Foreword:
I have read R.H. Bruck's A Survey of binary systems, where the notion of halfoperation is given. A halfoperation $\ast$ differs from a (binary) operation since $a\ast b$ may not be defined for all ordered pairs $(a,b)$. For example, from the Cayley table
$$\begin{array}{c|cc}\ast & 0 & 1\\ \hline 0 & 0 & 1\\1 & & 0\end{array}$$
we deduce that $0\ast 1=1$ while $1\ast 0$ is not defined ($(0,1)$ belongs to the range of the halfoperation, $(1,0)$ does not).
Let us now consider two well known operations over $\mathbb{R}^3$:
- vector product: ${\mathbb{R}^3\times\mathbb{R}^3\to\mathbb{R}^3}\qquad{(a,b)\mapsto a\wedge b}$,
- scalar product: ${\mathbb{R}^3\times\mathbb{R}^3\to\mathbb{R}}\qquad{(a,b)\mapsto a\cdot b}$.
In italian (my native language) we call both of them binary operation:
- operazione binaria interna one like vector product (interna: "inner" or "internal"),
- operazione binaria esterna one like scalar product (esterna: "outer" or "external").
Question:
Does such distinction exist in english? I found no traces of it in literature and, reading papers, forums and SE, I have experienced that this matter may sometimes give raise to ambiguities or misunderstanding.
How should we properly call an operation like subtraction $-$ over $\mathbb{N}$, since $a-b$ only belongs to $\mathbb{N}$ if $a\ge b$? Halfoperation over $\mathbb{N}$?
How should we properly call an operation like scalar product $\cdot$ over $\mathbb{R}^3$, since $a\cdot b$ exists for all $(a,b)\in\mathbb{R}^3\times\mathbb{R}^3$ but never belongs to $\mathbb{R}^3$? Is this not a binary operation while vector product is?
And, above all:
In the end the word binary, alone, means that the operation is a function of two variables (its arity is $2$), but does not deal with closure...
...Right?
