How do you multiply this

Question

How can you multiply these ordinal numbers: $(\omega+1)(\omega+1)(\omega2+2)$

I tried and have gotten to this: $(\omega^2+1)(\omega2+2)$ Is that the correct way, or did i made a mistake?

Answer 1

Ordinal multiplication is distributive in the rightmost element. Therefore

$$(\omega+1)(\omega+1)=(\omega+1)\omega+(\omega+1)=\omega^2+\omega+1$$

Therefore:

\begin{align*} (\omega+1)(\omega+1)(\omega\cdot2+2) &=(\omega+1)(\omega+1)\omega\cdot2+(\omega+1)(\omega+1)2 \\ &=(\omega+1)(\omega+1)\omega+(\omega+1)(\omega+1)\omega+(\omega+1)(\omega+1)2 \\ &= (\omega^2+\omega+1)\omega+(\omega^2+\omega+1)\omega+(\omega^2+\omega+1)+(\omega^2+\omega+1) \\ &= \omega^3+\omega^3+\omega^2+\omega+1+\omega^2+\omega+1 \\ &=\omega^3 \cdot 2+ \omega^2 \cdot 2 + \omega + 1 \end{align*}

Answer 2

Your result is correct. Have you tried to compute the Cantor normal form? If I am not mistaken, it should be $\omega^3 2 + \omega^2 2 + 1$.

edit: my bad, I made the same mistake ($(\omega+1)^2 = \omega^2 +1$) as you. As written above, the correct answer is $\omega^3 2 + \omega^2 2 + \omega + 1$.