Surcomplex Transfinite Maths

Question

So just curious about the surcomplex numbers... Do they have transfinites like the surreal numbers? For example:

ω*i = ?

Would that even make sense?

Thanks!

Answer 1

Surcomplex numbers are defined to be numbers of the form $a + bi$ for $a,b$ surreals. In particular, $b$ is allowed to be $0$, so the surcomplex numbers contain the surreals inside them. That means, since there are transfinite surreals, there are also transfinite surcomplex numbers. It also means - since the arithmetic operations are well-defined everywhere on the surcomplex numbers - that you can do any arithmetic you like with them. For example, $\omega \cdot i$ is just $0 + \omega i$ (or, written more simply, $\omega i$). I realize that's not very satisfying, but it's exactly what you get when you multiply $2 \cdot i$: you just get $2i$.

You can do more: say you want to know what $(\omega + 3i)(\omega - i)$ is. Using elementary algebra, $(\omega + 3i)(\omega - i) = \omega\cdot\omega + 3i\omega - i\omega -3i^2$. As usual, $i^2 = -1$, and $\omega\cdot\omega$ can most easily be written $\omega^2$. Simplifying, $(\omega + 3i)(\omega - i) = (\omega^2 + 3) + 2\omega i$.