2
$\begingroup$

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 $=$?

$\endgroup$
4
  • 9
    $\begingroup$ = is a relation: en.wikipedia.org/wiki/Finitary_relation $\endgroup$ Commented Nov 6, 2014 at 19:01
  • 3
    $\begingroup$ = can be an operator, in some computer languages, that returns "True" or "False". In mathematics, however, it is a relation. $\endgroup$
    – vadim123
    Commented Nov 6, 2014 at 19:07
  • $\begingroup$ If you want to think of $=$ as an operator here, you have to then think of $5*5=25$ as an expression (the result of applying the $=$ operator). If you want to think of $5*5=25$ as a sentence that is either true or false, then consider $=$ as a relation. $\endgroup$
    – Steve Kass
    Commented Nov 6, 2014 at 20:21
  • 1
    $\begingroup$ They're all relations. A map or operator is a special relation, that is it must be coinjective and cosurjective. $\endgroup$ Commented Nov 6, 2014 at 20:44

3 Answers 3

5
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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!! $\endgroup$ Commented Nov 7, 2014 at 11:03
  • $\begingroup$ 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.) $\endgroup$ Commented Dec 1, 2014 at 9:39
1
$\begingroup$

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)

$\endgroup$
2
  • 2
    $\begingroup$ The operator is the Iverson bracket, not the equals sign. $\endgroup$
    – egreg
    Commented Nov 6, 2014 at 21:04
  • $\begingroup$ @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 $b$. $\endgroup$
    – Artem
    Commented Nov 6, 2014 at 21:48
-1
$\begingroup$

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.)

$\endgroup$
3
  • $\begingroup$ How is an operator a map? $\endgroup$ Commented Nov 6, 2014 at 20:54
  • $\begingroup$ Could I have someone verify this before I accept? $\endgroup$
    – warspyking
    Commented Nov 6, 2014 at 20:54
  • $\begingroup$ Operator and map are synonymous! Operator however is usually reserved for maps that additionally respect algebraic structure. $\endgroup$ Commented Nov 6, 2014 at 20:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .