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\begin{document}
\begin{center}
\textbf{\Large Lecture notes for Math 510}\\
{\small 12/6/10}
\end{center}

\vspace{5ex}

These notes grew out of Prof.\ David Whitman's at the San Diego
campus of SDSU. I have made some changes, added in some details,
expanded some arguments, and added the section titled ``Why axiomatic
geometry?''

This is the instructor version. The student version has some of the
proofs missing.

\section{Linear order}

We are going to use the following terms without defining them: point, line,
lie on (as in a point lies on a line, or a line contains a point),
between, congruent. We will instead give a set of axioms that describe
their behavior.

We will use capital letters to denote points and lower-case letters to
denote lines.

\begin{axiom}
\label{point-and-line}
There exist at least one point and one line.
\end{axiom}

%Actually, one line is enough, it already implies two points.

\begin{axiom}
\label{two-points}
Every line contains at least two distinct points.
\end{axiom}

\begin{axiom}
\label{euclid-1}
Any two distinct points have a unique line containing them.
\end{axiom}

Let $ABC$ mean that the point $B$ is between the points
$A$ and $C$. Remember that between is an undefined primitive. We
did not define what it means here, only introduced a convenient
notation for it.

\begin{axiom}
\label{abc-cba}
If $ABC$ then $CBA$.
\end{axiom}

\begin{axiom}
\label{abc-bac}
If $ABC$ is true, then $BAC$ is not true.
\end{axiom}

We will want to restrict $ABC$ to points that lie on the same line.

\begin{defn}
We say that two or more points are \emph{collinear} if there is a
line that they all lie on.
\end{defn}

Note that this definition does not require that the points all be
distinct. But for the purposes of betweenness, we will want points to
be distinct.

\begin{axiom}
\label{distinct-collinear}
If $ABC$ then $A$, $B$, and $C$ are distinct and collinear.
\end{axiom}

A technical note: all our statements (axioms, theorems, etc) are
universally quantified. This means that when we talk about points $A,
B, C$, we really mean any three points. For example, Axiom
\ref{abc-cba} really means ``For any points $A, B, C$, if $ABC$ then
$CBA$.'' The reason we have left out the universal quantifiers is that
they would make our sentences unnecessarily cumbersome. This has not
caused any problems, as we have understood that $A, B, C$ referred to
any three points, not three particular points called $A, B, C$.

We are ready to prove our first theorem. It will be convenient to
start with a small lemma.

\begin{lemma}
\label{ax4-contra}
If $ABC$ is false then $CBA$ is false.
\end{lemma}

\begin{proof}
Axiom \ref{abc-bac} applied to $CBA$ gives if $CBA$ then $ABC$. Now
take the contrapositive.
\end{proof}

\begin{thm}
\label{badperm}
If $ABC$ is true than the following are false: $BAC$, $CAB$, $ACB$,
and $BCA$.
\end{thm}

\begin{proof}
The first of these follows directly from Axiom \ref{abc-bac}.

Now apply Lemma \ref{ax4-contra} to not $BAC$ to get not $CBA$.

Using Axiom \ref{abc-cba}, we get $CBA$. Applying Axiom \ref{abc-bac}
to $CBA$, we find that $BCA$ must be false. Now apply Lemma \ref{ax4-contra} to not $BCA$ to get not $ACB$.
\end{proof}

\begin{thm}
Let $A, B \in k$ and $A, B \in l$. If $k \neq l$, then $A=B$.
\end{thm}

\begin{proof}
This is for you to prove.
\end{proof}

\begin{axiom}
\label{oneinthemiddle}
If $A$, $B$, and $C$ are distinct and collinear then $ABC$ or $BAC$ or
$ACB$ must be true.
\end{axiom}

\begin{thm}
\label{4-collinear}
\mbox{}
\begin{enumerate}
\item
If $ABC$ and $ACD$, then $A, B, C, D$ are distinct and collinear.

\item
If $ABC$ and $BCD$, then $A, B, C, D$ are distinct and collinear.

\item
If $ABC$ and $ABD$ and $C \neq D$, then $A, B, C, D$ are distinct and collinear.
\item
If $ABC$ and $ADC$ and $B \neq D$, then $A, B, C, D$ are distinct and collinear.
\end{enumerate}
\end{thm}

\begin{proof}
This is for you to prove.
\end{proof}

We will now add two more axioms dealing with the idea of
betweenness. Think about what these mean and why we want them to
hold. (You may want to draw a picture.) While it may be quite obvious
to you that these should be true when you think about lines and points
in the Euclidean plane, they do not follow from the axioms we already
stated.

\begin{axiom}
\label{extendmiddle}
If $ABC$ and $ACD$ then $ABD$ and $BCD$.
\end{axiom}

\begin{axiom}
\label{insertmiddle}
If $ABC$ and $ADC$ and $B \neq D$, then $ABD$ or $ADB$.
\end{axiom}

\begin{thm}
\label{abc-bcd}
If $ABC$ and $BCD$ then $ACD$ and $ABD$.
\end{thm}

\begin{proof}
This is for you to prove.
\end{proof}

\begin{thm}
\label{thm5}
If $ABC$ and $ABD$ and $C \neq D$ then $ACD$ or $ADC$.
\end{thm}

\begin{proof}
This is for you to prove.
\end{proof}

\begin{thm}
\label{thm6}
If $ABC$ and $ABD$ and $C \neq D$ then $BCD$ or $BDC$.
\end{thm}

\begin{proof}
This is for you to prove.
\end{proof}

\begin{thm}
If $ABC$, $BCD$, and $ADE$, then $BCE$ and $ACE$.
\end{thm}

\begin{proof}
This is for you to prove.
\end{proof}

\begin{defn}
\label{def-segment}
Let $A \neq B$. The \emph{line segment} $\overline{AB}$ is the set
\[ \overline{AB} = \left\{ X \mid X=A\ \text{or}\ X=B\ \text{or}\ AXB
\right\}. \]
\end{defn}

\begin{defn}
How would you define ray?
\end{defn}

\begin{thm}
\label{thm8}
Let $A \neq B, C$. If $B \in \overrightarrow{AC}$ then $\overrightarrow{AB} =
\overrightarrow{AC}$.
\end{thm}

\begin{proof}
This is for you to prove.
\end{proof}

\section{Line segment congruence}

In this section, we will set up an axiomatic system to allow us to
compare line segments. Recall that we already have already defined
what a line segment is (see Definition \ref{def-segment}).

Let $\overline{AB} \equiv \overline{CD}$ mean that $\overline{AB}$ and
$\overline{CD}$ are congruent. Just like in the previous section, we
will not define what congruence means, but we will describe how it
behaves with a set of axioms. We will then use these axioms to prove
theorems about congruence. To make writing easier, we will usually
drop the line from above $AB$ to denote the line segment
$\overline{AB}$. This does not usually cause a problem because we have
nothing else that we would denote by $AB$.

First a quick observation about the line segment $AB$:

\begin{lemma}
\label{AB=BA}
If $A \neq B$ then $AB=BA$.
\end{lemma}

\begin{proof}
We proved this in class.
\end{proof}

This has some notable consequences for congruence. For example, if $AB
\equiv CD$ then also $BA \equiv CD$, $AB \equiv DC$, and $BA \equiv
DC$.

\begin{axiom}
\label{cong-reflexive}
$AB \equiv AB$
\end{axiom}

\begin{axiom}
\label{cong-transitive}
If $AB \equiv CD$ and $CD \equiv EF$ then $AB \equiv EF$
\end{axiom}

\begin{axiom}
\label{same-length}
If $ABC$, $DEF$, $AB \equiv DE$, and $BC \equiv EF$ then $AC \equiv DF$.
\end{axiom}

\begin{axiom}
\label{extend-segment}
If $A \neq B$ and $C \neq D$ then there exists a point $E$ such that
$CDE$ and $AB \equiv DE$.
\end{axiom}

\begin{axiom}
\label{same-point}
If $ABC$, $ABD$, and $EF \equiv BC$, and $EF \equiv BD$ then $C=D$.
\end{axiom}

\begin{thm}
\label{same-point-simpler}
If $ABC$, $ABD$, and $BC \equiv BD$ then $C=D$.
\end{thm}

\begin{proof}
We proved this in class.
\end{proof}

\begin{thm}
\label{point-before}
If $A \neq B$ then there exists a point $C$ such that $CAB$.
\end{thm}

\begin{proof}
We proved this in class.
\end{proof}

\begin{thm}
\label{ab-not-equiv-ac}
If $ABC$ then $AB \not\equiv AC$.
\end{thm}

\begin{proof}
This is for you to prove.
\end{proof}

\begin{thm}
\label{cong-symmetric}
If $AB \equiv CD$ then $CD \equiv AB$.
\end{thm}

\begin{proof}
This is for you to prove.
\end{proof}

We now know that segment congruence is symmetric. It will make our
subsequent proofs smoother to refer to this property as symmetry,
rather then referencing this theorem every time.

\begin{thm}
If $ABC$, $DEF$, $AB \equiv DE$, and $AC \equiv DF$ then $BC \equiv
EF$.
\end{thm}

\begin{proof}
We proved this in class.
\end{proof}

\begin{thm}
\label{split-segment}
If $ABC$ and $AC \equiv DF$ then there exists a point $E$ such that
$DEF$ and $AB \equiv DE$ and $BC \equiv EF$.
\end{thm}

\begin{proof}
This is for you to prove.
\end{proof}

\begin{thm}
If $A, B, C$ are distinct and collinear and $AB \equiv BC$ then $ABC$.
\end{thm}

\begin{proof}
This is for you to prove.
\end{proof}

\section{Comparing line segments}

The intuitive idea behind congruence of line segments is that two line
segments should be congruent if they are the same size. Of course, in
our axiomatic system, we don't really have a notion of size and we
cannot measure the length of a line segment. In $\R^n$, we could
define the distance of two points, so we could just say that the
length of the line segment $AB$ is $d(A,B)$. But this is specific to
$\R^n$ and may not be available in another geometric system where our
axioms still hold. We will instead use congruence to compare the sizes
of line segments, as in the next definition.

\begin{defn}
We say $AB<CD$ if there exists a point $E$ such that $CED$ and $AB
\equiv CE$.
\end{defn}

\begin{thm}
\label{comparison}
\mbox{}
\begin{enumerate}
\item
If $AB<CD$ and $CD \equiv EF$ then $AB<EF$.

\item
If $AB<CD$ and $CD<EF$ then $AB<EF$.

\item
If $AB \equiv CD$ and $CD<EF$ then $AB<EF$.
\end{enumerate}
\end{thm}

\begin{proof}
We proved part (a) in class. Parts (b) and (c) are yours to prove.
\end{proof}

\begin{thm}
If $AB$ and $CD$ are segments then $AB<CD$ or $AB \equiv CD$ or
$CD<AB$.
\end{thm}

\begin{proof}
We proved this in class.
\end{proof}

\begin{thm}
\label{consistent-order}
If $AB<CD$ then $AB \not\equiv CD$ and $CD \not < AB$.
\end{thm}

\begin{proof}
This is for you to prove.
\end{proof}

\begin{thm}
If $ABC$, $ADC$, $AB \equiv BC$, and $AD \equiv DC$ then $B=D$.
\end{thm}

\begin{proof}
This is for you to prove.
\end{proof}

\section{Why axiomatic geometry?}

The motivation for studying geometry through an axiom system as this
is that our axioms can fit many different kinds of geometries. Points
and lines can mean different things in different geometries. We have
seen such things when we constructed and tested models for incidence
geometry earlier in the course.

The great thing about axiomatic geometry is that in any interpretation
where all of our axioms hold, all of our theorems must also
hold because we only relied on the axioms to prove them. This becomes
particularly important when you have to work with a geometry that is
quite different from the Euclidean geometry we are used to. In fact,
there are many such strange geometries and they do play important
roles in various branches of mathematics and physics. For example,
hyperbolic geometry is central in general relativity, and geometries
with ultrametric distances are at the heart of an area of number
theory called $p$-adic analysis. Intuition based on familiar drawings
of points and lines does not get you very far in such geometries. If
you want to prove theorems, you need to start with the
axioms. Playing with the axioms and proving theorems should help you
develop an intuition about these other geometries too.

We are now going to give a familiar interpretation of our axiomatic
geometry and prove that it is a model.

Recall that
\[ \R^n = \{ (x_1, x_2, \ldots, x_n) \mid x_1, x_2, \ldots, x_n \in \R \}
\]
where $(x_1, \ldots, x_n)$ is an ordered $n$-tuple. We call $x_i$ the
$i$-th \emph{coordinate} of $(x_1, \ldots, x_n)$. We will call the elements
of the set $\R^n$ \emph{points}. We will continue to denote them by
capital letters, but we will use lower-case letters for the
coordinates. For two points $A=(a_1, \ldots, a_n)$ and
$B=(b_1, \ldots, b_n)$, we say $A=B$ if $a_i=b_i$ for all $i=1,
\ldots, n$.

We will now define what a line is. In order to do so, we will need two
distinct points. Let $A=(a_1, \ldots, a_n)$ and $B=(b_1, \ldots, b_n)$ be
distinct points in $\R^n$. We say that a \emph{line through $A$ and
  $B$} is the set of points
\[ l = \{(x_1, \ldots, x_n) \mid x_i = a_i + t(b_i-a_i) \text{\ where\
} t \in \R \}. \]
Note that this is the usual equation of a line from analytic
geometry. Also, it is clear that $A,B \in l$ with $t=0$ and $t=1$. We
will see that we can write down other equations of lines that go
through $A$ and $B$. When we are not interested in naming two points
that the line goes through, we will simply talk about the line, rather
than the line through $A$ and $B$.

You can now verify that Axioms 1--3 hold.

We need to define our interpretation of betweennes. For three distinct
points $A, B, C$ let $ABC$ hold if and only if $d(A,C)=d(A,B)+d(B,C)$,
where $d$ is the usual Euclidean distance on $\R^n$:

\begin{defn}
\label{euclidean-distance}
For $A, B \in \R^n$, the \emph{distance} of $A$ and $B$ is
\[ d(A,B) = \sqrt{\sum_{i=1}^n(a_i-b_i)^2}. \]
\end{defn}

Here are a few useful properties of distance.
\begin{thm}
\label{distance-props}
Let $A,B,C$ be points.
\begin{enumerate}
\item
Nonnegativity: $d(A,B) \geq 0$ with $d(A,B)=0$ if and only if $A=B$.

\item
Symmetry: $d(A,B) = d(B,A)$.

\item
Triangle Inequality: $d(A,C) \leq d(A,B)+d(B,C)$ with equality if and
only if $B \in \overline{AC}$.
\end{enumerate}
\end{thm}

You can now verify that Axioms \ref {abc-cba} and \ref{abc-bac} hold.

We verified Axiom \ref{distinct-collinear} in class. You can now
verify Axioms \ref{oneinthemiddle}--\ref{insertmiddle}.

We are ready to define line segment congruence in our
interpretation. We will now really need to use distance. We will say
$AB \equiv CD$ if $d(A,B) = d(C,D)$.

We verified Axioms \ref{cong-reflexive} and \ref{cong-transitive} in
class. You should prove that Axioms
\ref{same-length}--\ref{same-point} also hold.

\end{document}
