# Surreal Numbers and Normal Forms

On the way to figuring out how to do sign-expanded forms of infinite and infinitesimal numbers, we need to look at yet another way of writing surreals that have infinite or infinitesimal parts. This new notation is called the normal form of a surreal
number, and what it does is create a canonical notation that separates the parts of a number that fit into different commensurate classes.

What we’re trying to capture here is the idea that a number can have multiple parts that are separated by exponents of ω. For example, think of a number like (3ω+π): it’s not equal to 3ω; but there’s no real multiplier that you can apply to 3ω that captures the difference between the two.

Diving in to the meat of the subject: suppose we’ve got an arbitrary positive surreal number, N.

From the last post, we know that it’s part of a commensurate equivalence class C(x), where x is an ordinal, and that that class has a leader ωx, which is simplest (youngest) number in that class. So what we can do is write N in terms of its commensurate class leader. By the definition of the commensurate class and the nature of the real numbers, we know that there is some real-number multiplier r such that: ¬(∃z∈C(x) : | (ωx*r)-N < z)); that is, that ωx*r lies arbitrarily close to N in the commensurate class of N. So, we can almost write N as ωx*r.

The almost leads us to the catch. The restriction that we used above, that z is in the commensurate class C(x) of N is the problem. As I said in the beginning of the post, there are numbers that have parts that span multiple commensurate classes: things like 3+(1/ω). 1 (ω0) is the leader for the class of 3; and 3 is the closest multiplier, but 3*ω0≠3+(1/ω). So in fact, what we really need to write an arbitary N in this form is ωx*rx+Nx, where Nx is the simplest number in a commensurate class smaller than x.

Now, we have a new number, Nx. Suppose we repeat that process, to come up with a normal form for Nx. We wind up with N=ωx*rxx-1*rx-1+Nx-1. Now, suppose we keep repeating it until we get to someplace where Nx-α=0. Then we’ve got a full normal form for N. Of course, this being surreal-number land, there’s no guarantee that the normal form for N will terminate – that is, we can’t be sure that we’ll ever find an α for which Nx-α=0. But that’s OK: we have surreal numbers whose {L|R} form are infinite, and whose sign expansion forms are infinite; it’s OK if their normal forms are infinite too. But it does involve a trick or two.

But, given that I’ve had this much of this post written since 8:30 this morning, and it’s now almost 9:30pm, I think I’ll save the trick of infinite normal forms for tomorrow.