Revised 27jan13
\begin{document} \maketitle \textbf{ $\C$ 2010, Prof. George K. Francis, Mathematics Department, University of Illinois} \section{Introduction}
For some people following the textbook on this subject is very difficult.
It seems like a collection of "story problems on steroids."
The context and common speech gets in the way of understanding the content.
For such and similar reasons mathematicians have found that expressing such
arguments in terms of logical symbols, and possibly adding figures,
greatly reduces this difficulty. Translated into the common tongue,
a symbol might be read in several different ways. For example,
$\exists \epsilon > 0$ can be read as saying "for at least one positive epsilon", "for some positive epsilons", "for almost every but not for all positive epsilons" etc.
While the logical meaning of a symbolic expression is totally unambiguous,
its reading in common speech is not, not even in legal or legislative writing.
Recall that ``If it rains then it pours." written $ R \implies P $, could
also be read as ``It is raining, therefore it is pouring.", and even ``It is pouring
because it is raining." If you express this as ``Either it's not raining or it's pouring"
you would write it differently, so: $ \neg R \or P $.
Next, recall the quantifiers from MA347. The
\textit{existential quantifer}, $\exists$, can be read to say ``there exists'', occasionally,
``for some" works better in English. Similarly, the
\textit{universal quantifier}, $\forall$, is usually read ``for all'', but sometimes
``for any" is clearer. Just don't make the common error of equating ``for any" to
"for some".
The $\exists! $ abbreviates ``there exists an unique". Thus, for
\begin{itemize}
\item $(\forall n \in \mathbb{N})(\exists m \in \mathbb{N})( m > n) $ \\
\item $(\forall n \in \mathbb{N})(\exists! \ m \in \mathbb{N})( m > n) $ \\
\item $(\forall n \in \mathbb{N})(\exists! \ m \in \mathbb{N})( m = n +1) $ \\
\end{itemize}
the first and third and true, and the second one is false.
\section{Hvidsten starting Page 20}
We begin with the example on page 20 of Hvidsten. The students get capital letters $A,B,C, ... $
and the school classes they are in get lower case $k,\ell,m ... $.
And we abbreviate ``student $A$ is in class $k$" by just `` $Ak$''
We can now state the axioms precisely, without the distraction of thinking about
kids running around their classrooms, and what the teacher is wearing today.
\begin{eqnarray*}
A1 & \{P_1,P_2,P_3\} & \mbox{ Given 3 capital letters } \\
A2 & (\forall A \ne B)(\exists ! \ k)(Ak \and Bk) & \mbox{Euclid's 2nd postulate} \\
A3 & (\forall k)(\exists A)( \neg Ak) & \mbox{ Dimension > 1 } \\
A4 & (\forall k \ne \ell)(\exists ! \ A)(Ak \and A\ell) & \mbox{Dual of A2} \\
\end{eqnarray*}
Suppose we interpret the capital letters as naming points, and the
lower case letters as line. For the first axiom we might draw three points
$ ... $ that line up, or a point above two points $\therefore$.
The second axiom, which is Euclid's second postulate, still doesn't distinguish
between the two models: the line might pass through all three points, or there
are three lines forming a triangle.
The third axiom eliminates the "3 pts on a line" case. You might think that
we're done, we've got a model that completely captures the first 3 axioms and
we don't need any more. But wait! there could be any number of lines that don't
have any points on it. Even though classes that don't have any students in them
don't make much sense, it does happend in college. Of course the classes are
cancelled. But you can't read more from the context than you're told.
To eliminate this case, we need the fourth axiom. It takes a little reflectionto agree that with all four axioms, we really can't draw a different picture.
We will take this issue up next.
\subsection{Models of Geometry}
So we have a uniquely possible picture which is consistent with each of the four
axioms. In a few pages we shall see that this is an exremely important achievement.
We have created a unique \textit{model} for the axiom system. Each of the axioms
is true in this model. Moreover, any other such model would be \texit{isomorphic}
to one we have. For two models to be isomorphic to each other there has to
be a \textit{bijection} (one-to-one and onto correspondence) between two sets
of points, and the two sets of lines, and, also require, that the only
\textit{relation} between points and lines, called \textit{incidence} is
preserved.
The nice thing about a model, and from now on you may not use the word ``model"
in any other sense in this course, is that you can determine the truth or falsity of propositions in the original axiomatic system by using properties of
points, lines and incidence, as they occur in the model.
For example, that there are exactly three classes, and no student is taking
all three of them.
\subsection{Projective Geometries}
The \textit{dual} of a sentence about points and lines is the same sentence
but with the roles of points and lines reversed. Thus we have just proved
the dual of A1, because we can truthfully say ``There exactly three lines."
The sentence ``No student is taking all three of them" can be written
more precisely as follows, and rewritten
\begin{eqnarray*}
A'3 & \neg (\exists A)(\forall k)( Ak) & \equiv & (\forall A)(\exists k)(\neg Ak) \\
\end{eqnarray*}
which is exactly the dual of axiom A3. Thus we can predict that for every
theorem in this axiom system, its dual is also a theorem. Or, for every
theorem we prove we get another true one for free. Recall that a "theorem"
is a proposition which follows logically from the axioms. But since we have
a unique model, we can infer theorems from the model.
You may also recall that two lines that do not meet are said
to be \textit{ parallel }. What should we call two points that have
no common line, \textit{ strangers}, perhaps?
At any event, we see that this toy geometry has no parallel lines. Euclid's second postulate also says that there are no strangers, i.e. every pair of points
is joined by a line.
\section{Icecream Exercise 1.4.3 Explained}
Those of you who figured this one out using a model probably got a very pretty
picture for it, a pentagram (5-pointed star) inscribed in a pentagon. The
followers of Pythagoras were very fond of this figure, and it has deep geometrical
meaning. To do that, you would have to \textit{interpret} the children as
points and flavors of icecream as line. But let's be naive and see what the
other interpretation gets us.
The children are $A,B,C ... $ and the flavors of icecream are $h,k,\ell ... $.
And a child $A$ liking flavor $h$ would say $Ah$. Note that Hvidsten uses
the letters A1, A2, for all axioms and this makes it hard to keep them straight.
So let's use a different letter, maybe I for icecream.
\begin{eqnarray*}
I1 & \{ k_1, k_2, k_3, k_4, k_5 \} & \mbox{exactly 5 flavors} \\
I2 & (\forall k \ne \ell)(\exists !\ A)(Ak \and A\ell) & \mbox{dual Euclid 2} \\
I3 & (\forall A)(\exists h \ne k)(hA \and kA \and (\forall \ell)( \ell A \implies (\ell = h) \or (\ell = k))) & \mbox{valence 2}\\
\end{eqnarray*}
Starting with I1 there are very many ways we could draw pictures of just 5 lines.
Try it with just two lines first. ( One line is dull.) They could cross at a point, or not.
(Deep breath, this was a profound statement for this course!) Already for 3 lines,
they could all be parallel like a Roman III, two cross like a
Roman XI, an asterisk, or they form a triangle. We hope for the last. For 4 and
especially 5 lines there are very many more possibilities for just I1. (How many?)
The second axiom says that every pair of lines cross at a unique point. Try
drawing this. You'll soon realize that you want to bend some lines into curves,
wiggly ones at that. Can you do it without curving? More than one way?
(How about a pentagram again?)
And, of course, there could be all sorts of stray points not on any of these lines.
The third axiom, I3, forbids this, saying every point lies on exactly one pair
of lines. Since there are only 5 choose 2 = 10 such pairs, there are a most
10 points. The third axiom is powerful.
Study your picture, is there any other way of drawing it? Well, yes there is.
But to deduce true theorems from the axioms its fine to have just one model in
which all the axioms are true. And whichever way you draw it, there will be
exaclty 10 crossing of the 5 lines.
I think instead of the \textit{Icecream Geometery} we should call this
\textit{Pythagoras' Geometry of Five} whose model is the Pythagorean
five-pointed star, or \textit{Pentagram}.
\section{The Four Point Geometry}
The four-point geometry has these axioms [Hvidsten p32]
\begin{itemize}
\item P1: There are exactly four Points.
\item P2: Two distinct Points belong to one and only one Line.
\item P3: Each Line has exactly two Points belonging to it.
\end{itemize}
Note how we have adopted the convention of capitalizing primitives in an
axiomatic system to keep from confusing them with the common names of
geometrical objects in their models.
Comments on Toy Axiom Systems
\begin{document} \maketitle \textbf{ $\C$ 2010, Prof. George K. Francis, Mathematics Department, University of Illinois} \section{Introduction}
For some people following the textbook on this subject is very difficult.
It seems like a collection of "story problems on steroids."
The context and common speech gets in the way of understanding the content.
For such and similar reasons mathematicians have found that expressing such
arguments in terms of logical symbols, and possibly adding figures,
greatly reduces this difficulty. Translated into the common tongue,
a symbol might be read in several different ways. For example,
$\exists \epsilon > 0$ can be read as saying "for at least one positive epsilon", "for some positive epsilons", "for almost every but not for all positive epsilons" etc.
While the logical meaning of a symbolic expression is totally unambiguous,
its reading in common speech is not, not even in legal or legislative writing.
Recall that ``If it rains then it pours." written $ R \implies P $, could
also be read as ``It is raining, therefore it is pouring.", and even ``It is pouring
because it is raining." If you express this as ``Either it's not raining or it's pouring"
you would write it differently, so: $ \neg R \or P $.
Next, recall the quantifiers from MA347. The
\textit{existential quantifer}, $\exists$, can be read to say ``there exists'', occasionally,
``for some" works better in English. Similarly, the
\textit{universal quantifier}, $\forall$, is usually read ``for all'', but sometimes
``for any" is clearer. Just don't make the common error of equating ``for any" to
"for some".
The $\exists! $ abbreviates ``there exists an unique". Thus, for
\begin{itemize}
\item $(\forall n \in \mathbb{N})(\exists m \in \mathbb{N})( m > n) $ \\
\item $(\forall n \in \mathbb{N})(\exists! \ m \in \mathbb{N})( m > n) $ \\
\item $(\forall n \in \mathbb{N})(\exists! \ m \in \mathbb{N})( m = n +1) $ \\
\end{itemize}
the first and third and true, and the second one is false.
\section{Hvidsten starting Page 20}
We begin with the example on page 20 of Hvidsten. The students get capital letters $A,B,C, ... $
and the school classes they are in get lower case $k,\ell,m ... $.
And we abbreviate ``student $A$ is in class $k$" by just `` $Ak$''
We can now state the axioms precisely, without the distraction of thinking about
kids running around their classrooms, and what the teacher is wearing today.
\begin{eqnarray*}
A1 & \{P_1,P_2,P_3\} & \mbox{ Given 3 capital letters } \\
A2 & (\forall A \ne B)(\exists ! \ k)(Ak \and Bk) & \mbox{Euclid's 2nd postulate} \\
A3 & (\forall k)(\exists A)( \neg Ak) & \mbox{ Dimension > 1 } \\
A4 & (\forall k \ne \ell)(\exists ! \ A)(Ak \and A\ell) & \mbox{Dual of A2} \\
\end{eqnarray*}
Suppose we interpret the capital letters as naming points, and the
lower case letters as line. For the first axiom we might draw three points
$ ... $ that line up, or a point above two points $\therefore$.
The second axiom, which is Euclid's second postulate, still doesn't distinguish
between the two models: the line might pass through all three points, or there
are three lines forming a triangle.
The third axiom eliminates the "3 pts on a line" case. You might think that
we're done, we've got a model that completely captures the first 3 axioms and
we don't need any more. But wait! there could be any number of lines that don't
have any points on it. Even though classes that don't have any students in them
don't make much sense, it does happend in college. Of course the classes are
cancelled. But you can't read more from the context than you're told.
To eliminate this case, we need the fourth axiom. It takes a little reflectionto agree that with all four axioms, we really can't draw a different picture.
We will take this issue up next.
\subsection{Models of Geometry}
So we have a uniquely possible picture which is consistent with each of the four
axioms. In a few pages we shall see that this is an exremely important achievement.
We have created a unique \textit{model} for the axiom system. Each of the axioms
is true in this model. Moreover, any other such model would be \texit{isomorphic}
to one we have. For two models to be isomorphic to each other there has to
be a \textit{bijection} (one-to-one and onto correspondence) between two sets
of points, and the two sets of lines, and, also require, that the only
\textit{relation} between points and lines, called \textit{incidence} is
preserved.
The nice thing about a model, and from now on you may not use the word ``model"
in any other sense in this course, is that you can determine the truth or falsity of propositions in the original axiomatic system by using properties of
points, lines and incidence, as they occur in the model.
For example, that there are exactly three classes, and no student is taking
all three of them.
\subsection{Projective Geometries}
The \textit{dual} of a sentence about points and lines is the same sentence
but with the roles of points and lines reversed. Thus we have just proved
the dual of A1, because we can truthfully say ``There exactly three lines."
The sentence ``No student is taking all three of them" can be written
more precisely as follows, and rewritten
\begin{eqnarray*}
A'3 & \neg (\exists A)(\forall k)( Ak) & \equiv & (\forall A)(\exists k)(\neg Ak) \\
\end{eqnarray*}
which is exactly the dual of axiom A3. Thus we can predict that for every
theorem in this axiom system, its dual is also a theorem. Or, for every
theorem we prove we get another true one for free. Recall that a "theorem"
is a proposition which follows logically from the axioms. But since we have
a unique model, we can infer theorems from the model.
You may also recall that two lines that do not meet are said
to be \textit{ parallel }. What should we call two points that have
no common line, \textit{ strangers}, perhaps?
At any event, we see that this toy geometry has no parallel lines. Euclid's second postulate also says that there are no strangers, i.e. every pair of points
is joined by a line.
\section{Icecream Exercise 1.4.3 Explained}
Those of you who figured this one out using a model probably got a very pretty
picture for it, a pentagram (5-pointed star) inscribed in a pentagon. The
followers of Pythagoras were very fond of this figure, and it has deep geometrical
meaning. To do that, you would have to \textit{interpret} the children as
points and flavors of icecream as line. But let's be naive and see what the
other interpretation gets us.
The children are $A,B,C ... $ and the flavors of icecream are $h,k,\ell ... $.
And a child $A$ liking flavor $h$ would say $Ah$. Note that Hvidsten uses
the letters A1, A2, for all axioms and this makes it hard to keep them straight.
So let's use a different letter, maybe I for icecream.
\begin{eqnarray*}
I1 & \{ k_1, k_2, k_3, k_4, k_5 \} & \mbox{exactly 5 flavors} \\
I2 & (\forall k \ne \ell)(\exists !\ A)(Ak \and A\ell) & \mbox{dual Euclid 2} \\
I3 & (\forall A)(\exists h \ne k)(hA \and kA \and (\forall \ell)( \ell A \implies (\ell = h) \or (\ell = k))) & \mbox{valence 2}\\
\end{eqnarray*}
Starting with I1 there are very many ways we could draw pictures of just 5 lines.
Try it with just two lines first. ( One line is dull.) They could cross at a point, or not.
(Deep breath, this was a profound statement for this course!) Already for 3 lines,
they could all be parallel like a Roman III, two cross like a
Roman XI, an asterisk, or they form a triangle. We hope for the last. For 4 and
especially 5 lines there are very many more possibilities for just I1. (How many?)
The second axiom says that every pair of lines cross at a unique point. Try
drawing this. You'll soon realize that you want to bend some lines into curves,
wiggly ones at that. Can you do it without curving? More than one way?
(How about a pentagram again?)
And, of course, there could be all sorts of stray points not on any of these lines.
The third axiom, I3, forbids this, saying every point lies on exactly one pair
of lines. Since there are only 5 choose 2 = 10 such pairs, there are a most
10 points. The third axiom is powerful.
Study your picture, is there any other way of drawing it? Well, yes there is.
But to deduce true theorems from the axioms its fine to have just one model in
which all the axioms are true. And whichever way you draw it, there will be
exaclty 10 crossing of the 5 lines.
I think instead of the \textit{Icecream Geometery} we should call this
\textit{Pythagoras' Geometry of Five} whose model is the Pythagorean
five-pointed star, or \textit{Pentagram}.
\section{The Four Point Geometry}
The four-point geometry has these axioms [Hvidsten p32]
\begin{itemize}
\item P1: There are exactly four Points.
\item P2: Two distinct Points belong to one and only one Line.
\item P3: Each Line has exactly two Points belonging to it.
\end{itemize}
Note how we have adopted the convention of capitalizing primitives in an
axiomatic system to keep from confusing them with the common names of
geometrical objects in their models.
Question 1.
Translate Axiom P1 of the 4-point geometry into symbolic logic just as
we did for the Pythagoras's Geometry of Five (Icecream Geometry)
Question 2.
Translate Axiom P2 of the 4-point geometry into symbolic logic just as
we did for the Pythagoras's Geometry of Five (Icecream Geometry)
Question 3.
Translate Axiom P3 of the 4-point geometry into symbolic logic just as
we did for the Pythagoras's Geometry of Five (Icecream Geometry)
\subsection{The Model}
Next, build a model for the 4-Point geometry. Start by placing 4 points
in the plane and use P2 (which is Euclid's 1st and 2nd Postulate) to
add Lines. But note that P3 places some restrictions of the segments
which can be Lines in this geometry.
Question 4.
Look around this lesson. Which figure represents a model for 4-Point
Geometry?
Note that this model could really not be constructed in any other way.
One says that such an axiom system is \textit{categorical} if all
of its models are \textif{isomorphic}, meaning that they are essentially
the same. Categorical axiom systems have the interesting property that
any fact you can read off their models are automatically theorems in
the axiomatic system.
\subsection{Theorems in the 4-Point Geometry}
By now you really are ready to tackle the 2 exercises on the 4-point
geometry (1.5.4 - 1.5.5). One way is to simply look at the model and
count. The other way is to write out a logical proof without references
to a model. Which do you think is easier?
\subsection{Categoricity}
It is actually non-trivial to prove that a model for an axiom system is
categorical. For instance, reconsider five-point geometry and look at
the figure at the beginning of this lesson. And go back to the discussion
of I3. If all we wanted to do is keep points strictly on lines, we might
have said so: I3* Every point has to be on two lines. But then the complete
pentagram, consisting of a pentagon with five more secants. This geometry
is not the same, it has two non-isomorphic models: one has 5 Lines, and the
other has 10 Lines. In this much weaker geometry, the proposition
"There are exactly five Lines" is not a theorem. There are two models, and
the proposition is true in one and false in the other. This situation is
hugely important for this course, and will be treated under its proper
title, namele \textit{independence}.
On the other hand, suppose we reconsider the 3 point geometry. And
perversely, we choose circles to be our Lines. Then the very first
interpretation $...$ is out. On the other hand, for $\therefore$ has
a circle through it, so those three Points are Collinear in our
geometry. To continue, we might take three mutally tangent circles, and
their tangencies to be our Lines and Points. But these two models are
isomorphic because the bijection which takes each Point to itself, and
the corresponding circle to the lines preserves incidence. So it is the
bijection we want.
\end{document}