I'm looking through some notes on the surreal numbers, and have noticed that Dyadic rationals can be constructed in finitely many number of days, since they are products of integers with some power of $1/2$. I want to try prove this but I am not sure how to go about this. Any suggestions on how to start this, or any useful resources?


1 Answer 1


$\newcommand{surr}[2]{\left\{ #1 \mathbin{\big\vert} #2 \right\}}$If you know about multiplication of surreal numbers, this is easy. Let $D$ be the set of surreal numbers constructed in finitely many days, or equivalently surreal numbers represented by games with finitely many positions. It is then clear from the recursive definitions that $D$ is closed under addition, multiplication, and negation. So to show $D$ contains all dyadic rationals, it suffices to show that $1/2\in D$, which is easy because we can just write down the game $\surr{0}{1}$ and check that $\surr{0}{1}+\surr{0}{1}=1$.

Alternatively, here is a proof using just the structure of the surreals as an ordered abelian group. Since $D$ is closed under addition and negation, it suffices to show that $1/2^n\in D$ for all $n$.

To show that, the idea is that $1/2=\surr{0}{1}$, $1/4=\surr{0}{1/2}$, $1/8=\surr{0}{1/4}$, and so on. To prove this, recursively define $x_0=1$ and $x_{n+1}=\surr{0}{x_n}$. You can then prove by induction that $2x_{n+1}=x_n$ for all $n$, from which it follows that $x_n=1/2^n$ for all $n$.

The details of how to carry out this induction are hidden below:

Before starting the induction, note that by definition we have $0<x_{n+1}<x_n$ for all $n$ (and these inequalities guarantee that $x_n$ is indeed a number for all $n$). The base case $n=0$ is a simple computation (or alternatively, it is essentially the same as the induction step below; the case of the argument which uses the induction hypothesis can be skipped since $x_{n-1}$ does not exist).

Now suppose that $n>0$ and $2x_n=x_{n-1}$; we will must prove $2x_{n+1}=x_n$. To do so, we show that the second player can always win the game $$x_{n+1}+x_{n+1}-x_n=\surr{0}{x_n}+\surr{0}{x_n}+\surr{-x_{n-1}}{0}.$$ Suppose Left goes first. If they move in either copy of $x_{n+1}$ they reach the position $x_{n+1}-x_n<0$ and so Right wins. If Left instead moves in $-x_n$ then Right can move in one copy of $x_{n+1}$ to reach the position $x_{n+1}+x_n-x_{n-1}$. By the induction hypothesis, $x_{n+1}+x_n-x_{n-1}=x_{n+1}-x_n<0$ so again Right wins.

Now suppose Right goes first. If they move in either copy of $x_{n+1}$ they reach the position $x_{n+1}+x_n-x_n=x_{n+1}>0$ so Left wins. If they instead move in $-x_n$ they reach the position $x_{n+1}+x_{n+1}>0$ and again Left wins.

The converse is also true: every element of $D$ is a dyadic rational, so $D$ is exactly the set of dyadic rationals. The converse is harder to prove; see Prove that a surreal number is born in a finite stage if and only if it is of the form $\frac m{2^n}$. for one approach. More generally, you can find proofs of all the basic properties of surreal numbers in Conway's book On Numbers and Games.


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