the $=$ operator). If you want to think of $55=25$ as a sentence that is either true or false, then consider $=$ as a relation. – Steve Kass Nov 06 '14 at 20:21
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    They're all relations. A map or operator is a special relation, that is it must be coinjective and cosurjective. – C-star-W-star Nov 06 '14 at 20:44

3 Answers3

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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.

MJD
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  • What you refer to is that every relation can be sort of lifted up to another relation. This way you can turn eeeverything into a function. But that is not what is desired!! – C-star-W-star Nov 07 '14 at 11:03
  • Besides, there's a precise definition of what an operator is not only mathematical jargon - that is the equal sign is an operator!! (I'm not bothering as I wrote an answer, too, but as this one is still being here.) – C-star-W-star Dec 01 '14 at 09:39
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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)

Artem
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    The operator is the Iverson bracket, not the equals sign. – egreg Nov 06 '14 at 21:04
  • @egreg Why are Iverson brackets an operator? Isn't the equal sign a binary operator that operates on two values, $a$ and $b$? Inside the brackets we have a statement that is either true or false based on input $a$ and $b$. – Artem Nov 06 '14 at 21:48
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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.)

C-star-W-star
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