I'm trying to recall the term for an operator/function where f(a, b) = f(b, a). For example, a + b = b + a for all values of a and b. However a - b != b - a unless a = b.

4 Answers 4


I think the word you are looking for is commutative:

(mathematics, of a binary operation) Such that the order in which the operands are taken does not affect their image under the operation.

Addition on the real numbers is commutative because for any real numbers s , t, it is true that s + t = t + s.

Addition and multiplication are commutative operations but subtraction and division are not.

  • 13
    Just a comment. Commutative is largely isolated to describe binary operators that usually take an infix notation---such as addition and multiplication. "Binary operator" is itself a jargon term meaning that the function sends two elements from some set and spits out a new thing in the same set. If you're dealing with a function that takes two things and spits out something not necessarily from that same set, mathematicians would preferably say "symmetric". Symmetric also extends to functions with more than two arguments. If f(a,b,c)=f(b,c,a)=f(c,b,a)=..., we'd call f symmetric.
    – user70564
    Nov 1, 2018 at 22:04
  • Yes - Peter's answer addresses that (and was originally a comment similar to yours). Nov 2, 2018 at 8:26
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    This answer is incorrect, and should not be upvoted or accepted. @RobertWolfe has explained it. The correct term is symmetric. Commutative is an acceptable term in very specific cases, and f(a,b) in the OP's post is not one of them.
    – Myridium
    Nov 5, 2018 at 1:45

You seem to be asking for the word for operations and functions. The word mathematicians generally use is different in the two cases.

A function which does not depend on the order of the arguments is usually called a symmetric function. So f(a,b,c) = f(b,a,c) = f(b,c,a), although the word commutative is occasionally used.

And as the other answers say, an operation which does not depend on the order of the operands is called a commutative operation. So a·b=b·a.

  • 2
    Good point about the distinct terminology for functions vs. operations. On some level, functions and operations are equivalent (ask any Lisp programmer), but by convention they are usually treated differently. You never hear someone actually say "that's a commutative function".
    – John Y
    Nov 1, 2018 at 21:47
  • 5
    This is a better answer than the currently-accepted one. Nov 3, 2018 at 0:46

Commutative Operation

Any operation ⊕ for which a⊕b = b⊕a for all values of a and b. Addition and multiplication are both commutative. Subtraction, division, and composition of functions are not. For example, 5 + 6 = 6 + 5 but 5 – 6 ≠ 6 – 5.


As others have noted, the common term for this is "commutative". For a brief etymology explaining why the word for this is "commutative", see https://math.stackexchange.com/questions/3045924/why-is-it-called-commutative-property

You might also hear "Abelian", named after the mathematician Abel, who studied mathematical structures called "groups" with commutative operators in the 19th century.

FYI there's a hilarious joke about that which will help you remember this fact:

What's purple and commutes? An Abelian grape.

See https://en.wikipedia.org/wiki/Abelian_group for more details.

  • 1
    I've never heard of "Abelian operators". Abelian groups are those whose group operation is commutative. Nov 3, 2018 at 0:45
  • @R abelian is abstract algebra's term for commutative. I think the only time you will ever hear it used would be only in the context of abstract algebra. Nov 4, 2018 at 12:42
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    @MikeR it is true in abstract algebra that the adjective abelian is generally applied only to the word "group", not to an operator.
    – ajd
    Nov 5, 2018 at 23:22
  • All the comments above are correct and I do not believe I said anything to contradict them. I said that the original poster may hear the word "Abelian" used in a context where "commutative" might be expected, which is true, and why that is the case, which is of historical interest. Plus, a hilarious joke! Nov 5, 2018 at 23:47

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