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This is a very basic / fundamental question about arithmetic.

Multiplication of natural numbers can be thought of as repeated addition, which amounts to "copying" a given number of times. For example, $2 \times 3$ can thought of as the number $3$ taken $2$ times, that is, $3 + 3$. But it can also be viewed as the number $2$ taken $3$ times, that is $2 + 2 + 2$. Visually, we can look at these two ways of multiplying as

$$\begin{array}{ccc} \bigcirc & \bigcirc & \bigcirc\\ \bigcirc & \bigcirc & \bigcirc \end{array} \quad\quad \textrm{or} \quad \quad \begin{array}{cc} \bigcirc & \bigcirc \\ \bigcirc & \bigcirc \\ \bigcirc & \bigcirc \end{array}$$

Of course, the result is $6$ in both cases, but it seems to me that fundamentally, this is not exactly the same operation. When considering the "structure" of the repeated addition, multiplication appears to be asymmetric, although it is commutative when considering only the integer result. (Actually, I think commutativity becomes a meaningful property only if we first consider this asymmetry.)

Given this asymmetry, is there any convention on the order of arguments for integer multiplication? That is, should we write $2 \times 3 = 2 + 2 + 2$, or is $2 \times 3 = 3 + 3$ preferable?

Roland
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  • I prefer to write $2\times3=6$ :-) – Angina Seng May 14 '17 at 18:49
  • They're the same precisely because we don't care about the "rectangular structure" this builds up, but instead the count of the things making up the rectangle. These counts will always coincide. – Mark Schultz-Wu May 14 '17 at 18:55
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    For integers it doesn't matter. But for multiplication of ordinal numbers (which isn't commutative), the convention is that $\omega \cdot 2 = \omega + \omega$ (i.e., it means “count to $\omega$ twice”, which is different from “count to two $\omega$ times” = “count to $\omega$”). Personally, I find that this feels a bit backwards, since for integers I tend to think of $a \cdot b$ as “count to $b$, $a$ times), but it's pretty well established nowaday (although I seem to remember reading somewhere that Cantor's original definition was the other way around). – Hans Lundmark May 14 '17 at 19:53
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    Also, see here: https://matheducators.stackexchange.com/questions/9842/repeated-addition-standard-notation/ – Hans Lundmark May 14 '17 at 19:56
  • @HansLundmark, thanks for the link, that page contained a lot of useful info! It seems from the answers on that page that multiplication has indeed been considered asymmetrical in this sense since Euclid, and commutativity is then proved, not stated axiomatically. You might want to write up an answer based on you comment and I can accept it. – Roland May 14 '17 at 21:24
  • For both $2 + 2 + 2$ and $3 + 3$, it is a decomposition of $6$. One of them first group elements in pairs and then group the pairs. The other one first group elements in triplets and then group the triplets. If one look at the process backwards in time, they transform to each other. The two ways of interpreting multiplication are related by a time reversal symmetry. – achille hui May 15 '17 at 07:12

2 Answers2

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There is a well-established convention for multiplication of ordinal numbers in set theory (where it matters): $\alpha \cdot \beta$ means “count to $\alpha$, $\beta$ times”, or in other words, “$\alpha + \dots + \alpha$”. This follows from taking the definition to be inductive in the factor on the right: $$ \alpha \cdot (\beta+1)=\alpha\cdot \beta+\alpha . $$ But for integers (where it doesn't matter), opinions differ. Of course, while developing the theory, one has to choose one way or the other, until the commutativity has been proved.

See also this question at the Mathematics Educators StackExchange.

Hans Lundmark
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One can interpret this as a demonstration that multiplication is commutative: $a \times b = b \times a$, since one produces the horizontal orientation of the rectangle, the other the vertical.

(This relies on choosing a specification for which of $2+2+2$ and $3+3$ you mean by $2 \times 3$: which choice we make is unimportant, but only because the operation does turn out to be commutative (in particular, this is false for other types of object).)

Chappers
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  • Thanks, I understand this, My question was if there is any convention on choosing that "specification". (It looks like there isn't.) – Roland May 15 '17 at 05:21