| 1 | \newpage |
| 2 | \section{Notes: Chapter 3} |
| 3 | |
| 4 | \subsection{Review} |
| 5 | |
| 6 | In the third chapter, Knuth begins to embrace similarity exactly as we did in |
| 7 | Definition \autoref{defi:similar}, noting that different forms exist which are |
| 8 | similar to each other. He also more carefully tracks through the relations in |
| 9 | $\surreal{}{0} \leq \surreal{}{} \leq \surreal{0}{}$. |
| 10 | |
| 11 | Finally, at the end of the chapter, Knuth begins speculating about the form of |
| 12 | numbers to come and their relation to each other, pointing out, via an |
| 13 | erroneous categorization of positive number forms, that categorization isn't as |
| 14 | simple as it first appears |
| 15 | |
| 16 | |
| 17 | \subsection{Exploration} |
| 18 | |
| 19 | My goal is that of assigning meaningful names to newly generated numbers under |
| 20 | the assumption that Knuth's three names ($-1$, $0$ and $1$) were naturally |
| 21 | meaningful. Since that implies some notion of `distance from zero' (after all, |
| 22 | I'm attempting to build a numberline), defining some form of addition and |
| 23 | subtraction seems like a natural starting point. |
| 24 | |
| 25 | Remember that we're still only considering finite generations for our universe. |
| 26 | That means our addition operation will never be closed since our finite |
| 27 | universe always has a largest element and nobody can stop us from adding $1$ to |
| 28 | it. However, as such closure-violating elements are members of our universe in |
| 29 | a later generation, only when considering closure, we will mentally think of |
| 30 | our universe as something like $\mathbb{U} \equiv \bigcup_{n \in \mathbb{N}} |
| 31 | \mathbb{U}_n$ and otherwise ignore the problem. |
| 32 | |
| 33 | Thus, updating my caveat from my Chapter 1 notes: |
| 34 | |
| 35 | \begin{framed} |
| 36 | \noindent All definitions and theorems only consider the case of finite left |
| 37 | and right sets. In these finite cases, we're pretending that addition is |
| 38 | closed, but it's not. |
| 39 | \end{framed} |
| 40 | |
| 41 | Based on my Chapter 2 notes, when considering my numbers in the line defined by |
| 42 | my total order, I want every reduced form number that has both left and right |
| 43 | ancestors to be the geometric mean of those ancestors. And any reduced form |
| 44 | number missing one or both ancestors is similar to another, more suitable |
| 45 | reduced form number. |
| 46 | |
| 47 | Also in my Chapter 2 notes, I hand-wavingly defined such a geometric mean |
| 48 | recursively, just to build a mental picture. However, since I don't see a way |
| 49 | to define the same concept using only the left and right sets of the operands, |
| 50 | a recursive definition might be the right approach as long as it reduces the |
| 51 | overall age of the numbers involved, allowing me to use the same |
| 52 | sum-of-generations type argument used when proving transitivity. |
| 53 | |
| 54 | \begin{defi} \label{defi:addition} |
| 55 | For two numbers $x$ and $y$, define \emph{addition} as |
| 56 | $$ |
| 57 | x \sgkadd y = \surreal{X_L}{X_R} \sgkadd \surreal{Y_L}{Y_R} |
| 58 | \equiv \surreal{\set{X_L \sgkadd y} \cup \set{Y_L \sgkadd x}}{\set{X_R \sgkadd y} \cup \set{Y_R \sgkadd x}} |
| 59 | $$ |
| 60 | where $\set{A \sgkadd b}$ means the set of numbers $a \sgkadd b$ for all $a \in A$. |
| 61 | |
| 62 | We are using the symbol $\sgkadd$ in order to keep our addition distinct |
| 63 | from whatever Knuth eventually defines. |
| 64 | \end{defi} |
| 65 | |
| 66 | Given the obvious symbolic symmetry, we won't bother explicitly proving this |
| 67 | operation is commutative. |
| 68 | |
| 69 | Keep in mind that this is only guaranteed to produce a valid number (per Axiom |
| 70 | \autoref{ax:number-definition}) subject to our caveat regarding closure. |
| 71 | |
| 72 | Since we have defined this operation in terms of specific forms, we must ensure |
| 73 | the operation behaves identically with respect to all similar forms. It would |
| 74 | be a shame if, for example, $0+0=0$ only held for certain values of $0$. |
| 75 | |
| 76 | \begin{theorem} \label{thm:sgkadd-welldefined} |
| 77 | The binary operation $\sgkadd$ on $\mathbb{U}$ is well defined. That is, for |
| 78 | numbers $x, x', y, z \in \mathbb{U}$ such that $x \sgkadd y = z$ and $x |
| 79 | \similar x'$, $\exists z' \in \mathbb{U}$ such that $x' \sgkadd y = z' |
| 80 | \similar z$. |
| 81 | \end{theorem} |
| 82 | |
| 83 | \begin{proof} |
| 84 | TODO |
| 85 | \end{proof} |
| 86 | |
| 87 | \begin{theorem} \label{thm:sgkadd-identity} |
| 88 | The number $0 = \surreal{}{}$ is the identity element for the binary |
| 89 | operation $\sgkadd$ on $\mathbb{U}$. That is, for any number $x \in |
| 90 | \mathbb{U}$, $x \sgkadd 0 \similar 0 \sgkadd x \similar x$. |
| 91 | In this behavior, the number $0$ is unique up to similarity. |
| 92 | \end{theorem} |
| 93 | |
| 94 | \begin{proof} |
| 95 | TODO |
| 96 | \end{proof} |
| 97 | |
| 98 | \begin{theorem} \label{thm:sgkadd-associative} |
| 99 | For all $x, y, z \in \mathbb{U}$, it holds that |
| 100 | $$(x \sgkadd y) \sgkadd z \similar x \sgkadd (y \sgkadd z).$$ |
| 101 | \end{theorem} |
| 102 | |
| 103 | \begin{proof} |
| 104 | TODO |
| 105 | \end{proof} |
| 106 | |
| 107 | \begin{defi} \label{defi:inverse} |
| 108 | For a number $x$, let \emph{negation} be defined as |
| 109 | $$ |
| 110 | -x = -\surreal{X_L}{X_R} \equiv \surreal{-X_R}{-X_L} |
| 111 | $$ |
| 112 | where $-A$ means the set of numbers $-a$ for all $a \in A$. |
| 113 | \end{defi} |
| 114 | |
| 115 | \begin{theorem} \label{thm:sgkadd-inverse} |
| 116 | For every number $x \in \mathbb{U}$, there exists a number $-x \in |
| 117 | \mathbb{U}$ such that $x \sgkadd -x = 0$. |
| 118 | In this behavior, the number $-x$ is unique up to similarity. |
| 119 | \end{theorem} |
| 120 | |
| 121 | \begin{proof} |
| 122 | TODO |
| 123 | \end{proof} |
| 124 | |
| 125 | Putting that all together, $(\mathbb{U},\sgkadd)$ is well defined, closed, and |
| 126 | respects the three group axioms. It's a group. Let's name it |
| 127 | $\mathbb{U}_{\sgkadd}$. It's also commutative. |
| 128 | |
| 129 | |
| 130 | \subsection{Conjecture} |
| 131 | |
| 132 | $\mathbb{U}_{\sgkadd}$ really is a group. |
| 133 | |