Topic: Surreal Numbers, and my slightly nervous connection with them.

[Quantum Burrito]

Actually, I might as well tell you the formula for multiplication, so you can check your answers:

ab = {A_Lb+aB_L−A_LB_L, A_Rb+aB_R−A_RB_R | A_Lb+aB_R−A_LB_R, A_Rb+aB_L−A_RB_L}

(where A_L represents the left set of a, etc.)

[stuck]

Well, yeah, if you do anything but science and maths this is rather trivial. But Quantum Mechanics has been taking a lot from pure maths, specifically linear algebra and abstract algebra, and knowledge of quantum mechanics without that is extremely shallow.

I suppose we can show 1/3 like so. *This is a preliminary idea, so it is probably not very rigorous, though I think I've done a good job* Similar to the way you defined ε, we define a series greater than n, N_G, such that its tail tends toward n. Note, however, that n is not in the series, it just gets infinitesimally  close. Then we construct another series, N_L, less than n such that its tail becomes infinitesimally close. Again, n is not in N_L.

Using these two series, we construct the surreal number {N_L| N_G}. Since n is between N_L and N_G, the surreal number must be n.

Still, there must be a "cleaner" way to define the N series. Since we know all numbers of the form 1/2^k*a can be generated by QB's previous construction of ε, there are two numbers of this set that must be infinitesimally above and below n, meaning the surreal number is n.

I'll work on that multiplication.

[Quantum Burrito]

I'l like to drop in something else I have found (well, reconstructed from drips and drabs on the internet):

Surcomplex Numbers

The Surcomplex numbers are a proper superset (i.e., they contain) the Complex numbers. Also, 'surcomplex' is an awesome word.
Anyway, as you may or may not know, the complexes are formally defined as an ordered pair of reals, (r,i).

These ordered pairs are defined in set theory as {{r},{r,i}} (as are all ordered pairs). However, the arithmetic operations on them must be defined.

Addition:
(a,b)+(c,d)=(a+c,b+d)

Multiplication:
(a,b)*(c,d)=(ac-bd,bc+ad)

The inverses can be derived from these equations. As can exponentiation, etc.

Now, although the Surreals on their own cannot construct the square root of -1 (try it from the definition of multiplication I gave, you'll end up needing a set that can multiply with the empty set without becoming the empty set, which is impossible), neither can the reals. Bit using the same ordered pair methodology, but with Surreal numbers, a set of Surcomplex numbers can be defined:

(0,1)*(0,1)=(0*0-1*1,1*0+0*1)=(-1,0)

Any ordered pair with 0 as the right projection is a Surreal number, so (0,1) times itself in the Surcomplexes is -1. Just as in the Complexes, and so we shall label this number i.

Best thing about this, is we now have complex infinites and infinitesimals.

(ε,0)*(0,ω)=(ε*0-0*ω,0*0+εω)=(0,1)=i
(ε,4ω)*(1,2)=(ε-8ω,4ω+2ε)=(-8ω+ε,4ω+2ε)=(-8ω+ε)+(4ω+2ε)i
(0,ω)*(ω,0)=(0*ω-ω*0,ω*ω+0*0)=(0,ω²)=ω²i

And, Seeing as the complexes remain unchanged as a subset of the Surcomplexes, e^πi+1=0. So, now all we need to do is find an identity that includes ω and ε as well!