Reason for order of multiplication of ordinals

Question

I understand that addition and multiplication on ordinal numbers cannot be commutative and I know why $1+\omega$ and $2\times\omega$ must be different from $\omega+1$ and $\omega\times2$, respectively. For addition, I see why $1+\omega$ is just $\omega$, because it represents a position after $1$ and then $\aleph_0$ elements, and I can include the first element into the rest and the position wouldn't change.

What I am intrigued about is why $\omega+\omega$ is equal to $\omega\times2$ and not $2\times\omega$, as I would think.

How I interpret an expression like $5×2$ is "five-times two", like two seen five-times, as in $2+2+2+2+2$. In the context of ordinals, I would also imagine the expression $\omega+\omega$ (as I see omega two-times) to be $2\times\omega$. In contrast, I would interpret an expression like $\omega\times2$ to be $\underbrace{2+2+2+...}_\omega$, which also corresponds to a set of $\aleph_0$ elements and thus be equal to $\omega$. However, the usual definition is the exact opposite.

I understand that notation is not the underlying mathematics and one may freely redefine the operator in terms of changing the order of operands, but this definition seems to me arbitrary and inconsistent. Is there any particular reason why it was defined so?

Answer 1

Well, it is quite arbitrary. But one soft reason to define it this way is that it makes ordinal arithmetic left distributive, i.e. for all ordinals $\alpha, \beta, \gamma$

$$\alpha \cdot (\beta + \gamma) = \alpha \cdot \beta + \alpha \cdot \gamma$$

and allows for left cancellation $$ \alpha > 0 \wedge \alpha \cdot \beta = \alpha \cdot \gamma \implies \beta = \gamma. $$

Now, since historically these (and similar) laws have been stated predominantly for 'left variants' rather than their 'right' counterparts, it feels natural to me to define ordinal arithmetic the way it is.

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edit: I actually also see an argument for the other position. If we consider $\alpha \cdot \beta$ it's $\beta$ that behaves more like a scalar in that, for every $\alpha > 0$ $$\beta \mapsto \alpha \cdot \beta$$ is strictly increasing and, as above, $\alpha \cdot (\beta + \gamma) = \alpha \cdot \beta + \alpha \cdot \gamma$. Since modules usually have their scalars to the left (and I consider your 'beer example' to be a natural module), this might serve as an argument to define at least ordinal multiplication the other way.

Answer 2

Personally, I accepted this order of multiplications by noting that $\omega^{(\omega + 1)} = \omega^\omega \cdot \omega.$

More generally, $\alpha^{(\beta + \gamma)} = \alpha^\beta \cdot \alpha^\gamma$.