# Is the definition of ordinal exponentiation for successor ordinals commutative?

The definition of ordinal exponentiation for successor ordinals is:

$\alpha^{\beta+1} = \alpha^\beta \cdot \alpha$.

In general (for all ordinals $\beta$, also limit ordinals), is it true that

$\alpha \cdot \alpha^\beta = \alpha^\beta \cdot \alpha$?

And if so, what is the best way to prove it?

• FYI, here's a summary of some basic ordinal arithmetic results. – Dave L. Renfro Mar 16 '17 at 19:44
• This version of my post on basic ordinal arithmetic results is better, because the original spacing that I used (to add clarity) is used. – Dave L. Renfro Mar 16 '17 at 19:59

No.

$$\omega\cdot\omega^\omega = \omega^{1+\omega} = \omega^\omega < \omega^{\omega + 1} = \omega^\omega\cdot\omega$$

• Thanks. What if $\beta$ is a successor ordinal? – user386617 Mar 16 '17 at 20:03
• Take $\beta=\omega+1$, or any infinite ordinal really. – Asaf Karagila Mar 16 '17 at 20:11

No -- $\alpha\cdot \beta$ is the order type of $\beta$ many copies of $\alpha$ after each other.

This means that $2\cdot\omega$ is just $\omega$: an $\omega$-sequence of elements that come two at a time is the same as an $\omega$-sequence of single elements.

$$2\cdot \omega = 2+2+2+\cdots = (1+1)+(1+1)+\cdots = 1+1+1+1+\cdots = \omega$$

On the other hand $\omega\cdot 2$ is an entirely new ordinal, the order type of two copies of the natural numbers after each other.

$$\omega\cdot 2 = (1+1+1+\cdots) + (1+1+1+\cdots)$$