Are all operations functions?

Question

I have looked at Wikipedia(I know it's not completely reliable) but on it an operation is formally defined as: "A function ω is a function of the form $ω : V → Y$, where $V ⊂ X_1 × … × X_k$." and I have also heard one of my professors mention in passing that an operation is a "special" kind of function.

Here's the thing, functions have only one output and therefore operations should too, however the indefinite integration operation has infinitely many outputs if the antiderivative exists. Can you please clear this up for me, perhaps the definition of an operation is wrong? And if so what is exactly an operation then? Thank you very much.

Answer 1

Usually, like in algebra, an operation is a function.

See :

• Ethan Bloch, Proofs and Fundamentals : A First Course in Abstract Mathematics (2nd ed - 2011), page 251 :

Definition 7.1.1. Let $A$ be a set. A binary operation on $A$ is a function $A \times A \to A$. A unary operation on $A$ is a function $A \to A$.

See also :

• Saunders Mac Lane & Garrett Birkhoff, Algebra, page 12.

In other fields of mathematics, like linear algebra and analysis, we have also Operators and Functionals.

Answer 2

In general, "operation" is an informal term, without a precise definition. Usually operations are functions, but as you point out with the example of antiderivatives, there are things that are sometimes referred to as operations which are not functions. If I had to give a generally applicable definition, I would say that an operation is a relation which is being thought of as if it were a function (i.e., something that takes an input and gives an output), even if it may not actually be a function (because it can actually give more than one output, or because it is not defined on some values that you might expect to be in its domain). In some contexts, though, an "operation" definitely does refer to a function. For instance, the term "binary operation" (or more generally, "$n$-ary operation") refers to a kind of function, as in Mauro Allegranza's answer.