**Defining an Order**

Now that we have verified the table of possibilities for adding games (at the end of II.1), and understand *P*-positions as “0”, we can note a couple things: An *L*-position plus an *L*-position is another *L*-position. The negative of an *L*-position is an *R*-position (so not an *L*-position), and vice versa. This makes the *L*-positions kind of like positive numbers: When you add two positives you get a positive, and the negative of a positive is never positive. For this reason, we say that a Conway position is “positive” if it’s an *L*-position, and “negative” if it’s an *R*-position. Why not the other way around? There are good reasons for both choices, but this is the one that stuck (a later post will have a better reason for this one).

With the concept of positive in hand, we can go on to define > and ≥, etc. G>H when G-H is an *L*-position.G≥H when G>H or G=H (equivalently, when Left has a winning strategy when *Right* goes first in G-H.) Under these definitions, all of the regular inequalities with integers like “-2>-3” hold for the Conway position versions of them. Also, most of the basic facts about inequalities hold: G>H implies -G<-H, etc. Basically, G>H means something like “G is better for Left than H no matter who goes first”.

**The order isn’t ideal**

However, be warned: there are two things that “go wrong” compared to common inequalities you’re used to. Firstly, not every Conway position is positive (*L*-position), negative (*R*-position) or 0 (*P*-positions), so not every pair of positions is comparable. For instance, since * is an *N*-position, “*>0”, “*=0”, and “*<0" are all false! We write "*‖0" and say "star is incomparable with zero" or "star is fuzzy (with respect to zero)" or "star is confused with zero". Slightly confusingly for some programmers, the notation "*<>0″ is used sometimes for this “confused” relationship. The other main thing that goes wrong is that there are games like * and {1+*|-1} which aren’t nonnegative to begin with, but become nonnegative (or even positive, in the latter case) when you add them to themself: *+*=0, {1+*|-1}+ {1+*|-1}>0. In technical language, these Conway positions form a “perforated partially ordered abelian `GROUP`

“.

**More Confusion**

To become a little more comfortable with this “confused” idea, let’s look more carefully at some familiar positions. *‖0, but how does * compare with 1? Well, in *-1, if Left goes first they move to -1, and then Right moves to 0 and wins (eventually, if Left happens to have moves in the particular representation of 0, like {-1|1}=0). If Right goes first then they move to -1 and win (eventually). Therefore *-1 is an *R*-position so *<1. Since -*=*, we have -1<* for free. So somehow, * occupies a rough position between -1 and 1, but it's confused with 0.

With ±1={1|-1}, there's even more confusion. If Left goes first in (±1)-1, they move to 1-1=0, and win. If Right goes first, they move to -1-1=-2, and win. Therefore, ±1-1‖0, so ±1‖1 (similarly, ±1‖-1). However, let's see how ±1 compares to 2: If Left goes first in ±1-2, their only move (up to equivalence) is to 1-2=-1, and they lose. If Right goes first, they move to -1-2=-3 and win. Either way Right wins, so ±1-2). You might visualize this as ±1’s location on a number line being unclear: like a cloud over the interval [-1,1], but definitely between -2 and 2. Analogously, ±n‖m for -n≤m≤n but ±n<m for m>n and ±n>m for m<-n.