TITLE: Is a binary function a set of ordered pairs?
QUESTION [1 upvotes]: The definition of a function, according to the book I have, is: a function is a set of ordered pairs, no two of which have the same first component. Thus an ordered pair in this set can be expressed as $\pmb{(a,b)}$ or $\pmb{(a, f(a))}$.
However, a binary function, from what I've read, implies a ternary relation, expressed as $\pmb{(a, b, f(a,b))}$, which is not an ordered pair but a triplet. Thus does not satisfy the definition of a function? Shouldn't it be expressed as ($\pmb{(a,b)}$, $\pmb{f(a,b)}$)? Because only then will it become an ordered pair. Makes sense because of the definition. If we defined the relation of sets $\pmb{A}$, $\pmb{B}$ and, let's say, $\pmb{C}$ as the cartesian product, then we get $\pmb{A \times B \times C}$ which implies taking the product of a set of ordered pairs, $\pmb{A \times B}$, and a set, $\pmb{C}$, whose elements aren't ordered pairs, so the possible combinations will be of the form ($\pmb{(a,b)}$, $\pmb{f(a,b)}$) or ($\pmb{(a,b)}$, $\pmb{c}$) where $\pmb{a}$, $\pmb{b}$ and $\pmb{c}$ are elements of $\pmb{A}$, $\pmb{B}$ and $\pmb{C}$ respectively, satisfying the definition previously stated and not ($\pmb{a}$, $\pmb{b}$, $\pmb{c}$), or is it a mere notation meant to simplify ($\pmb{(a,b)}$, $\pmb{f(a,b)}$)?

REPLY [4 votes]: Yes.
To be properly formal a binary function $f: A \times B \to C$ is a collection of pairs $(x,y)$ with $x \in A \times B$ and $y \in C$ such that each $x$ is the first element of exactly one pair. We write $f(x)$ for the second element of the pair.
Since elements of $A \times B$ have the form $(a,b)$ the function is a collection of pairs $((a,b),y)$. Then to be really REALLY formal you should write the outputs as $f((a,b))$ rather than $f(a,b)$.
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