Is $3 \times 2 = 3+3$ or $2+2+2$?

Question

$3\times 2$ is $3+3$ or $2+2+2$. We know both are correct as multiplication is commutative for whole numbers. But which one is mathematically accurate?

Answer 1

In every ring and even in every semiring, $3 \times 2$ is by definition $(1+1+1)\times (1+1)$ . Since multiplication distributes over addition, you get on the one hand $$ (1+1+1) \times (1+1) = (1+1+1) + (1+1+1) = 3 +3 $$ and on the other hand $$ (1+1+1) \times (1+1) = (1+1) + (1+1) + (1+1) = 2 + 2 + 2 $$

Answer 2

I think this is an interesting and non-trivial question, but essentially impossible to answer. Let me explain why I think this.

The usual (I think?) way to define multiplication in the natural numbers is recursively: we define, for each $n \in \mathbb N$, a function $Mult_n : \mathbb N \to \mathbb N$ as follows:

• First we define the base of the recursion to be $Mult_n(1) = n$.
• Then for all $m$ we define $Mult_n(m + 1) = Mult_n(m) + n$.

(Alternatively, if one is using the convention that the smallest natural number is $0$, we can begin the recursion with $Mult_n(0) = 0$. It doesn't really matter.)

With this definition, $$Mult_3(2) = Mult_3(1) + 3 = 3 + 3$$

whereas $$Mult_2(3) = Mult_2(2) + 2 = (Mult_2(1) + 2) + 2 = (2 + 2) + 2$$

At this point you do one of the following things:

Option 1:

• We introduce the notational shorthand $m \times n = Mult_m(n)$
• We then prove the important (and not entirely obvious!) theorem that $Mult_n(m) = Mult_m(n)$ for any natural numbers $m, n$.

Option 2:

• We prove the important (and not entirely obvious!) theorem that $Mult_n(m) = Mult_m(n)$ for any natural numbers $m, n$.
• We introduce the notational shorthand that $m \times n$ means the common value of $Mult_m(n)$ and $Mult_n(m)$, which have been proven to be equal.

The crucial point here is that it does not matter which order we do it in. If we introduce the notational shorthand first, then $3 \times 2$ is by definition equal to $3 + 3$, and the fact that this is equivalent to $2 + 2 + 2$ is a consequence of the theorem.

On the other hand, if prove the theorem first, before introducing the notational shorthand, then $m \times n$ is by definition equal to both $3 + 3$ and $2 + 2 + 2$, which are equal by the theorem.

At the end of the day the difference between these two approaches has to do not with the content of the mathematics but with how you choose to narrate the exposition of the mathematics, which is not really a mathematical question but more of a stylistic choice.

Answer 3

The answer to this question depends on context. Strictly you need also addition to be associative even to know that $2+2+2$ is well-defined. Addition is usually taken to be a binary operation, and if you want to be very strict you should be expressing it as such.

Both answers are arithmetically correct. Any useful definition will have both of them (or equivalent statements) as consequences.

The foundational question is not whether these statements are correct or true, but whether there are natural axioms from which they flow as consequences. If we are intending to model arithmetic, we would normally say that the axioms were flawed if they did not imply both statements (or equivalents with particular notations for functions and brackets and the like).

As indicated above there are notational issues as well as foundational ones. In some definitions, because addition is binary, $2+2+2$ would not be counted as a well-formed expression. There are good, as well as bad, notational conventions. Actually defining the conventions (notations) by which apparently natural arithmetical expressions and equations are validly written is somewhat trickier than first appears. But when it is done, the informal expressions we use can generally be seen to be equivalent to valid expressions in appropriate formal systems.

Answer 4

It might be worth mentioning that in the context of ordinal arithmetic, where the order actually matters, the currently accepted convention is that $\alpha \cdot \beta$ means “repeat the operation of ‘counting to $\alpha$’ $\beta$ times”, so that $$\omega \cdot 2 = \omega + \omega$$ (“count $1,2,3,\dots$, and then do it again”) and $$2 \cdot \omega = 2 + 2 + 2 + \cdots = \omega$$ (“count $1,2,1,2,1,2,\dots$”).

Cantor himself originally used the opposite convention, but later switched (in Beiträge zur Begründung der transfiniten Mengenlehre, where at the end of §17 he writes “Die scheinbare Abweichung der Formeln [...] hängt nur mit der veränderten Schreibweise des Produktes zweier Zahlen zusammen, da wir nun den Multiplikandus links, den Multiplikator rechts setzen, [...]”; see this HSM.SE question.