Are the operators of the four basic operations of arithmetic binary operators?

Question

1. Supposing I have a subtraction (multiplication and division) of 3 or more numbers, e.g., 20 - 3 - 2 - 3 - 1, will the subtraction end up being done in pairs, i.e., the - operator is binary and therefore only supports 2 operands at a time?

2. Would the multiplication (x) and division (/) operators also be binary operators, i.e., will they operate only on 2 operands at the same time?

For example:

9 x 3 x 2 x 2 x 4 -> Will the operation be performed with only 2 operands at a time?

100 / 2 / 5 / 2 -> Will the operation be performed with only 2 operands at a time? I am new to algebra.

My question is just to understand what happens "in detail" when we have more than 2 operands in these 3 basic arithmetic operations, I am new to algebra

Answer 1

Formally speaking for instance in the context of Peano Axioms, one begins with the natural numbers. In this case addition is a binary operation $+ : \mathbb{N} \times \mathbb{N} \to \mathbb{N}$ defined as repeated successor ($S : \mathbb{N} \to \mathbb{N}$). In particular one defines the function "add $n$ to an arbitrary $m$" recursively for all "$n$" and then a new function which simply takes in two inputs $n$ and $m$; i.e., effectively a binary operation.

Multiplication is defined recursively as repeated addition and thus is also binary.

Subtraction is usually defined for the integers and bigger number systems in terms of addition between natural numbers and is thus also binary.

Division is usually defined for even bigger numbers systems (although nothing prevents one from dividng in the integers as long as one knows that not all results will be integers themselves) in terms of multiplication and is thus also binary.

Having said this, it is a matter of conventions. I suppose one could come up with different definitions for these operations which are not necessarily binary. However, for most conventional practical purposes there seem to be no obvious advantage in doing so.

PS: Note, as already pointed out in the comments, that thanks to the commutativity and associativity of these operations, their binary nature is not particularly important.

Answer 2

In answer to the question in your title, yes, each of them is a binary operator. But most of the questions you pose in the body of your post are of the form “Will it be performed…?” And the answer to that question depends on context. In most contexts, the canonical order of operations is implied.

In any event, the reason they’re called binary operators is that they always (always) have exactly two operands. So consider the general form

$x_0 \star_1 x_1 \star_2 \cdots \star_n x_n$.

Syntactically, it is an alternating sequence of operators $\star_i$ interleaved with operands $x_j$. In principle, the $n$ operations could be evaluated in any of $n!$ orders. When any particular operation $\star_i$ is evaluated, its first operand is either $x_{i-1}$ or the value of a (previously evaluated) subexpression involving $x_{i-1}$. Likewise, $\star_i$’s other operand is either $x_i$ or the value of a subexpression involving it.

The evaluation always involves successively applying binary operations to pairs of operands (each of which may itself be the value of a nested subexpression).