So, today we’re going to play a bit more with nimbers – in particular, we’re
\ngoing to take the basic nimbers and operations over nimbers that we defined last time, and
\ntake a look at their formal properties. This can lead to some simpler definitions, and
\nit can make clear some of the stranger properties that nimbers have.<\/p>\n
<\/p>\n
The first thing we did with nimbers was define nimber addition. The way that we did it was iterative: take a set of existing nimbers that exist at ordinal stage N, and use a simple procedure to try to create as many nimbers as we could by adding pairs of stage-N nimbers. Then we ran out, of new numbers we could create, we’d move to the next ordinal by figuring out the smallest<\/em> number that hadn’t yet been included into the nimbers – and then we’d repeat the generation of the stage N+1 nimbers by addition within the set.<\/p>\n So – being formal, and distinct from that process, we’ll repeat the definition of nimber addition:<\/p>\n α+β = the least ordinal distinct from all α’+β and α+β’ What odd properties does addition have in nimbers? Well, the most obvious odd Remember that nimbers are related to the game of nim. In nim, you’ve got a set of piles of stones. In each move, a player can remove any number of stones from one<\/em> pile. The first<\/em> player who can’t pick up any stones because there are none left loses. Each nimber is, essentially, a model of a game with a certain sized pile of stones. So given a single nimber N, it’s a model for game of nim with a single pile of N stones. The addition expression using nimbers N amd M, N+M is a model of a game of Nim with two<\/em> piles: one of size N, and one of size M. Doing the addition is basically asking what’s the nimber that models a single pile<\/b> game that has exactly the same winners (assuming perfect players) as the two-pile game N+M?<\/em><\/p>\n So suppose you’ve got a game of nim with two piles of size N – the game N+N. If the player who moves second plays well, they’re guaranteed<\/em> to win. Why? because all they need to do is mimic<\/em> the first player. If the player one takes X stones from pile 1, then player two takes X stones from pile 2. If they just continue to mimic player one using the opposite pile, then they’re guaranteed to win – because eventually player one will have to take the last stone from one pile; and then player two will take the last stone from the other pile, which will leave player one with no move: player two wins.<\/p>\n So it’s the same outcome as a gave that started with no<\/em> stones at all: player one can’t move, so player 2 wins.<\/p>\n To get to the next interesting bit, we need to toss in another definition. In the definition of addition, we used α’ to refer to nimbers smaller than α. We need another similar<\/em> notion. We’ve define the number building process, and the meaning of addition in terms of the minimum excluded ordinal<\/em> or mex<\/em>. Similarly to the way that we define α’ in terms of addition, we can define α*<\/sup> in terms of pure mex: if we’ve got a set of nimbers, S, and we know that the nimber α is the minimum excluded ordinal of S – that is, α=mex(S), then we’ll define α*<\/sup> as a variable that ranges over the members of S. So α*<\/sup> can represent any nimber in a set which has α as its mex. The difference between α*<\/sup> and α’ is that α’ is by definition less than α – but α*<\/sup> can be larger<\/em> than α – it just needs to be part of an ordinal set that excludes<\/em> α <\/p>\n Ok. So – addition obviously needs to satisfy the field axioms, because the whole point of nimbers is that they’re a new kind of number field. So we know that α+β=α+γ if and only if β=γ – that’s just standard field stuff. But also<\/em>, in the nimbers, this means that α+β = mex(α*<\/sup>+β ∪ α+β*<\/sup>). This is a major break from our intuitive idea of addition. The α’ based definition is nice, because it This is getting long, so I think I’ll stop here. Next time, we’ll look in more detail at nimber multiplication.<\/p>\n","protected":false},"excerpt":{"rendered":" So, today we’re going to play a bit more with nimbers – in particular, we’re going to take the basic nimbers and operations over nimbers that we defined last time, and take a look at their formal properties. 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\nwhere α'<α and β'<β.<\/p>\n
\nproperty we’ve already seen: for any nimber N, N+N=0. That’s pretty strange. Why does that make sense?<\/p>\n
\nbuilds on the intuitive idea that you can add two big things by breaking them into pieces, and then adding up the pieces. But now, we’re saying that we can define the addition of two small numbers by somehow mashing together to big<\/em> numbers. It’s a strange idea – but once you’ve absorbed the notion that for two numbers α and β both larger than zero, α+β is not<\/em> necessarily larger than α then it should make sense. And once you have that new definition of addition, then some more field axioms – like associativity – just fall right out.<\/p>\n