# II.5a: Arithmetic of the Surreal Numbers

Knowledge about the ordinals isn’t required for this post or the next one, but it would definitely add to your understanding. I’ll be talking about the Knuth-named “Surreal Numbers”, which for our purposes are just the Numbers (no “short” restriction) from the previous post. Conway denoted the proper class (bunch that’s too big to be a set) of surreal numbers “No“, but nowadays “$\mathbb{S}$” (for “Surreals”) is pretty popular, in analogy to ℝ for the reals. They’re the Surreals because they include the reals but go further. In fact, in some sense, the Surreals contain everything you might want to think of as a number (including the ordinals). Since the previous post contained the facts about Surreals that are relevant to CGT, and it would take pages and pages to prove all the basic assertions about the Surreals, let alone of the non-basic ones, this post will omit most of the proofs (which can generally be found in On Numbers and Games, the primary source for this post and the next).

Where unspecified, Gℓ will range over the left options of G, etc. Remember that Numbers can be added (inductively) via the formula G+H={Gℓ+H,G+Hℓ|Gr+H,G+Hr} and they can be subtracted by adding the negative: -G={ -Gr | -Gℓ }. It turns out we can do more with Numbers, though.

Numbers can be multiplied.
With all of these Numbers that act like dyadic fractions and integers, it would be nice if we had a procedure that let us multiply 2={1|}={{0|}|}={{{|}|}|} by 1/2={0|1}={{|}|{{|}|}} to get 1={{|}|}. To get a formula for G*H, we just need a bunch of things less than it for the left options, and a bunch of things more than it for the right options.

A common first guess (perhaps even of Conway’s?) is to define it recursively by G*H={G*ℓ*H,G*Hℓ|Gr*H,G*Hr}. However, a moment’s thought will show that this doesn’t actually make any sense for negative numbers, and another moment’s thought will show that this is literally the same definition as that for G+H! With a lot of thought, Conway came up with a janky definition that required G and H to be in some simplified form. With a lot more thought (months), Conway found the “right” definition.

The key idea is that we have four inequalities (coming from property 4. from the previous post and intuition about multiplication) (G-Gℓ)*(H-Hℓ)>0, (Gr-G)*(H-Hℓ)>0, etc. If we expand these out and manipulate, we get things like G*H>Gℓ*H+G*Hℓ-Gℓ*Hℓ, G*H<Gr*H+G*Hℓ-Gr*Hℓ, etc. This leads to the recursive definition G*H={Gℓ*H+G*Hℓ-Gℓ*Hℓ, Gr*H+G*Hr-Gr*Hr | Gℓ*H+G*Hr-G*ℓ*Hr, Gr*H+G*Hℓ-Gr*Hℓ}. All the properties that you expect of multiplication hold up to equivalence (the product of positive numbers is positive, 1 times anything is itself, multiplication distributes over addition, etc.).

Numbers can be divided.
This is a little surprising since if we imposed the finiteness condition, there’d be no Number corresponding to 1/3 (even though we have 1/2={0|1} and 1/4={0|1/2}). To find H/G, we really just need to multiply H by 1/G. To find 1/G, we really just need to be be able to find 1/G for positive G (we can take the negative afterwards if necessary). Even then, it’s tricky to write down a formula (and took a rather long while to find).

If G is positive, then we don’t need any of the negative left options (one of the left options has to be nonnegative to make G positive), so we’ll throw them out. Also, we’ll throw in 0 to the left options of G (G was positive so this is equal by the simplicity rule from the previous post). Then we can define H=1/G recursively by the following: H={0,(1+(Gr-G)*Hℓ)/Gr, (1+(Gℓ-G)*Hr)/Gℓ | (1+(Gℓ-G)*Hℓ)/Gℓ, (1+(Gr-G)*Hr)/Gr}. Now this looks problematic because it has options of H in the formula, but you can build the left and right sets with induction. Start with 0 as an option of H and use all of G’s options. Then go through all of the options of H you just got, etc. (Technical note: This transfinite induction doesn’t need more steps than the cardinality of the set of G’s options to the ℵ0.)

If you don’t want to use a crazy double-inductive definition, Kruskal solved the H part of the recursion: you can have the options (both left and right) of 1/G be (1-Π(1-G/Gi))/G where Gi ranges over the positive options of G, the products are always finite (the empty product of no terms is 1) and the dividing by G in the formula isn’t actually a problem since there’s a G in the expansion of the numerator that can cancel. That expression is a left option of 1/G exactly when an even number of the Gi are left options of G.

In technical language, the Surreal Numbers are an ordered `FIELD` (they’re too big to be a field), and they contain an isomorphic copy of every ordered field.

You can take square roots (and beyond)!
There’s a vaguely similar doubly-recursive formula to the one for 1/G found by Clive Bach for the square root of a nonnegative Number. And a similar singly-recursive version, they’re both kinda complicated. To take other roots and do other things, there’s lots of complex machinery (you can get a different definition of square root using infinite series stuff, for instance).

Similarity to arithmetic for On2.
I talked about the “ordinal Nimbers” On2 in a previous post. As far as I know, these multiplication, division, and square root formulas were found for the Surreals first, and then they just tried the same formulas on the impartial games and they all magically worked. (There’s a lot of simplification that comes from the fact that G=-G for impartial G.) I admit I don’t know of a good reason why the same definitions should work for such a different class of positions; they break down horribly for general positions.

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