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 ω^{x}, 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) : | (ω^{x}*r)-N < z)); that is, that ω^{x}*r lies arbitrarily close to N in the commensurate class of N. So, we can *almost* write N as ω^{x}*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 (ω^{0}) is the leader for the class of 3; and 3 is the closest multiplier, but 3*ω^{0}≠3+(1/ω). So in fact, what we really need to write an arbitary N in this form is ω^{x}*r_{x}+N_{x}, where N_{x} is *the simplest number* in a commensurate class smaller than x.

Now, we have a new number, N_{x}. Suppose we repeat that process, to come up with a normal form for N_{x}. We wind up with N=ω^{x}*r_{x}+ω^{x-1}*r_{x-1}+N_{x-1}. Now, suppose we keep repeating it until we get to someplace where N_{x-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 N_{x-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.