X-Git-Url: http://git.subgeniuskitty.com/surreal-numbers/.git/blobdiff_plain/8484c2fd2ad8d6a4bcc64f16142185ad71ffbd5f..46905e173537fd1c9e8590beafd42b2dbf79612c:/notes/chapter-1.tex diff --git a/notes/chapter-1.tex b/notes/chapter-1.tex index 265fd80..78834af 100644 --- a/notes/chapter-1.tex +++ b/notes/chapter-1.tex @@ -1,4 +1,357 @@ \newpage \section{Notes: Chapter 1} -Hello from chapter 1. +\subsection{Review} + +In the first chapter, Knuth provides an algorithm for generating a set of +symbols, a `little old ladies at the country club' rule for selecting a subset +of those symbols, and a binary relation on that selected subset. + +\begin{axiom} \label{ax:symbol-generation} +Every symbol is comprised of a left and right set of other symbols. The initial +symbol is named $0$ and is comprised of the empty set for both its left and +right sets. All further symbols are generated by repeatedly selecting +(potentially empty) subsets of the set of all existing symbols, using these +selections as left and right sets to define a new symbol, and then adding that +symbol to the set of all existing symbols. +\end{axiom} + +From this set of all symbols, growing with each iteration, we select the subset +of symbols which satisfy the next axiom, calling them \emph{numbers}. This +subset forms our \emph{universe}, also denoted $\mathbb{U}$. + +\begin{axiom} \label{ax:number-definition} +Every number corresponds to two sets of previously created numbers, such that +no member of the left set is greater than or equal to any member of the right +set. +\end{axiom} + +Since this selection rule implies a binary relation, Knuth provides it. + +\begin{axiom} \label{ax:leq-relation} +One number is less than or equal to another number if and only if no member of +the first number's left set is greater than or equal to the second number, and +no member of the second number's right set is less than or equal to the first +number. +\end{axiom} + +As his final trick, Knuth applies the algorithm of Axiom +\autoref{ax:symbol-generation} to the hand-crafted initial input ($0 \equiv +\surreal{}{}$), retains only those symbols which are valid numbers per Axiom +\autoref{ax:number-definition}, and applies Axiom \autoref{ax:leq-relation} to +establish the following relation on the newly expanded universe. + +$$-1 \equiv \surreal{}{0} \leq 0 \equiv \surreal{}{} \leq 1 \equiv \surreal{0}{}$$ + + +\subsection{Exploration} + +I attempted to create a program which applies these axioms, generating numbers +and organizing them by relation. Along the way I realized that I was assuming +at least a total order and maybe a well order. If those assumptions are +incorrect, the program's data types and visualization algorithms change +significantly. Consequently, I began poking my new universe to see what type of +order Axiom \autoref{ax:leq-relation} actually defines. + +\begin{framed} + \noindent Note that the following definitions and theorems only consider + the case of finite left and right sets. +\end{framed} + +\begin{defi} \label{defi:number-in-number} +Given two numbers $x$ and $y = \surreal{Y_L}{Y_R}$, we say that $x$ is +\emph{in} $y$ iff $x \in Y_L \cup Y_R$. We also write this as $x \in y$. +\end{defi} + +\begin{defi} \label{defi:ancestor} +Given two numbers $x$ and $y$, we say that $x$ is an \emph{ancestor} of $y$ iff +there exists a relationship $x = a_1 \in ... \in a_n = y$ with finite $n$. +\end{defi} + +Although I think I can make everything work while only using the notion of +`ancestor', generating numbers in a stream with no other age-based +distinguishing attributes, I include the following definition to match Knuth's +notion of generating new numbers in a day-by-day or power-set like fashion. + +\begin{defi} \label{defi:generation} +A \emph{generation} shall refer to the numbers created by applying Axiom +\autoref{ax:number-definition} to all possible combinations of extant numbers. +Generations are numbered sequentially such that generation-0 consists of the +number $0$, generation-1 consists of the numbers $-1$ and $1$, etc. +\end{defi} + +\begin{defi} \label{defi:generation-function} +For a surreal number $x$, the \emph{generation function} $g\colon \mathbb{U} +\rightarrow\mathbb{N}$ returns the generation in which $x$ was created. For +example, $g(0) = 0$ and $g(-1) = 1$. +\end{defi} + +\begin{defi} \label{defi:younger} +Given two sets of surreal numbers $X$ and $Y$ with equal cardinality, we say +that $X$ is \emph{younger} than $Y$ iff, $\forall x_i \in X$ and $\forall y_j +\in Y$ it holds that $\sum g(x_i) < \sum g(y_j)$. Similarly, we say that $Y$ is +\emph{older} than $X$. +\end{defi} + +With these definitions in place, let's start pursuing an order. + +\begin{theorem} \label{thm:transitive} +Axiom \autoref{ax:leq-relation} is transitive. That is, if $x \leq y$ and $y +\leq z$, then $x \leq z$. +\end{theorem} + +\begin{proof} +Seeking a contradiction, let $x = \surreal{X_L}{X_R}$, $y = +\surreal{Y_L}{Y_R}$, and $z = \surreal{Z_L}{Z_R}$ be the youngest set of +numbers satisfying $x \leq y$, $y \leq z$ and $x \nleq z$. Note by inspection +that generation-0 and generation-1 obey transitivity. + +By Axiom \autoref{ax:leq-relation}, this creates two possible cases. There +could exist an $x_L \in X_L$ such that $x_L \geq z$ or there could exist a $z_R +\in Z_R$ such that $z_R \leq x$. + +In either case, we end up with a set of three numbers which do not obey +transitivity, namely \set{y, z, x_L} or \set{z_R, x, y}. But, since $x_L +\in X_L$ and $z_R \in Z_R$, we know that $g(x_L) < g(x)$ and $g(z_R) < g(z)$. +In other words, if we consider the sum $g(x) + g(y) + g(z)$, it must be greater +than both the sum $g(y) + g(z) + g(x_L)$ and the sum $g(z_R) + g(x) + g(y)$. + +Since this means a younger set of numbers exist for which transitivity fails, +we have reached the desired contradiction. +\end{proof} + +\begin{theorem} \label{thm:reflexive} +Axiom \autoref{ax:leq-relation} is reflexive. That is, $x \leq x$. +\end{theorem} + +\begin{proof} +Let $x$ be the youngest number satisfying $x \nleq x$. Note that $0 \leq 0$ by +inspection. + +By Axiom \autoref{ax:leq-relation}, this creates two possible cases. Either +$\exists x_L \in X_L$ such that $x \leq x_L$ or $\exists x_R \in X_R$ such that +$x_R \leq x$. + +In the first case, by Axiom \autoref{ax:leq-relation}, we find that $X_L \ngeq +x_L$, implying that $x_L \ngeq x_L$, a contradiction since $x_L \in X_L$ from +the start. We arrive at a similar contradiction for the second case. +\end{proof} + +Per \autoref{thm:transitive} and \autoref{thm:reflexive}, we find that Axiom +\autoref{ax:leq-relation} is at least a preorder on our universe. Normally we +would move on to proving that Axiom \autoref{ax:leq-relation} is antisymmetric, +however, there is a problem. Working by hand with Axioms +\autoref{ax:symbol-generation} and \autoref{ax:number-definition}, generation-2 +contains the following numbers. + +$$\surreal{-1}{}, \surreal{1}{}, \surreal{-1,0}{}, \surreal{0,1}{},$$ +$$\surreal{-1,1}{}, \surreal{-1,0,1}{}, \surreal{-1}{1}, \surreal{0}{1},$$ +$$\surreal{-1,0}{1}, \surreal{}{1}, \surreal{-1}{0}, \surreal{}{-1},$$ +$$\surreal{-1}{0,1}, \surreal{}{0,1}, \surreal{}{-1,0}, \surreal{}{-1,1},$$ +$$\surreal{}{-1,0,1}$$ + +By application of Axiom \autoref{ax:leq-relation}, note that $\surreal{}{} \leq +\surreal{-1}{1}$ and also $\surreal{-1}{1} \leq \surreal{}{}$, yet clearly +$\surreal{}{} \neq \surreal{-1}{1}$. It turns out that Axiom +\autoref{ax:leq-relation} fails to define even a partial order. + +For now, let's simply define antisymmetry into existence, give it a special +name and symbol so we know where we've strayed from Knuth, and move on with +creating our program. Worst case, it'll turn out that we're generating +something embedded in a larger space, like the linearly-ordered $\mathbb{R}$ +embedded in the not-linearly-ordered $\mathbb{C}$. + +\begin{defi} \label{defi:similar} +Two numbers $x$ and $y$ are \emph{similar} iff, per Axiom +\autoref{ax:leq-relation}, $x \leq y$ and $y \leq x$. This is denoted by $x +\similar y$. +\end{defi} + +Applying this new definition to the twenty numbers from generations 0-2, we +find that they start to collapse into equivalency classes like the +aforementioned $\surreal{}{} \similar \surreal{-1}{1}$. More on this later. + +Returning to our investigation of order, now that we have `proven' +antisymmetry, our universe has a partial order. The only thing required to +upgrade to a total order is proof that every number is comparable, but first a +lemma. + +\begin{theorem} \label{thm:weak-self-relation} +For $x = \surreal{X_L}{X_R}$, it holds that $X_L \leq x \leq X_R$. +\end{theorem} + +\begin{proof} +Seeking a contradiction, assume $\exists x_L \in X_L$ such that $x_L \nleq x$. + +By Axiom \autoref{ax:leq-relation}, this creates two possible cases. Either +$\exists (x_L)_L \in (X_L)_L$ such that $(x_L)_L \geq x$ or $\exists x_R \in +X_R$ such that $x_R \leq x_L$. By Axiom \autoref{ax:number-definition}, the +latter case is impossible. + +We address the remaining case by inductively noting that $(x_L)_L \leq x_L$, +thus $x \leq (x_L)_L$. However, since this means that $x \leq (x_L)_L \leq +x_L$, by \autoref{thm:transitive}, $x \leq x_L$. By definition, this means $X_L +\nleq x_L$, a contradiction since $x_L \in X_L$. + +The proof of $x \leq X_R$ follows similarly. +\end{proof} + +\begin{theorem} \label{thm:all-comparable} +For every $x$ and $y$ in the universe, if $y \nleq x$, then $x \leq y$. +\end{theorem} + +\begin{proof} +Seeking a contradiction, let $x$ and $y$ be numbers such that $y \nleq x$ and +$x \nleq y$. + +By Axiom \autoref{ax:leq-relation}, this creates two possible cases. Either +$\exists x_L \in X_L$ such that $x_L \geq y$ or $\exists y_R \in Y_R$ such that +$x \geq y_R$. In the first case, by \autoref{thm:weak-self-relation}, $x_L \leq +x$, so by \autoref{thm:transitive}, $y \leq x$, a contradiction. A similar +contradiction is reached in the other case. +\end{proof} + +Well, that's it. We have a total order, but we're no longer in the same +universe as Knuth. We can even go one step further and note that, by only +considering a finite number of generations, we are guaranteed to only generate +a finite quantity of numbers, and a finite totally ordered set is automatically +well ordered. + +With a total order in hand, we can strengthen $x \nleq y$ into $x > y$, +motivating the following theorem. + +\begin{theorem} \label{thm:strong-self-relation} +For any number $x = \surreal{X_L}{X_R}$, it holds that $X_L < x < X_R$. +\end{theorem} + +\begin{proof} +Simply recognize why `linear order' is a synonym for `total order'. +\end{proof} + +Moving away from considering order, note that creating Definition +\autoref{defi:similar} imposed order on our universe at the cost of some +uniqueness. Let's examine that loss of uniqueness a bit further. + +The twenty numbers from generations 0-2 break down into seven similarity-based +equivalence classes, as shown below. + +$$\surreal{}{-1} \similar \surreal{}{-1,0} \similar \surreal{}{-1,1} \similar \surreal{}{-1,0,1}$$ +$$\surreal{}{0} \similar \surreal{}{0,1}$$ +$$\surreal{-1}{0} \similar \surreal{-1}{0,1} \similar \surreal{-1}{}$$ +$$\surreal{}{} \similar \surreal{-1}{1}$$ +$$\surreal{0}{1} \similar \surreal{-1,0}{1}$$ +$$\surreal{0}{} \similar \surreal{-1,0}{} \similar \surreal{}{1}$$ +$$\surreal{1}{} \similar \surreal{0,1}{} \similar \surreal{-1,1}{} \similar \surreal{-1,0,1}{}$$ + +Thus, whereas Knuth's universe has 20 numbers by generation-2, our universe +only has seven numbers. For convenience when writing our program which creates +a (finite set based) surreal numberline, we desire a unique representation for +each equivalence class. Toward that end, we introduce the following +definition and theorem. + +\begin{theorem} \label{thm:reduced-form} +Given an arbitrary number $x$, there exists a number $x' = +\surreal{X'_L}{X'_R}$ which is similar to $x$ and has the property that the +cardinality of both $X'_L$ and $X'_R$ are one or zero. +\end{theorem} + +\begin{proof} +Let $x = \surreal{X_L}{X_R}$. Since a finite, total ordered set contains a +maximum and minimum element, it follows that we can define $X'_L$ as containing +only the maximum element of $X_L$ and define $X'_R$ as containing only the +minimum element of $X_R$. If either $X_L$ or $X_R$ are the empty set, the +corresponding $X'_L$ or $X'_R$ are also the empty set. + +Use these two newly defined sets to construct $x' = \surreal{X'_L}{X'_R}$. This +clearly satisfies Axiom \autoref{ax:number-definition} by inheritance. Then, +to show that $x \similar x'$, we apply Axiom \autoref{ax:leq-relation} in both +directions. + +In the forward direction, $x \leq x'$ requires that, $\forall x_L \in X_L$, it +must hold that $x_L \ngeq x'$. Indeed, from \autoref{thm:strong-self-relation}, +\autoref{thm:transitive}, and our explicit construction of $X'_L$ as the +maximum element of $X_L$, the desired relation holds since $X_L \leq X'_L < +x'$. + +Also in the forward direction, $x \leq x'$ requires that, $\forall x'_R \in +X'_R$, it holds that $x'_R \nleq x$. By the same theorems/construction as +before, the desired relation holds since $x < X_R \leq X'_R$. + +The reverse direction follows via a similar pair of arguments. +\end{proof} + +\begin{defi} \label{defi:reduced-form} +Given an arbitrary number $x$, the $x'$ guaranteed to exist by +\autoref{thm:reduced-form} is termed the \emph{reduced form} of $x$. +\end{defi} + +Note that reduced forms are not unique since $\surreal{}{} \similar +\surreal{-1}{1}$ but $\surreal{}{} \neq \surreal{-1}{1}$. + +Note also that uniqueness is not trivially achieved by applying the generation +function to an equivalence class. Although all elements are comparable, the set +is finite, and the natural relation inherits the reflexive and transitive +properties from $\mathbb{N}$, the relation is not antisymmetric since +$\surreal{-1}{0} \neq \surreal{-1}{}$ and yet both are from generation-2. Thus, +stealing well ordering from $\mathbb{N}$ and using it to define a unique +element to represent the equivalence class turns out to be harder than I had +hoped. + +I suspect the first time a reduced form number from a given equivalence class +is generated, it will be alone. I base that on viewing the growing set as being +comprised of two types of numbers, first a primary set forming a sparse line +bookended by the empty set, making the set longer with every generation. + +$$\set{}$$ +$$\set{} < 0 < \set{}$$ +$$\set{} < -1 < 0 < 1 < \set{}$$ + +Within this set a secondary set is forming, sitting directly in the middle +between every existing number, making the set denser with every generation. +Although harder to whip up in \LaTeX, I visualize this like a house of cards +viewed end-on, a honeycomb lattice linking two parents to each child, adding a +full spatial dimension to the honeycomb with each generation. + +If this visualization is true, then uniqueness of the reduced form representing +an equivalence class in the first generation said class is encountered, follows +naturally, even though this result doesn't hold for any following generations. + +Regardless, for the purpose of writing a program that generates numbers and +orders them, the truth of this conjecture is irrelevant. We can simply identify +the first number encountered in an equivalence class as the canonical +representation of that class. The naturally serial nature of the program will +serialize creation of numbers, adding a finer grained age relation and making +it trivial to choose a `first' element to represent each class. + +All this started because I wanted a program to generate some numbers and +investigate the relation. We finally have the requisite tools to return to that +task, namely the strong self relation in \autoref{thm:strong-self-relation} +which solves our comparison problem and the uniqueness of reduced forms in the +subset of numbers(/equivalence classes) our program will actually generate. + + +\subsection{Conjecture} + +If we can build an addition operation which holds for $1 + (-1) = 0$, then we +could start trying to assign meaningful names to some of the elements from +generation-2. It appears that numbers of the form \surreal{n}{} behave like the +number $n+1$ and similarly, \surreal{}{n} behaves like $n-1$. That seems to +build the integers. + +If we write a program to generate a bunch of new surreal numbers and graph them +as ``generation vs magnitude'', perhaps we can assign some meaning to numbers +which don't fit the pattern mentioned in the previous paragraph. Maybe these +behave like $1/n$? It sort of feels like surreal numbers constructed via finite +repetitions of our current process will end up building something vaguely like +the dyadic rationals. + +Are numbers just the average of their left and right sets (in reduced form)? +For example, $\surreal{-1}{1} \similar \surreal{}{}$. If so, then generation-2 +includes numbers like $\pm \frac{1}{2}$. + +If my suspicions about symmetry are correct, anytime the number of equivalence +classes is even, there are at least two equivalence classes that are similar. + +Since the universe is growing in both direction, I don't think we can impose a +useful well ordering on the whole set unless considering a finite number of +generations.