\newpage
\section{Chapter 4: Bad Numbers}

\subsection{Review}

This was an interesting chapter, much busier than the previous two. It was nice
to see that we haven't veered too far into nonsense.

Knuth generates all the elements of generation three and they match our
explorations from the Chapter 1 notes. He also proves a transitive law holds
for Axiom \autoref{ax:leq-relation} using the same argument I used, showing
that younger (simpler) sums reach a contradiction established by the original
numbers. Then, with transitivity, he notes that all his numbers are similar to
our Definition \autoref{defi:reduced-form} reduced form numbers. Knuth ends the
chapter with a wink for meeting him at this point.


\subsection{Exploration}

I continue my attempt to define an addition operation which respects the
implications of the names assigned by Knuth ($-1$, $0$ and $1$) and which
extends those names in the pattern I think is developing.

From Definition \autoref{defi:addition}, using the symbol $\sgkadd$, we are defining addition as
$$
x \sgkadd y = \surreal{X_L}{X_R} \sgkadd \surreal{Y_L}{Y_R}
\equiv \surreal{\set{X_L \sgkadd y} \cup \set{Y_L \sgkadd x}}{\set{X_R \sgkadd y} \cup \set{Y_R \sgkadd x}}
.
$$

Since we have defined this operation in terms of specific forms, we must ensure
the operation behaves identically with respect to all similar forms. It would
be a shame if, for example, $0+0=0$ only held for certain values of $0$.

We already know the outputs aren't well behaved in this manner, but we patched
that up by only considering closure in terms of future generations. Now we
examine the inputs.

\begin{theorem} \label{thm:sgkadd-welldefined}
    The binary operation $\sgkadd$ on $\mathbb{U}$ is well defined. That is, for
    numbers $x, x', y, z \in \mathbb{U}$ such that $x \sgkadd y = z$ and $x
    \similar x'$, $\exists z' \in \mathbb{U}$ such that $x' \sgkadd y = z'
    \similar z$.
\end{theorem}

\begin{proof}
    TODO
\end{proof}

With the inputs and outputs behaving as needed, we now apply our operation to
the set, checking if we meet the group axioms.

\begin{theorem} \label{thm:sgkadd-identity}
    The number $0 = \surreal{}{}$ is the identity element for the binary
    operation $\sgkadd$ on $\mathbb{U}$. That is, for any number $x \in
    \mathbb{U}$, $x \sgkadd 0 \similar 0 \sgkadd x \similar x$.
    In this behavior, the number $0$ is unique up to similarity.
\end{theorem}

\begin{proof}
    TODO
\end{proof}

\begin{theorem} \label{thm:sgkadd-associative}
    For all $x, y, z \in \mathbb{U}$, it holds that
    $$(x \sgkadd y) \sgkadd z \similar x \sgkadd (y \sgkadd z).$$
\end{theorem}

\begin{proof}
    TODO
\end{proof}

\begin{defi} \label{defi:inverse}
    For a number $x$, let \emph{negation} be defined as
    $$
    -x = -\surreal{X_L}{X_R} \equiv \surreal{-X_R}{-X_L}
    $$
    where $-A$ means the set of numbers $-a$ for all $a \in A$.
\end{defi}

\begin{theorem} \label{thm:sgkadd-inverse}
    For every number $x \in \mathbb{U}$, there exists a number $-x \in
    \mathbb{U}$ such that $x \sgkadd -x = 0$.
    In this behavior, the number $-x$ is unique up to similarity.
\end{theorem}

\begin{proof}
    TODO
\end{proof}

Putting that all together, our operation $(\mathbb{U},\sgkadd)$ is well
defined, closed, and respects the three group axioms. Our universe $\mathbb{U}$
is a group under $\sgkadd$, our addition operation. Let's name this group
$\mathbb{U}_{\sgkadd}$.

Note: This group is also commutative, seen by running transposed symbols
through Definition \autoref{defi:addition} and noting that sets are inherently
unordered.


\subsection{Conjecture}

Given our definition of addition, it seems natural to consider defining
multiplication in the same recursively-younger manner. Then see if it's also a
group and if it relates to $\mathbb{U}_{\sgkadd}$ in the usual manner.
