How come $\omega^3=2$ for infinite nimber $\omega$?

Question

The Details:

I don't know much about nimbers. It is my understanding that their multiplication is different than ordinary numbers. I'm not sure how to multiply infinite nimbers together; it's given recursively as

$$ab = \text{mex}\{a'b+ab'+a'b'\mid a'<a, b'<b\},$$

where $\text{mex}$ is the minimum excluded element and $+$ is nimber addition. I don't have any experience with $\text{mex}$.

In the section Nimbers and the Game of Nim of Conway & Guy's, "The Book of Numbers", it is stated that

$$\omega^3=2.$$

The Question:

How come $\omega^3=2$ for the infinite nimber $\omega$ with respect to nimber multiplication?

Thoughts:

My guess is that

$$\begin{align} \omega\omega^2&=\text{mex}\{a'\omega^2+\omega b'+a'b'\mid a'<\omega, b'<\omega^2\}\\ &\stackrel{?}{=}\text{mex}\{a'\omega^2+\omega b'+a'b'\mid a', b'\text{ finite}\}\tag{1}\\ &\approx\text{mex}\{ c\mid c<2\}\tag{2}\\ &=2, \end{align}$$

where $(1)$ is because $\omega$ is infinite and $\omega^2$ is also infinite (I presume!).

But $(2)$ baffles me.

Background:

Despite a cursory reading of some of Conway's, "On Numbers and Games" back in 2019 and this question about $\ast$, I know next to nothing about combinatorial game theory. Perhaps I'm trying to run before I can walk . . .

Please help :)

Answer 1

I'm maybe a little out of my experience, so please take things with a bit of grain of salt, but despite any sketchiness I hope this helps a bit.

We should establish that the nimber product $\omega \cdot \omega$ gives the ordinal $\omega^2$, but I'll skip that for the moment.

Then in considering the nimber product $\omega \cdot \omega^2$, your $(1)$ is not correct; in general $b'$ can range over any ordinal less than $\omega^2$ including ordinals of the form $c\omega+d$ with $c$ and $d$ finite.

If we are trying to prove $\omega^3$ is finite, we only care about cases when $a'\omega^2 + \omega b' + a'b'$ is finite. When $a'$ and $b'$ are both finite and non-zero, this term will be infinite. To get something finite, we need to first cancel the $\omega^2$ which means (taking for granted the niceness of nimber algebra) $c=a'$ as $a'\omega^2 + (a'\omega+d)\omega + a'(a'\omega + d) = a'\omega^2 + a'\omega^2 + d\omega + a'^2\omega + a'd = (d+a'^2)\omega + a'd$. Then we must cancel the $\omega$ which means $d=a'^2$, so our term is simply $a'^3$; in other words we can only get finite nimbers that are perfect nimber-cubes. So, our set contains $0$ (from $a'=b'=0$) and $1$ (from $a'=1$, $b'=\omega+1$). It would remain to show that $2$ is not the cube of a finite nimber (though, of course it is apparently the cube of the infinite nimber $\omega$). (I'm not going to show this, sorry).

To get comfortable, it may help to explore some simpler cases of infinite nimber arithmetic.

With finite $n$, the nimber sum $\omega + n$ is the ordinal $\omega + n$. Any finite nimber is in the set $\{a+n|a<\omega\}$. Infinite ordinals $\omega + x$ with $x<n$ simply come from the $\{\omega+b|b<n\}$ (inductively).

With finite $n>0$, the nimber product $n\cdot \omega$ is the ordinal $n\omega$: to get $c\omega+d$ with $c<n$, we take $a'=c$ and $b'=d/(n+c)$.

In the nimber product $\omega \cdot \omega$, we can get $\{a'\omega + \omega b' + a'b'\} = \{(a'+b')\omega + a'b'\}$. To prove the result is infinite, we must have all finite ordinals in this set. We get finite ordinals when $a'=b'$, so this is true only if every finite nimber is the square of a finite nimber (which, apparently, is true). More generally, we would get $c\omega+d$ if $a'+b'=c$ and $a'b'=d$, for which I suppose it is relevant to relate this to writing $x^2+cx+d=(x+a')(x+b')$ (ie if every finite quadratic polynomial can be factored with finite roots then $\omega \cdot \omega = \omega^2$). (Also, not going to show this.)

Answer 2

The hidden question is why we need an infinite nimber to define a cube root of $2$ in the first place. We use the field theory of finite nimbers to show that these cannot include a cube root of $2$.

As readers of this thread probably know, the finite nimbers are made up of an infinite sequence of subfields, each having $0$ as the minimum element and $M(k)=2^{2^k}-1$ as the maximum element, with $k$ a nonnegative whole number. Subfields for smaller $k$ are embedded in those with larger $k$ like Matryoshka dolls. For any value of $k$, a nonzero nimber $N$ within the subfield will always satisfy $N^{M(k)}\overset{nim}{=}1$.

Now suppose that a finite nimber $N$ satisfies $N^3\overset{nim}{=}2$. Let $k$ be the smallest $k$ for which $M(k)\ge N$, thus identifying the minimal subfield containing $N$.

Cubing the proposed equality gives $N^9\overset{nim}{=}1$. But for $k=0$ the only nimbers, and thus the only cubes there of, are $0$ and $1$; and for $k\ge1$ we also have $N^{M(k)}\overset{nim}{=}1$ where $M(k)$ is divisible by $3$ without being divisible by $9$! So, by using linear combinations of exponents, we prove that $N^3\overset{nim}{=}1$, contradicting the hypothesis. We must go outside the finite-nimber Matryoshka doll, to infinite nimbers, to satisfy $N^3\overset{nim}{=}2$.

We can extend the argument. Since $2\overset {nim}{=}3×3$ the noncubic character of $2$ over the finite nimbers implies a similar noncubic character for $3$. Then, using the closure properties of the subfields, we can prove that every nimber from $0$ up to $M(k)$ is $\in\{1,2,3\}$ times a cube in the same subfield, and then only $0$ and nimbers with $1$ as the multiplier can ever be finite-nimber cubes. For instance, when all the nimbers from $0$ through $M(2)=15$ are cubed the set of distinct results is $\{0,1,8,10,13,14\}$; all other nimbers in this subfield are $2$ or $3$ times one of the nonzero elements of this set and thus never can be additional cubes of finite nimbers.