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\begin{document}

\title{History of mathematics}
\subtitle{Log of a course}
\author{David Pierce}
\date{June 20, 2011\\
(last edited \today)}
\publishers{Mathematics Department\\
Mimar Sinan Fine Arts University\\
Istanbul\\
\url{http://math.msgsu.edu.tr/~dpierce/}}

\uppertitleback{\center{
This work is licensed under the\\
 Creative Commons Attribution--Noncommercial--Share-Alike
License.\\
 To view a copy of this license, visit\\
  \url{http://creativecommons.org/licenses/by-nc-sa/3.0/}\\
\mbox{}\\
\cc \ccby David Pierce \ccnc \ccsa\\
\mbox{}\\
Matematik B\"ol\"um\"u\\
Mimar Sinan G\"uzel Sanatlar \"Universitesi\\
Bomonti, \c Si\c sli\\
\.Istanbul 34380 Turkey\\
\url{http://math.msgsu.edu.tr/~dpierce/}\\
\url{dpierce@msgsu.edu.tr}
}
}

\maketitle%[-1]

\chapter*{Preface}

This is about a course given in Ankara in which third- and fourth-year undergraduates read original sources of mathematics.  Now I live in Istanbul, where I am involved in a course in which first-year undergraduates read Book I of Euclid's \emph{Elements.}  I occasionally consult the present document, and in so doing, I may make minor adjustments, or I add more diagrams, so that the discussion of Euclid becomes more self-contained.

Web pages in the domain \url{metu.edu.tr} are obsolete; but this
should now be replaceable with \url{mat.msgsu.edu.tr}.
\tableofcontents

\listoffigures

\addchap{Prolegomena}

\addsec{What is here}

This book is a record of a course in the history of
mathematics, held at Middle East Technical University, Ankara, Turkey,
during the 2009/10 academic year.  Officially the course 
was 
\begin{compactenum}[(1)]
  \item
Math 303, History of Mathematical Concepts I, in the fall
semester; 
\item
Math 304, History of Mathematical Concepts II, in the
spring.  
\end{compactenum}
There were about twenty students in each semester; but only
four students took both semesters.  The two semesters correspond to
the two numbered parts of this book.  
According to the catalogue, the course content is thus:
\begin{quote}\sloppy
  [Math 303:] Mathematics in Egypt and Mesopotamia, Ionia and Pythagoreans,
  paradoxes of Zeno and the heroic age. Mathematical works of Plato,
  Aristotle, Euclid of Alexandria, Archimedes, Apollonius and
  Diophantus. Mathematics in China and India.  [Math 304:] 
Mathematics of the Renaissance, Islamic contributions. Solution of the
cubic equation and consequences. Invention of logarithms. Time of
Fermat and Descartes. Development of the limit concept. Newton and
Leibniz. The age of Euler. Contributions of Gauss and
Cauchy. Non-Euclidean geometries. The arithmetization of analysis. The
rise of abstract algebra. Aspects of the twentieth century.  
\end{quote}
Most parts of this description correspond to chapter titles in the
suggested textbook by Boyer~\cite{MR996888}.  But I did not use a modern
textbook.  My way of teaching the course was inspired
by my experience at St John's College,\label{SJC} with campuses in
Annapolis, Maryland, and Santa Fe, New Mexico, USA.  As a student at
St John's, I 
learned mathematics by reading, presenting, and discussing the works
of Euclid, Apollonius, Descartes, Newton, and others.  In teaching
Math 303--304, I hoped my own students could learn in the same way.  So my
course had no textbook other than the works (in English translation)
of the mathematicians that we studied.  In class, students presented
the content of these works at the blackboard. 

My notes\label{Johnnies} in
Part~\ref{part:one} below
started out as emails to a discussion group, the `J-list', composed of
St John's alumni. 
The dates used as section heads in this part are the original dates of
composition of these emails; but I have done some editing and added
some diagrams (though not yet as many as might be added for the
convenience of the reader). 

In the spring semester, the
conversion of emails into \LaTeX\ (so that they could be incorporated
in a book such as this one) became too tedious; also I wanted to
use diagrams immediately; so I started 
composing my notes directly in \LaTeX.  In
Part~\ref{part:two} of this book, section titles are simply dates of classes.

A big difference between courses at St John's College and courses at METU is
that the latter have written examinations.  The exams that I wrote
for Math 303--304 are in Appendix~\ref{app:exams}.

Whether the course was a success might be judged from student
comments, which I invited on the final exams; these are in
Appendix~\ref{app:comments}. 

On the other hand, students are not necessarily the best judges of
their own progress.  It is also the case that one of the best and most
enthusiastic students, Mehmet D., did not write me any comments;
below I shall mention some of what he told me face to face.
Meanwhile, 
I judge the course to have been successful, at least insofar as it taught
students that they \emph{could} read some of the great works of
mathematics.  As can be seen from their comments, some of the students
wished I had just 
\emph{told} them what was in those books.  If the course had been simply a
mathematics course, I could have done that.  But the course was a \emph{history}
course, and the whole point of history is to understand what people in
the past have \emph{thought.}  In saying this (and I shall say more
about it below), I am following the Oxford 
philosopher R.G. Collingwood (1889--1943), some of whose remarks on
history are in Appendix~\ref{app:Col}.

My attempts to communicate to my department what I was doing with the
course are in Appendix~\ref{app:cor}, along with the responses of the
sole person who did respond.

Appendix~\ref{app:background} consists of some notes on ancient Greek
mathematics that I put on the webpage of Math 303 at the beginning of
the semester. 

Appendix~\ref{app:Gk} gives the Greek alphabet (many Greek words are quoted in the main text).

\addsec{Apology}

If I were to teach Math 303--304 again (which I should like to do),
then I should 
certainly make some changes.  But the practice of reading and
presenting original sources, especially older ones, ought to be
maintained, for reasons including the following. 

\minisec{Scientific history}

Studying history does not mean learning to express \emph{opinions}
about what people of the past thought; it is learning what they
thought.  In saying this, I have in mind the distinction between
opinion and knowledge expressed by the character of Socrates in
Plato's \emph{Republic} \cite[II, p.~92; 506C]{Shorey}: 
\begin{quote}
Have you not observed that opinions (\Gk{d'oxai}) divorced from
knowledge (\Gk{>epist'hmh}) are ugly things?\footnote{\Gk{o>uk
    >'h|sjhsai t`as >'aneu >epist'hmhs d'oxas, <ws p~asai a>isqra'i?}} 
\end{quote}
A teacher can \emph{tell}
students what he believes Euclid thought, and the students can learn
to repeat these teachings; but the teachings are only opinions for the
students, if not for the teacher, unless the students test the
opinions against what Euclid actually wrote.   

A teacher's lectures on
math history may be useful for students' \emph{mathematics.}  In
\emph{A Comprehensive Introduction to Differential
  Geometry}~\cite[p.~vi]{MR82g:53003a}, Spivak writes, 
\begin{quote}
  Of course, I do not think that one should follow all the intricacies
  of the historical process, with its inevitable duplications and
  false leads.  What is intended, rather, is a presentation of the
  subject along the lines which its development \emph{might} have
  followed; as Bernard Morin said to me, there is no reason, in
  mathematics any more than in biology, why ontogeny must recapitulate
  phylogeny.  When modern terminology finally is introduced, it should
  be as an outgrowth of this (mythical) historical development.
\end{quote}
Spivak here is getting ready to teach mathematics, not history.  It is
useful for him and his readers to look at the 
history of the mathematics; but then that history will be adapted to
the needs of the mathematics.  In this case, as Spivak suggests,
history becomes a myth, a kind of story.  It may be an enjoyable or
useful story.  The story may be based on historical \emph{knowledge}
on the part of the storyteller-mathematician.  But then the story is not
designed to share all of that knowledge.  For the listener or reader
then, the  story---the myth---can only be a kind of \emph{opinion,} in
the sense of Plato.  It is no longer history.  

In \emph{The Principles of History}~\cite[pp.~12 f.]{Collingwood-PH},
Collingwood derides what he calls 
`scissors-and-paste' history:
\begin{quote}
There is a kind of history which depends altogether upon the testimony
of authorities\dots it is not really history at all, but we have no
other name for it\dots  History constructed by excerpting and
combining the testimonies of different authorities I call
scissors-and-paste history. 
\end{quote}
By contrast, the scientific historian will pay
attention to the latest research \cite[p.~35]{Collingwood-PH}:
\begin{quote}
\dots whereas the books mentioned in a bibliography for use of a
scissors-and-paste historian will be, roughly speaking, valuable in
direct proportion to their antiquity, those mentioned in a
bibliography for the use of a scientific historian will be, roughly
speaking, valuable in direct proportion to their newness. 
\end{quote}
What this means for math history, I
think, is that we must not treat Euclid's \emph{Elements,} say, as
the word of God or even the unaltered word of Euclid.  We may well pay
attention to Russo's argument in `The First Few Definitions in the
\emph{Elements}' \cite[10.15, pp.~320--7]{MR2038833} that the obscure
definition of straight line now found in the \emph{Elements} is the
work, not of Euclid, but of a careless copyist.  Still, there is
little point in reading Russo without reading the text associated with
Euclid's name. 

\minisec{Experience}

Most of our students will not be professional mathematicians.  The experience of
making sense of a difficult text, getting up in front of an audience,
and talking about their understanding, will be more useful to our students
than any particular piece of mathematical knowledge.  Indeed, I think this is
so, even for the students who \emph{will} be mathematicians.  At any
rate, as I said, my own undergraduate education consisted entirely of
this kind of learning.  Any ability I have now as a teacher was
nurtured by this experience. 

\minisec{Tradition}

Many people derive satisfaction from their membership in a group.  The
group might be a political party, a nation, humanity, or the
supporters of a football team.  If one is studying mathematics, I
suppose the best group to feel oneself a member of is the group of
\emph{mathematicians,} if not just the group of \emph{thinkers.} 
By actually \emph{reading} Euclid and his successors, we come to know that we
are part of a tradition that dates back thousands of 
years.  This point is reinforced when we consider that much of the
mathematics that our 
undergraduates learn was created by mathematicians who had read
Euclid.  Most of the course Elementary Number Theory I (Math 365) at METU, for
example, can be found in Gauss's \emph{Disquisitiones Arithmeticae}
(1801), of which
Wikipedia\footnote{\url{http://en.wikipedia.org/wiki/Disquisitiones_Arithmeticae},
  accessed June 19, 2010.} says: 
\begin{quote}
The logical structure of the \emph{Disquisitiones} (theorem statement
followed by proof, followed by corollaries) set a standard for later
texts.  
\end{quote}
This claim is not sourced, but it seems short-sighted: the
statement--proof, statement--proof style of mathematical writing is
found in Euclid, whom Gauss implicitly credits in his preface
\cite[p.~xvii]{Gauss}: 
\begin{quote}
Included under the heading ``Higher Arithmetic'' are those topics
which Euclid treated in Book VIIff.\ with the elegance and rigor
customary among the ancients\dots\footnote{The continuation of the
  sentence is, `but they are limited to the rudiments of the science.'
  There has indeed been progress since Euclid.} 
\end{quote}
We may not expect our students to write as well as Gauss, even if he
was only their age when he was writing; but they would do well to have
Euclid as a model (and Gauss).\footnote{In my experience, the best
  mathematical writers among students at METU grew up in the
  former Soviet Union.  I don't know if something in the Soviet
  tradition should be credited.  On page \pageref{Soviet} I quote a
  Soviet textbook that I used in high school.} 

\minisec{Changes}

Although mathematics has an age-old tradition, the subject has
changed since Euclid; but this can be difficult to 
see.  Obviously we have more theorems now; less obviously, the
\emph{spirit} of mathematics has changed.
In \emph{The Foundations of Geometry} \cite{MR0116216}, David Hilbert
appears to think that, in 
axiomatizing geometry, he is only refining the work of Euclid.  If
so, Hilbert is wrong.  We think today that Euclid's five postulates
are not in fact sufficient to justify all of his propositions; rather,
there are hidden assumptions, overlooked by Euclid, which Hilbert
uncovers.  Even in Proposition 1 of Book I of Euclid's
\emph{Elements,} there is an implicit assumption that two
circumferences, each containing the center of the other, must
intersect; this assumption is justified by no postulate.  But we have
no 
reason to think that Euclid is \emph{trying} to uncover all of his
`hidden assumptions'.  He is just writing down what is
true.  

One may say further the intersection of the circles in Euclid's I.1 is not a
\emph{hidden} assumption; the intersection is evident from the
diagram.  Today we think nonetheless that the existence the point of
intersection should still be noted separately in words or
symbols.  Evidently Euclid did not think the same way.
(See pp.~\pageref{one-one} and~\pageref{axioms} below.)  

Euclid also has no
notion of `non-Euclidean' geometry, so he has no need 
to distinguish his geometry logically from any other.
His postulates, along with his demonstrations, serve as a sort of
\emph{explanation} of why his propositions are true; but there is no
reason to expect the postulates and the demonstrations to provide a
\emph{complete} explanation,---if the notion of completeness even
makes sense in this context.  
Now, although he does not seem to say so clearly, it may well be that
\emph{Hilbert's} goal was a complete set of axioms for
geometry; but what could this have meant?  Since Hilbert's
\emph{Foundations,} several different 
notions of logical completeness have been defined.  Hilbert in fact
succeeded in writing down a \emph{categorical} set of axioms,
in that any two geometries in which the axioms are true
must be isomorphic to one another.  But we can hardly say that Euclid
aimed to do the 
same, if for him there was only one geometry in the first place.

At the beginning of `On teaching mathematics' \cite{MR1618209},
V.I. Arnold says:\footnote{I took the text
  from \url{http://pauli.uni-muenster.de/~munsteg/arnold.html} (July 30,
  2010), but the original source \cite{MR1618209} is not given there;
  I found it later.}    
  \begin{quote}
    Mathematics is a part of physics. Physics is an experimental
    science, a part of natural science. Mathematics is the part of
    physics where experiments are cheap. 

The Jacobi identity (which forces the heights of a triangle to cross
at one point) is an experimental fact in the same way as that the
Earth is round (that is, homeomorphic to a ball). But it can be
discovered with less expense. 

\sloppy
In the middle of the twentieth century it was attempted to divide
physics and mathematics.  The consequences turned out to be
catastrophic. Whole generations of mathematicians grew up without
knowing half of their science and, of course, in total ignorance of
any other sciences. They first began teaching their ugly scholastic
pseudo-mathematics to their students, then to schoolchildren
(forgetting Hardy's warning that ugly mathematics has no permanent
place under the Sun).  
  \end{quote}
I can't say that Arnold is right about mathematics in general.  He
may be right to say that \emph{Euclid's} mathematics is physics (as
physics is understood today).  Again, Euclid \emph{explains} why many
geometric propositions are true.  If one finds the explanations
inadequate, one may probe further; this does not make Euclid wrong.
Likewise, we have many explanations
of the motions of the heavens---explanations by Ptolemy, Copernicus,
Kepler, Newton, and Einstein.  None of these explanations is wrong.
Each new explanation builds on the preceding, as Hilbert built on
Euclid; on the other hand, each physicist had a different project:
each was looking for a different \emph{kind} of answer to the question,
`Why do the heavens appear as they do?'\footnote{For example, Ptolemy
  wanted to know what configurations of circular
  motions could account for the dance of the planets in the sky.  From
  Kepler, Newton understood 
  that the planets moved in ellipses about the sun; Newton sought a
  different kind of account of this, namely a law of force.}

In the preceding paragraphs, I have expressed opinions about Euclid,
Hilbert, and others.  I might 
express these opinions to students; but the students should question the
opinions while consulting Euclid himself (and 
Hilbert, and the others).  It may well be that a modern mathematician
misunderstands his ancient predecessors, because his main business is
to be a mathematician and not an historian.  If one just wants to
learn mathematics from the mathematician, that's fine; if one also
wants to learn history, one should go to the source.

\minisec{Proof}

Many of Euclid's propositions are propositions that I learned to prove
in high school, albeit from a modern textbook.\footnote{I didn't much like
the textbook, which was \cite{Weeks-Adkins}.  I wanted to read Euclid, and did so, first on my own,
and then at St John's.}  As I understood it, the
purpose of my high-school course was not so much to learn those
geometrical results themselves, but to learn the
\emph{possibility} of proving those results.  Unfortunately students
at METU seem never to get such a course, either in high
school or at METU (see p.~\pageref{proofs-new}).  We do teach proof;
but at the same time we are teaching modern mathematics, and
this complicates
things.  I count Descartes as modern.  Descartes gives us a
method of great power, which students begin learning in
their first semester at METU, in
Analytic Geometry (Math 115).  However, it is difficult to understand 
the method's power of \emph{proof.}\footnote{See
  \S\ref{sect:exam-I-1} for an exam that required application of
  Descartes's analytic geometry, as well as Newton's conception of
  quadrature.  Most students performed very poorly on this exam; later
  I discuss what to do about this.} 

Using analytic methods, how would we prove that the
base angles of an isosceles triangle are equal?  Given such a
triangle, we can set up a rectangular coordinate system in
which the vertices $A$, $B$, and $C$ of the triangle are respectively
$(0,a)$, $(-b,0)$, and $(c,0)$, where $a$, $b$, and $c$ are all
positive (Fig.~\ref{fig:DC}). 
\begin{figure}[ht]
\centering
\psset{unit=7mm}
  \begin{pspicture}(-4,-1)(4,5)
    \psline{->}(-4,0)(4,0)
    \psline{->}(0,-1)(0,5)
    \psline(-3,0)(0,4)(3,0)
    \psdots(-3,0)(0,4)(3,0)
    \uput[ul](0,4){$A$}
    \uput[ur](0,4){$(0,a)$}
    \uput[d](-3,0){$B$}
    \uput[ul](-3,0){$(-b,0)$}
    \uput[d](3,0){$C$}
    \uput[ur](3,0){$(c,0)$}
    \uput[u](-1.5,0){$b$}
    \uput[r](0,2){$a$}
    \uput[u](1.5,0){$c$}
  \end{pspicture}
  \caption{`Analytic' proof of Euclid's I.5}\label{fig:DC}
\end{figure}
Then $AB=AC$ if and only if
$a^2+b^2=a^2+c^2$, that is, $b=c$ (since both are positive).  In this
case, the angles at $B$ and $C$ 
have the same cosine, namely $b/\sqrt{a^2+b^2}$, so the angles
are equal.  Fine; but this argument uses notions not found till page
133 of the analytic geometry text~\cite{Karakas} used at METU; even then, the text just assumes familiarity with cosines, when full
knowledge of these will not come till a later course of mathematical analysis.

By contrast, for Euclid, the equality of the base angles of an
isosceles triangle is only Proposition 5 of Book I of the
thirteen books that make up the \emph{Elements.}  Notwithstanding the `hidden assumptions' mentioned above,
I don't know anything better than Euclid's `synthetic' geometry for
giving students a notion of what is a sound proof.

\minisec{Thrills}

I just mentioned the power of Descartes's analytic geometry.  It is a
thrill to learn this geometry from Descartes himself.  The thrill is
worth sharing with students; but it does not come 
cheap (my colleague's dismissive comments on p.~\pageref{fun} notwithstanding).  One needs to
have read Descartes's predecessors, and to have
read them faithfully---\emph{not} translated into modern, symbolic,
\emph{Cartesian} language.  But textbooks like Boyer \cite{MR996888} present
the old work in just this anachronistic way.

\minisec{Discoveries}

It may happen that new mathematics comes out of taking old mathematics
seriously.  I can only offer my own example: a paper about the
logic of vector spaces \cite{MR2505433},
directly
inspired by reading Euclid and Descartes.%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\footnote{The main 
  algebraic result is that if we have a vector space of dimension
  greater than $n$, then we can enlarge the scalar field so that the
  dimension of the space is reduced to $n$, while every set of $n$
  vectors that are linearly independent over the original scalar
  field remain independent over the new field.  
  Logically then, model-theoretically, the theory of $n$-dimensional vector spaces over algebraically closed fields is, in the appropriate signature, model-complete.
Casting all humility aside, I quote from the anonymous referee:
`The paper is well-written and very interesting. The structures are
indeed basic, yet I found several results which surprised me, and the
technical proficiency with which things are handled makes publishing
the presentation worth-while. For example realizing the geometric idea
of Descartes, while taking care to make all formulae existential, is
an example of the added value of the paper. To me personally even the
fact that the scalar field can be recovered from the parallelism
predicate was new.'} 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\minisec{Coverage}

The wonderful new \emph{Princeton Companion to
  Mathematics}~\cite{MR2467561} contains short biographies of 96
mathematicians, in chronological order.  The first 14 mathematicians
listed are: 
%\begin{quote}
\begin{multicols}{2}
  \begin{compactenum}[(1)]
\item
Pythagoras,
\item
Euclid,
\item
Archimedes,
\item
Apollonius,
\item
Abu Ja'far Muhammad ibn M\=us\=a al-Khw\=arizm\=\i,
\item
Leonardo of Pisa (known as Fibonacci),
\item
Girolamo Cardano,
\item
Rafael Bombelli,
\item
Fran\c cois Vi\`ete,
\item
Simon Stevin,
\item
Ren\'e Descartes,
\item
Pierre Fermat,
\item
Blaise Pascal,
\item
Isaac Newton.
\end{compactenum}
\end{multicols}
%\end{quote}
In my course, we read works of seven of the mathematicians on this
list.  (There is no extant text by Pythagoras.  In one class I
lectured additionally on Archimedes.) 
We also read two
other mathematicians, namely Th\=abit ibn Qurra and Omar Khayy\'am;
but we could have dropped the former, in line with the suggestion of Ali in
\S\ref{sect:comments-spring} (p.~\pageref{B2Ali}).  It is indeed a shame not to read any
of the remaining 82 mathematicians on the \emph{Princeton Companion's}
list.  But there just isn't time to read many more.  One could read
the work of many mathematicians in a source book like
Smith's~\cite{MR0106139} or Struik's~\cite{MR858706}, but I think the
coverage would be too superficial to be of much value. 

If it is desired, then Newton's contemporaries (such as Leibniz) and
successors can be studied in the courses that 
cover their work.  About courses I have taught at METU, I can say that Gauss
can be read in Elementary Number Theory I (Math 365), while Set Theory (Math 320) and Introduction
to Mathematical Logic and Model Theory (Math 406) can make use of van
Heijenoort's anthology of matheamatical logic~\cite{MR1890980}.  It just does not seem fair
to me to use a course like Math 303--304 to teach students about
mathematicians whose work they do not have time to \emph{know.}   

If one wants a royal road to a view of the grand sweep of mathematical
history, one can read Struik's \emph{Concise History of Mathematics}.  However, one might be uneasy with the author's materialistic
approach.  Struik writes for example: 
\begin{quote}
The main line of mathematical advance passed through the growing mercantile cities under the direct influence of trade, navigation, astronomy, and surveying.  The townspeople were interested in counting, in arithmetic, in computation.  Sombart labeled this interest of the fifteenth- and sixteenth-century burgher his \emph{Rechenhaftigkeit.}%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\footnote{[Struik's note:] W. Sombart, \emph{Der Bourgeois} (Munich and Leipzig, 1913), p.~164.  The term \emph{Rechenhaftigkeit} indicates a willingness to compute, a belief in the usefulness of arithmetical work.}  
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Leaders in the love for practical mathematics were the ``reckon masters,'' only very occasionally joined by a university man, able, through his study of astronomy, to understand the importance of improving computational methods. 
\hfill\cite[Ch.~V, \S6, p.~82]{Struik}

\enlargethispage{\baselineskip}
The rapid development of mathematics during the Renaissance was due
not only to the \emph{Rechenhaftigkeit}
of the commercial classes but
also to the productive use and further perfection of machines.\\ \mbox{}\hfill\mbox{\cite[Ch.~VI, \S1, p.~93]{Struik}}
\end{quote}
This explanation of the development of mathematics is perhaps correct; but it is hardly complete.
The development of mathematics is due, first of all, to
\emph{mathematicians} (who may or may not be `university men').
They may take their inspiration from various sources; but those sources do not \emph{cause} the mathematics to be created.  This is a point worth making in a course, and
it is a point made by the practice of reading \emph{mathematicians.} 
Struik may not disagree.  In the introduction of his history, he writes:
\begin{quote}
  The selection of material was, of course,  not based exclusively on
  objective factors, but was influenced by the author's likes and
  dislikes, his knowledge and his ignorance.  As to his ignorance, it
  was not always possible to consult all sources first-hand; too
  often, second- or even third-hand sources had to be used.  It is
  therefore good advice, not only with respect to this book, but with
  respect to all such histories, to check the statements as much as
  possible with the original sources.  This is a good principle for
  more than one reason.  Our knowledge of authors such as Euclid,
  Diophantus, Descartes, Laplace, Gauss, or Riemann should not be
  obtained exclusively from quotations or histories describing their
  works.  There is the same invigorating power in the original Euclid
  or Gauss as there is in the original Shakespeare, and there are
  places in Archi\-medes, in Fermat, or in Jacobi which are as beautiful
  as Horace or Emerson.
\end{quote}

\addsec{Possibilities for the future}

I would make some changes in teaching Math 303--304 again.  Here are
some notes about what might be done. 

If students are going to make presentations, they must prepare for these
conscientiously, with the understanding that a poor presentation will
disappoint not only their teacher.  \emph{Classmates} must challenge
students who try to fake their way through a proof.  Such challenges
happened occasionally in my class (see for example
p.~\pageref{confused}); I wish I could encourage students 
to make more of them, or (better) convince the speakers not to try to
fake their way.  (See pp.~\pageref{scolding}, \pageref{Besmir-silent},
and~\pageref{Melis-copying}, and \S\ref{sect:frust} 
[p.~\pageref{Emir-reading}], for some problematic days.)  

Some 
formal measures might be of help.  I did learn all 
of my students' names; but I found out too late that they didn't
always know \emph{one another's} names.  I sometimes tried arranging
the desks in a semicircle (see pp.~\pageref{desks}
and~\pageref{desks-again}).  I am told by Mehmet (whom I mentioned
above) that what I really must do is \emph{grade} the students on
their individual presentations.  Mehmet is not a student who needs
such a goad, but (if I understand him) other students do need prodding
by the threat of low marks.  In this case, I can only
hope that what students first do for marks, they may later do for
their own satisfaction.  As it was, I did tell students that they got
credit for \emph{attending} class; I did not say that students would
be graded on the \emph{quality} of their attendence and participation.

  Also (suggests Mehmet), students should know many weeks in advance
  what they will be presenting.  This should be possible, now that I
  know (from this very log) at what pace the course can proceed.
  Mehmet did think the practice of reading original sources like Newton
  should be continued.\footnote{Mehmet took both semesters
    of Math 303--304 and is now [June 2010] going to study for a doctorate in
    physics at Yale University.}

Classes proceeded more slowly than I expected, sometimes because
students had indeed not conscientiously prepared for them.  If one wants to
cover more material, one can skip some 
propositions in class, while holding the students responsible for
learning them independently.  Students might still work together, as Ece
suggested; see \S\ref{sect:comments-spring} (p.~\pageref{B2Ece}).  Still, it should be
noted that, though at the beginning of Math 304 I assigned
presentations to pairs or triples of students, the students
generally didn't work together.

The teacher could compromise his principles and make 
some presentations himself.  Indeed, as I noted above, I did this with
Archimedes.  I did it too with Book V of the 
\emph{Elements,} on proportion (see p.~\pageref{proportion}), and I
should have done it more; here, understanding the mathematics is hard 
enough, even if one is not trying to learn the mathematics straight
from Euclid.  The final exam of Math 303 showed that students had
\emph{not} generally learned Euclid's definition of proportion (the
one that must have inspired Dedekind's definition of the real numbers
\cite{MR0159773}).\footnote{Russo~\cite[pp.~46 f.]{MR2038833}\ ridicules historians like Heath, who are impressed that Euclid could
  have `anticipated' Dedekind's theory of irrational numbers.  Euclid
  didn't anticipate Dedekind; he \emph{taught} Dedekind, who read him
  in school.} 

It was hard for the students not to have much sense of what would be
on exams.  I didn't have much sense myself, when I started the course.
Nonetheless, in the first semester, students generally impressed me by
their understanding on exams; but in the second semester, they
disappointed me.  I was quite pleased with the problems I wrote on the
last two exams of Math 304; but in another year, some such problems
should be worked out with students in class.  

In lectures in the second semester, I compromised and \emph{stated}
for the students the results from Apollonius that we would need for
Newton.  Proofs of some of these results did however end up as the
exam problems just mentioned. 
Again, it would be better to make proofs of \emph{all} of these
results more clearly a part of the class, either in lectures or in
homework.  Proofs can use the streamlining that Descartes makes
possible, at least if the point is to be able to read Newton's
\emph{Principia.}  (In Math 304, I told the students out loud that
problems like those on the second exam could show up on the final; but
the students seemed not to do anything with this warning.) 

Anthologies like
Katz~\cite{Katz} are useful for identifying the old works of
mathematics that may be worth reading.  However, it may be misleading
to see a brief excerpt out of context.  It would be desirable (where
possible) to consider an anthology's selections in the context of the
larger works from which they have been taken.\footnote{I have
  therefore asked the METU library to order some of these larger works.}

Unfortunately most students of Math 304 probably will not have taken
Math 303.  Therefore it may be better to divide the contents of the
course not chronologically, but thematically, perhaps with geometry
and analysis in one semester, number theory and 
algebra in the other.  The former could start with Book I of Euclid's
\emph{Elements;} the latter could leave this book as background
reading, but start seriously with Book II. 

I had originally
thought of finishing Math 304 with Lobachevsky, but there was no
time, and anyway most of the students had not read Euclid, because
they had not taken Math 303.  In a
rearrangement of the course, Lobachevsky could be accommodated
somehow.  On the other hand, Lobachevsky is number 31 on the
(chronological) list in the \emph{Princeton Companion;} we would skip
a lot of great names to get to him.
\vspace{1cm}

\newlength{\sig}
\settowidth{\sig}{(Minor editing, June 20, 2011)}
\mbox{}\hfill
\begin{minipage}[t]{\sig}
  D. P.\\
Ankara\\
July 30, 2010\footnotemark
\end{minipage}\footnotetext{Minor editing, June 20, 2011, and later.}


%\addtocontents{toc}{\enlargethispage{1\baselineskip}}
%\tableofcontents

\part{Fall semester}\label{part:one}

%\input{math-303-log}

\chapter{Euclid}


\section{Sunday, October 4}

This is about a course I am teaching now: a course in which students
read and present Euclid in more or less the St John's style.  We have
had three hours of class so far, and I am excited to think that the
course may work out as I have hoped.

My own memories of the first-year math tutorial at St John's are dim.
Possibly Johnnies\footnote{As noted on p.~\pageref{Johnnies}, these
  remarks were originally addressed to St John's College
  alumni, who call themselves Johnnies.} are better prepared to read
Euclid, precisely 
because they haven't spent two or three years studying modern
mathematics as my students now have---and because Johnnies have come
to college expecting to read old books.

I am teaching at Middle East Technical University, in the capital of
Turkey.  The course I am writing about now is a third-year course that
comes with the rather pretentious title `History of Mathematical
Concepts I'.  I didn't ask to teach it.  But students wanted the
course to be offered, and a few weeks ago, one of the assistant chairs
of our department asked Ay\c se (my colleague and spouse) if she would
teach the course.  She didn't want to, but said that I might.  
Eventually I was offered the course.

I didn't want to teach the course as my colleagues had done in the
past: from a textbook like Boyer's \emph{History of
  Mathematics}~\cite{MR996888}. But I 
realized that the course could be an opportunity to read original
texts with students.  I decided to take it on.

Unfortunately our students are used to skipping class.  I think they
may pick up this idea from high school.  High school in Turkey does
not prepare students for the national university entrance exam; the
students take special lessons on evenings and weekends for this.
There the students learn all the tricks that their regular teachers
don't tell them.  So what's the point of spending time with a regular
teacher at all?

In my other courses at METU, I have not required attendance.  If
students want to study on their own, that has been fine; all that
matters is their performance on exams.  But in a course where the
whole point is to read and discuss Euclid, this won't do.

Students started registering for courses this semester on Wednesday,
September 23.  On that day, my course was open to third-year students,
with a capacity of 30.  In the afternoon, 23 students had registered.
Then I sent out an email to all math students, warning them of the
unusual nature of the course.  I threatened them with failure if they
did not come to class.  The authorities had not limited course
capacity as severely as I had wanted, so I tried to scare away
uninterested students.  On Thursday, the course was open to
fourth-year students, and the capacity was raised to 40.  This
capacity was 
reached, but now only 17 of those students were third-year.  It seems
I had driven off six students.

In the following week---last week---I met my class twice.  On the
first day, Tuesday, 24 students showed up.  On the second day, Friday,
only 18 showed up, though four of them had not come on Tuesday.  So I
have seen a total of 28 students, out of 40 who had registered (plus
three more who couldn't, but still wanted to come).  This is probably
typical.  Students don't have to commit to courses till the coming
week, `add-drop week' (when I shall make sure that those three extra
students can register, if they still want to).

On Tuesday, the first day of class, at first I didn't speak of
definitions, postulates, and common notions.  I just proved Euclid's
Proposition I.1 (to construct an equilateral triangle), and then I
asked what we had assumed in constructing the triangle.  Thus we
recognized a need for Postulates 1 and 3.  One student observed what
is famously missing from Euclid: we need to know that the two circles
in the construction intersect.  (I don't hold this to be a flaw in
Euclid.  As I think I have learned from Mr Thomas on the J-list, the flaw is to
think that Euclid is trying to present an axiomatic system as we
understand such things today.)

We went on to prove and discuss Propositions I.2 and 3.

Unfortunately here in Ankara one cannot order textbooks and expect
students to buy them.  One reason is that our department does not tell
us what we are teaching until it is too late to order books.  Another
reason is that books are expensive, and as long as the library has a
copy, students will have it photocopied.  As for Euclid, the library
seems to have lost some volumes of the Dover edition of Heath, and the
library hasn't bought the Green Lion edition yet.  So I have pointed
the students to several web editions of Euclid.  I could make my own
copies of the Green Lion or Dover edition available for photocopying,
but I won't.

Perhaps the recent Fitzpatrick translation of Euclid is the most
useful, if only because the author has put a \url{pdf} file on the
web.\footnote{\url{http://farside.ph.utexas.edu/euclid.html}}
However, I recall that Mr Thomas had some criticism of this edition,
or at least was dubious about the reliability of the Greek text that
accompanies the translation.  My perusal suggests that Fitzgerald is
more literal than Heath, but his footnotes may be misguided.

On Tuesday, on the web\footnote{\url{http://metu.edu.tr/~dpierce/Courses/303/}}
as well as in class, I hoped I had been clear enough about what was
expected from students.  On Friday however, it appeared that few
students had got around to actually reading Euclid; or perhaps they
were shy about admitting it.  One student agreed to present
Proposition 4 (what English-speakers may learn as SAS); but then it
transpired that she had read only \textbf{my} account of this proposition,
which I had also put on the web, perhaps by mistake.

For Proposition 5, nobody was initially forthcoming.  I drew a
triangle $ABC$ on the board, with $AB = AC$, and invited somebody to try
to prove the equality of angles $ABC$ and $ACB$.  This was a useful
exercise, for me at least.  One student came up and drew a circle
whose center was A and whose circumference contained $B$ and $C$.  Tolgay
tried to argue that the base angles of the triangle subtended equal
arcs of the circle, or were inscribed in equal arcs, or something like
that.

The state of Tolgay's mathematics may be like what I imagine
mathematics to have been before Euclid.  Tolgay understands that we
can prove some propositions from other propositions.  But he has no
clear notion of a systematic development, from a few basic principles,
of a whole body of mathematics.

When I visited St John's as a prospective student, my guide took me to
his dorm room, where he and his friends told me excitedly that whereas
in high school you were \textbf{told} that things were true, at St John's you
\textbf{proved} them.  So these were my people.  In fact my own high-school
geometry course had been rigorous,---so much so that I understood the
whole point to be not isosceles triangles and parallelograms, but
proof itself.  Still, during that course, I obtained a copy of Euclid, and I wished we
could read this instead of our regular textbook.

After geometry, I had a two-year course of calculus, where we proved
everything from the axioms for a complete ordered field (the so-called
real numbers).  As I understood it, this was what mathematics was all
about.

Our students at METU are among the best in Turkey, and they have
learned to do some math problems that I haven't a clue how to solve.
But apparently it's hard to ask about proofs on a multiple-choice
university entrance exam.  In any case, our students don't seem to
come to us with much notion of proof.  We have a first-year course
that is supposed to instil such a notion; but it is also supposed to
teach about `linear orderings' and `equivalence classes' and various
other modern abstract notions.  I have thought that students might be
better served by a course of reading Euclid.

The student Tolgay at the board, trying to prove I.5 with a circle,---
he can apparently think creatively, but if after two years of
university mathematics he can't catch on to what Euclid is about, even
just from attending an hour or so of my class, then I think there may
be something wrong with our department's program.

Maybe my criticism is premature.  In any case, I suggested that Tolgay
was trying to use some propositions that were indeed correct, but that
we had not proved yet.

Another student [name forgotten] came forward and tried to prove I.5 by drawing through
$A$ a straight line parallel to $BC$.  I pointed out that as yet we knew
nothing about parallel lines.

Finally it appeared that somebody had read Euclid.  Proposition 5 was
presented faithfully by Ali, who on Tuesday had transferred the \url{pdf} file of
Fitzgerald's Euclid from my flash drive to his.

I gave Pappus's proof of I.5, which is much easier to write down than
Euclid's, but perhaps harder to believe.  (Triangles $ABC$ and $ACB$ are
equal in all respects, by I.4.)  Students seemed to like this proof,
including Ali.  I asked whether Euclid might have known the proof.

Proposition 6 was also presented \emph{\`a la} Euclid, this time by a young
woman who had dropped out of my set-theory class last semester because
her father was dying.  I learned about the death after the course,
when Elif sent me an email thanking me for letting her pass anyway.  I
had been quite lenient that semester, because of another student, who
had had to undergo treatment for leukemia.  Elif wrote that she hoped
to do better in another course of mine; perhaps this semester she
intends to fulfill that pledge.

For Proposition 7, Cihan came forward to give the proof.  He was one
of the three students who had asked me to enlarge the capacity of the
course so that he could join.  Cihan seemed to have read the Euclid,
though he got confused.  I pointed out that Euclid's proof covered
only one case: another arrangement of the points was possible for
which the proof wouldn't work.  I left consideration of this case as an
exercise.

I think I myself did Proposition 8.

For Proposition 9 (to bisect an angle), Cihan eventually came forward
again, though seemingly without any notion of Euclid's construction.
He gave his own argument, assuming that a straight line could be
bisected (as in Proposition 10).  Ahmet came forward with a correct
method of bisecting a straight line, but he had trouble proving it
until Ali came up to help.

First Ali gave some advice from his seat, in Turkish.  I let the
Turkish discussion go on for a bit, then pointed out (in Turkish) that
not everybody knew Turkish.  I didn't mean only myself.  There was a
British student in the class, here just for the semester.  There's
also an Albanian student, though he may have learned Turkish.  There
is a student from Azerbaijan, but Azeri and Turkish are mutually
comprehensible.

By this time it was late on Friday afternoon, and our two hours were
up.  I had pointed out to the sleepy students that I hadn't chosen the
schedule.  I decided reluctantly to take volunteers for the next few
propositions, to be presented next Tuesday.

\section{Thursday, October 8}

Tuesday's class is from 16.40 to 17.30.  On Tuesday of this
week, I went to the classroom ten minutes early, to be able
to get on with the business of learning people's names.
Two students were there.  Yunus asked how many exams there
would be, and I repeated what I had written on the webpage:
one midterm and one final.  He asked what would be on the
exams, and I said I would ask for proofs, perhaps of
propositions from Euclid, perhaps of others.  He asked what
they (the students) could use in the proofs, and I said they
could use whatever Euclid used.  He asked whether that meant
they could bring a list of Euclid's propositions to the
exam, and I said no.  I said students should know the Greek
alphabet.  Yunus asked where they could get the alphabet,
and I repeated what I had added to the webpage that day:
They could get it from Wikipedia for example, or they could
download a page prepared by me.  I had brought a printout to
class, so I gave it to Yunus.

It is tedious to read the last paragraph, but it was tedious
to go through this dialogue with Yunus in the first place.
The system trains students to ask such questions; and anyway
I \textbf{am} going to have to assign letter grades at the end of
the semester.

By the time class was supposed to start, one other student
had shown up.  She said many students were coming from
another class, which was then being held in a building far
away, because of the ongoing renovations in our department's
building.  I asked about that other class, and then I
realized it was my own spouse's class!  Ay\c se assured me
later that she had ended class on time.  But you know,
students don't feel like rushing from one class to another.

\begin{sloppypar}
Soon more students came, and the presentations of
propositions started.
\end{sloppypar}

As I recall my own freshman math tutorial with Mr Kutler in
Annapolis, propositions were not preassigned to students.
I don't recall any problem finding volunteers on the spot,
although this may be because people like me were prepared to
volunteer if nobody else was.  The year before, when I
visited a math tutorial as a prospective student, a
volunteer was \textbf{not} forthcoming for Proposition $N$.  The
tutor then closed his eyes and brought his pencil down on
the list of students.  The student so picked asked
nervously, `Could I do Proposition $N+1$ instead?'  I suppose
he knew he had to present something, and he couldn't be
prepared for everything, so he prepared for that
proposition.

I had hoped the class I am teaching now could be like the
one I was a student in, or at least like the one I was
a prospective student in.  But as I suggested in my report on last
Friday's class, I gave up on that idea pretty quickly.  I
took volunteers on Friday for the following Tuesday's
propositions.

On Tuesday, therefore, the exchange-student Jeremy came
forth with I.11 all prepared.  He started writing out the
statement of the proposition, in what seemed to be a direct
quote.  I worried that he was just going to quote the proof
as well, but he didn't.  I raised several questions, during
the proof and afterwards.  `How do you know point $E$ exists?'
`Why doesn't Euclid just draw a circle to find $E$, rather
than appealing to Proposition~2?'

Things continued in this way.  I raised questions, trying to
suggest the kind of critical approach that I hoped the
students themselves would take.  But I suppose it's hard for
them to get critical about the seemingly basic propositions
we are going through.  

Besmir proved I.12.  Then Yunus proved I.13, that a straight
line set up on another makes angles equal to two right
angles.  He did it more tersely than Euclid, and I wondered
why Euclid's approach had an extra complication.  Yunus
argued that, if the straight line $AB$ is set up on $CD$ [so
that $B$ lies on $CD$], and $EB$ is set up at right angles to $CD$,
[and $E$ is on the same side of $CD$ that $A$ is on], then
\begin{equation*}
       CBA  + ABD = CBE + EBD = \text{2 right angles}.
\end{equation*}
(See Figure~\ref{fig:I.12}.)
\begin{figure}[ht]
\psset{unit=1.5cm}
  \centering
  \begin{pspicture}(-1,-0.3)(1,1)
%\psgrid
\psline(-1,0)(1,0)
\psline(0,0)(0,1)
\psline(0,0)(-0.5,1)
\uput[d](-1,0){$C$}
\uput[d](0,0){$B$}
\uput[d](1,0){$D$}
\uput[r](0,1){$E$}
\uput[l](-0.5,1){$A$}
  \end{pspicture}
  \caption{Euclid's I.12}\label{fig:I.12}  
\end{figure}
But Euclid argues (in prose that one might rewrite as
follows):
\begin{align*}
       CBA + ABE\phantom{{}+EBD}       &= CBE,\\
       CBA + ABE + EBD &= CBE + EBD,\\
             ABE + EBD &=\phantom{CBA+{}}       ABD,\\
       CBA + ABE + EBD &= CBA + ABD,\\
             CBA + ABD &= CBE + EBD = \text{2 right angles}.
\end{align*}
Why such length?  I don't know, unless, for Euclid, the sum
of angles $CBA$ and $ABD$ is not itself an angle, so it cannot
be immediately identified with the sum of $CBE$ and $EBD$.

Indeed, I have seen it said that one sign of Euclid's
greatness is his \textbf{not} trying to treat angles as if they
were the same sort of magnitudes as straight lines.  Today
one might say that the sum of right angles $CBE$ and $EBD$ is a
`straight angle', whose measurement is 180 degrees.  Then
`obviously' the sum of $CBA$ and $ABD$ is the same.  But this is
not obvious for Euclid, and rightly so.

It is a failing that Heath does not comment on I.13, except
for a remark on translating one clause.  I don't know if
this is Heath's failing, or a failing of other commentators
whose work he reviews in his own notes.

Damla, Friday's volunteer for I.14, did not show up for
class on Tuesday.  Cihan stepped up to prove the
proposition.  In the proof by contradiction, he established
\begin{gather*}
       CBA + ABE = \text{2 right angles},
\\
       CBA + ABD = \text{2 right angles};
\end{gather*}
then he concluded
\begin{equation*}
       CBA + ABE = CBA + ABD.
\end{equation*}
I asked how this was justified, and he mentioned Common
Notion 1:  Things equal to the same are equal to each other.
I pointed out that we don't know that \textbf{these} two right
angles are equal to \textbf{those} two right angles.  Somebody
pointed out that we had Postulate~4.

It has been two days since the course, and I don't
remember who brought up Postulate~4.  Se\c cil presented I.15 (`vertical
angles are equal'), and then I took volunteers for Friday's
class.  I spent half an hour after class talking with
one student and then another, about the Heath translation
versus the Fitzgerald, and about how ancient mathematics may
differ from our own.

\section{Friday, October 9}

My course is an `elective', and students who don't like it
can take another.  (For several electives, such as Ay\c se's on
graph theory, enrollment is maxed out.  But a third-year
elective on Lebesgue integration has only 18 students
registered;  a couple of fourth-year electives have less
than 15 students; another course will be closed for lack of
interest.)

On this last day of `add-drop' week, I am down to 22
registered students.  Of those 22, there are 6 whom I have
never seen in class.  Another came to the first class only.
Of the remaining 15, all but two have come to every class;
the other two came only to the most recent class, but seem
to be serious about the course: they are among the
volunteers to present propositions today.

In short, I seem now to have 15 interested students.  After
today's class, each of these~15 will have been up to the
board at least once to present a proposition.  Those other
students who didn't want to do this will have dropped out.

In my very first seminar at St John's, I was a bit surprised
when a tutor launched into an opening question without any
preliminary remarks about how things would be done.  But
that was fine of course, and I guess I'm following that
model today.

I plan to cover all of Book I, and then pick and choose (in
St John's fashion, as I recall it).  We should cover the
theory of proportion.  The students all know the `Euclidean
algorithm' for finding greatest common divisors; it would be
good for them to see this in its original presentation.\footnote{We
  eventually skipped this in class, however.} 

After that, I don't know.  We should do something of
Archimedes and Apollonius.  (I note that Euclid, Archimedes,
and Apollonius seem to be the main sources of examples in
Reviel Netz, \emph{The Shaping of Deduction in Greek Mathematics}~\cite{MR1683176}, the
book that Mr Thomas recommended.)   

At the Nesin Mathematics Village
in the summer of 2008, I presented much of Book I of
Apollonius in 12 hours of lectures.  This would translate
into four weeks of my present class; but it would be too
fast (as it was in 2008), and not in the right spirit.
Really students should be presenting propositions, even if
this seems to slow things down.

Next semester, I am likely to be assigned `History of
Mathematical Concepts II', which is supposed to start with
the Renaissance.  Ay\c se suggested I could spend all of the
present semester on Euclid, and all of next on Apollonius,
regardless of what the course catalogue says; but I don't
think I'll do this.


\section{Saturday, October 10}

On Friday, October 9, Euclid class began with Ahmet's
presentation of I.16: in a triangle, an exterior angle is
greater than either of the opposite interior angles.  Not
all students were present at the beginning of class.  I had
come to class with various things to say, but I could say
them any time.  I let Ahmet be the first speaker, so he
could have the experience of seeing latecomers walk in while
he talked (and so that they would see that they were
interrupting one of their classmates).

Ahmet and I are old friends: he took model
theory with me last fall and set theory last spring, and he
used to ask challenging questions after class.  He is a
double major in math and philosophy.  He and another
undergraduate named Burak inspired me to offer a reading
course this semester, in addition to the two courses I am
normally assigned.  (We intend to read together the late
Paul Cohen's book~\cite{MR0232676}---based on his lectures at Harvard---on
his proof of the independence of the Continuum Hypothesis.)

Ahmet expressed the equality of two lines by writing
\begin{equation*}
       \lvert AE\rvert = \lvert CE\rvert.
\end{equation*}
Other students had used this notation before.  I asked what
the vertical bars meant, and of course Ahmet said that they
denoted taking the \textbf{lengths} of the lines.  If I understood
his point, he said that we couldn't do the math unless we
had the abstract notion of length.  I observed that, as far
as I knew, Euclid didn't refer to length as such; he just
used ordinary language, saying $AE$ was equal to $CE$.  Nobody,
including myself, recalled that, in Definition 2, a line is
`breadthless length'.  Now, there is an argument (by Lucio
Russo, in \emph{The Forgotten Revolution}~\cite{MR2038833}) that Euclid didn't
write this or any other definition of line, but in any case,
Definition 2 says, not that a line \textbf{has} a length, but that a
line \textbf{is} a length.

In the book recommended by Mr Thomas called \emph{The Shaping of
Deduction in Greek Mathematics}~\cite{MR1683176}, Reviel Netz observes that
the \textbf{words} of Euclid and other ancient mathematicians do
not completely determine the diagrams.  
I assume now that
the interested reader can look at the diagram in Heath's
translation (now reproduced as Figure~\ref{fig:I.16}).
\begin{figure}
  \centering
  \begin{pspicture}(0,-1.5)(5,3)
    \psline(3.5,-1.5)(2,3)(0,0)(4.5,0)
\psline(0,0)(5,3)(3,0)
\uput[ul](2,3){$A$}
\uput[d](0,0){$B$}
\uput[dl](3,0){$C$}
\uput[r](4.5,0){$D$}
\uput[75](2.5,1.5){$E$}
\uput[ur](5,3){$F$}
\uput[l](3.5,-1.5){$G$}
  \end{pspicture}
  \caption{Euclid's I.16}\label{fig:I.16} 
\end{figure}
After Ahmet had proved angle $ACD$ greater than
angle $BAC$, he observed that, by the same construction, $BCG$
is greater than $ABC$, while $BCG$ is equal to $ACD$ by I.15.  I
said it wasn't necessary to repeat the construction, but
nobody seemed to get the point until I spelled it out: since
we have proved $ACD > BAC$, we have proved the general
statement of the proposition, another instance of which is
the inequality $BCG > ABC$.  No further proof is necessary.

When Ahmet was finished, before I could make any of the
general remarks I had prepared, Mehmet stood up to continue
with I.17: any two angles of a triangle are together less
than two right angles.  I let him proceed.  (Mehmet, Ahmet,
and Burak were by far the best students in set theory last
semester.  Mehmet is majoring in physics as well as math.)

It may have been during Mehmet's presentation that a student
whom I hadn't seen before, Rashad, mentioned 180 degrees.
Perhaps he thought I.17 was obvious, since all three angles
of a triangle add up to 180 degrees.  There was some
laughter when I pointed out that we didn't know anything
about degrees; perhaps the other students had got used to
hearing me say such things.

After Mehmet, I introduced the parts of a proposition that
are spelled out by Proclus in his commentary~\cite{MR1200456} on Book I of
the \emph{Elements.}  The relevant section of Proclus is quoted in
the introduction to the Green Lion edition of Heath's Euclid
(though unfortunately without a page number---it's 159 in
the cited translation).  So recent Johnnies should know the
parts of a proposition.  Netz gives them in his book as
well, where they are called
\begin{compactenum}[(1)]
\item\label{item:enunciation}
Enunciation (\Gk{pr'otasic}: what is to be proved in general terms);
\item
Setting out (\Gk{>'ekjesic}: the `givens' as labelled in the diagram);
\item
Definition of goal (\Gk{diorism'oc}: the `to prove');
\item
Construction (\Gk{kataskeu'h}: additional straight lines and so forth
that are needed in
the proof);
\item\label{item:apodeixis}
Proof (\Gk{>ap'odeixic});
\item
Conclusion (\Gk{sump'erasma}: a repetition of the enunciation, and what
Heath replaced with `Therefore etc.').
\end{compactenum}
In writing~\eqref{item:apodeixis} on the board, I asked Ahmet whether he had
encountered the word `apodictic' in a philosophy course; he
seemed to find the word vaguely familiar.

In listing the six parts, I just wanted to be clear that the
things we call `propositions' have a definite form, a form
which, for the sake of the reader, the writer might choose
to follow.

By the way, Netz in effect points out that our use of the
word `proposition' is an instance of metonymy.\label{page:metonym}  Properly the
proposition is only the enunciation: part~\eqref{item:enunciation} above.  Netz
argues that, for the Greeks, the `metonym' for the whole
six-part package was not the enunciation, but the diagram.
Now, the diagram is not one of those six parts.  One
might think that the diagram is like the \textbf{sight} or \textbf{look}
of a person, while the six written parts are the \textbf{voice} of
the person.  In any case, Netz's argument is
tenuous, or else I am reading too much into it.\footnote{Editing these
  remarks later, I don't
  remember why I thought Netz's argument tenuous.}  He observes
that, even when the same diagram \textbf{could} be used for two
propositions, it almost never is.  In `translating' the
\emph{Conics,} Heath~\cite{Heath-Apollonius} mutilates Apollonius
precisely by making one diagram fit many propositions.

In class, later presentations of propositions seemed to be
influenced a bit by Proclus's list of parts.  But I saw then
that I had a task for the future: to convince students not
to write down the `definition of goal' without being clear
that it hasn't actually been proved yet.  The students tend
to write formulas without writing any words to explain their
interrelations.  Proofs should be persuasive prose
compositions; but the students get little or no practice in
writing in school.  (Remember, the university entrance exam
is all multiple choice.)

In particular, in her proof of I.18 (in a triangle, the
greater side subtends the greater angle), \"Ozge used some of
the terminology from Proclus.  Next up was M\"ursel, who had
sat in only on the previous class before deciding to
register for the course.  His argument was quite detailed in
a good way.  But he kept looking at me, sitting at the side
of the room, until I reminded him that I wasn't the only
student in the class.  He had a soft voice, and I think it
was he whom I asked, `Do you think your classmates can hear
you?'  Those classmates said `No!'

Break time was coming up.  I stepped up to show that, as I
gather from Netz, Euclid tends to be mistranslated in
English.  The `setting out' of I.1\label{one-one} is not `Let $AB$ be the
given straight line' but rather `Let the given straight line
be $AB$':
\begin{quote}\centering
  \Gk{>'Estw <h doje~isa peperasm'enh <h AB.}
\end{quote}
Netz argues that Euclid is not \textbf{creating} a
straight line whose endpoints are defined to be $A$ and $B$;
rather, there is already a straight line; it is `given'; it
is there on the diagram (Figure~\ref{fig:I.1})
\begin{figure}[h]
\centering
\psset{unit=1.5cm}
\begin{pspicture}(3,2)
\pscircle(1,1)1
\pscircle(2,1)1
\uput[l](1,1){$A$}
\uput[r](2,1){$B$}
\uput[u](1.5,1.866){$C$}
\uput[l](0,1){$D$}
\uput[r](3,1){$E$}
\pspolygon(1,1)(2,1)(1.5,1.866)
\end{pspicture}
\caption{Euclid's I.1}\label{fig:I.1}
\end{figure}
and its endpoints are those that
you see near the letters $A$ and $B$.

What this means to me is that the recent translator
Fitzpatrick is wrong to say in a footnote to I.1, `The
assumption that the circles do indeed cut one another [at $C$]
should be counted as an additional postulate.'  That the
circles do cut one another is too obvious to need
postulating: you can see right there in the diagram that the
circles cross.  In our modern ideal, a proof proceeds by
mechanical application of formal rules to strings of written
symbols.  A proof is a list of such strings, with nothing to
do with any diagram.  But Euclid was not trying to write
such proofs.

So Fitzpatrick, like Heath, seems to have overlooked one
feature of the Greek.  But perhaps he did well to translate
Greek perfects as English perfects:  `Let the circle
$BCD$\dots\emph{have been} drawn.'  The diagram is not being
constructed as the reader reads; the diagram is already
there, it already \textbf{has been} constructed.  (And since the
reader is probably reading a scroll, the reader never needs
to flip a page.  Again I owe these observations to Netz.
But in the Green Lion edition~\cite{MR1932864} of Heath's Euclid, diagrams
are repeated on every spread where they are needed.  Is
anybody here young enough to have benefitted from this
feature?  The luxury of a scroll, the convenience of a
codex!  How's that for an advertising slogan?)

During the break, I noticed a sticker on the new classroom
windows that said \emph{\i s\i\ yal\i t\i ml\i\ \c cift cam.}  This meant
`heat \underline{\qquad} double glass', and one can guess the meaning that
fills in the blank, but I didn't recognize \emph{yal\i t\i ml\i.}  I
guessed that it came from a verb \emph{yal\i tmak}, which might in
turn be the causative form of a verb \emph{${}^*$yalmak.}  Melis and
Ali told me I was right on the former point, wrong on the
latter.  \emph{Yal\i tmak,} said Ali, meant `insulate' or
`isolate'.  Knowing that even my spouse confuses these two
English words, I wrote them on the board.  Jeremy from the
UK explained the distinction.  I observed that \emph{insula} was
island in Latin.  I recalled that, in ancient times, way out
on the tip of what is now Turkey's Dat\c ca Peninsula, there
was a city called Knidos.  We once talked on the J-list
about the Aphrodite of Knidos.  The Knidians tried to
`isolate' or `insulate' themselves by cutting a canal across
their isthmus, making their home an island; but they failed.
(The story is in Herodotus.  In class I observed that
\emph{insula} appeared in `peninsula', but forgot that the
Turkish word, \emph{yar\i mada,} was also literally `half-island'.)

We were still in the break, and not all students had
returned to the classroom, but I couldn't wait to talk about
Greek imperatives, particularly in the third person.  As I
had reviewed in Smyth's \emph{Greek Grammar}~\cite{Smyth} in the morning, there are
three kinds of Greek imperatives: present, aorist, and
perfect.  Moreover, the personal endings have different
forms in active and passive voice.  Turkish has just one
kind of imperative, and distinctions of voice are handled in
a different part of the verb.  (I didn't get into the middle
voice, but Turkish as well as Greek might be said to have
one.)  Actually, classical Greek apparently forms its active
perfect imperative periphrastically, as a participle + `let
it be' (\Gk{>estw}).  Turkish uses a similar construction for a
perfect imperative (as in \emph{Ge\c cmi\c s olsun} `may [your trouble]
have passed').  As in Greek, the Turkish for `let it be' is
one word (\emph{olsun}).

By this time class was officially going, and I emphasized my
main point: not that students should know all of the Greek
imperatives, but that Euclid used one of them in particular,
the perfect, to talk about things that had already been
constructed.  Nonetheless, Elif asked me to clarify the
distinction between the present and aorist imperatives.  I
hadn't actually used those terms, but had just written down
a sort of paradigm (Figure \ref{fig:imp})%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\footnote{My Greek is so rusty that I am not
  entirely confident of these forms, though Smyth does give the
  perfect middle/passive of \Gk{gr'afw} as a paradigm in his \P~406.}
  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
based on \Gk{gr'afw}. 
\begin{figure}[ht]\centering
\renewcommand{\arraystretch}{1.2}
  \begin{tabular}{|r|l|l|l|}\hline
        & active                & middle & passive \\\hline
present & \Gk{graf'etw}         & \multicolumn2{c|}{\Gk{graf'esjw}}\\\hline
aorist  & \Gk{gray'atw}         & \Gk{gray'asjw} & \Gk{graf'htw} \\\hline
perfect & \Gk{gegraf`wc >'estw} & \multicolumn2{c|}{\Gk{gegr'afjw}} \\\hline
  \end{tabular}
  \caption{Greek imperatives}\label{fig:imp}
\end{figure}
As for the
distinction, I just repeated briefly what I had gathered
from Smyth:
\begin{compactenum}[(1)]
\item 
\Gk{graf'esjw} let [it] be drawn [generally]
\item
\Gk{gray'asjw} let [it] be drawn [now]
\item
\Gk{g'egrafjw} let [it] have been drawn, \emph{\c cizilmi\c s olsun}
\end{compactenum}

Still, Taner\label{Taner} asked what the point of learning Greek was,
since math was hard enough in one's own language, and harder
still in English.  I said I just wanted him to know the
alphabet and to recognize the Greek words, like \Gk{gr'afw},
that are the source of our mathematical vocabulary.  I\label{DHH}
passed along the rumor that I had heard on the J-list from
Mr Billington, who had heard it from a British woman [Deborah Hughes
  Hallett] who
once taught in my department here in Ankara (though well
before my time): math students do better if they have
learned the Greek alphabet.  I pointed out in Turkish that,
unfortunately, I didn't know much Turkish, but anyway,
English was spoken at METU.  In English I said it would be
good if somebody would translate Euclid into Turkish; but
the translator should work from the original Greek, not the
English.\footnote{Ahmet Arslan, translator of Aristotles's
  \emph{Metaphysics} into Turkish~\cite{Metafizik}, confesses in his
  preface that he does not know Greek: he used French and English
  translations for his own work.} 

Taner was supposed to present I.20 (two sides of a triangle
are together greater than the third), so perhaps he was
nervous.  He had come to my office earlier in the day, not
having consulted the course webpage or actually looked at
Euclid yet.  He asked what the point of the course was, and
I told him as best I could.

In class, his presentation showed understanding, but was
rushed, and people besides me raised questions.  Ali asked
Taner to write down what he was proving.  
(See Figure~\ref{fig:I.20}.)
\begin{figure}
  \centering
  \begin{pspicture}(3,2.5)
    \psline(1,1)(3,0)(2.5,2.5)(0,0)(3,0)
\uput[ul](1,1){$A$}
\uput[l](0,0){$B$}
\uput[r](3,0){$C$}
\uput[ul](2.5,2.5){$D$}
  \end{pspicture}
  \caption{Euclid's I.20}\label{fig:I.20}
\end{figure}
In his haste he
got it wrong, instead of
\begin{equation*}
       BA + AC > BC
\end{equation*}
writing something like
\begin{equation*}
       BA > BC.
\end{equation*}
For a while I thought maybe he meant that we could \textbf{assume}
what he had written down, and this would have been correct,
as far as it went.  We got things straightened out in the
end.  As Taner returned to his seat, he apologized to me for
his poor English; actually his English was good.

At some point in the class, I went back to talk about I.16
(Figure~\ref{fig:I.16}).
Some commentators say it must use a hidden assumption, since
it doesn't use the parallel postulate, and yet it fails on
the surface of a sphere.  I observed that the proposition
fails if the point $F$ ends up below $BD$ (that is, on the other
side from $A$).  Ali said that couldn't happen, because then
$BF$ would not be straight---that is, it would not `lie evenly
with the points of itself'.  I suggested that, by the
`obvious' continuity principle, $BF$ and $BD$ would have two
points in common, in violation of a principle that we read
into the first postulate.

\c Ca\u gda\c s finished the presentations with I.21.  Then I observed
that somebody---it was the Epicureans, according to Proclus,
but I had forgotten this---somebody had ridiculed Euclid
for proving propositions like I.21 and especially I.20
(which it depends on), when they are obvious even to an ass, a
donkey.  This got some laughs, perhaps more so because I had
given the Turkish word for donkey, \emph{e\c sek}; the English might not
have been in their vocabulary.

Of those registered students whom I had never seen,
three---Rashad, Tu\u gba, and Nur---were in class finally.
They claimed they knew what they were getting in for by
registering for the course.  Everybody else had presented
propositions; so the newcomers became first in line for next
Tuesday.  After they agreed to this, other students were
eager to sign up for propositions, so I wrote down their
names too.

I ended the class by raising the question of whether we
could now improve the proof of I.8, the SSS rule for
congruence of triangles.  Euclid's proof involves `applying'
one triangle to another; I asked whether we could avoid
this.  (I had apparently taken up this issue in an essay I
wrote at St John's.  I dug up the essay this August when I
visited my mother and went through those old things of mine
that remained in her house.)

\section{Tuesday, October 13}

Today, in fifty minutes, we covered four
propositions.  Tu\u gba was first up, for I.22 (construct a
triangle, given the sides).  Or that's what I thought; she
seemed to have prepared I.20 instead.  Well, last class was
the first one she actually attended; but Taner proved I.20
then.  I asked Tu\u gba if she could prove I.22 anyway, but she
preferred to present a proposition next time.

I thought Rashad volunteered to prove I.22.  He came to the
board and started writing down---Proposition I.23.  This is
the one he had signed up for.  I said we couldn't do 23
before 22.  He wasn't prepared to prove 22 (even though 23
uses it); so he sat back down.

Cihan volunteered to prove I.22, and did it.  If the three
given sides are $A$, $B$, and $C$, then as Euclid points out, we
must have
\begin{align*}
       A + B &> C,&
       A + C &> B,&
       B + C &> A.
\end{align*}
But how do these requirements come into the picture?  I
discussed this with Cihan, and we drew three pictures, with
two non-intersecting circles each, showing what goes wrong
when any of the three inequalities above is violated.

But \"Ozge at least was brave enough to say, in effect, that
she didn't get it.  She came up to the board, and we
discussed the matter some more, until she was satisfied.

Rashad then proved I.23: to construct a given angle on a
given straight line and given point on it.  One does this
by constructing a triangle with the desired angle.  I
observed that I.22 hadn't spoken about \textbf{where} one could
construct the triangle.  I invited Rashad to show how to
construct the desired triangle in the \textbf{place} where it was
desired.  He did this, easily.

Proposition I.24 was Nur's: If two triangles have two sides
equal to two sides respectively, but the one included angle
is greater than the other, then the side opposite the one is
greater than the side opposite the other.  In Heath's
diagram (Fig.~\ref{fig:I.24}),
\begin{figure}[ht]
\mbox{}\hfill
\begin{pspicture}(0,-0.8)(2,2.5)
\pspolygon(0,0)(2,0.5)(1,2)
\uput[u](1,2){$A$}
\uput[r](2,0.5){$B$}
\uput[dl](0,0){$C$}
\end{pspicture}
\hfill
\begin{pspicture}(3,-0.74)(5,2.5)
\pspolygon(3,0)(5,0.5)(4,2)
\psline(3,0)(4,-0.24)(5,0.5)
\psline(4,2)(4,-0.24)
\uput[u](4,2){$D$}
\uput[r](5,0.5){$E$}
\uput[dl](3,0){$G$}
\uput[d](4,-0.24){$F$}
\end{pspicture}
\hfill\mbox{}
\caption{Euclid's I.24}\label{fig:I.24}
\end{figure}
 Nur chose the point $G$ merely to satisfy $DG = AC$.
When I asked if there were any other condition, she said No.
Eventually she just saw, or remembered, or saw in the
notebook that she had set aside, that angles $EDG$ and $BAC$
should be equal.

Nur completed the proof as Euclid gives it.  Then I asked:
What if the point $F$ happens to fall inside triangle $DEG$?  I
suppose I'm glad I hadn't consulted Heath's commentary on
this proposition; Heath does discuss this other case, and he
gives a simple proof---which I had overlooked.\footnote{Editing in December, 2012, I am not sure what `simple proof' I meant.  Probably it was the proof using I.21, which I did mention later.  Heath also points out that we may assume $AC\geq AB$; but that in this case we should prove that the point $F$ falls below $EG$.  He refers to De Morgan's proposed lemma, that every straight line drawn from the vertex of a triangle to the base is less than the greater of the two sides.  De Morgan reportedly proves this from his corollary to I.21, that among straight lines from a point to a give straight line, the perpendicular is the shortest, and they get longer as they move away from the perpendicular.  As an alternative to De Morgan's argument, Heath suggests `the method employed by Pfleiderer, Lardner, and Todhunter', whereby, in the original diagram for I.24, we let $DF$ (extended if
 necessary) meet $EG$ at $H$ (see the figure at the end of this note).
  Then $DHG>DEG$ by I.16, and $DEG\geq DGE$ by I.18, so $DHG>DGE$, and therefore $DG>DH$ by I.19, so $DF>DH$, and $F$ must fall below $EG$, as desired.  Finally, Heath suggests a `modern alternative proof' of the proposition, all cases at once; this involves the bisector of angle $FDG$, meeting $EG$ at $H$.
  \begin{center}
  \mbox{}\hfill
  \psset{unit=8mm,labelsep=4pt}
\begin{pspicture}(3,-0.74)(5,2.5)
\pspolygon(3,0)(5,0.5)(4,2)
\psline(3,0)(4,-0.24)(5,0.5)
\psline(4,2)(4,-0.24)
\uput[u](4,2){$D$}
\uput[r](5,0.5){$E$}
\uput[dl](3,0){$G$}
\uput[d](4,-0.24){$F$}
\uput{2pt}[ul](4,0.25){$H$}
\end{pspicture}
\hfill
\begin{pspicture}(3,-0.74)(5,2.5)
\pspolygon(3,0)(5,0.5)(4,2)
\psline(3,0)(4,-0.24)(5,0.5)
\psline(4,2)(4,-0.24)
\psline(4,2)(3.56,0.14)(4,-0.24)
\uput[u](4,2){$D$}
\uput[r](5,0.5){$E$}
\uput[dl](3,0){$G$}
\uput[d](4,-0.24){$F$}
\uput{1pt}[ul](3.56,0.14){$H$}
\end{pspicture}
\hfill\mbox{}
%\caption{Euclid's I.24}\label{fig:I.24}
\end{center}
}

Nur and others claimed that, in the picture I had drawn (as in Fig.~\ref{fig:I.24alt}),
\begin{figure}[ht]
\mbox{}\hfill
\begin{pspicture}(0,-0.5)(2,2.5)
\pspolygon(-0.1,0)(2,0)(0,2)
\psline(0,2)(0.7,0.12)(-0.1,0)
\psline(0.7,0.12)(2,0)
\uput[u](0,2){$D$}
\uput[dr](2,0){$E$}
\uput[ur](0.7,0.12){$F$}
\uput[dl](-0.1,0){$G$}
\end{pspicture}
\hfill
\begin{pspicture}(0,-0.5)(2,2.5)
\pspolygon(-0.1,0)(2,0)(0,2)
\psline(0,2)(0.7,0.12)(-0.1,0)
\psline(0.7,0.12)(2,0)
\uput[u](0,2){$D$}
\uput[dr](2,0){$E$}
\uput[ur](0.7,0.12){$F$}
\uput[ul](-0.1,0){$G$}
\psline(-0.1,0)(-0.14,-0.8)
\psline(0.7,0.12)(1.05,-0.76)
\uput[ur](1.05,-0.76){$H$}
\uput[ul](-0.14,-0.8){$K$}
\end{pspicture}
\hfill\mbox{}
\caption{Alternative diagram for I.24}\label{fig:I.24alt}
\end{figure}
with $F$ inside $DEG$ (but $DF = DG$), it is obvious that $EF$ is
shorter than $EG$.  Could they prove it?  Well, they didn't
think it needed proof.  I claimed it did.

Somebody could have cited I.21, whereby $EF + FD < EG + GD$,
which yields the claim.  But nobody did.  I suggested
extending $DG$ and $DF$, so that the \textbf{exterior} angles at the
base of isosceles triangle $DFG$ are equal.  Thus the second
part of I.5 could be used, in the same way that the first
part is used for the case that Euclid does give.  (I'm glad
I hadn't consulted Heath, because I might not have noticed
this argument if I had.)

Checking Heath now at I.7, I learn that Proclus similarly
proved the omitted case of that proposition.  (I have the
Proclus, but have not been reading him systematically;
maybe I should.)

By this time, the class was almost over.  I don't know if
the students were happy that we covered so few propositions.
In the remaining minutes, Cihan proved I.25 (the converse of I.24), which has a
purely logical proof, I would say.  There was no time to
observe that I.6 (equal angles are subtended by equal
sides) could have postponed till after I.18 (the greater
side subtends the greater angle); then this with I.5 (equal
sides subtend equal angles) would have allowed a purely
logical proof of I.6.

As I say, there was no time to discuss the alternative proof
of I.6.  Maybe next time, or not.  I don't mind going
slowly, if there are things to say; but I do want to get to
mathematics that is more difficult in a conventional
sense---such as the theory of proportion.

Perhaps next time I'll mention the Steiner--Lehmus
Theorem
without naming it---so the students can't just look up the
proof, although that's what I did.  Sam Kutler told us about
this theorem when I was in his math tutorial, but we didn't
discuss the proof.  Conway's argument---which I read
today---that there can be no \textbf{direct} proof is
intriguing.

\vspace{\baselineskip}
{\footnotesize 
\emph{Note:}
The Wikipedia article on the Steiner--Lehmus Theorem provides a
  link to\\
\mbox{}\hfill
\url{http://www.mathematik.uni-bielefeld.de/~sillke/PUZZLES/steiner-lehmus}
\hfill\mbox{}\\
  (accessed December 6, 2012), which discusses a number of proofs.
  Here is an adaptation of the first of these proofs (said to be taken
  from Coxeter's \emph{Introduction to Geometry,} which cites a letter
  from H.G. Forder): 
\psset{labelsep=3pt}
\begin{center}
\psset{unit=2cm}
\begin{pspicture}(1,1)
\pspolygon(0,0)(1,0)(1,1)
\psline(0,0)(1,0.42)
\psline(1,0)(0.5,0.5)
\psline(1,0)(0.71,0.71)
\psline(0.54,0.54)(0.65,0.27)
\uput[r](1,1){$A$}
\uput[l](0,0){$B$}
\uput[r](1,0){$C$}
\uput[r](1,0.42){$D$}
\uput[l](0.5,0.5){$E$}
\uput[u](0.71,0.71){$F$}
\uput[u](0.54,0.54){$G$}
\uput[d](0.65,0.27){$H$}
\end{pspicture}
\end{center}
Assume triangle $ABC$ is not isosceles.  We may assume the angle at
$B$ is smaller than the angle at $C$ [I.18].  The straight line $BD$
bisects the angle at $B$, and $CE$ bisects the angle at $C$ [I.9].  We
construct angle $ECF$ to be equal to $DBA$ [I.23], we let $BG=CF$
[I.3], and we let angle $BGH$ be equal to angle $BFC$ [I.23].  Then
triangle $BGH$ is equal in all respects to triangle $CFE$ [I.26].  In
particular, $BH=CE$.  Now, $GH\parallel FC$ [I.28].  Also $BG<BF$,
since $BF>FC$ [I.19].  Thus $BH<BD$.  Therefore $CE<BD$.  The
contrapositive is that if $CE=BD$, then the triangle \emph{is}
isosceles.  We have not used the Fifth Postulate. 

An algebraic argument runs as follows.  We first establish Stewart's Theorem:
\begin{center}
\psset{unit=1.5cm}
\mbox{}\hfill
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\pspolygon(-1.2,0)(0.8,0)(0,1)
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\uput[u](0,1){$A$}
\uput[d](-1.2,0){$C$}
\uput[d](0.8,0){$B$}
\uput[d](-0.4,0){$D$}
\uput[ul](-0.6,0.5){$b$}
\uput[ur](0.4,0.5){$c$}
\uput[l](-0.2,0.5){$d$}
\uput[u](-0.8,0){$n$}
\uput[u](0.2,0){$m$}
%\uput[d](-0.2,0){$a$}
\end{pspicture}
\hfill
\begin{pspicture}(-1.2,-0.1)(1,1.1)
\pspolygon(-1.2,0)(0.8,0)(0,1)
\psline(0,1)(0,0)
\psline(0,1)(-0.4,0)
\uput[u](0,1){$A$}
\uput[d](-1.2,0){$C$}
\uput[d](0.8,0){$B$}
\uput[d](0,0){$E$}
\uput[d](-0.4,0){$D$}
\uput[ul](-0.6,0.5){$b$}
\uput[ur](0.4,0.5){$c$}
\uput[l](-0.2,0.5){$d$}
\uput[u](-0.8,0){$n$}
\uput[u](0.35,0){$m-x$}
\uput[u](-0.2,0){$x$}
%\uput[d](-0.2,0){$a$}
\end{pspicture}
\hfill\mbox{}
\end{center}
For now, $b$, $c$, and $d$ are simply lengths; in particular, they are
positive elements of $\R$.  We let $a$ be the length of $CB$.  But we
conceive of $n$ as a \emph{vector} in $\R$ from $C$ to $D$, and $m$ as
a vector from $D$ to $B$, so that 
\begin{equation*}
a=n+m.
\end{equation*}
We may allow $D$ to fall to the left of $C$, in which case $n$ is
negative; if to the right of $B$, then $m$ is negative.  Similarly $x$
is a vector from $D$ to $E$, where $E$ is the foot of the
perpendicular dropped from $A$.  Now we compute: 
\begin{gather*}
d^2-x^2=b^2-(n+x)^2,\\
	d^2=b^2-n^2-2nx,\\
	d^2=c^2-m^2+2mx,\\
	ad^2=b^2m+c^2n-mn^2-nm^2,\\
	a(d^2+mn)=b^2m+c^2n.
\end{gather*}
In case the cevian $AD$ is an angle bisector, we have $b/c=n/m$; this
will be true, even if $AD$ bisects an \emph{external} angle at $A$,
provided we then allow $b/c$ to be negative (because $n/m$ will be
negative): 
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\uput[u](2,2){$A$}
\uput[l](0,0){$C$}
\uput[dr](3,1){$B$}
\uput[dr](2,0.67){$D$}
\uput[r](6,2){$D'$}
\uput[ul](1,1){$b$}
\uput[ur](2.5,1.5){$c$}
\uput[dr](2.5,0.833){$m$}
\uput[dr](4.5,1.5){$m'$}
\uput[dr](1,0.33){$n$}
\end{pspicture}
\end{center}
In this case then,
\begin{align*}
m&=a-n=a-\frac{bm}c,&
m&=\frac{ac}{b+c},&
n&=\frac{ab}{b+c},
\end{align*}
so Stewart's Theorem yields
\begin{equation*}
	d^2=bc\biggl(1-\Bigl(\frac a{b+c}\Bigr)^2\biggr)
	=bc\cdot\frac{(b+c)^2-a^2}{(b+c)^2}
	=\frac{bc(b+c+a)(b+c-a)}{(b+c)^2}.
\end{equation*}
Here, if $AD$ is the bisector of the \emph{external} angle at $A$,
then $b+c$ will be the \emph{difference} between the lengths of $AC$
and $AB$, and this is less than $A$ [I.20]; but also $bc<0$, so $d^2$
is still positive.  In any case, $a+b+c$ is positive. 

Suppose the (internal) bisector of the angle at $B$ also has length $d$.  Then
\begin{equation*}
0=\frac{a(c+a-b)}{(a+c)^2}-\frac{b(c+b-a)}{(b+c)^2}.
\end{equation*}
Here, if $AD$ is the bisector of the \emph{external} angle at $A$,
then we must take $b$ to be negative, while $a$ and $c$ are positive.
In any case, the right-hand member here has $a-b$ as a factor.
Indeed, the right-hand member is equal to the following. 
\begin{gather*}
\frac{c\bigl(a(b+c)^2-b(a+c)^2\bigr)+(a-b)\bigl(a(b+c)^2+b(a+c)^2\bigr)}{(a+c)^2(b+c)^2},\\
(a-b)\cdot\frac{c(c^2-ab)+(a+b)(c^2+ab)+4abc}{(a+c)^2(b+c)^2},\\
(a-b)\cdot\frac{c^2(a+b+c)+ab(a+b-c)+4abc}{(a+c)^2(b+c)^2}.
\end{gather*}
In case $b>0$, the other factor is positive, so $a-b=0$, and the
triangle $ABC$ is isosceles. 

John Conway (as quoted at the link above) argues that this is not a
direct proof, since our computations alone do not yield $a=b$, but we
need also that $b$ is positive.  If $b<0$, then we may have for
example $a=c=1$, in which case 
\begin{align*}
b^2+5b+2&=0,&
b&=\frac{-5\pm\surd17}2,&
b&\approx -0.438
\end{align*}
(the other root is too large in absolute value).
} 

\section{Friday, October 16}

Today,  Propositions I.26--33 were presented by
Tolgay, Besmir, Jeremy, Se\c cil, Elif, \"Ozge, Taner, and Yunus,
respectively.

I had an evil thought today: that some students might be
making long presentations in order to make sure we don't
cover much material.  It's probably false, but Ay\c se has
learned from students that they have tricks for slowing down
her classes by getting her to talk about irrelevant things.

Tolgay proved I.26: the triangle-congruence theorem that we
might call ASA and AAS.  He got a bit confused at the board.
I'm not sure how many of his classmates were paying close
attention on a warm Friday afternoon in October.  Did they
see clearly where he needed to go, as he did not?  At least
Tolgay was learning: what you think you understand when you
are by yourself may become strange when you are standing up
in front of others.

\emph{I} dragged things out after Tolgay wrote:
\begin{gather*}
       \angle AHB = \angle DFE;
\\
       \angle ACB = \angle DFE;
\\
       \angle AHB > \angle ACB, 
\end{gather*}       
which is impossible.  (See Fig.~\ref{fig:I.26}.)
\begin{figure}[ht]
\mbox{}\hfill
\begin{pspicture}(0,-0.5)(2,2.5)
\pspolygon(0,0)(2,0)(0.5,2)
\psline(0.5,2)(1.5,0)
\psline(2,0)(0.375,1.5)
\uput[u](0.5,2){$A$}
\uput[dl](0,0){$B$}
\uput[dr](2,0){$C$}
\uput[ul](0.375,1.5){$G$}
\uput[d](1.5,0){$H$}
\end{pspicture}
\hfill
\begin{pspicture}(0,-0.5)(2,2.5)
\pspolygon(0,0)(2,0)(0.5,2)
\uput[u](0.5,2){$D$}
\uput[dl](0,0){$E$}
\uput[dr](2,0){$F$}
\end{pspicture}
\hfill\mbox{}
\caption{Euclid's I.26}\label{fig:I.26}
\end{figure}
I wanted him to spell out the intermediate conclusion that
$\angle AHB = \angle ACB$.

Talking to me once after class, Jeremy heard me say that I
thought students should be able to express Euclid's
enunciations in their own words, while being reasonably
faithful to Euclid's style.  In presenting I.28, Jeremy did
write down the enunciation in his own words; and he remarked
that he was doing so.  But really, I didn't think he needed
to take the time to do that in writing; like most of his
classmates, he could have drawn a picture and just worked
with that.  But Jeremy does seem to aim at giving crisp
presentations; so perhaps he sets a good example.

After \"Ozge proved I.29 (parallel lines make equal alternate
angles, etc.), we took a break.  During the break, Cihan
asked me if this was the first time we had used the Fifth
Postulate, and if so, didn't that mean that everything
before I.29 was true without the Fifth Postulate.  Yes, I
said, except that (as I had suggested at the time) I.16
raises a question for some commentators.

Ali wasn't sure that we had really needed Postulate 5 in I.29.
Here, finally, I said we should talk about this in class.
Ali seems to be one of the most attentive students; if he is
confused at this point, others must be.

In class then, I had Ali raise his question.  I reviewed a
case of a `logical' theorem, such as I.19 [`In any triangle the greater angle is subtended by the greater side'], which follows
immediately from I.5 [equal sides subtend equal angles] and I.18 [the greater side subtends the greater angle].  I asked whether I.29 could
follow in such a way from its converse, I.27.  Ali, at
least, agreed that it couldn't.  I couldn't resist
mentioning Lobachevsky and his working out the consequences
of the negation of I.29.

After Taner presented I.32 (an exterior angle is equal to
the two opposite interior angles, and all interior angles
are equal to two right angles), I asked him whether he had
been familiar with the fact before.\label{proofs-new}  Yes, he said, but he
had never proved it.  I had been wondering just how familiar
all these propositions seemed to the students.  They
confirmed for me my understanding that, on the university
entrance exam, they just have to be able to compute numbers
(perhaps angle measures in a geometrical figure).

I told Taner I was sorry he hadn't proved these propositions
before.  He said he was sorry too.

At some point I asked who was in the geometry class of one
of my colleagues, Cem \emph{Bey.}  Cihan, Taner, and maybe some others said
they were.  

What are you doing? I asked.  

Proofs! said
Taner.  He didn't know the English, he said, but they were
studying things like the \emph{a\u g\i r merkezi.}

Oh, the center of gravity, I said.

Yes, and the Nine Point Circle, said Cihan.

I wrote down the statement of the Steiner--Lehmus Theorem
without naming it; but Cihan knew the name---from that
Geometry class, apparently.  Anyway, I need to talk to Cem \emph{Bey.}

On other days, there have been only just enough desks\label{desks} in
our classroom.  If that had been true today, I was going to
push them all to the edge.  In fact, I told this to some
students before class, and one of them asked, `Why?'

I didn't say it was because of talk of Harkness Tables and
so forth on the J-list.  I just said everybody was a sort of
teacher in class, and I didn't want people sitting behind
others and chit-chatting---as had happened a little bit
before.

Well, there were a lot more desks in the room this time,
and I didn't feel like pushing them all around, but maybe I
should have.  Rashad sat in back playing with his cell
phone.  Taner and Se\c cil, conferring over a desk, said they
were working on the mathematics; I said we should all talk
about it together, but nothing really came of this.

I was going to say here that my class was not quite St
John's, but then I remembered a Johnny classmate who used to
sit at the seminar table reading science fiction novels in
her lap.

\section{Thursday, October 22}

On Tuesday, October 20, I did arrange the classroom desks in a
sort of semicircle.  When some students walked in late and tried to go
behind, their classmates indicated that they should find a seat in the
circle.  Students moved their desks to accommodate the newcomers.

Propositions I.34--39 were presented by Ahmet, M\"ursel, Tu\u gba, Melis, Nur,
and Rashad.

I.35 (`parallelograms which are on the same base and in the same parallels
are equal to one another')  is supposedly the place where the meaning of
equality changes.  Indeed, M\"ursel wrote the statement on the board as I
have quoted it, but after `equal' he inserted a parenthetical comment:
`(equal in area)'.  After his presentation, I asked what `area' was,
claiming that we had no notion of area as a number (even with units, as
$5\ \mathrm{cm}^2$).  Euclid just says the two parallelograms are equal, and I don't
see any difference in meaning from I.4, where the statement is:
\begin{quote}
If two triangles have the two sides equal to two sides respectively, and
have the angles contained by the equal straight lines equal, they will
also have the base equal to the base, the triangle will be equal to the
triangle, and the remaining angles will be equal to the remaining angles
respectively, namely those which the equal sides subtend.
\end{quote}
If by saying the two triangles were equal here, Euclid had meant they were
congruent, then he would not have bothered to mention the further
equalities of sides and of angles.

It could have been after I.36, but I think it was after I.38 when Jeremy
asked: why didn't Euclid just use the method of `application', as in I.4
and I.8?  After all, I.36 is the variant of I.35 where the parallelograms
are on \textbf{equal} bases.  Then I.37 is like I.35, but is about \textbf{triangles} on
the same base; and I.38 is about triangles on equal bases.

So in proving I.36 and I.38, why didn't Euclid just say, Apply one figure
to the other so that they have the same base?  I just suggested that
Euclid preferred to avoid using the method of application whenever
possible.  That's perhaps an inadequate answer, since he could have
avoided the method in proving I.8, but didn't.

It occurs to me only now that a better answer to Jeremy's question would
be that, after applying one figure to the other, one can no longer be sure
without proof that the two figures are in the same parallels.

Jeremy addressed his question to me, calling me `Sir'.  If it happens
again, I'll ask him at least to use the Turkish \emph{Hocam} (`My
teacher'; but even bus drivers are \emph{Hocam} at our university).  Really Jeremy
should have addressed the whole class with his question; but I don't know what more I can
do than I have already done to raise interest in general discussion.

After their presentations, students \textbf{are} uttering a \emph{pro forma} `Any
questions?' to their classmates.

It is interesting to see the different styles of students at the board.
Some make their arguments almost entirely out loud, and I ask them to
write down some of the details.  Others try to write down everything, and
I suggest that they can leave some things out (especially the general
enunciation).

I hope I'm not micromanaging.  From freshman mathematics at
St John's College, I do recall a time when one student was drawing a straight
line from $A$ to $B$, and Mr Kutler suggested it would be better drawn from $B$
to $A$.

Back in Ankara, at the end of class, I took volunteers for the remaining
propositions in Book I.  Already \c Ca\u gda\c s had been lined up for I.40; but
since that proposition seems not to be a Euclid original, I asked \c Ca\u gda\c s
if he would do I.41 instead.  He agreed, mentioning his awareness that it
was Heiberg who had declared I.40 an interpolation.

After Book I, I plan to lecture on the theory of proportion.\label{proportion}  There are
two challenges to reading Euclid: the mathematics, and Euclid's style of
doing mathematics.  For proportion at least, I shall try to mitigate the
latter challenge through the use of symbolism.  I am aware that written
symbolism is subject to the same criticism that Socrates levelled at
writing in general:  It allows the brain to get lazy.

In any case, I want to challenge the students with Apollonius as soon as
possible.

I talked some time on Tuesday morning with Cem \emph{Bey,} who seems quite
pleased to learn about my course.  He asked if I would cover non-Euclidean
geometry in my own course, since, for himself, the discovery of
non-Euclidean geometry was more important than landing on the moon.  That
sounded naive to me: of course the geometry was more important! In any
case, I said I would do non-Euclidean geometry in the spring semester, if
I was assigned the modern math history course.

Cem \emph{Bey} confirmed that there was no Turkish translation of Euclid; there
had just been an Ottoman translation of a geometry text used at Sandhurst.
(Cem \emph{Bey} is, among other things, a scholar of Ottoman mathematics, and he detests the Turkish
language reforms, which have deprived him of the expressiveness of Ottoman
Turkish.)

Regarding Euclid, Cem \emph{Bey} mentioned that we did have a colleague whose
mother tongue was Greek; but I already knew this.  (She was born in
Rhodes, but took Turkish citizenship some time after coming to Ankara to
study; she now needs a visa to visit her family back in Rhodes.)

Cem \emph{Bey} was impressed to learn that there were `still' colleges like St
John's in the West.  Though he considered German almost as his mother
tongue, he regretted not having been able to read Kant.

I don't really need more projects, but now I want to translate Euclid into
Turkish.  At least Turkish has a better way than English does for
expressing Euclid's passive third-person perfect imperatives.

\section{Saturday, October 24}

My next installment is fairly rushed; but as usual I want to keep a record
while the memory is fresh.

On Friday, October 24, we finished Book I.  I don't know how well the
students appreciated the \textbf{poetry} of Euclid, but I think they enjoy the
class reasonably well.  At least, I never have any trouble getting
volunteers for presenting propositions.

I have wondered if it is tedious for some students to see their classmates
work out laborious proofs of `obvious' facts.  Even if the students accept
that proofs are necessary, maybe they get frustrated to see a classmate
struggle with what should be an `easy' proof.  But I don't have any real
evidence of this, and anyway the proofs are getting harder now.

Before class, Ali asked if we would sit in a semicircle again.  He didn't
mind sitting in the front row, he said, but maybe other students did.  I
decided just to leave the chairs as they were, in a rectangular array.  I
didn't notice any chatting in the back this time.

Moreover, I had a couple of guests, one of whom I knew from the Nesin Mathematics Village in
the summer of 2008.  She's only now a first-year student in our
department.  I don't know what she had heard about my class, but she came
and listened and took notes.  She had a male companion, but I don't know
if he was equally interested.  Indeed, my spouse noticed a guest in her
linear algebra that day: he seemed to be the boyfriend of one of the
students.

Ali also asked me whether we had Euclid's text only because of the Arabs.
I said there were some texts of which this was the case, but I didn't
think it was so for Euclid.

Ali also asked why we were skipping I.40.  He didn't understand how
Heiberg could decide that it was an interpolation.  I didn't know how, but
recalled something about a papyrus fragment mentioned by Heath.  Anyway,
Ali agreed that I.40 was not particularly surprising or important.

After \c Ca\u gda\c s has proved I.41, when Ali started proving I.42 (construction
of a parallelogram in a given angle equal to a given triangle), he asked
if his triangle was too small.  Nobody complained, so he continued.  But
he stood right in front of his picture, facing and talking to the
blackboard.  I suggested that many people couldn't see, but he just said
`I already asked if my picture was too small.'

Tolga made his first presentation with I.43 (the complements of
parallelograms about the diameter of a parallelogram are equal).  The
previous class had been the first that he attended.  It wasn't too clear that he understood
what he was proving.  He may have been confused by Euclid's convention of
writing $EG$ and $HF$ to designate \textbf{parallelograms.}  He seemed surprised when
I drew attention to the convention, although he had supposedly finished
his proof.  I tried to get him to shade the two equal parallelograms $EG$
and $HF$, but he didn't understand the request until somebody else explained
in Turkish.

I.44 (construction of a parallelogram in a given angle, on a given side,
equal to a given triangle) was pleasant.  It's nice how Euclid proves that
$HB$ and $FE$ must actually meet when produced.  But \"Ozge was vague about what
the point $L$ was, so I inquired about this.  It turns out that here, as
elsewhere, Euclid does not describe the construction of the figure, but
talks about a figure that has already been constructed.  \"Ozge drew $KL$
parallel \textbf{and equal to} $FH$.  Euclid just draws it parallel, and $L$ is
determined because this is where it meets $HA$ extended; but Euclid names $KL$
before he has even referred to the extension of $HA$.  I seem to recall
being a bit disconcerted, as a student, by this habit; but I don't recall
considering the reason for the habit.  On Friday, I went to the board to
ask why $HA$ and (the line that ends up being) $KL$ can't be parallel.  I
thought of using the Fifth Postulate again, but the students told me that
if $HA$ and $KL$ were parallel, then the intersecting lines $FE$ and $FG$ would be
parallel.

Ahmet proved I.45 (construction of a parallelogram equal to a given
rectilinear figure).  He still wanted to say that the two figures had
equal \textbf{areas,} not that the figures were themselves equal.  But it was
break time, so I postponed my complaint till after that.

Then I drew three lines on the board: one straight, two curved.  I asked:
Do they have lengths?  I asked for a show of hands; most people said the
lines \textbf{did} have lengths; a couple said No.  

I asked, If you think this
curved line has a length, what is it?  

Jeremy said, We need a unit.  

I drew a unit.  He said, We need a smaller unit.

I acknowledged that with calculus one could \textbf{define} the length of a
curved line; but we couldn't do it with the tools at hand.  I wanted to
argue that it was meaningless to abstract a `length' from a line unless
you could compare two lines (for example) and say they were equal; then
you might define `length' as `that by virtue of which two equal lines are
equal.'  But Euclid gives us no way to compare curved and straight lines;
so it is meaningless to talk about the length of a curved line.  Rashad
wanted to be able to pull a curved line straight; I observed that we had
no such postulate.

Jeremy argued that we should still allow a concept of length, for the sake
of philosophy, or something like that.  I said we should avoid talking
nonsense, to keep philosophy from getting a bad name.

With plane figures, the matter is different.  Euclid does give us the
means to compare them.  Now we know that triangles can be `equal' to
parallelograms and other figures.  OK then, what makes them equal is their
`areas', if you like; but we are still far from having `area' as a number.

Today we may approximate areas by of figures by dividing them into little
squares.  Euclid turns rectilinear figures into parallelograms by I.45;
but the parallelograms need not even be rectangles.  Here I uttered my
complaint that we today were obsessed with \textbf{right} angles: that every
angle on our campus, in fact, was right.

Back during the break, \c Ca\u gda\c s had looked at my Green Lion edition~\cite{MR1932864} of
Euclid and wondered about the claim on the back that Euclid was the most
celebrated mathematician of all time.  Did I agree with that? he asked me.
He seemed to think that \emph{somebody} was going to write some such compendium
as Euclid's; Euclid just happened to be the one who did it.  Maybe I
didn't understand his idea.

After Besmir constructed a square in I.46, Rashad gave a careful proof of
I.47.  Jeremy asked why we call it Pythagorean, if Euclid proved it.
Rashad said that Euclid's theorem was different: Pythagoras's theorem was
\begin{equation*}
a^2 + b^2 = c^2.
\end{equation*}
  I told Ay\c se about this later.  I thought Rashad
meant that Pythagoras was interested in identifying `Pythagorean triples',
like $(3,4,5)$ or $(7,12,13)$; but Ay\c se suggested that possibly Rashad didn't
actually see the connection between a geometrical square and the square of
a number.

Time was running out, but I thought we should fit in I.48 to complete Book
I, and Se\c cil was ready.  It's a nice proof:  In proving a converse, Euclid
often uses the method of contradiction; he could do the same for I.48, but
he avoids this and gives a direct proof.

Next time, a few propositions from Book II.  I wonder what the volunteers
will make of them.

\section{Wednesday, October 28}

On Tuesday, October 27, I had
five students lined up for the first five propositions of Book II.  Jeremy
was number 1; but Ali told me that Jeremy couldn't come: something about
seeing the police concerning his residence permit.  Jeremy hadn't asked
Ali to present II.1 in his place, but I asked Ali if he would do it
anyway, and he agreed.

Class hadn't started yet, so Ali went on to ask me whether Euclid
does define `area' somewhere.  It seems these kids are obsessed with
assigning numbers to geometrical figures.  Without numbers, it's not math!
That's what the system seems to teach them.

I mentioned that Euclid was surely aware of the desirability of assigning
numerical areas to plots of land; it just wasn't his interest in the
\emph{Elements.}  The \emph{Elements} ends with the construction of the five Platonic
solids, I said; Ali seemed to find this exciting.

Five minutes into class, there were still only about five students.  One
of them speculated that the others had gone home for the holiday:
Thursday is Republic Day.  Friday is not an official holiday, but I
cancelled class anyway so that Ay\c se and I can take a long weekend down by
the Mediterranean, and the students can have a break too.  But I didn't
mean for them to take the whole week off.

Some time during Ali's presentation, several other students walked in.

Ali used the notation $(A,B)$ for the rectangle contained by $A$ and $B$.  When
Mehmet presented II.2, he used $\operatorname R(A,B)$ for this rectangle, and $\operatorname S(A)$ for the
square on $A$.  I invited him to write a modern algebraic formulation of the
proposition, and he did.

Tolga's presentation of II.3 was confused,\label{confused} but it was Ali and somebody
else (Cihan?) who got him to straighten things out, not I.

Yunus took a long time with II.4, though in the end he seemed to be able
to recover a rigorous demonstration in the manner of Euclid.  When I
asked him, he admitted that he didn't see the point of proving the
proposition, since it is obvious that $(x+y)^2 = x^2 + 2xy + y^2$.  I
suggested that this equation is just symbols, but the geometry is the
`real thing.'  I think I mentioned that Descartes thought the Greeks must
have had some sort of algebra.  I don't know, myself.  When I worked
through Book I of Apollonius in 2008, I had the impression that he did
\textbf{not} have the sort of algebraic point of view that I could adopt.

I asked \"Ozge to postpone her presentation of II.5 till next week.  I used
the remaining few minutes to give a preview of proportion.  I stated VI.1
in words and also in the form
\begin{equation*}
       ABC : ACD :: BC : CD.
\end{equation*}
But what does this \textbf{mean?} I asked.  I wrote it in the form that the
students would expect:
\begin{equation*}
\frac{\operatorname{area}(ABC)}{\operatorname{area}(ACD)}=  \frac{\lvert BC\rvert}{\lvert CD\rvert},
\end{equation*}
but argued that we didn't know what this meant.  Actually, Ali and others
might argue that they do know what this means, with calculus.

A thought about Book II: Heath suggests that Euclid proves the first ten
propositions independently because he is mainly interested in establishing
a \textbf{method.}  He could derive II.2 from II.1, for example, but that's not
the point.  In Book~I though, it \textbf{is} the point, or \textbf{a} point:  I mean,
Euclid's bisection of an angle in Proposition 9 is not the most efficient; but
it relies on Proposition 1, and perhaps for this reason Euclid prefers the
construction he gives to an independent construction.  It's as in a
joke:\footnote{The joke is based on a section of Smullyan's book
  \emph{What is the Name of This Book?}~\cite{Smullyan}.}  
\begin{quote}
How does a mathematician boil water?  By filling the kettle, putting it on
the stove, and turning on the flame.

What if the flame is already on?  Then the mathematician turns it off,
thus reducing the problem to the previous problem.
\end{quote}

\section{Friday, November 6}

Class on Tuesday, November 3, was a lecture by me.  \"Ozge was going to
present II.5, but she wasn't there on time, so I just jumped ahead and
presented II.10.  In algebraic formulation, this is
\begin{equation*}
       (2x + y)^2 + y^2 = 2x^2 + 2(x + y)^2,
\end{equation*}
which can be rearranged to form the equation
\begin{equation*}
       (2x + y)^2 - 2(x + y)^2 = y^2 - 2x^2.
\end{equation*}
In particular, if $(a, b)$ is a solution to
\begin{equation*}
       2x^2 - y^2 = 1,
\end{equation*}
then $(a + b, 2a + b)$ is a solution to
\begin{equation*}
       y^2 - 2x^2 = 1.
\end{equation*}
So we can generate a sequence
\begin{equation*}
       (1, 1), (2, 3), (5, 7), (12, 17), (29, 41), \dots, (a_n, b_n), \dots
\end{equation*}
where $b_n/a_n$ tends towards the square root of $2$.  This is what we `know'
today; but what does it \textbf{mean?}  What is `the square root of $2$'?

I proved VI.2, having assumed VI.1 without proving it or even defining
proportions exactly.  Then I went back to discuss the definitions in Book
V of having a ratio and having the same ratio.  Then I proved VI.1.

I saw a lot of sleepy faces as I stood at the blackboard.  This reminded
me that it had been good when the \textbf{students} were doing the presenting at
the blackboard.

Well, today I give an exam on Book I, which is why I didn't want to
trouble them to give presentations on Tuesday.  The exam problems are [see also \S\ref{sect:exam-1-1}]:
\begin{compactenum}[1.]
\item
  To find the error in a proof that all triangles are isosceles.  (I
don't know if some students will have seen this in some popular book.)
\item
To translate some Greek words (like \Gk{je'wrhma} and \Gk{pol'ugwnon}) into English.
\item
To write down the Greek alphabet.
\item
To give a proof of I.6, analyzed into the six parts described by Proclus
(enunciation, exposition, specification, construction, proof, and
conclusion).  A confusing point here is that Euclid's proof is by
contradiction, so the `construction' step is based on a hypothesis that
turns out to be false.  So what part does this false hypothesis lie in?
I don't know whether Proclus contemplated this question.  One doesn't
really \textbf{need} the false hypothesis though, one can just construct the
point $D$, which in the end turns out to be the same as $A$.
\item
To prove I.8 (`SSS') without using Euclid's method of application.  I
had invited the students once or twice to consider this problem.
\item
Something new: In triangle $ABC$, suppose $BC$ is bisected at $D$, and
straight line $AD$ is drawn.  Assuming $AB$ is greater than $AC$, prove that
angle $BAD$ is less than $DAC$.  It's possible that few will get this, but I
want to find out.
\end{compactenum}
Meanwhile, at the end of class on Tuesday, I lined up volunteers for next
Tuesday to present some propositions about circles from Book III:  3, 20,
21, 22, and 31.  I chose these because they seem to be needed for
Apollonius, and I am keen to get to him.  (`Had we but world enough and
time,' we would just read all of Euclid.  `But at my back I always hear
time's winged chariot hurrying near.')  After Book III, I'll get students
to present from Book V, so they can deal with proportionality themselves.

\section{Friday, November 13}

Last Friday, November 6, I gave the students in `math history' class a
written exam.  I was depressed afterwards, because I had the impression
that the students had not done well.  I feared that my own enthusiasm had
blinded me to the difficulties that the students must have with Euclid.

After I read the exams on Saturday morning, I felt a lot better.  I saw
that I had not been wrong to put a `hard' problem on the exam.  Several
students found a better proof than the one I had thought of.

On another problem, a student introduced a novel method of proof.  Maybe
he didn't clearly see that he was doing this, but:  If you are given that
angle $ABC$ is greater than angle $DEF$, doesn't that mean that there is
\textbf{some} straight line $AD$ drawn inside angle $ABC$ so that angle $ABD$ is equal
to angle $DEF$?  It would seem so, except that Euclid insists on being able
to construct $AD$.

Unfortunately most students had not taken seriously my demand that they
learn the Greek alphabet.  But some had, including one, Taner, who had
once complained [p.~\pageref{Taner}] about doing math in English rather than Turkish.

When I met with my set theory study group on Saturday, our classroom had
the Greek alphabet on the board, with a few mistakes.  I recognized the
hand as that of Tolgay---who had not made those
mistakes on the exam.

I put a photo of that blackboard on the course webpage, along with
solutions and commentary on the exam.

On Tuesday, November 10, \"Ozge presented II.5.  I was glad I had asked her
to do this, because this provided an opportunity for discussing problems
like:  Given a straight line $AB$ and a square $C$, how can we find point $D$ on
it so that the rectangle contained by $AD$ and $DB$ is equal to $C$?  I worked
this out \textbf{analytically,} that is, by assuming that we already have D and
working backwards.  I gave the students the exercise of doing the same
thing, only with $D$ on $AB$ \textbf{extended.}

I also did II.11---to divide a straight line in extreme and mean ratio
(though the terminology is not available at this point yet)---again in the
analytic style, though not in Descartes's algebraic style.  It seems to me
quite plausible now that, as Mr Thomas reported on the J-list, Newton also thought about
his work in the ancient style, visually, rather than by symbol
manipulation.  Actually, Mr Thomas wrote:
\begin{quote}
One of the things that I \textbf{knew}
coming out of St John's was that Newton derived his results
`analytically' and then cast them into `synthetic' form \emph{\`a la} Euclid.
Cohen~\cite{MR1965137} tells us, however, that there is `no shred' of evidence that this
was so.  And Newton apparently never threw anything away, so the absence
of evidence is telling.
\end{quote}
So I wonder if `analytic' and `synthetic' are the right words here.  But I
haven't got Newton with me yet.

Meanwhile, today, November 13, half the students didn't show up.  I was
told that they had an exam in another class right after mine, and they
wanted to study.  Their teacher was a friend we went out with
last night, actually.  Anyway, students presented a few propositions about
circles from Book III that I had asked for, and then we moved on to Book
V.  But I haven't time to say much about that yet.  At the beginning of
class, I did discuss a common English error [at least by native
  Turkish speakers]:\label{page:let}   to write 
\begin{quote}\centering
Let $AB$ is the given straight line
\end{quote}
rather than
\begin{quote}\centering
Let $AB$ be the given straight line.
\end{quote}
I talked about the Turkish subjunctive and imperative verbs and noted the
English periphrastic equivalents.

\section{Wednesday, November 18}

On November 13, Elif began with III.3, namely:
In a circle, a diameter bisects a chord that is not a diameter if and only
if the two are at right angles.  (I note by the way that Euclid does not
seem to use the word `chord'.)

In Book III, I tried to select for presentation only those propositions
that would be needed for Apollonius.  To do this, I relied on the editors
of Euclid and Apollonius.  Proposition III.3 really relies on III.1:  To
find the center of a given circle.  But the Green Lion edition doesn't
indicate as much.  I didn't notice this until Elif presented III.3.  I
asked her if she could prove III.1, but she couldn't.  I asked if somebody
else could do it, and Cihan said he could.  He did it too, by taking the
intersection of the perpendicular bisectors of two chords with a common
endpoint.  Euclid doesn't do this: he take the midpoint of the
perpendicular bisector of one chord.

Is III.1 really required for III.3?  The former is a `problem', the latter
a `theorem'.  The latter simply needs to know that the center of a circle
exists; but it does exist, by definition of a circle.  Possibly this is
why the Green Lion editors, or Heath before them, left off a reference to
III.1.

But Euclid does not seem to rely on the existence of something unless he
can actually construct it.  Later I shall mention an exception to this, in
Book V.  Meanwhile, I think that, in Euclid's own terms, III.3 relies on
III.1.

Jeremy was supposed to do III.20, but he was missing, so I did it.  The
theorem was no surprise to the students:  the angle at the center is
double the angle at the circumference.  Cihan raised the question of what
happens when the `angle at the circumference' is considered as drawn to
the smaller half of the circumference.

Nur proved III.21: angles in the same segment are equal.  I asked her when
she had first learned this.  Before high school, she said.

Tu\u gba presented III.22: the opposite angles in a quadrilateral inscribed
in a circle are equal to two right angles.

Taner presented III.31, but said he was confused about something.  I
discounted this, until I realized that there really was something strange.
The proposition is mainly that the angle \textbf{in} a semicircle is right; that
the angle in a segment that is greater than a semicircle is less than a
right angle; and less, greater.  Taner had no problem with this.

But then Euclid says the angle \textbf{of} a segment that is greater than a
semicircle is greater than a right angle.  He's talking about a
\textbf{curvilinear} angle.  The same sort of angle is mentioned in III.16, but
we had skipped this: `\dots further the angle of the semicircle is greater,
and the remaining angle less, than any acute rectilineal angle'.

In \emph{Euclid: The Creation of Mathematics}~\cite[p.~83]{MR1695164}, by
Benno Artmann, I 
read the claim that III.16 shows that Euclid was aware of
`non-Archimedean' orderings.  That's a strong claim.  In the language of
Book V, the angle between a tangent and a circle does not \textbf{have a ratio}
with any rectilineal angle, since no multiple of the former can exceed the
latter.  But Euclid does not seem to remark on magnitudes that do not have
a ratio.  Throughout Book V, there often needs to be an assumption that
certain magnitudes have a ratio; Euclid does not mention this assumption.
One might wonder whether different people compiled Book V and the earlier
books.

In class we moved on to Book V ourselves.  Cihan presented V.1; Yunus,
V.2.  But there was not much to present.  Unfortunately, again I don't
recall clearly what happened: it was five days ago.

Actually I do recall that Yunus misstated his proposition first, so we
corrected it.  Euclid states V.2 in terms of six magnitudes, a first through
a sixth.  If he was going to use algebraic notation, I thought Yunus might
call these magnitudes $A_1$, $A_2$, \dots, $A_6$.  But he didn't.

I must have got up to write my own algebraic formulation of these
propositions:
\begin{gather*}
       nA_1 + nA_2 + \dotsb + nA_m = n(A_1 + A_2 +\dotsb + A_m),
\\
       n_1A + n_2A + \dotsb + n_mA = (n_1 + n_2 +\dotsb + n_m)A.
\end{gather*}
Here capital letters are magnitudes, and minuscules are multipliers.

It does appear that Euclid himself treats multipliers in isolation in one
place, VII.39: `To find the number which is the least that will have given
parts.'  He means for example to find a number that will have a third
part, a fourth part, and a seventh part.  In this case we must find the
least number $A$ for which there are numbers $B$, $C$, and $D$ such that $A = 3B =
4C = 7D$.

But in the general situation, Euclid says, in effect, let $E$, $F$, and $G$ be
the given parts.  But then we can't just apply Proposition VII.36 to this,
taking the least number measured by $E$, $F$, and $G$.  No, first we have to
take \textbf{numbers} $H$, $K$, and $L$ that are `called by the same name as' $E$, $F$, and
$G$.  So in my example, $E$, $F$, and $G$ are not three, four, and seven; they are
third, fourth, and seventh.  But third, fourth, and seventh what?  I don't
know.  Proposition VII.39 is the last in Book VII, so maybe it was added
later.  Heath doesn't suggest this however; nor does he remark on its
strangeness.  He says only that VII.39 is `practically a restatement' of
VII.36.  If so, then we really should inquire why Euclid makes the
restatement.

Back to my class, and my equations above.  I suggested to the class that
magnitudes and their multipliers were like vectors and scalars in linear
algebra.  In particular, the two equations above are certainly not
`practically the same'.  This point comes out to us if we write (as I did
in class) V.3 as
\begin{equation*}
       k(mA) = (km)A.
\end{equation*}
Multipliers can multiply each other; magnitudes as such cannot.

I hadn't actually assigned V.3.  Tolga had volunteered for V.4, but was
not present.  I presented it.  It is the first proposition about
proportions:  Symbolically,
\begin{quote}\centering
       if  $A:B::C:D$, then  $kA:mB::kC:mD$.
\end{quote}
I should have assigned V.3, since V.4 uses it.

Mehmet was supposed to present V.7, but we were out of time.  That's good,
because he and I got to talk about his proposition before he presented it.
He asked me after class whether V.7 wasn't simply
\begin{quote}\centering
       if  $a = b$, then  $a/c = b/c$,
\end{quote}
and if so, isn't it obvious?  I said:
\begin{compactenum}[1.]
\item
If you think it is obvious, you can say so when you present it.
\item
Do not write fractions like $a/c$.  Today we think of a fraction as a
single thing, a number.  But in Euclid we have no justification for
thinking this way.  We can think the way we do only because people like
Euclid have done the groundwork.
\item
Do not use the equals sign between ratios.  Equality is a `common
notion'.  If $A = B$ and $B = C$, then we know $A = C$.  We don't know this with
ratios, but must prove it; in fact it is V.11.
\end{compactenum}
Unlike, say, `less than', proportion is not a relation between
two things; it is a relation between \textbf{four} things.  Such relations are
almost unheard of, unless we want to take an expression like `These are my
parents, and this is my sister' as signifying a single relation between four
people.  Moreover, proportion is something we \textbf{define.}  So we cannot just
`intuit' its properties; we have to prove them.

That's roughly what I said to Mehmet, except I didn't use the example of
familial relations.  I recall talking more, even bringing up what I
learned from Mr Thomas, that there's no evidence that Newton did \textbf{not}
think about mathematics the way the Ancients did.  Mehmet is majoring in
physics as well as mathematics.  I don't know if he found it attractive to
learn that, if I did get to teach the second semester of this course, I
wanted to read Newton.

\section{Friday, November 20}

On Tuesday, November 17, Mehmet started math history class by presenting
Euclid's Proposition V.7.  He stated it as
\begin{quote}\centering
       If $A = B$, then $A:C::B:C$ and $C:A::C:B$
\end{quote}
and then he declared that it was obvious.  As I noted in my last
entry, such a declaration is what I had suggested (if he thought it
correct).

I said maybe the proposition is obvious, \textbf{after} one observes that
\begin{quote}\centering
       $A:B::C:D$ if and only if $C:D::A:B$
\end{quote}
(the `$::$' relation is symmetric).  Mehmet said, correctly I think, that
all one really needs to observe is that
\begin{equation*}
       A:C::A:C
\end{equation*}
(the `$::$' relation is reflexive).  But then one needs a `substitution
principle': if two things are equal, then one of them can be substituted
for another in any mathematical statement.

Now, Euclid does not have a proposition to the effect that $A:C::A:C$.
Would he take such a proposition as obvious, or as pointless?

The Common Notions include 1, that things equal to the same are equal to
each other.  If we want to express this symbolically, we might write
\begin{quote}\centering
       If $A = C$ and $B = C$, then $A = B$.
\end{quote}
But then we should also observe that
\begin{quote}\centering
       If $A = B$, then $B = A$,
\end{quote}
so that, if we should find that $A = C$ and $C = B$, then $A = B$.  However,
there is no express Common Notion to the effect that, if a first thing is
equal to a second thing, then the second thing is equal to the first.
This is just tacitly understood.  One does need it in modern mathematical
arguments.

`A thing is equal to itself' is not a Common Notion either.  It may be
true, but, although devotees of Ayn Rand may worship the equation `$A = A$',
I can't think of an occasion where such an equation is used in
mathematics.  This is why I suggest that it might be pointless for Euclid
to prove $A:C::A:C$.

Again, from $A:C::A:C$, one could derive V.7; but again, this would be by a
`principle of substitution', and such a principle could not very well be
stated in Euclidean terms.  Euclid is not in the business of manipulating
formal expressions in some artificial `language'.  The best way for him to
prove V.7 is probably just as he does it.  In particular, V.7 is \textbf{not}
obvious.

But I can subject my students to only so much of this speculation.  Back
in class, we moved on to V.8.  Besmir was supposed to do it; but he
thought he was supposed to prove III.8.  Other presenters in class knew we
were in Book V; so the mistake must have been Besmir's.

I presented V.8 myself.  It brought to light yet another difficulty with
Book V.  The claim is this, symbolically:
\begin{quote}\centering
       If $A > B$, then $A:C > B:C$ and $C:B > C:A$.
\end{quote}
Assume $A > B$.  We want multiples of the various magnitudes so that
\begin{quote}\centering
       $kA > mC$, but $kB < mC$;
\end{quote}
we might write this as
\begin{quote}\centering
       $kA > mC > kB$.
\end{quote}
To achieve this, we should make the gap between $kA$ and $kB$ greater than $C$.
But
\begin{equation*}
       kA - kB = k(A - B),
\end{equation*}
by V.5.  So we just take $k$ large enough that
\begin{equation*}
       k(A - B) > C.
\end{equation*}
To do this, we assume that $A - B$ and $C$ \textbf{have a ratio} in the sense of
Definition V.4.  More on this presently.  Meanwhile, accepting this, we
let $mC$ be the first multiple of $C$ that exceeds $kB$.  Then $kA > mC > kB$, as
desired.

In Heath's translation, Proposition V.8 is:
\begin{quote}
Of unequal magnitudes, the greater has to the same a greater ratio than
the less has; and the same has to the less a greater ratio than it has to
the greater.
\end{quote}
There is an implicit assumption: each of the three magnitudes mentioned
does have a ratio to each of the rest.  That's fine.  But there's another
assumption, which the proof requires:  If two \textbf{different} magnitudes have
a ratio to a third ratio, then so does their difference.  This means, for
example, we can't take the sum of a square $A$ and a straight line $B$ and call
the result $C$: for then we should have, presumably, $C > A$, although
$C:A::A:A$.

Ali presented V.9; but he just said it followed immediately from V.8,
being its contrapositive.  I noted that it was the contrapositive,
provided one noted that, of two unequal magnitudes, exactly one is greater
than the other.

Ahmet presented V.11:
\begin{quote}\centering
       If $A:B::C:D$ and $C:D::E:F$, then $A:B::E:F$
\end{quote}
(the `$::$' relation is transitive).  He just said it followed from the
transitivity of implication:  If $A$ implies $B$ and $B$ implies $C$, then $A$
implies $C$.  He admitted it was slightly more complicated, because of the
quantifiers: we assume
\begin{compactenum}[(1)]
\item
for all $k$ and $m$, if $kA > mB$, then $kC > mD$,
\item
for all $k$ and $m$, if $kC > mD$, then $kE > mF$.
\end{compactenum}
We want to conclude
\begin{compactenum}[(1)]\setcounter{enumi}2
\item
for all $k$ and $m$, if $kA > mB$, then $kE > mF$.
\end{compactenum}
Actually, using formal logic might obscure the point.

\c Ca\u gda\c s presented V.12:
       If 
       \begin{equation*}
       A_1 : B_1 :: A_2 : B_2 :: \dots :: A_n : B_n,
\end{equation*}
       then 
       \begin{equation*}
       A_k : B_k :: A_1 + A_2 + \dotsb + A_n : B_1 + B_2 + \dotsb + B_n.
\end{equation*}
I think it was here, and I think it was Ali who pointed out the assumption
that each of these magnitudes should have a ratio to the others; otherwise
we might be adding squares and straight lines, which would lead to
problems as discussed above with V.8.

\section{Friday, November 20}

Mr Gorham asked on the J-list:
\begin{quote}
Isn't reciprocity included in the notion of `equal to'? Maybe I'm
thinking of it linguistically instead of mathematically but it seems to
me there's no need to spell out that if $A=B$ then $B=A$ because `$=$'
contains within it the idea of a two way street.
\end{quote}
What do you mean by `included in'?  When Euclid writes `Things equal to
the same are equal to one another,' well, indeed that reciprocal pronoun
`one another'---I guess one would call it reciprocal, or something like
that---that `one another' suggests the meaning, `$A$ is equal to $B$, and $B$
is equal to $A$.'

But we modern mathematicians recognize that equality has three distinct
properties:
\begin{compactdesc}
	\item[Reflexivity] $A = A$;
\item[Symmetry] if $A = B$ then $B = A$;
\item[Transitivity] if $A = B$ and $B = C$ then $A = C$.
\end{compactdesc}
It is of some interest that Euclid (or somebody writing under that name)
distinguished only the last (or some formulation of the last) as a Common
Notion (again, unless you want to read symmetry also into that Common
Notion).  However, the relation of `less than or equal to' is reflexive
and transitive, but not symmetric.  Other such examples show that no two
of the properties imply the third.

Eva Brann's friend Barry Mazur has an article on his
homepage\footnote{I can't find the article at
  \url{http://www.math.harvard.edu/~mazur/} anymore, but it seems
  to have been published as \cite{MR2452082}.}
called `When is one thing equal to some other thing?'  It's been a while
since I read it, but the theme (as I recall) is the mystery about what
equality is in mathematics.

\section{Saturday, November 21}

In the class of Friday, November 20, it was interesting to see the
different styles of different students in presenting propositions from
Book V.

Rashad began with V.14, using modern symbolism.  He was a bit late, and
before his arrival, I had ranted a bit about the irresponsibility of
agreeing to present a proposition and then not showing up.  Then Rashad
entered in a rush.

Melis continued with V.15, following Euclid's style exactly.  The
enunciation is, `Parts have the same ratio as the same multiples of them
taken in corresponding order.'  For convenience, I would write:
\begin{equation*}
       A:B::kA:kB.
\end{equation*}
Melis just wrote out the words, and gave the proof as Euclid does, with
a diagram like Euclid's, for the case where (in my notation above) $k = 3$.
Such `proof by example' is perhaps considered short of rigorous today; at
least, it's out of style.  But what really is the problem with it?

I asked Melis, `What if there were seventeen of the part in the whole,
instead of three?'

She said, `The proof would be the same.'  She's right.

Se\c cil presented V.16, which symbolically is
\begin{quote}\centering
       If $A:B::C:D$, then $A:C::B:D$.
\end{quote}
She also followed Euclid closely, but I had the feeling that this was
because she did not comprehend the proof very well.  Actually she confused
some letters, but had a bit of trouble correcting them when the mistake
was pointed out.  Well, I know one's brain can stop working well when one
is standing at the blackboard; it had happened to me earlier in the day
in our departmental algebra seminar.

M\"ursel was next with V.17:  `If magnitudes be proportional \emph{componendo,}
they will also be proportional \emph{separando.}'  But he didn't write out
the words, and I don't think many of the students are using Heath's
translation with those Latin expressions.  M\"ursel just gave a symbolic
statement and proof.

Talha, volunteer for V.18, was missing.  Actually he hadn't volunteered:
I \textbf{assigned} almost everybody a proposition from Book V.  Talha started
attending class late, and he has never presented a proposition.

I presented V.18 myself, noting what seems to be a first for us in Euclid:
an assertion of existence without construction.  I mean, Euclid says that
if $CD$ is \textbf{not} to $DF$ as $AB$ is to $BE$, then $CD$ must be to \textbf{some} $DG$ as $AB$ is
to $BE$.  Well, this seems to be a new postulate.  We are in no position yet
to \textbf{construct} a magnitude that has a given ratio to another.

In the book~\cite[Ch.~14, p.~134]{MR1695164} that I mentioned another day,
Benno Artmann passes on a claim that some propositions in Book V are
copied verbatim from Eudoxus, since nobody wanted to change the words
of the master.  Maybe the proof of V.18 is evidence for this.  [In
  fact Artmann was talking about V.8.]

Tolgay presented V.20, and \"Ozge, V.22.  Really, V.20 appears to be just a
lemma for use in proving V.22, which is
\begin{quote}\centering
       If $A:B::D:E$ and $B:C::E:F$, then $A:C::D:F$ \emph{ex aequali.}
\end{quote}
And that was all it seemed necessary to do from Book V.  [This was wrong; I turned out to want V.23 for the final exam.  I just gave it to the students then.]

I had done VI.1 and 2 on an earlier day; now Besmir did VI.4 (equiangular
triangles have proportional sides).  I asked how he knew that $BA$ and $ED$
met \textbf{beyond} $A$ and $D$.  The answer seemed to be that, if $ED$ met $BA$ between
$A$ and $B$, then $ED$ would cross $AC$; but these two lines are parallel.

Elif presented VI.6: triangles with one equal angle, and the sides about
it proportional, are equiangular.  Euclid's is another peculiar proof,
like that of I.48, where along one leg of a triangle, a new triangle is
constructed that turns out to be congruent to the first.  If the new
triangle were constructed on the same side as the first, then it would
coincide with the first; but Euclid wouldn't like this, so he would assume
(by way of contradiction) that the triangles were \textbf{not} congruent.  Thus
the fact that a straight line has two sides allows Euclid sometimes to
avoid proofs by contradiction.

Se\c cil was scheduled for V.8, but our time was almost up, and she was happy
enough to postpone her presentation till Tuesday.  Tolgay was scheduled
for V.11, but he had already left, apparently to collect his thoughts
before an exam immediately after my class.  I took volunteers for the
remaining propositions in Book VI.

\section{Friday, December 11}

I sent my last report on my `math history' class almost three weeks ago,
on the class of Friday, November 20.  Since then, there have been only
three classes, one hour each.  Friday, November 27, was the Feast of the
Sacrifice.  I have no classes on Thursday or Monday, so I got a five-day
weekend; but I spent it at my desk at home, working on various projects.
There was no real external compulsion to do this work, just my inner
drive.

One (but only one) of my projects was preparing to give a talk the
following weekend in Istanbul.  There wasn't really much to do, since I
could more or less repeat the talk I had given in France in the summer;
but I made a lot of adjustments.  The occasion was a 60th-birthday
conference.  I had met Oleg Belegradek when I was a student at Maryland.  Then he was
still working in Siberia; now we have both ended up in Turkey.  For the
birthday event, Oleg's former Kemerovo colleague, Boris Zil'ber, came from Oxford
where he now works.

I cancelled math history class on Friday, December 4, to take an afternoon
bus to Istanbul.  It was a bad time of day to go to Istanbul; evening
traffic held us up for an hour.  The driver said the delay the previous
evening had been two hours.

At the party on Saturday evening, somebody gave Oleg a present:
\emph{Logicomix}~\cite{Logicomix}.\footnote{\url{http://www.logicomix.com/en/}}

I noted that the book had been purchased at Robinson Crusoe Books, so on
Monday I went there myself and bought a copy.  I read it on the bus back
to Ankara that afternoon.  It was my first graphic novel, and I was
impressed; but why shouldn't I be impressed to see a
mathematician-philosopher made into a tragic hero like Orestes?

\asterism

In class on Tuesday, November 24, Se\c cil presented VI.8; Tolgay, VI.11;
Mehmet, VI.12; and Jeremy, VI.15.  Then I was moved to scold (some of) the
students for poor preparation.\label{scolding}  Se\c cil had looked at
her notes repeatedly. 
Jeremy was more polished, and he was able to write down the numbers of the
propositions that justified the steps of his proof; but he couldn't just
explain in words why the steps were justified.  I can't fault anybody for
having difficulty with the mathematics; but I fault Jeremy for trying to
fake his way through a proof.  I said to the class that notes were not
absolutely forbidden, since we regular teachers did use them ourselves in
teaching; still, I said, one ought be able to understand and reproduce the
general flow of one of Euclid's arguments without copying from a notebook.

Then Yunus got up and gave an exemplary exposition of VI.16, without notes
at all.  (He did take a glance at the proposition in my copy of Euclid
before proceeding to the blackboard.)

M\"ursel followed with VI.17, a special case of VI.16.  Then Elif finished
the day with VI.18: to construct on a given straight line a rectilineal
figure `similar and similarly situated' to a given one.  I was sorry she
just used a quadrilateral like Euclid, rather than drawing a more
outlandish figure to emphasize the generality of the proposition.

\asterism

For the holiday, Cihan was flying to Bosnia to see his Serbian girlfriend,
whom he had met in France.  Apparently he didn't get back in time to
present VI.19 on Tuesday, December 1; so I presented it.

Rashad did VI.21, which he said was immediate.  Nur presented VI.23; but I
recall going to the board myself to talk about what `compound ratio'
meant.  Ali finished our coverage of Book VI with Proposition 24.

You see I haven't too much to report here, in part because I am late in
making the report.  But I think the students have been bored; they may
think all of these propositions about proportion are obvious.  Students
have been cutting class too, perhaps to prepare for other classes and
their exams.  Perhaps they haven't understood that a necessary and nearly
sufficient condition for getting a good grade in my course is showing up
to class.  It's hard to believe, unfortunately, that they don't care about
their grade.

Mehmet finished the day with Proposition 1 of Book XI: 
\begin{quote}
A part of a
straight line cannot be in the plane of reference and a part in a plane
more elevated.
\end{quote}
  We discussed whether there was really anything to prove
here.  Euclid argues that, if the contrary does happen, then the part of
the straight line that \textbf{does} lie in the `plane of reference' can be
extended in that plane.  Then two straight lines will have a common
segment.  I should think this was obviously absurd; but Euclid \textbf{proves}
the absurdity by drawing a circle with two distinct diameters that have a
common endpoint.  Mehmet didn't repeat that argument, and indeed the
circle doesn't appear in Heath's diagram.

Postulate 2 justifies extending straight lines, but says nothing about
planes.  It is quite an exaggeration, this modern idea that Euclid builds
up his whole system from `axioms'.\label{axioms}  The Modern then has to say that Euclid
got it wrong, since propositions like XI.1 are `really' axioms too.

\asterism

On Tuesday, December 8, Besmir presented XI.2:  In Heath's translation:
\begin{quote}
If two straight lines cut one another, they are in one plane, and every
triangle is in one plane.
\end{quote}
  Besmir used Fitspatrick's version, which
inserts a parenthesis after `triangle':  `formed using segments of both
lines'.  Before starting, Besmir asked if he should write down the
proposition, and I told \textbf{him} to decide.  This was a mistake.  He wrote
all those words, then he started writing the words of the proof, without
saying anything.  He said he would explain the proof after he had written
it down.  I suggested he do these at the same time.  He tried; but after
some questions from me, he had to admit that he didn't understand the
proof.  Well, Heath has a note:  `It must be admitted that the `proof' of
this proposition is not of any value.'  There's really nothing that can be
proved here, in our sense at least.  Euclid's proof does suggest that
assumption that two intersecting straight lines must lie in one plane, at
least near the point of intersection; then XI.1 can be used to finish the
proof.

In any case, it may be of value to confront the students with weird
proofs; it may induce them to be more questioning of what they read.

When he was finished, Besmir asked to leave and go study for an exam.  I
said the door was unlocked.

Elif continued with XI.3: `If two planes cut one another, their common
section is a straight line'.  Despite several attempts, she just couldn't
get the diagram right.  This reminded me that a three-dimensional
imagination may be difficult to acquire.

But \c Ca\u gda\c s gave an accomplished presentation of XI.4.

Taner was supposed to present XI.5, but he was absent, so I did it:  If
three straight lines are at right angles, at the same point, to another
straight line, then the three are in one plane.  Both Cihan and Ali raised
questions about the argument.  I hope they would have raised them also if
Taner had been presenting.  I think Cihan had not quite understood that
(in Heath's diagram) $AB$, $AC$, and $AF$ are all in one plane, and angles $ABC$
and $ABF$ are right (which is absurd unless $BC$ and $BF$ coincide).  Indeed,
the claim cannot be seen from the figure alone.

At the end I set up the proposition explicitly as a converse of XI.4.
Together, 4 and 5 are that, if two straight lines meet a third at right
angles, and at the same point, and a third straight line meets the others
at that point, then this straight is in the plane of the first two if and
only if it is at right angles with the third.

With all of this talking, I used up the remaining time.

Today, if we can get through eight more propositions, then we shall be
finished with Euclid.  Apollonius is next.

\section{Tuesday, December 15}

In class on Friday, December 11, we finished with Euclid.  It's too bad,
because we stopped with Book XI; but now there are just ten hours left for
Apollonius.

Tolgay proved XI.6: straight lines at right angles to the same plane are
parallel.  Maybe this proposition sets a record for the most auxiliary
lines: to prove the parallism of two straight lines, \textbf{five} additional
straight lines are drawn.

We skipped XI.7, owing to a mistake of mine:  I picked the propositions in
Book XI by looking at what was needed in Book I of Apollonius, according
to the editors of the latter; but I forgot to check which propositions
the needed propositions themselves needed.

In fact XI.8 calls on XI.7: a straight line joining points on two parallel
straight lines is in the same plane as the parallels.  But the latter
hardly needs proof---or can hardly be proved, as opposed to being assumed.

Proposition XI.8 is the converse of XI.6, and Tu\u gba proved it.  When she
started drawing the diagram, I suggested that she could just use the one
left by Tolgay, since it is the same down to the lettering of the points.

[As I noted on p.~\pageref{page:metonym},] Reviel Netz suggests that
for the ancients, the diagram is a `metonym' for 
a proposition; the diagram `individuates' the proposition.  By contrast,
for us (he says), the \textbf{enunciation} of the proposition is the metonym:
this is what we quote when we want to specify which proposition we are
talking about.

However, XI.6 and 8 have identical diagrams.  But in fact, as they are
drawn in Heath at least, one is a mirror image of the other: it is
reversed.  So when I suggested that Tu\u gba use Tolgay's diagram, she looked
at it and decided she had better use her own.

\"Ozge proved XI.9: parallels to the same are parallel to each other.  Is it
obvious?  I don't know about the students, but I think it is not obvious
in three dimensions.  It's not a \textbf{surprising} proposition; but proving it
takes a bit of work, and it's `real' work: you take a plane to which the
straight line A is at right angles; if B and C are parallel to A, then
they are at right angles to the plane by XI.8; then they are parallel to
each other by XI.6.  And these are the record-breaking propositions in
terms of numbers of auxiliary straight lines needed in their proofs.  Book
XI has some of the same logical music as Book I.

Se\c cil did XI.10, then we skipped ahead to XI.14, which Yunus did; then
M\"ursel did XI.15, using and adapting Se\c cil's diagram from XI.10.  I don't
know if he was influenced by my earlier comments, or would have done this
anyway.

Mehmet did XI.16, and Rashad, XI.18, and that was it.

We had some time left.  There were 13 students present, and I wanted to
assign the first 14 propositions of Apollonius.  Tolga (who has not
attended many classes) said he would take the I.1 and 14.  Then I just
wrote down the rest of the students from left to right.  Tolgay said he
might miss next class, because of an exam.  I said he shouldn't, but I
moved him to a later proposition anyway.

I read out Kepler's warning, at the beginning of the Green Lion edition~\cite{MR1660991} of Apollonius,
to the effect that some work is inherently difficult, and Apollonius is an
example.
Now we'll see what happens!

\chapter{Apollonius and Archimedes}

\section{Tuesday, December 15}

Probably it's good that we started today with Proposition I.1 of
Apollonius, rather than skip ahead to something meatier.  Tolga proved it,
as it is proved in the text: I mean, he didn't prove it as if he had
thoroughly understood it and was passing on his understanding.  Not that
there's so much to understand: a straight line joining the vertex of a
conic surface and another point in the surface lies on that surface.

The original Taliaferro translation~\cite{MR0053831} in the Britannica
\emph{Great Books of the Western World} introduces a
small error, which is repeated in the Green Lion edition:  There are three
diagrams, showing three possible configurations.  In two of the diagrams,
$B$ is on straight line $AF$, not $AE$, and indeed the text would not make much
sense if $B$ were on $AE$.  But in the third diagram, $B$ \textbf{is} on $AE$ (extended).

The Heiberg edition does not feature such a
mistake.\footnote{I printed this out from
  \url{http://www.wilbourhall.org/} which has all sorts of old math texts.}

Actually, before Tolga started, I wrote down a bunch of Greek terms from
Apollonius that gave rise to English words (although the latter may not be
the words used to translate the former):  The English words were cycle,
periphery, parallepiped, epiphany, center, basis, scalene, diameter.

Tolga made the English mistake commonly made [and which I mentioned on p.~\pageref{page:let}]:  He wrote for example `Let $C$
is not on the surface...'  I asked him to replace `let' with `suppose'.
But I can't say that the grammatical difference between `let' and
`suppose' here is important.

After Tolga, Elif presented I.2.  She started sketching the figure, and
she said something about `vertically opposite points'.  She had evidently
been confused by the expression, `If on either of the two vertically
opposite surfaces two points are taken\dots'  I jumped up to try to clarify
matters with my own diagram.

Elif worked through Apollonius's proof that the straight line joining the
points lies within the surface.  I asked if the result was obvious.  She
said she had thought it was, but on the other hand the proof was a real
proof.

Ali proved I.3, that if a cone is cut by a plane through the vertex, the
section is a triangle.  

\c Ca\u gda\c s asked, Can't the section be just a straight
line, as when the cutting plane is tangent to the conic surface?  

Ali
said, But then the cone would not have been properly \textbf{cut} by the plane:
the cone is supposed to be cut into two pieces.

Ali asked whether the cone could be infinite, or something like that.  I
observed that the \textbf{cone} has a base, though the \textbf{conic surface} can be
extended indefinitely, and Yunus would be proving something involving this
fact with I.8.  (Yunus acknowledged this.)

Somehow I was moved to distinguish the conic surface from the cone by
saying the surface was two-dimensional.  Ali asked, What does that mean?
I think he was teasing me, alluding my own tendency to ask the students
what \textbf{length} is.  We laughed.

Tu\u gba proved I.4, that a plane parallel to the base cuts the cone in a
circle.  As she was drawing her figure, I asked whether the proposition
was obvious.  She said it was.  I got up and drew an extremely oblique
cone (hers was nearly right) and asked, Is the proposition still obvious?
She smiled and said it still was.  Nonetheless she did the proof.

One ends up proving $DG = GH = GE$, where $H$ was chosen arbitrarily on the
section, so that $DGE$ ends up being the diameter of a circle.  Rashad asked
whether the `last line' was really necessary; he was referring to the
straight line $AHK$, used to prove that $GH$ is equal to $DG$ and $GE$.  I looked
to others for an answer.  Ali said in effect that if we didn't have $H$,
then all we can prove is $DG = GE$; but this doesn't establish that $G$ is the
center of a circle.

I suggested that, if we just proved $DG = GH$, that would be enough to
establish that $G$ is the center of a circle.  But again $H$ is a random
point, and $E$ is not random: it is in a straight line with $DG$.  Special
cases do tend to get special treatment: so the term `ellipse' will not
cover the circle, presumably because any curve that is a circle should be
called just that.

There were five minutes left, but Se\c cil said they weren't enough for her
to prove I.5 (which may be the first non-obvious proposition).  So we
stopped.

There were just 13 students present, two who had not come on Friday:
\c Ca\u gda\c s and Nur.  \c Ca\u gda\c s had the text though; Nur did not.  The text is the
Green Lion edition, my copy being perhaps the only one in Ankara, except
the photocopied pages that the students now have.  I asked the library to
order a copy, but it isn't in.  I do see however that somebody other than
myself has asked the library to order:
`\emph{Apollonius de Perge, Coniques : texte grec et arabe,} \'etabli, traduit et
comment\'e sous la direction de Roshdi Rashed; text in Arabic and Greek with French translation of facing pages;
introductions, commentaries, and notes in French.'

\section{Saturday, December 19}

Last night in Apollonius class, I became sad and depressed about the
whole enterprise; but afterwards my belief was restored.

Se\c cil presented I.5 of the \emph{Conics.}  She went about constructing the
diagram, but she didn't explain what was special about points $G$ and
$K$.  When I enquired, she said there was nothing special about them.
So she had missed the whole point, namely that triangle $AGK$ is similar
to axial triangle $ABC$, but lies `subcontariwise'.  I let her continue
with the proof.  Eventually she asserted that triangles $DFG$ and $KFE$
were similar, and I pointed out that the missing hypothesis was
required for this.  I think it was Cihan or Ali who told Se\c cil where
the hypothesis was in the text.

When Se\c cil wrote down the equation
\begin{equation*}
       \operatorname{rect.} DF,FE = \operatorname{sq.} FH,
\end{equation*}
she also wrote the justification supplied by the editors:
\begin{quote}
       Eucl. III.31, VI.8 porism, and VI.17.
\end{quote}
This suggested that she didn't just \textbf{see} why the claim was so.  I
invited her to draw the circle in the plane of the blackboard---the
circle whose diameter was $DE$, to which $HF$ was dropped
perpendicularly.  She did this, but positioned $F$ as if it were the
center of the circle.

When the proposition was finished, so that, in priniciple, we knew
that an oblique cone had a circular section in two different
directions, I asked Se\c cil if this was surprising.  She said it was.

As I look back at the proposition, I see we didn't remark on the
importance of having the plane of the axial triangle be at right
angles to the base.  In general, students seem to be drawing their
cones as if they were right anyway.

Taner proved I.6 very confidently, but he seemed to have relied mainly
on the diagrams to tell him what the assertion \textbf{was.}  He got it
wrong.  He thought the cone was being cut by a plane through the
vertex, making the triangular section $AKL$; and he thought the base $KL$
was a diameter of the base of the cone.

Ali questioned how the cone was being cut.  Perhaps he had understood
that the cone was `really' being cut so as to make the axial triangle
$ABC$; or perhaps he was just trying to reconcile Taner's claims with
the text, and not succeeding.  Taner kept insisting that there was
only one cutting plane, making $AKL$.

Well, $AKL$ \textbf{can} be understood as the result of cutting the cone with a
plane through the vertex.  But that's not how it arises in the text.

I got up and tried to argue this point.  Eventually Taner agreed that
he had been confused.

While I was up at the board, I saw that many classmates were not
paying attention.

Besmir\label{Besmir-silent} was next with I.7.  He got up, drew a diagram, and started
writing down words without making a sound.  What should I do?  The
classmates are not Johnnies who will speak up if something isn't going
right.  They may think it is the \textbf{teacher} who should not allow time
to be wasted.  Eventually I asked Besmir if he had any teachers who
came to class and wrote silently on the board.  He just said he needed
to write everything down before talking about it.

Well, since (I think) he was working from memory, maybe he needed to
concentrate silently.  In that case, I would rather he used notes.

By the time Besmir was finished, the usual ten-minute break time was
almost over.

I suppose the Apollonius is harder than I think.  Unfortunately I can
recover no memory of the relevant mathematics tutorials at the
St John's.  How did I prepare and present propositions?  How did
others?  I can't remember.  But I do think that Johnnies were
\textbf{engaged} in class in a way that most of my students now seem not to
be.  I think Johnnies understand that they are supposed to be reading
every proposition themselves.  My students now may study only the
propositions that they are supposed to present.

It might be recalled that Johnnies have but one mathematics tutorial
at a time.  My students are taking other math classes.

By the time they come to Apollonius, Johnnies have spent some weeks or
months thinking three-dimensionally with Euclid and Ptolemy.  Perhaps
my students now have not had so much experience.  But Ay\c se pointed out
later that they \textbf{do} have such experience, from vector calculus.  From
our point of view, some students are just lazy.  If I were giving an
ordinary lecture class, Ay\c se reminded me, most students would not be
very engaged in the class, if they bothered to come to class at all.
They would cram before exams, and that would be it.  Why should I
expect things to be different now?

After a late ten-minute break, from which not everybody had returned,
Yunus gave a reasonably accomplished presentation of I.8.  He answered
somebody's question (Ali's, I think) about exactly what was being
proved.  I asked whether Apollonius was proving that the conic
section, which `increases indefinitely,' also opens indefinitely
wide.  Yunus said No, and I guess he's right; the fact will be a
consequence of later propositions.

M\"ursel was next with I.9.  During the break, he had asked me why
Apollonius could say `Therefore $GED$ is a straight line.'  I pointed
out the ensuing reference to `Eucl. XI.3'.  Since he didn't have a
copy, I handed him my volume III of the Dover edition~\cite{MR17:814b} of Heath's
Euclid.  Here I recalled the Green Lion remark on the usefulness of a
one-volume edition.  I had brought volumes II and III to class; but what
if volume I had been needed?  My Green Lion edition was at home.

Anyway, I don't know why M\"ursel was confused, since before the claim
in question, Apollonius says `therefore $D$, $E$, $G$ are points on the
common section of the planes.'

At the board, M\"ursel took a long time.  I gave up hope of getting to I.11 and the definition of the parabola.

Ahmet presented I.10 sheepishly, since it seemed so simple.  I asked
if he had a modern way to describe the result of the proposition.  He
didn't, but Cihan offered the word `convex'.  A conic section is
convex: any straight line drawn between two of its points lies
entirely inside the section.  I went to the board to suggest that the
diagram of the proposition was perhaps misleading, since points $G$ and
$H$ on the section were drawn on opposite sides of the axial triangle,
but they need not have been.  I recalled that Elif had proved the
proposition (namely I.2) that justified I.10; she agreed with my
memory and observation.

Time was just about up.  I already had students signed up for all
propositions through I.15; I took more names for the next five.  The
new names included the students who hadn't shown up to the previous
class.

I asked the class what they thought of Apollonius.  Was it
interesting?  Was it just hard?  Cihan said it was both.  He is one of
the better students.  He sits at the front and takes notes from his
classmates' presentations.

I started talking about how there was no point in doing `math history'
unless you read original works.  Secondary sources will `modernize'
the treatment.  If there is really only \textbf{one} mathematics, that may be
fine; but the unity of mathematics is not obvious.

I don't remember exactly how it all happened, but several of us ended
up sitting around for half an hour after class talking about
mathematics.  The active participants besides myself were Ali, Cihan,
and Mehmet, but Elif, M\"ursel, and Se\c cil also stayed around.  When the
rest of the class was still there, I said something about how
Apollonius was rigorous mathematics, whereas there had been periods,
as in the 18th century, when math was not about rigor, but was about
deriving equations with however-tenuous justification.  I had just
been reading about Euler's derivation of the value of the sum 
\begin{equation*}
\frac 11+\frac14+\frac19+\frac1{16}+\dotsb
\end{equation*}
of the
reciprocal squares.  Because of something he said, I directed a question at Ali:  `Do you want to see Euler's
derivation?'  How could he say no?

I got up to the board, but said class was officially over, so anybody
who had to leave should feel free.  Those who stayed around seemed to
be impressed by the derivation of the value of $\uppi^2/6$ for the sum,
though they agreed that it was not rigorous.  I reported the claim
that British mathematics in this period fell behind continental
mathematics precisely \textbf{because} British mathematicians had an
excessive devotion to rigor, which kept them doing mathematics in the
ancient style.  I did mathematics for its rigor, I said; but I had to
acknowledge that great advances in mathematics had been made by people
who didn't share my interest.

Mehmet made a distinction between mathematics and physics: the
mathematician wanted to prove things; the physicist, to discover
them.  I think those were the words he used.  He's majoring in physics
as well as mathematics.  He talked about physical laws whose
mathematicial derivations were not sound.  A physicist told him
this didn't matter, since the laws agreed with nature.  But Mehmet
wanted a rigorous derivation of the laws.

I suggested that physics and mathematics had been indistinguishable
for much or least some part of history.  Ali said that, in the
Renaissance, there had been \textbf{one} science, and \textbf{one} art, and  a
person like Da Vinci could do both.

We carried on for a while, as I said, until nobody seemed to have more
to say.  But then Mehmet asked me if there were another edition of
Apollonius he could look at, since he had not been able to make sense
of his proposition, I.15.  I showed him the Greek text of Heiberg with
facing Latin translation.  I said there was Heath's English version~\cite{MR17:814b},
which was not a proper translation; and I explained the origin of the
Taliaferro edition.  I had the original Britannica version of
Taliaferro with me, and Mehmet observed that the diagram for I.15 was
completely different there.  I don't know why Mehmet would have had a
problem with the Green Lion diagram, which is beautiful, albeit
anachronistic.  I forgot about the English edition by Rosenfeld, which
I had once found on line.  The English needed editing, and there were
no diagrams: the reader was supposed to consult Heiberg for the
diagrams.  The translator claimed that, as a mathematician, he could
correct the deficiencies of Taliaferro.\footnote{It seems Rosenfeld died last
year, but his translation is at
\url{http://www.math.psu.edu/katok_s/Apollonius.html}.}

Anyway, Mehmet is one of the brightest students I have had; if he is
struggling with Apollonius, I suppose that should tell me something.
I shouldn't feel disappointed that we are not likely to complete the
remaining fifty propositions of Book I of the \emph{Conics} in the remaining
seven hours of class!

\section{Friday, January 8}

There have been four hours of class (three sessions) that I couldn't
report on.  We spent those hours on Propositions 11--15 of Book I of Apollonius.

On Tuesday, December 22, Cihan presented 11, which introduces the parabola.
Everything was fine, as I recall.  Rashad was then supposed to present 12,
introducing the hyperbola, but he was missing, so I had Tolgay go ahead to
13, on the ellipse.  He admitted there were things he didn't understand,
but I said we would work them out together.

However, there wasn't time to finish in the one hour (rather, 50 minutes)
that we had.  So Tolgay started over on December 25.  He used colored
chalk to distinguish parts of the diagram: a good touch.  When Tolgay was
finished, I asked Rashad if he could draw a new figure, but keep Tolgay's
proof on the board, since it was basically the same as the one he needed
to give.  But no, Rashad had to start afresh.

Apollonius and Euclid may repeat statements in the course of a
demonstration.  They do not have the modern technique of writing an
equation (or rather, proportion), displayed by itself on one line, with a
number, so that it can be referred to later by that number.  But we today
can use this technique at the blackboard.  Indeed, if a proportion is
somewhere on the blackboard, and we want to use it, we can point to it and
say `By this we can conclude\dots'  I think Rashad was one, but not the only
one, of the students whom I tried to convince to use this technique,
rather than rewrite something that hasn't yet been erased.  He still
wanted to rewrite.  Such students are at the stage of following an
argument of Apollonius step by step, without seeing it as a whole.

Tolga presented 14, on the opposite sections, in the manner I have come to
expect: he sounds fairly polished, but he may not really know what is
going on.  Afterwards, I tried to emphasize the point of the proposition:
No matter how oblique your cone or rather your `opposite surfaces' are, no
matter whether your cutting plane cuts one surface near the vertex, and
the other far away, you get the same section from either surface.

On December 28, Mehmet presented 15, the finding of a second diameter of
the ellipse.  After his successful conclusion, I admitted that the
proposition was still mysterious to me, although it became unsurprising if
one wrote things out with `Cartesian' coordinates.  I had written out my
own rearranged and streamlined argument, but I didn't take the time to
show the students.  (The point is that most of the argument can be written
out as a chain of equal ratios, as in $A:B::C:D::E:F::G:H::\dots$)

I said that an ellipse by definition is a certain kind of conic section;
by demonstration, the ordinates of an ellipse have a certain relation to
the abscissas.  Proposition 15 shows that an ellipse has a second diameter, with
respect to which new ordinates have a similar relation to new abscissas.
But this does not show that there is a cone that would give us that second
diameter along with the ellipse: showing this would take a lot more time
and propositions.

There is a remarkable point in the demonstration where Apollonius takes
the difference between one area, say $A$, and another area, say $B$, although
$B$ is not actually a part of $A$.  We know that $B$ is \textbf{equal}
to a part of $A$;
but still, to speak of the difference between two disjoint areas suggests
the idea of an area as something \textbf{abstracted} from a figure.

But I hadn't yet checked the Greek text.  Apollonius doesn't speak of `$A$
minus $B$'; he says `$A$ exceeds $B$ by $C$.'

Time was up.  Unlike December 25, January 1 was a holiday, so our next
scheduled class was to be January 5.  However, I stayed home with a cold
that day.  It wasn't a matter of life or death, but I just didn't feel
like putting in the effort of making my way to the university.  In fact,
perhaps staying home didn't improve my rate of recovery; I still feel worn
out by the cold, though I am in the office getting ready for the last
class of the semester.

I did spend time over the weekend thinking of what might be done next
semester.  There are passages of Al-Khwarizmi, Al-Uqlidisi, Thabit ibn
Qurra, and Omar Khayyam that are worth reading, along with Cardano,
Descartes, Newton, and Lobachevski.  Unfortunately, if there is not enough
time, it is the first four, not the last, that should be jettisoned.

\section{Friday, January 8}

The last class of the semester is over.

On Tuesday, December 28, I got an email from Melis, who was scheduled to
prove Proposition 16.  She was however writing from her home in \.Izmir
(Smyrna), whither she had made a snap decision to go.  It is common for
students to make the holidays longer than they are officially scheduled to
be.

We did Proposition 15 on Tuesday, but there was no time for 16 anyway.  If
there had been, I may have proved it myself, more efficiently (in my view)
than Apollonius does.  The proposition is to me a rare example where a
proof by contradiction is better than a direct proof.

So today we opened with Melis's proof of 16: that an hyperbola has a
conjugate diameter.  But Melis\label{Melis-copying} didn't actually
give \textbf{her} 
proof.  She went up with her copy of the Green Lion text and started
copying its contents onto the board.  After a while I asked her what
she was trying to prove, since she hadn't made it clear.  

`I don't know' she said.

Another dilemma for Teacher.  I could be a disciplinarian and send Melis
to her seat, with a reminder that nobody should write down anything that
\emph{she herself} doesn't understand and believe.  But then I should have been
doing this throughout the semester.  In the event, I told Melis what she
was supposed to be proving.

However, when Melis continued copying things down from the text, I
questioned this practice, noting that we all had the text and could read
it for ourselves; she was supposed to be \textbf{explaining} to us, I said.  She
made some weak attempt at this, but it was more like explaining to
herself.

Near the end, Apollonius makes a leap that the editors justify with a
footnote.  Melis ignored the footnote, and Cihan asked why the leap was
justified.  I don't \textbf{think} he was testing Melis; he really wanted to
know.  Ali attempted an explanation, and then it appeared that \textbf{he} had
missed the point of the proposition: he thought $AK = BL$.  This is true, but
it is precisely what must be proved.

Melis finished somehow, and then I got up to offer my proof by
contradiction of the same result, as well as a general comment on the
import of the proposition.

Nur did 17, a simple but confusing proposition; and Nur indeed was
confused.  I went to the board and discussed the situation with her.
Actually I'm not sure how we know that point $C$ exists (under the
assumption that leads to contradiction).  I said this.

Ahmet did 18, and \c Ca\u gda\c s, 19.  Taner was supposed to do 20, the next
proposition after 16 that involves actual lengths.  But he was absent, so
I proved this, along with 21.  (The last had been unassigned, but I
thought we should finish with this rather than 20.)

Then, in a semester course on ancient mathematics, I spent half an hour
talking about Archimedes.  I gave his rigorous quadrature of the parabola,
then mentioned the non-rigorous version on the Archimedes Palimpsest
discovered in Istanbul by Heiberg a century ago.

Then I had to stop.  But before class I had written on the board the names
of the mathematicians I wanted to read next semester.  Ali at least was
interested and wanted to know how to get a hold of the texts for a friend.

In the break, \"Ozge asked about the final exam, and in particular whether
they still had to know the Greek alphabet.  I said Yes.  She complained
that she didn't want to memorize it again.  I said that most of her
classmates had not bothered to do this on the first exam, so I thought it
was fair to ask again; but I said I would re-memorize the Russian alphabet
(which I had learned for one of the two language exams for my
doctorate, though I forgot everything soon after the exam).


\part{Spring semester}\label{part:two}

%\input{math-304-log}

\addsec{About the course}

\emph{This is from the Math 304 webpage:}

This course is a continuation of Math 303, but that course is not a
prerequisite for this one. Practices will be as in Math 303: 
\begin{compactitem}
\item 
    attendence is required;
    \item
all students will spend time making presentations at the blackboard;
\item
 there is no `textbook'. 
\end{compactitem}

This course will make no attempt to fit the catalogue
description. Some phrases in that description are apparently based on
chapter titles in Boyer's \emph{History of Mathematics.}  But again, this
course will not follow a textbook; we shall read original sources
(albeit in translation, from Arabic, Latin, French, \dots). This approach
is slower, but more honest to the title of the course. Why? 
\begin{compactitem}
\item 
 I accept the conclusion of the philosopher R. G. Collingwood [see
   Appendix~\ref{app:Col}]  that
 history is the history of thought.
This means, in particular, that
 doing history of mathematics means thinking the mathematical thoughts
 of past mathematicians. 
 \begin{compactitem}
 \item 
This is difficult work, but nobody else can do it for us.
\item
This work can hardly be done without looking directly at what these
mathematicians actually wrote.  
 \end{compactitem}
\item
Second-hand accounts of past mathematics may give a misleading view,
as for example by translating everything into modern algebraic terms.  
\end{compactitem}

Anybody who is interested can read a conventional `history of
mathematics' on their own. But there is no substitute for working
together, as a group, to understand some old piece of original
mathematics. 

Some students took Math 303 in hope of learning some history in the
sense of stories. The words `history' and `story' are indeed
cognate, coming through French from the Latin \emph{historia,} which is from
the Greek \Gk{<istor'ia}. However, we know almost nothing about the
personal lives of ancient mathematicians. About more recent
mathematicians, more is known. For example, there is this interesting
piece of information: 
\begin{quote}
    After his death, Newton's body was discovered to have had massive
    amounts of mercury in it, probably resulting from his alchemical
    pursuits. Mercury poisoning could explain Newton's eccentricity in
    late
    life.\footnote{\url{http://en.wikipedia.org/wiki/Isaac_Newton},
      accessed February 17, 2010}   
\end{quote}
This is irrelevant to the understanding of Newton's mathematics
(though it might be used as an excuse for not understanding Newton). 

Some students in Math 303 were disappointed in the quality of some of
their classmates' presentations. However, student presentations are
essential to this course. You don't really understand something unless
you can stand up and talk about it. Also, in this course, everybody
should have read what is being presented at the blackboard, and
everybody should be prepared to criticize a faulty presentation, or to
raise questions.  

\chapter{Al-Khw\=arizm\=\i, Th\=abit ibn Qurra, Omar Khayy\'am}
 
\section{Thursday, February 18}

There were 20 students in class; three of
them were not
among the 37 registered students.  I discussed what is on the
webpage.\footnote{\url{http://metu.edu.tr/~dpierce/Courses/304/}}  I
stated Propositions
5 and 6 of Book II of Euclid's \emph{Elements} \cite{MR1932864} and I
drew 
the diagrams that prove the propositions.  In algebraic notation, the
propositions are:
\begin{align*}
        (x + y)(x - y) + y^2&= x^2&   \text{ if } y &< x,\\
        (x + y)(y - x) + x^2& = y^2&   \text{ if } x &< y.
\end{align*}
See Figure~\ref{fig:0}.
Written algebraically, the propositions become the `same' if we switch
$x$ and $y$ 
in the second line.
\begin{figure}[ht]
\psset{unit=9mm}
%\mbox{}\hfill
\begin{pspicture}(-1.1,-0.5)(4.5,2)
%\psgrid
\psline(2,0)(2,2)(4,2)(4,0)(0,0)(0,1.4)(2.6,1.4)(2.6,0)
\psline(2.6,1.4)(2.6,2)
\uput[u](1,0){$x$}
\uput[l](0,0.7){$x-y$}
\uput[u](2.3,0){$y$}
\uput[u](3.3,0){$x-y$}
\uput[r](4,1){$x$}
\uput[d](0,0){$A$}
\uput[d](4,0){$B$}
\uput[d](2,0){$C$}
\uput[d](2.6,0){$D$}
\end{pspicture}
\hfill
\begin{pspicture}(0,-0.5)(5.8,3.1)
%\psgrid
\psline(2,0)(2,2.6)(4.6,2.6)(4.6,0)(0,0)(0,0.6)(4.6,0.6)
\psline(4,0.6)(4,2.6)
\uput[u](1,0){$x$}
\uput[u](3.3,0){$y$}
\uput[r](4.6,0.3){$y-x$}
\uput[r](4.6,1.6){$x$}
\uput[u](3,2.6){$x$}
\uput[d](0,0){$A$}
\uput[dr](4.6,0){$D$}
\uput[d](4,0){$B$}
\uput[d](2,0){$C$}
\end{pspicture}
%\hfill\mbox{}
\caption{Quadratic equations as in Euclid}\label{fig:0}
\end{figure}

But Euclid doesn't write things this way.  I introduced the propositions
by asking: 
\begin{quote}
If a straight line is to be divided in two, where should the
point of division be chosen so as to maximize the area of the rectangle
bounded by the two pieces?
\end{quote}
Ali answered (he was one of the best students in last semester's course):
He said the straight line should be bisected.  

Why? I asked.  He observed
that if the point of division approached one of the ends of the line, then
the rectangle would become small.

This was a reasonable way to think of the problem, I thought.  But then I
have a memory of thinking this way as a child:  I was playing with a
rubber band, and I wondered if the area enclosed by the band remained
constant through all possible contortions of the band (in a plane).  The
answer was obviously No, if one observed that the band could be
straightened out so as to enclose nothing.

My maximization question in class on Thursday was one that may come up in
a calculus class.  But I don't think anybody should be impressed at the
ability of calculus to answer the question, since the answer is so easily
found without calculus.  Indeed, if you divide the line equally and
unequally, then Euclid's II.5 shows by \emph{how much} the rectangle
bounded by the
equal parts exceeds the rectangle bounded by the unequal parts: it exceeds
by the square on the line between the two points of section.

Euclid's II.6 is about what happens when the line is divided `externally'.
Euclid doesn't use this language, and I don't know whether he thought of
it.  Me, I am delighted to find that two propositions are just instances
of one idea; but I can only guess whether Euclid sought such delight.
(Presumably he saw that II.5 and II.6 were intimately related; but I don't
know what he thought the relation was.)

Again, Proposition II.5 is that, if straight line $AB$ is bisected at $C$, and
$D$ is chosen elsewhere on $AB$, then
\begin{equation*}
        \rect{AD, DB} + \sq{CD} = \sq{AC}.
\end{equation*}
Proposition II.6 is about what happens when $D$ is chosen on the
\emph{extension} 
of $AB$ beyond $B$.  Then
\begin{equation*}
        \rect{AD, BD} + \sq{AC} = \sq{CD}.
\end{equation*}
These become the same proposition if we use `directed' lines and allow
`negative' areas, so that $\rect{AD, DB}$ is the `negative' of
$\rect{AD, BD}$. 
But I don't know of any reason to think that Euclid considered this
possibility.

In the remainder of class, I started to state what we would need to know
about conic sections in order to understand Omar Khayy\'am's solution of
cubic equations by means of conics.

Specifically, I said that a parabola has an \emph{axis} and a \emph{parameter.}
Suppose the parameter is $AB$, and the axis is $AD$, drawn at right angles.
If $C$ is chosen on the parabola itself, and $CD$ is drawn at right angles to
the axis, then
\begin{equation*}
         \sq{CD} = \rect{AB, AD}.
\end{equation*}
An ellipse or an hyperbola has an axis $AB$ and a parameter $BC$ so
that, if $D$ is chosen on the curve, and $DE$ is dropped at right
angles to the axis (or the axis extended, in the case of the hyperbola),
then
\begin{equation*}
        \sq{DE} : \rect{AE, EB} :: BC : AB.
\end{equation*}
I postponed till another time the definition of proportion.  After class a
student (possibly Mehmet Arif \c Sekercio\u glu) asked for
clarification of the definitions 
of conic sections.  I sketched a cone as in Apollonius
\cite{MR1660991} and said that he 
proved that, if you cut the cone, the sections had the properties I
described.  Maybe I'll say this to the whole class later.

I did intend to be a bit intimidating in the first class, trying to ensure
that only committed students stayed with me beyond Add-Drop Week.  
\begin{comment}
There
is another third-year course, Galois Theory, which supposedly has only
four students in it.  One student told Ay\c se that students don't want to
take that course, since the teacher is hard to understand and gives low
grades.  Well, I don't want to have to accommodate more students owing to
my colleague's failure to offer an attractive course of his own.
\end{comment}

 \section{Tuesday, February 23} 

Perhaps I shall not have a problem.
Twenty-three students 
showed up---more than last time, but not many more.  (Seven from Thursday
did not return.)

I had told the students on Thursday to read the selections from
al-Khw\=arizm\=\i{} and Th\=abit ibn Qurra (taken from Katz
\cite{Katz}) that I had put on the
webpage.  I said the students 
should be prepared either to explain these passages or say why they didn't
make sense.  But when I came to class on Tuesday, it appeared that only
one student (Zhala) had actually printed out the selections.  Another
student (O\u guzhan) had read the selections on the computer screen and taken
notes; he said he could expound their contents.  Perhaps others had done
something similar; but to find out, I should have had to interrogate
them one by one.

Instead of doing that, I gave the book to the nearest student (Dilber)
and asked her to read the first paragraph of al-Khw\=arizm\=\i; we discussed this, then another
student read the next paragraph, and so on.

The first paragraph (after the preface invoking the blessings of the
deity) seems to allude to `Arabic' numerals.  That's what one student
said, and I agreed, saying that if we had more time it could be fun to
read Al-Khw\=arizm\=\i's exposition of the Hindu base-ten numeration
system: I 
gather this exposition is the reason why we call them Arabic numbers.  I
wrote on the board
\begin{quote}\centering
        1 2 3 4 5 6 7 8 9
\end{quote}
and asked what one calls these in Turkish; the students said \emph{rakam.}
Then I wrote
\begin{quote}\centering
        I II III IV V VI \dots
\end{quote}
They told me these were \emph{Romen rakamlar\ i.}  There seemed to be some
awareness that English uses the term `Arabic numerals' for the former; but in
Turkish they are just numerals.

Al-Khw\=arizm\=\i{} introduces squares, roots, and numbers.  But they are all
numbers.  His first example is
\begin{quote}\centering
        Square is equal to five roots of the same.
\end{quote}
With student approval, I wrote this as
\begin{equation*}
        x^2 = 5x.
\end{equation*}
(Let me just say once for all that when I write such things, I
periodically recall that our authors do \emph{not} use such language.)
Al-Khw\=arizm\=\i{} then concludes
\begin{equation*}
        x = 5.
\end{equation*}
I asked if there was any problem here.  Somebody said $x = 0$ was another
solution; but it seemed to be agreed that this was of no interest.

When al-Khw\=arizm\=\i{} got to the more complicated example---
\begin{quote}%\centering
        one square, and ten roots of the same, amount to thirty-nine
        dirhams
         [$x^2 + 10x = 39$]
\end{quote}
---I had O\u guzhan go to the board and present al-Khw\=arizm\=\i{}'s cookbook
solution.  It is a solution that in my opinion is not self-justifying: it
arrives at the answer $3$, and one can check that this is correct, since
three squared plus ten times three is indeed thirty-nine; but one does not
know \emph{why} this should be correct.

\begin{sloppypar}
More on this later.  The students seemed to understand that
al-Khw\=arizm\=\i{}'s `dirham' just meant a unit.  Ali knew that it
had been in particular a unit of weight.  I observed that it was still
the monetary unit of Morocco and that it derived from the Greek
\Gk{draqm'h}. 
\end{sloppypar}

Meanwhile I had Zhala go to the board to write out the solution to
\begin{quote}%\centering
        square and twenty-one in numbers are equal to ten roots of the
        same square          
        [$x^2 + 21 = 10x$];
\end{quote}
Yasemin read out the steps of the solution as necessary.  Here two
solutions arise.  Why?

Well, al-Khw\=arizm\=\i{} does go on to give a geometrical justification.  For
this, I had Murat go to the board to draw the diagrams, while
somebody else---Salih Kanl\i da\u g, I think---read out the steps.

Murat's full name is Murat Yasar Kurt, but he told me likes to be called
MuYaKu (`like Japanese' he said).  He turned out to have a
printout of the text as well.  He was not particularly prepared to draw
al-Khw\=arizm\=\i{}'s diagram; but he worked it out.

So now we had two solutions of $x^2 + 21 = 10x$ on the board: the
`arithmetic' solution that Zhala had written, and the geometric one that
MuYaKu had written.  Some students agreed with me that the geometric
solution was at the same time a \emph{proof} that it was a solution.  But
MuYaKu said they were both proofs, just done in different styles.

I wrote out the geometric argument more quickly, arriving at the answer $3$.
What you do is draw a square, then extend one of the sides to have total
length $10$ units; see Figure~\ref{fig:1}, left.  You complete a
rectangle next to the square and on the
extension of the side.  The rectangle is supposed to have area $21$ units;
and this with the square makes $10$ `roots'.  Now bisect the line of length
$10$.  This has already been divided unequally, and the rectangle formed
from the two pieces has area $21$, as we said.  By Euclid II.5, the square
on $5$ exceeds $21$ by the square on the line between the two points of
division.  So this line is $2$ units long, and the original square has side
$3$ units long.
\begin{figure}[ht]
%\mbox{}\hfill
\begin{pspicture}(5,2.5)
\psset{unit=5mm}
\psline(0,0)(10,0)(10,3)(0,3)(0,0)
\psline(5,0)(5,5)(0,5)(0,3)
\psline(3,0)(3,5)
\end{pspicture}
\hfill
\begin{pspicture}(5,3.5)
\psset{unit=5mm}
\psline(0,0)(10,0)(10,7)(0,7)(0,0)
\psline(5,0)(5,5)(10,5)
\psline(7,0)(7,7)
\psline(5,3)(7,3)
\end{pspicture}
%\hfill\mbox{}
\caption{A quadratic equation as in al-Khw\=arizm\=\i}\label{fig:1}
\end{figure}

I didn't actually refer to Euclid; we in effect reproved the proposition.
Anyway, $7$ is also a solution to the original problem: why didn't this come
directly out of the geometric argument?

O\u guzhan knew the answer: al-Khw\=arizm\=\i{}'s drawing assumes that the midpoint
of the line of length $10$ lies beyond the side of the original square.  If
it lies inside, we get $7$.  See Figure~\ref{fig:1}, right.

Time was about up.  Al-Khw\=arizm\=\i{} considers three kinds of problems:
\begin{compactenum}
\item 
square and roots equal a number,
\item
square and number equal roots,
\item
square equals roots and number.
\end{compactenum}
As an \textbf{exercise,} I suggested working out geometric solutions to the
remaining cases, as for example in the following instances:
\begin{gather*}
        x^2 + 10x = 39,\\
        x^2 = 4x + 21.
\end{gather*}
Probably Al-Kharizmi does this himself in the full text (which I linked to
on the webpage; I didn't want to use those versions in class though,
because they are full of footnotes explaining things in symbolic terms).

O\u guzhan had indicated that al-Khw\=arizm\=\i{} was solving equations
\begin{equation*}
        ax^2 + bx = c.
\end{equation*}
I agreed, but observed that he didn't use unspecified coefficients like
$a$ and $b$.

I think Th\=abit ibn Qurra does in effect use general (unspecified)
coefficients; 
but this will be our topic for Thursday's class. 

\section{Thursday, February 25}


Indeed, Th\=abit ibn Qurra gives a geometric solution to the problem:
\begin{quote}
\emph{m\=al} and roots equal a number.
\end{quote}
A note says \emph{m\=al} is the Arabic for \emph{asset.}  Indeed I'm
embarrassed to recall only now that Turkish has borrowed the word with
this meaning.  No student pointed this out; is that because the point
is obvious to them, or because they didn't notice it? 

In any case, the meaning here is `square', that is, `square of the
root'.  Th\=abit ibn Qurra draws a square $ABDG$ (but he calls it
$ABGC$) and extends it by a rectangle that represents the `roots'; the
whole rectangle then is the `number'.

Say one of the long side of the large rectangle is $AE$; this contains
$B$.  Let $BE$ be bisected at $W$.
(So ibn Qurra introduces letters in the order $ABGDEW$, at least in
translation; was he using Greek letters, including the digamma?  Ali
suggested it might be so, though I don't know if he knew about the
digamma.  Would Arabic letters be transliterated thus too?  No student
claimed knowledge of these letters.)

By II.6 of Euclid's \emph{Elements,} to which ibn Qurra refers explicitly,
\begin{equation*}
     \rect{EA, AB} + \sq{BW} = \sq{AW}.
\end{equation*}
But $\rect{EA, AB}$ is the given `number', and $BW$ is half the given
number of roots, so $\sq{BW}$ is known; hence $\sq{AW}$ is known; hence $AW$
is known.  The claims about what is `known' allude to Euclid's
\emph{Data} \cite{MR1989796},
though only the editor's footnote makes this explicit.  Finally, $AW$
minus $BW$ is known; but this is the desired root.

So the original equation is soluble in principle.  And this claim
holds generally.  Th\=abit ibn Qurra's alternative to using literal constants
in an equation like
\begin{equation*}
     x^2 + bx = c
\end{equation*}
is to make the equation into a picture.  We just somehow understand
that one picture can stand for many cases; to suggest otherwise is to
suggest that, even if we know how to solve $x^2 + bx = c$, we are not
sure we can solve $x^2 + dx = e$.

It is worth noting that Th\=abit ibn Qurra does not actually give a
construction for solving the equation; he just shows that it \emph{can be
done.}

Again with the passage of time, I've forgotten who presented the above
solution in class.
In the excerpt in the book we've been using, Th\=abit ibn Qurra goes on to
solve the equation
\begin{quote}
     \emph{m\=al} and number are equal to roots.
\end{quote}
I decided to skip doing this in class, in order to review conic
sections again, as they would be needed for Omar Khayy\'am's solution of
cubic equations next time.

\section{Tuesday, March 2} 

I asked if somebody
could present Omar Khayy\'am's 
solution (also in Katz \cite{Katz}) of an equation of the form
\begin{quote}\centering
     cube and number are equal to sides.
\end{quote}
(So `side' is what we called `root' before.)  Several students said
they hadn't been able to follow the argument.  Mehmet volunteered to
go to the board; but first I got G\"ok\c cen to read the selections from
Khayy\'am's introduction that are included in the text.  Some key
points: 
\begin{compactenum}\sloppy
\item 
Khayy\'am says you gotta know Euclid's \emph{Elements} and \emph{Data,} along
with the first two books of Apollonius's \emph{Conics;} but that's enough.
\item
There are four geometric `degrees': (absolute) numbers, sides,
squares, and cubes; you can talk about square-squares, but only
`metaphorically'.
\item
Only equations involving numbers, sides and square can be solved
\emph{numerically,} so far: perhaps somebody in future can do more.
Khayy\'am's solution of cubic equations will be \emph{geometric.}
\item
The numeric/geometric distinction was recognized by Euclid; why
else would he develop a theory of ratios of \emph{magnitudes} in Book V,
then an independent theory of ratios of \emph{numbers} in Book VII?
\end{compactenum}
Mehmet then worked out Khayy\'am's solution of the equation above.  It
involves a parabola and an hyperbola: their point of intersection
determines the solution.  Mehmet rewrote the equation symbolically as
\begin{equation*}
     x^3 + a = bx.
\end{equation*}
During the course of things, I asked: Why must the parabola and
hyperbola intersect?  Somebody, I think Fuad, said they need not.

Indeed, Khayy\'am notes that the curves might be tangent, or meet in two
points.  But he doesn't give conditions for tangency.  I suggested
this as an \textbf{exercise} for the students.

It is too bad most of the students were not with me last semester to
read Apollonius.  I just told them that Apollonius shows how conics
can be found with given axes and parameters, and this justifies what
Khayy\'am does.  But it's not a ruler-and-compass construction; indeed,
one needs a third dimension for the cones themselves.

I observed that if $x^2 + a = bx$, then $x$ is  half of $b$ plus (or minus)
the square root of the sum of $a$ and the square of half of $b$.  I did
this geometrically, but got confused, so the students helped me out.
I asked how we could \emph{construct} a square root, and Fuad came to the
board to do this with a circle, though he was a bit hesitant.  In any
case, there are algorithms for extracting square roots numerically.
(The anthology of texts has an algorithm for fifth roots, but I
skipped it.)

I observed that we didn't have a way to convert Khayy\'am's solution of
the cubic into a similar construction and method of computation.

After the break, I proposed another way to symbolize Khayy\'am's work, a
way closer to what he does.  Khayy\'am introduces lines of lengths $a$ and
$b$; then he in effect solves
\begin{equation*}
     x^3 + a^2b = a^2x.
\end{equation*}
He does this by contructing a parabola
\begin{equation*}
     x^2 = ay,
\end{equation*}
then a hyperbola
\begin{equation*}
     y^2 = x(x-b).
\end{equation*}
Eliminating $y$ shows that $x$ is as desired.  Indeed, from these two
equations we get
\begin{gather*}
     \frac ax = \frac xy = \frac y{x-b},\\
     \frac{a^2}{x^2} = \frac x{x-b},\\
     x^3 = a^2(x-b),\\
     x^3 + a^2b = a^2x.
\end{gather*}
This derivation follows Khayy\'am's verbal description pretty closely, I
think.  But I have no intuition for actually coming up with this
solution.  Well, that's what I said in class anyway; but now that I
think of it, I see that the steps of my algebraic derivation are
pretty easily reversible.

But did Khayy\'am think this way?  I don't know.  I also don't know
whether Khayy\'am's solution is original with him; I think Greek
mathematicians knew how to find cube roots with conic sections,
anyway.  But again, Khayy\'am refers explicitly to Euclid and
Apollonius; if he were using additional old work, he might have said
so.

In the reading, Khayy\'am also solves
\begin{quote}\centering
     cube and sides equal squares and number.
\end{quote}
I worked through the solution myself (it uses an hyperbola drawn with
respect to given asymptotes, and a circle).  Then I took volunteers
for presenting the several sections of our next reading:  Chapters I,
II, VI, XI, XXXVII, and XXXIX of Gerolama Cardano's \emph{Ars Magna} or \emph{De
Regulis Algebraicis.}

The book opens with an attribution of the invention of the art of
algebra to Muhammad the son of Moses the Arab,---that is, Al-Khw\=arizm\=\i.
It gives a `numerical' solution to cubic equations.  Anthologies
include this, but it's not much fun to read out of context.

At least Struik's anthology \cite{MR858706} has a fairly literal
translation.  (I 
don't remember what Smith's \cite{MR0106139} is like.)  The whole of
Cardano's book was
translated by Witmer in 1968 \cite{MR0250842}, but Witmer freely uses
modern notation. 
This helps one read, but is misleading.  The original Latin can be
found on the web: I found it through Wikipedia.  Unfortunately this is
the text from a posthumous edition of Cardano's complete works from
1663.  Unfortunately, as Witmer says, each edition of the \emph{Ars Magna}
kept the old mistakes and introduced new ones.

Over the weekend I started typing out some sections of the Latin, with
parallel translation: Witmer's or Struik's translation, with some
adjustments by me.  But this job became too daunting as the amount I
wanted to include grew.  So I just gave students copies of sections of
Witmer's translation, along with the Latin from the \verb+pdf+ file on the
web, in case they want to look at that (and Mehmet D., at least, said he
did). 

\chapter{Cardano}

\section{Thursday, March 4}

Ali and Emir presented sections 2--5 of Chapter 1 of Cardano's \emph{Ars
Magna.}  The two didn't know each other before signing up for this
assignment on Tuesday.  I think that, rather than working together, they
just decided to alternate sections.  Ali began, talking about sections 1
and 2; then Emir continued with 3, and so on.

When Ali started speaking, there was still a bit of chitchat in the room.
I asked him if people were paying attention to him.  He didn't seem too
concerned; but then I shut people up.

During Ali's and Emir's presentations, I occasionally went to the board to
make a point.  I could see that students were asleep or had blank
expressions.  At the end of class I said they all looked like Zombies, and
they laughed.

But then MuYaKu came to me and asked, `What exactly did we do today?'  I
said, `You should have asked this earlier!'

I thought the reading was fascinating, though on the surface it appears
trivial.  Superficially, the reading is a discussion of negative numbers:
for example, $9$ has two square roots, namely $3$ and $-3$.

But this is a misleading account, which is unfortunately encouraged by
Witmer's translation.  There is no `$-3$' for Cardano; there is $3$,
considered as minus.  Something like that.  There is no symbol $-3$; there
is just `$\mns 3.$' or `$3. \mns{}$' (Possibly the periods are just
thrown in by 
the typesetters.  The tildes are presumably to indicate an
abbreviation.)

Later in the book, Cardano will suggest the possibility of taking the
square root of a negative number.  Here he ignores this.  Hence for
example $81$ has only two fourth roots, namely $3$ and minus-$3$; but there
could be two more, namely the square roots of minus-$9$.

To be continued\dots

\section{Excursus}\label{sect:Klein}

By a Johnnie reading the foregoing, I was asked:
\begin{quote}
Do you have any
suggestions [for a seminal text on negative numbers]?   
\end{quote}
I answer: Texts other than what we read at St John's are new for me
as well.  In 
his \emph{Mathematical Thought from Ancient to Modern Times}
\cite{MR0472307}, Kline says, 
\begin{quote}
One of the first algebraists to accept negative numbers was Thomas
Harriot (1560--1621), who occasionally placed a negative number by
itself on one side of an equation.  But he did not accept negative
roots.
\end{quote}
What does this mean?  Did Harriot actually write equations in
the modern symbolic sense?  This seems to be one more example of why a
math history book is not of much value in isolation from the original
texts.

In any case, Cardano was dead before Harriot was born, but Cardano had
given some recognition to negative numbers, as I have said.

Kline makes another strange comment earlier in his book (page 192): 
\begin{quote}
In
arithmetic the Arabs took one step backward.  Though they were
familiar with negative numbers and the rules for operating with them,
through the work of the Hindus, they rejected negative numbers.
\end{quote}

By the way, leafing through Kline's book, I notice something that is
apparently wrong.  (I find it worthwhile to collect such examples, in
case one of my colleagues in future still wants to teach the history
course out of a modern textbook.)  Kline says (p.~194), 
\begin{quote}
As for the
general cubic, Omar Khayyam believed this could be solved only
geometrically, by using conic sections.
\end{quote}
But Khayyam says (in the translation in the Katz book), 
\begin{quote}\label{OK}
But, as for
the proof concerning these kinds [of equations], if the subject of the
question is simply a number, neither we nor any of the algebraists
have been able to do it except in the first three degrees: number,
thing, and \emph{m\=al}.  But perhaps someone else, who will come after us,
will know [how to do] it.
\end{quote}
Maybe Khayyam contradicts himself somewhere else on what the future
may hold, but I don't know why he would.

My correspondent replied:
\begin{quote}
Perhaps Klein is right.  I believe the case is that the general cubic
cannot be solved by an algorithm when there are 3 real roots, the
so-called `irreducible cubic.'  I have not checked this lately, but if
it is so, then Khayy\'am may be intending to say that some, not all
cubics can be solved numerically.  It is these irreducible cubics,
by the way, that Vi\`ete solves geometrically in props 16-18 of the
\emph{Supplementum Geometriae.}  
\end{quote}
I answered this as in \S\ref{sect:answer}.


\section{Thursday, March 4, again}

For Cardano (still) there are three kinds of quadratic equations:
\begin{compactenum}
\item \label{item:i}
square and roots equal to a number,
\item\label{item:ii}
 square and number equal to roots,
\item\label{item:iii}
 square equal to roots and number.
\end{compactenum}
(`Root' here is \emph{res} `thing', though Witmer doesn't like the
translation `thing'; he just uses $x$, though Struik uses `unknown'.)

Type~\eqref{item:i} has one solution (\emph{\ae stimatio,} I
believe)---that is, one 
\emph{positive} solution.  For us there is also a negative solution, but
not, apparently, for Cardano: for him the equation
\begin{equation}\label{eqn:*}
     x^2 + 3x = 28
\end{equation}
has just the solution $4$.  But the funny thing is, this means for
Cardano that the equation
\begin{equation}\label{eqn:**}
     x^4 + 3x^2 = 28
\end{equation}
has two solutions, $2$ and minus-$2$: but these are also `equal to each
other' for Cardano, rather as in `equal and opposite,' I suppose---and
this is a reason not to write `minus-$2$' as `$-2$', since $-2$ is obviously
different from $2$.

Cardano doesn't write out equation~\eqref{eqn:*}, only~\eqref{eqn:**};
so this is all Emir 
wrote in his presentation.  I asked Emir how he knew that~\eqref{eqn:**} had the
solutions $2$ and minus-$2$; in reply, he wrote out~\eqref{eqn:*} (with
$t$ instead of $x$, since he let $t = x^2$); then he solved it.  But
he did this by 
transforming it into
\begin{equation*}
     t^2 + 3t - 28 = 0
\end{equation*}
and then observing $28 = (4)(7)$ and $3 = 7 - 4$, so that the roots are $4$
and $-7$.

I said, `So you're factorizing,' and Emir agreed.  In other words,
Emir was finding what I would have written as
\begin{equation*}
     t^2 + 3t - 28 = (t + 7)(t - 4).
\end{equation*}
But he didn't actually write out the factorization this way; he just
wrote down the $4$ and the $7$.  To write more would have been against his
training to find answers as quickly as possible and fill in the right
circle on the multiple-choice answer form supplied with the national
university entrance exam.

Apparently Emir has not picked up on the geometric solutions we have
worked out, whereby one finds that t is the square root of the sum of
$28$ and the square of half of $3$, minus half of $3$.  This raises for me
the question of whether to encourage the students more to try to think
in the old-fashioned way.

Cardano works out similar examples with types~\eqref{item:ii}
and~\eqref{item:iii}:  Type~\eqref{item:ii} has either two or no
solution, so the corresponding quartic (with 
$x$ replaced with $x^2$) has either four or no roots.
Type~\eqref{item:iii} has one 
[positive] solution, so the quartic has two solutions.

How do we know that these solutions exist?  Ali observed at some point
that Cardano seemed to be making some sort of continuity assumption.
I said that we had a geometric construction of solutions of quadratic
equations.  But Ali seemed to understand `geometric' as `physical': we
could obtain a line segment as the solution, but our measurement of
this segment would yield a rational number, even though the correct
solution might be irrational.  Ali mentioned that the Pythagoreans
knew about the irrational, and that this caused a crisis for them; I
passed on what I had learned from Mr Thomas on the J-list, that there
was no evidence of such a crisis.

I also observed that Cardano was going to be using cube roots, even
though there is no ruler-and-compass construction of these.  But I
asked whether anybody knew an algorithm for extracting square roots.
Nobody did.  My father had once told me that he had learned such an
algorithm, and a couple of years ago I derived an algorithm for myself
for some reason, while teaching a number-theory course.  One of the
Arabic readings that I skipped in the Katz book concerns extraction of
a fifth root.  So I suppose Cardano believed in roots because they
could be calculated (albeit only approximately).

But Cardano observes further (and Ali presented this part) that if any
number (`even a thousand') of odd powers are `compared with' (that is,
equated to) a number, then there will be one `true' solution, but no
`fictitious' [negative] solution.  This is the most remarkable
statement in the reading.  Ali understood its import, but I don't know
if everybody else did. (As I said, they were zombies at this time of
day, this late in the week.  Maybe I should make tea for them, as Ayse
and I did one year when each of us was teaching a Saturday class, to
mostly the same students.)  If we have the equation
\begin{equation}\label{eqn:***}
     ax + bx^3 + cx^5 + dx^7 + \dotsb = N,
\end{equation}
then the left side increases from $0$ as $x$ increases from $0$; also the
left side grows without bound as $x$ does; `therefore,' for just the
right value of $x$, the left side will be exactly $N$.

Perhaps it's not hard to accept this.  There's a puzzle that goes
something like, If you drive your car at varying speeds $300$ miles in $5$
hours, must there be a $60$-mile stretch that you cover in exactly one
hour?  The answer is supposed to be Yes, because if you let $f(x)$ be
the time required to travel between the $x$-mile and $(x+60)$-mile points,
then $f(x)$ will sometimes be above, sometimes below one hour, so for
`some' $x$ it will be exactly one hour.  But this makes an unjustified
continuity assumption.

In \emph{Mathematical Thought from Ancient to Modern Times}
\cite{MR0472307}, Kline writes (p.~198):  
\begin{quote}
Perhaps most interesting is the Hindus' and Arabs' self-
contradictory concept of mathematics.  Both worked freely in
arithmetic and algebra and yet did not concern themselves at all with
the notion of proof.
\end{quote}
We may just as well refer to Cardano's `self-contradictory concept of
mathematics.'  What is the real proof that~\eqref{eqn:***} has a solution?
Cardano shows no sign that this is a reasonable question.

Cardano discusses also the signs of the roots of cubic equations.  He
states without proof that the equation
\begin{equation*}
     x^3 + a = bx
\end{equation*}
has no, two, or three roots, depending on whether two-thirds of $b$
times the square root of one third of $b$ is less than $a$, equal to $a$, or
greater than $a$.  Emir just reported the rule, giving no indication of
having thought about where it came from.  Cardano gives no indication
of its origins either.  (I should say that Emir was reporting the
symbolic formulation of things, as recorded in Witmer's footnotes, and
not Cardano's verbal formulation.  This is another reason not to like
Witmer's edition.)

If there are two roots, then one is negative and is `twice' the other
(that is, it is minus-$2$ times the other).  If there are three roots,
then one is negative and the sum of the other two.  Witmer has a long
footnote about whether Cardano understood the meaning of this: in the
two-root case, there are `really' three roots, but the two positive
roots are identical.

I pointed out to the class that we could understand the situation from
looking back at Khayyam's geometric solution.  He solved this case of
cubic with a parabola and an hyperbola with axes at right angles to
each other.  If these curves do not intersect, there is no solution;
if they are tangent, there is one (positive) solution; in the last
case, the curves intersect twice, giving two positive solutions.

Time was up.

As I said, MuYaKu came to me after class, asking what exactly we had
accomplished.  I don't remember exactly what I said, but I talked
about Khayyam's solution, which was still there on the board.  MuYaKu
said this wasn't a solution.  I think he meant that we didn't really
`have' the line that, according to Khayyam, is the solution.  I asked
him whether we `had' the square root of $2$.

Before MuYaKu talked to me, but after class was over, Ece asked about
what student presentations were supposed to be like.  She evidently
\emph{had} been reading Cardano and had observed all of these unjustified
statements; was she supposed to find justifications for them?

Good for her!  I said the readings were not the word of God.  Not
everything needed presentation; the presenter should decide what was
important.  If there were unjustified claims, the presenter should say
so.  I went on in this vein, and Ece nodded enthusiastically and said
she got the idea. 


\section{Excursus, continued}\label{sect:answer}

I answered the comment at the end of \S\ref{sect:Klein}:
Should I emphasize that was referring to the book by Morris Kline, not a
book by (for example) Jacob Klein \cite{MR1215482}?

The method given by Cardano, applied to
\begin{equation*}
        x^3 -15x - 4 = 0,
\end{equation*}
will I believe give as a solution the sum of the cube roots of $2 + 11\mi$ and
$2 - 11\mi$, where $\mi^2 + 1 = 0$.  The method doesn't tell us, however, that
these cube roots are $2 + \mi$ and $2 - \mi$, so that $4$ is a root of
the cubic. 

Are you suggesting that, 450 years before Cardano---who apparently thought
he was publishing the first numerical solution of a cubic---, Khayy\'am
already knew about such solutions?

Under \emph{cubic function,} Wikipedia says,
\begin{quote}
In the 11th century, the Persian poet-mathematician, Omar Khayy\'am
(1048--1131), made significant progress in the theory of cubic equations.
In an early paper he wrote regarding cubic equations, he discovered that a
cubic equation can have more than one solution, that it cannot be solved
using earlier compass and straightedge constructions, and found a
geometric solution which could be used to get a numerical answer by
consulting trigonometric tables. In his later work, the Treatise on
Demonstration of Problems of Algebra, he wrote a complete classification
of cubic equations with general geometric solutions found by means of
intersecting conic sections.
\end{quote}
The information about Khayy\'am's `early paper' seems to be second-hand;
there is no direct reference to such a paper, but to a
webpage\footnote{\url{http://www-groups.dcs.st-and.ac.uk/~history/Biographies/Khayyam.html}} that also uses the Khayy\'am quote that I gave (p.~\pageref{OK}):
\begin{quote}
Another achievement in the algebra text is Khayy\'am's realisation that a
cubic equation can have more than one solution. He demonstrated the
existence of equations having two solutions, but unfortunately he does not
appear to have found that a cubic can have three solutions. He did hope
that `arithmetic solutions' might be found one day when he wrote:---
\begin{quote}
    Perhaps someone else who comes after us may find it out in the case,
when there are not only the first three classes of known powers, namely
the number, the thing and the square.
\end{quote}
\end{quote}
I don't know how to read this as other than admission that somebody in
future may succeed where Khayy\'am and others have failed.

Again, Kline said,
\begin{quote}
Omar Khayyam believed [the cubic] could be solved only geometrically, by
using conic sections\dots
\end{quote}
Perhaps he meant to say, Khayy\'am believed the only geometric solution was
by using conic sections (and not straightedge and compass alone).

\section{Tuesday, March 9}

I wonder if it is a bad idea to read mathematicians like Euclid and
Newton without reading mathematicians like Cardano.  Euclid is a model
of exposition.  In reading him, you may not be sure where you are
going, but at least you know how you got where you are.  Newton
follows this model, more or less.  Cardano does not.  And yet, as I
just told a student, Cardano's work is the direct ancestor of the
mathematics being taught in another course in our department: Galois
theory.

Burhan and Fuad presented sections 6 and 7 of Chapter VI of
Cardano's \emph{Ars Magna.}  This is where Cardano establishes the rule
that---said Burhan---every Turkish student learns today as
\begin{equation*}
            (a + b)^3 = a^3 + 3a^2 b + 3a b^2 + b^3.
\end{equation*}
Cardano proves it geometrically, by dividing a line $AC$ at $B$, drawing a
square on $AC$, then erecting a cube on this square.  The square is
$ACEF$.  Cardano draws a line perpendicular to $AC$ at $B$ and marks on it
$BD$ equal to $BC$; through $D$ a line is drawn parallel to $AC$.  So the
square is divided into regions $DA$, $DC$, $DE$, and $DF$, and Cardano
considers what happens when you multiply their sum by the sum of $AB$
and $BC$.

Burhan worked out the product in detail, as Cardano does.  I thought
his classmates would get bored and start chatting amongst themselves;
but they remained mostly quiet.  When Burhan was finished, I asked if
he was now more confident in the truth of the identity he had learned
years ago; he said yes.

Again, Cardano writes out the cube of $AC$ in terms of $AB$, $BC$, and the
regions $DA$, $DC$, $DE$, and $DF$.  So he gets volumes like $AB.DA$
and $BC.DF$, 
which he observes to be equal to $AB^2.BC$, and so on. That is, he
introduces the new letters of the diagram, only to reduce everything
in the end to $AB$ and $BC$.

I went to the board and asked:  We didn't Cardano just compute
directly,
\begin{align*}
   (AB + BC)^3& = (AB + BC)(AB + BC)^2\\
              & = (AB + BC)(AB^2 + 2AB.BC + BC^2)
\end{align*}
and so on?  I think it was Fuad who said this wasn't acceptible,
because it was not geometric; it was just symbol manipulation.  He had
earlier named `distributivity' as the algebraic rule at work in such
an argument.

As Cardano does, Fuad then went to the board to derive the rule that
\begin{equation*}
     AB^3 + 3AB.BC^2
\end{equation*}
exceeds
\begin{equation*}
     BC^3 + 3(BC.AB^2)
\end{equation*}
by
\begin{equation*}
     (AB - BC)^3.
\end{equation*}
Today we can obtain this by replacing $BC$ with $-BC$ in the earlier
identity.  But Fuad gave Cardano's geometric argument.  Cardano takes
$G$ on $AC$ so that $AG = BC$.  Then we know
\begin{equation*}
     AB^3 = (AG + GB)^3 = AG^3 + GB^3 + 3AG.GB^2 + 3GB.AG^2.
\end{equation*}
Now  add $3AB.BC^2$ to both sides, and replace $AG$ with $BC$, getting
\begin{multline*}
     AB^3 + 3AB.BC^2 \\
= BC^3 + GB^3 + 3BC.(GB^2 + GB.BC + AB.BC).
\end{multline*}
One must show that the parenthetical quantity, $GB^2 + GB.BC + AB.BC$, is
equal to $AB^2$.  Fuad drew a picture and tried to fit the pieces
together to make the desired square; but he just couldn't do it.  \c Sule
seemed to explain adequately what to do; it involved orienting one of
the rectangles vertically rather than horizontally; but Fuad couldn't
see it.  I tried to get \c Sule to go to the board; but Oguzhan was the
one who did it.

I thought probably Fuad could see things if he worked by himself;
being at the blackboard can gum up the thought process, as new
teachers may soon learn.

Anyway, now we had Cardano's Corollary 1.  There's a Corollary 2,
which \emph{does} seem to be, for Cardano, the result of replacing $BC$ with
$-BC$ (that is, minus-$BC$).  Fuad hadn't really understood what Cardano
was doing; I hadn't either, except in the way I just said.  I don't
really know what a negative number is for Cardano, if it is anything
at all.

When I was reading Cardano early this morning, I thought:  Homer
continues to be great poetry; but Cardano does not continue to be
great mathematics.  If you want to see where our mathematics comes
from, you must read people like Cardano; but otherwise there is no
point.

Unless the point is that important mathematics need not be well
written.  Students of mathematics today should read Euclid as a model
for the logical development of mathematics.  But there is a claim that
the best mathematicians do not write like Euclid: they are to busy
proving things to polish their work.  Maybe students should be aware
of this too.

In class, it was break time.  Again MuYaKu came up to ask me
something.  This
time it was, Why didn't Cardano just draw the full cube?  I thought
somebody had raised this question in class.  I suggested that printing
the necessary diagram might have been too much of a challenge.  But
spatial intuition itself was probably not a barrier for Cardano.
One thousand, eight hundred years before Cardano, Euclid had had an
outstanding spacial intuition.  To show this, I drew on the board the
diagram that Euclid uses in the construction of the dodecahedron.

This was in the break; but then Seray asked if she and her partners
Makbule and Salih (Acar) could make their presentation on Thursday.  I
said OK; how could I not?  I did have some things to talk about.

Indeed, I started the next hour by talking about my diagram from
Euclid.  (For independent reasons, I had spent the weekend studying
`polytopes': analogues of polygons and polyhedrons in higher
dimensions.  This is why Euclid's diagram was fresh in mind.)

Seray and her partners will present Cardano's solution of a cubic
equation.  Today then I reviewed Khayyam's solution.  I \emph{derived}
Khayyam's solution to
\begin{equation*}
     x^3 + a^2 b = a^2 x
\end{equation*}
in the manner I suggested in my log entry written on March 3: rewrite
as
\begin{align*}
     x^3 &= a^2(x - b),&
     \frac{x^2}{a^2} &= \frac{x - b}x;
\end{align*}
now introduce y so that
\begin{equation*}
     \frac xa = \frac yx = \frac{x - b}y,
\end{equation*}
and find the intersection of the curves given by
\begin{align*}
     \frac xa &= \frac yx, &\frac yx& = \frac{x - b}y,
\end{align*}
that is,
\begin{align*}
     ay& = x^2 & y^2 &= x(x - b)
\end{align*}
---a parabola and hyperbola, respectively.

As an \textbf{exercise,} I suggested solving all other cubics in this style:
for example, cube and sides equal number---which I solved myself; the
curves are a parabola and circle.

I recalled that MuYaKu doubted that such `solutions' were really
solutions.  But I repeated something Ali had presented last time.  I
wrote it thus:  The equation
\begin{equation*}
     a_1x + a_3x^3 + a_5x^5 + \dotsb + a_{2n+1}x^{2n+1} = b
\end{equation*}
(all numbers positive) definitely has a solution, for Cardano,
presumably because, as $x$ grows from $0$ without bound, so does the left
hand side.  I drew a graph of this.  (I also wrote the word
`anachronistic' on the board, to make sure they knew the word I used
to describe my algebraic treatment; nobody admitted to knowing the
word when I said it out loud.)

But how do we know that the left hand side of the equation passes
through every value?  It seems to me that we can be more confident
that Khayy\'am's solution of a cubic really \emph{does} establish the
existence of a solution.  I said we could accept that the parabola and
the hyperbola in one of Khayyam's solutions really did intersect.
Actually Ali reminded me that they might \emph{not} intersect, in certain
cases of the parameters of the equation.

I ended class about five minutes early.  Salih, Seray, and Makbule
asked about the reading they were supposed to present; for example,
what about these numbers 1570 and 1663 in the footnotes?  I explained
that those were dates of later editions of the \emph{Ars Magma.}  What did
`binomium' and `apotome' mean?  I tried to give the Euclidean
definition first, but I had only the Green Lion \emph{Bones} summary
\cite{bones}, which 
doesn't have definitions; I forgot however that the terms are defined
in \emph{Propositions} X.36 and X.73.  Anyway, I said that, for Cardano
apparently, they are expressions like $A$ plus or minus the square root of
$B$.  We had a bit more discussion; for example, Cardano solves the
cubic
\begin{equation*}
     x^3 + 6x = 20.
\end{equation*}
One can see that this has the root $x = 2$.  But Cardano's method (or
perhaps
rather Tartaglia's method) gives the difference of cube roots of a
binomium and an apotome.  Are they the same? Salih asked.  What is the
use of Cardano's complicated solution?  I observed that one could
compute that solution approximately, then show that it must be $2$,
though I didn't think there was an algorithm for general
simplification of solutions.  I mentioned the Galois theory course, as
I said.  I suggested that Salih and his partners could look up `cubic
equation' on Wikipedia for ideas, if they wanted.  Cardano himself is
obscure. 

\section{Thursday, March 11}

Salih, Seray, and Makbule presented Chapter XI of Cardano: `On
the Cube and First Power Equal to a Number'.  It's about cubes, as the
title says; but Cardano's diagram is of two-dimensional regions.
Salih started class by trying to draw a real cube, divided into
sections; but he couldn't get it right, so Seray came to do it.  Then
Salih proceeded with his demonstration, in which he claimed to show
what I shall express algebraically as:
\begin{quote}
If  $u^3 - v^3 = 20$, and $3uv = 6$, and $x = u - v$, then $x^3 +6x = 20$.
\end{quote}
Cardano's apparent purpose is to \emph{solve} the equation $x^3 + 6x = 20$.
So $u - v$ is a solution, except Cardano doesn't say till later how he
gets $u$ and $v$.  For him they are lines $AC$ and $BC$, with $B$ lying between
$A$ and $C$.  Salih just said, more or less with Cardano,
\begin{quote}
let $AC^3 - BC^3 = 20$, and $AC.BC = 2$.
\end{quote}
How can we just let it be so?  I asked.  Salih apparently hadn't
considered this question, because Cardano doesn't.  Well then, Cardano
is a bad expositor; I said this, and the students chuckled.

Seray took over at the point where Cardano says, `Now assume that $BC$
is negative.'  She went through the calculations that Cardano
apparently does, but she couldn't say clearly what the point was.  I'm
not sure what the point is either.  Seray seemed to suggest that Salih
did the positive case, and she the negative.

I think rather that Cardano just has a long-winded way of arguing that
$AB^3 + 6 AB = 20$; assuming $BC$ is negative means \emph{subtracting} $BC$ from
$AC$ to get $AB$.

Makbule went to the board to work out Cardano's `rule' for finding a
numerical solution to $x^3 + 6x = 20$: she went through the stated
manipulations of $6$ and $20$ to obtain the solution
\begin{equation*}
  x=\sqrt[3]{\sqrt{108}+10}-\sqrt[3]{\sqrt{108}-10}.
%x = cube root (square root 108 + 10) - cube root (square root 108 - 10).
\end{equation*}
She couldn't explain \emph{why} $x$ could be so found.  Apparently she hadn't
actually checked by substitution that this $x$ worked.  I think she
agreed with me that this value of $x$ must be equal to $2$.  But then
Seray said it is \emph{approximately} equal to $2$.  She must have
misunderstood what I said to her and Salih at the end of the last
class (and reported in my notes here).

I argued that the equation $x^3 + 6x = 20$ can have only one positive
solution.  Since $2$ is obviously a solution, and the complicated thing
above is a solution (Ali reported that it really was: he checked it),
then they must be equal.

But then Ece said a cubic equation could have three solutions?  Ali
said there were \emph{at most} three solutions, but some of them might be
repeated.  But Ece couldn't give a reason why there should be 3
solutions; she had just heard it somewhere.

Students seemed familiar with the idea of multiple roots.  How many
cube roots has $1$? I asked.  Just one, they said, but it was a multiple
root.  I showed this was wrong: $x^3 - 1$ factorizes as
\begin{equation*}
(x - 1)(x^2 + x + 1),
\end{equation*}
and we can solve $x^2 + x + 1 = 0$; the solutions are not $x = 1$.  I try
to stress that this was not the sort of quadratic equation that our
writers had considered, since it has no positive root.  However, Zhala
knew that Cardano would be considering square roots of negative
numbers in the next reading (which she and her friends will present).

Nobody admitted to knowing what Cardano was really doing in his
solution of a cubic.  I wrote on the board what I wrote above, but in
more general terms:
\begin{quote}\centering
If  $u^3 - v^3 = b$, and $3uv = a$, and $x = u - v$, then $x^3 +ax = b$.
\end{quote}
I checked it by working out the cube of $u - v$.  Meanwhile, I left the
class with the \textbf{exercise:}\label{ex:u,v}
\begin{quote}\centering
If $u^3 - v^3 = b$, and $3uv = a$, then what are $u$ and $v$?
\end{quote}
Cardano knows how to find $u$ and $v$, since his `rule' requires it.  But
unless I am missing something, Cardano does not explain to the reader
how his rule can be derived.

The derivation is pretty easy for us now:  We get
\begin{align*}
u^6 + u^3v^3 &= 6 u^3,
&
u^6 + 3 a^3 &= 6 u^3,
\end{align*}
and the latter equation is quadratic in $u^3$.  But did Cardano know how
to do this?  He must have, in some sense.  But in Chapter 1, when he
looks at equations like $x^4 + 3 x^2 = 28$, which are quadratic in $x^2$,
he doesn't give an example that is quadratic in a cube.

Also, Cardano wasn't the first to solve the cubic.    In the preface
to his translation, Witmer writes:
\begin{quote}
It was [Cardano] who developed the proof that the formula or formulae
that he received from del Ferro and Tartaglia are correct; found the
method for reducing the more complex forms of the cubic\dots to one or
another of the simple forms\dots
\end{quote}
I see no suggestion that Cardano could \emph{derive} the formulas in
question.  Some scholarship is needed here, I think. 

\section{Tuesday, March 16}

In today's two-hour class we discussed chapter XXXVII of Cardano's
\emph{Ars Magna,} called (in Witmer's translation) `On the Rule for
Postulating a Negative'.  Yasemin, Zhala, and Duygu presented the
material.

First we waited for Duygu to show up; Zhala called her.  I asked the
students why they thought I wanted them to make presentations.  Melis
said, `To give us experience in lecturing.'  Ece said, `To make sure
we follow the material' (or something to that effect).  I agreed with
such reasons, but said that I also wanted to learn from the students.
Again I pointed out that Cardano's book was not the Quran or the
Bible; it was just written by some guy, who could make mistakes or be
confusing.

Still, Yasemin began her presentation by reciting from memory the
beginning of the chapter:
\begin{quote}
The rule is threefold, for one either assumes a negative, or seeks a
negative square root, or seeks what is not\dots
\end{quote}
I said I hoped she would explain what this all meant.

As I see it, the gist of the chapter is this:  There are some problems
whose solutions are normally positive numbers; if you change the
parameters, the solutions may become negative, but the same general
method of solution works.  In other cases, the solutions may involve
square roots of negative numbers; such solutions may not make sense,
but they still work in a `formal' sense.  (Note that `formal' here
does not refer, as it could, to the \emph{highest} level of reality, but to
one of the lowest.)

Cardano observes, for example, that the equation
\begin{equation*}
x^2 + 4 x = 32
\end{equation*}
has the solution $8$, while the related equation
\begin{equation*}
x^2 = 4 x + 32
\end{equation*}
has the solution $4$; this means the former equation has also the
solution minus-$4$.

Now, I should think that Cardano would compute the solution to the
first equation as
\begin{quote}
half of $4$ plus the square root of the sum of $32$ and the square of half
of $4$.
\end{quote}
He also knows that a number has two square roots, one being negative;
why does he not then observe that, in the last computation, if the
negative square root is used, one indeed gets minus-$4$?  Why does he
instead convert to the second equation above?  Does he think this
conversion makes the solution minus-$4$ more plausible?

Cardano illustrates with a word problem:  Francis's wife's dowry is
$100$ gold pieces more than Francis's own wealth; and the square of the
dowry is $400$ more than the square of Francis's wealth.  Cardano
doesn't make the reasoning explicit, but it follows that Francis must
be in debt: his wealth is `minus-$x$'.  Then one gets the equation
\begin{equation*}
x^2 + 400 = (100 - x)^2,
\end{equation*}
which one solves to find $x = 48$; so Francis is $48$ gold pieces in debt,
and his wife's dowry is $52$.

I had a lot to say about all of this, and the students had comments as
well.  Oguzhan asked why Cardano doesn't just let Francis's wealth be
$x$; then we would just find $x = -48$.  Zhala drew a vertical number
line, with Francis's wealth below $0$, and his wife's dowry above.

I observed: Cardano says at one point that the difference between the
squares is $400$ \emph{gold pieces} (Witmer leaves Cardano's \emph{aurei}
untranslated).  But the difference is $400$ \emph{squares} of pieces---which
however has no physical meaning that I know of.  I suggested that
nobody would ever be interested in the situation of Cardano's problem.

Finally, in solving the equation above, I woule proceed with something
like
\begin{gather*}
x^2 + 400 = 10000 - 200 x + x^2,
\\
200 x = 9600,
\\
x = 48.
\end{gather*}
This is what Yasemin wrote, except that, like Cardano, she wrote the
middle line here as
\begin{equation*}
9600 = 200 x.
\end{equation*}
That's fine, but in my mind it involves an extra step, either to
switch members of an equation, or to cancel a minus sign.  For me, I
suppose, an equation is a spatial entity, with definite left and right
sides.  For Cardano, perhaps the left and right are not so distinct,
and he can interchange equations $A = B$ and $B = A$ as easily as he might
interchange the two expressions $CD$ and $DC$ for the same line segment.
(Again, Cardano doesn't actually \emph{have} equations in our sense; he
just writes in words, This equals that.)

I told Yasemin she should decide whether Cardano's next two examples
were worth going through; she decided they weren't.

Zhala worked through Cardano's problem of dividing $10$ into two parts
whose product is $40$.  (This problem is not numbered; it is just the
first illustration of `Rule 2'.)  Cardano says, It can't be done, but
do it anyway; you get that the parts are
\begin{quote}\centering
$5$ plus the square root of minus-$15$, and $5$ minus the square root of
minus-$15$.
\end{quote}
He checks this by performing the multiplication in a box in the text:
I reproduce it as follows, using $\Rx$ for the `Rx' symbol that
Cardano (or his printers in 1663) use for a square-root sign.  
\begin{equation*}
  \begin{array}{|l}
5. \pls \Rx \mns 15\\
5. \mns \Rx \mns 15\\\hline
25 \mns \mns 15. \text{ quad.\ est } 40.\\\hline
  \end{array}
\end{equation*}
But he calls this `as subtle as it is useless'.

Oguzhan drew my attention to Cardano's comment, 
\begin{quote}
Yet the nature of AD
[a square] is not the same as that of $40$ or of AB [a line]\dots
\end{quote}
I
don't know that it sheds any light on square roots of negative
numbers.  Cardano does go on to observe that, whereas minus-$15$ is $5$
squared minus $40$, one could try taking the sum of $5$ squared and $40$
instead.  This doesn't give the right answer.  It does however suggest
that Cardano may in other cases \emph{guess} solutions to problems, and
then verify them by substition, rather than actually deriving them.
(I raised this issue in my last log entry.)

In problem 4, Cardano proposes to divide $6$ into two pieces, the sum of
whose squares is $50$.  Cardano gets that the pieces are $7$ and minus-$1$.
Zhala worked this out after the break, just following Cardano's
recipe, which is:  Take half of $6$, and add or subtract the square root
of the difference of the $25$ from the square of half of $6$:
\begin{align*}
&\frac62 \pls \Rx 25 \mns\sq{\frac62},&
&\frac62 \mns \Rx 25 \mns\sq{\frac62}.
\end{align*}
Fuad asked why this worked; Zhala didn't know.  I pointed out that the
recipe differs from the recipe for solving a quadratic equation: in
the latter case, under the radical sign, the square of half the number
of roots is always \emph{added,} never subtracted.

I proposed the rule that I thought Cardano was following.  Maybe he
gives it earlier in the book, in a part we didn't read.  I drew
pictures for this rule, but algebraically it is:
\begin{equation*}
(a - x)^2 + (a + x)^2 = 2(a^2 + x^2).
\end{equation*}
In our problem, $a = 6/2$, and $6$ is divided unequally into pieces $a - x$
and $a + x$; the sum of their squares is $50$, and therefore
\begin{gather*}
a^2 + x^2 = 25,
\\
x = \Rx (25 \mns 9) = 4;
\end{gather*}
so the pieces are $\mns 1$ and $7$.

Even if Cardano does work out this kind of solution earlier in the
book, maybe it's better not to read the solution, but rather come up
with it on one's own.  But it took me a long time to find the solution
myself.

Cardano's Rule 3 seems to be about numbers that involve both negatives
and square roots of negatives.  Duygu showed that Cardano's only
example under this rule is simply wrong.  Cardano seeks three numbers,
which Duygu labelled $I$, $II$, and $III$.  We want then
\begin{equation*}
\frac I{II} =\frac{II}{III}
\end{equation*}
and also further conditions.  (I thought it wonderful that Duygu used
Roman numerals as variables.)  If $I$ is a square, as $x^2$, then the
conditions are:
\begin{align*}
II &= x^2 - x,
&
III &= x^2 - x - \Rx (x^2 - x).
\end{align*}
Cardano asserts that $I$, $II$, $III$ are
\begin{align*}
&\frac14,&  
&\mns\frac 14,&
&\mns \frac14\mns \Rx\mns\frac14.
\end{align*}
Indeed, Cardano takes the product of $I$ and $III$, claiming it is
\begin{equation*}
\mns\frac1{16}\pls\Rx\frac1{64},
\end{equation*}
which is $1/16$, the square of $II$; but the product is really
\begin{equation*}
\mns\frac1{16}\pls\Rx\mns\frac1{64}.
\end{equation*}
Either Cardano forgot the minus-sign, or he is confused about its
importance.  I remarked that the translator didn't note a problem,
although he had caught an error earlier on the page.

We had ten minutes left, but 
\c Sule wanted to start presenting Chapter
XXXIX, section 3.  She gave a preliminary algebraic result, presented
geometrically, needed for solving quartic equations (equations
involving squares of squares).  She didn't make clear why the result
was needed though, until I questioned her at the end. 

\section{Thursday, March 18}

The preliminary result presented by \c Sule is needed in the form
\begin{equation*}
(x^2+6)^2+2(x^2+6)t+t^2=(x^2+6+t)^2.
\end{equation*}
Before \c Sule made use of this today, I made
assignments for our next readings, in Vi\`ete and Descartes.  MuYaKu
asked for one of these assignments, though I thought he was going to
talk, with \c Sule, about Cardano; but he said he hadn't been able to
understand Cardano. 

The assigned readings were:
\begin{compactenum}
\item 
 Chapters
I--III of Vi\`ete's \emph{Introduction to the Analytic Art}
\cite[Appendix]{MR1215482}, along with the fifth of the `laws of
zetetics' in Ch.~V.   
\item
Book I of Descartes's \emph{Geometry} \cite{Descartes-Geometry},
divided into these 
sections:
\begin{compactenum}
\item 
pp.~2--6,
\item
6--11,
\item
11--17,
\item
17--25,
\item
25--37.
\end{compactenum}
\end{compactenum}

\c Sule stated the problem that Cardano takes up:  to find three
numbers in proportion whose sum is $10$ and the product of the first
two of which is $6$.  If the middle number is $x$, then the first is
$6/x$, and the third is $x^3/6$, so  
\begin{gather*}
\frac 6x+x+\frac{x^3}6=10,\\
36+6x^2+x^4=60x,\\
36+12x^2+x^4=6x^2+60x.
\end{gather*}
The left-hand side is now a square, $(x^2+6)^2$.  This is still a
square if we add $2(x^2+6)t+t^2$ as above, getting 
\begin{equation}\label{eqn:Car-4}
(x^2+6+t)^2=(2t+6)x^2+60x+t^2+12t.
\end{equation}
The right-hand side is now a square if and only if
\begin{gather}\notag
(2t+6)\cdot(t^2+12t)=\Bigl(\frac{60}2\Bigr)^2=900,\\\label{eqn:Car-r3}
t^3+15t^2+36t=450.
\end{gather}
\c Sule did all this, following Witmer's translation pretty closely
(she used Witmer's $b$ instead of my $t$; again, Cardano himself uses
no such letters).  Then \c Sule didn't know what to do.  I pointed out
that if~\eqref{eqn:Car-r3} holds, then we can take square roots
in~\eqref{eqn:Car-4}, getting 
\begin{equation*}
x^2+6+t=\sqrt{2t+6}+\frac{30}{\sqrt{2t+6}},
\end{equation*}
a quadratic in $x$.

From Cardano, we had learned to solve cubics only if there was no term
in $t^2$.  I left it as an \textbf{exercise} to eliminate this term
from~\eqref{eqn:Car-r3} by letting $t$ be $s$ minus something. 

Ali observed that we could solve quartics now only if there was no
$x^3$ term; I left it as another \textbf{exercise} to eliminate such
terms, when present. 

Again, from the equation
\begin{equation*}
x^4+12x^2+36=6x^2+60x,
\end{equation*}
Cardano derives
\begin{equation*}
t^3+15t^2+36t=450.
\end{equation*}
He then asserts a general rule, which, as \c Sule had apparently
observed, was wrong:  The coefficient of $t^2$ in the second equation
is always five-fourths the coefficient of $12$ in the first.  Neither
Witmer nor Struik appears to notice this mistake.  It's hard to see
why Cardano would make the mistake, unless one remembers that Cardano
is reasoning with ordinary words, not our algebraic symbolism.  (He
also seems to be less advanced than Euclid in his concern for proof.)
If we do use algebraic symbolism, then from  
\begin{equation}\label{eqn:x^4}
x^4+4ax^2+4a^2=
2bx^2+4cx,
\end{equation}
that is,
\begin{equation*}
(x^2+2a)^2=
2bx^2+4cx,
\end{equation*}
we get
\begin{align*}
(x^2+2a+t)^2
&=2bx^2+4cx+2tx^2+4at+t^2\\
&=(2t+2b)x^2+4cx+t^2+4at,
\end{align*}
and we require
\begin{gather}\notag
\begin{aligned}
2c^2&=(2t+2b)(t^2+4at)\\
&=2t^3+(8a+2b)t^2+4abt,
\end{aligned}\\\label{eqn:t^3}
t^3+(4a+b)t^2+4abt=2c^2.
\end{gather}
Cardano seems to have compared~\eqref{eqn:x^4} and~\eqref{eqn:t^3} in
a special case and drawn the wrong conclusion.

After class, G\"ok\c cen asked me about higher dimensions.
Now, Omar Khayy\'am had written,
\begin{quote}
  If the algebraist uses the `square-square' in problems of geometry
  it is only metaphorically, not properly, for it is impossible that
  the `square-square' be counted as a magnitude. \cite[p.~557]{Katz}
\end{quote}
Why then was G\"ok\c cen's topology class (I think it was topology)
talking about higher
dimensions?  I talked to her a bit about hypercubes, but then I had to
run for the departmental seminar. 


\chapter{Vi\`ete and Descartes}

\section{Tuesday, March 23}

To today's class though, I brought a (projection into three
dimensions of a) hypercube, made with my old Ramagon\texttrademark{}
pieces.  I came early to class and found \c Sule and Mehmet Arif \c
Sekercio\u glu standing outside.  Both of them were curious about the
model I was carrying, but Mehmet was the one who took it from my
hands.  It was however \c Sule who recognized that the model indeed
consisted of two connected cubes (just as a cube itself consists of two
connected squares). 

Then G\"ok\c cen came.  In the classroom I talked more about the
model and higher dimensions in general, pointing out for example that
the vertices of the hypercube can be given, in $\R^4$, the 16 coordinates 
\begin{equation*}
(\pm1,\pm1,\pm1,\pm1),
\end{equation*}
while the 4 vertices of usual 3-dimensional tetrahedron can be given
the coordinates 
\begin{align*}
&(1,0,0,0),&&(0,1,0,0),&&(0,0,1,0),&&(0,0,0,1).
\end{align*}
Then I gave way to Melis and Ece for their presentation of the Vi\`ete
reading. 

In fact Ece and Melis had come to my office two hours before class,
saying they
had not picked up the reading till yesterday, and it was too long for
two people.  I showed them that it wasn't so long, and I talked to
them about it.  I discussed the Greek meanings of \Gk{zhthtik'h},
\Gk{poristik'h}, and \Gk{<rhtik'h}; I think I skipped
\Gk{>exhghtik'h}, apparently used as a synonym for \Gk{<rhtik'h}.  I
mentioned that 
the earlier name of Beyo\u glu in Istanbul, namely Pera (\Gk{P'era}), meant
`beyond,' that is, beyond the Golden Horn; this word was apparently
related to \Gk{poristik'h}.  I pointed out that, according to a note
that wasn't in their Vi\`ete photocopy, the distinction between zetetic and
poristic may have corresponded to the distinction between theorem and
problem that we had discussed in Math 303.  Melis recalled this, but
Ece had not been in that class.  In any case, I didn't claim to
understand just what Vi\`ete meant by zetetic, poristic, and
exegetic. 

Nonetheless, in her presentation, Melis spoke as if she understood the
distinctions between these words.  On the board, she made a diagram:
Figure~\ref{fig:Viete}.
\begin{figure}[ht]
\centering
\psset{arrows=->,levelsep=4cm,nodesep=2mm}
\newcommand{\ic}[1]{\TR[tnpos=r]{#1}}
\pstree[treemode=R]
{\TR{analysis}}{
\ic{zetetic (\Gk{zhthtik'h})}
\ic{poristic (\Gk{poristik\'h})}
\ic{exegetic (\Gk{>exhghtik'h})}
}
\caption{Vi\`ete's analysis of analysis}\label{fig:Viete}
\end{figure}
She said something about how zetetic 
involves \emph{finding} a solution: for example, 
\begin{equation}\label{eqn:Melis}
x^2+21=10x,
\end{equation}
finding
$x=3$ or $x=-13$.  But poristic for her seemed to involve a case where
you just \emph{see} a solution.  The exegetic was where the solution
just \emph{comes out;} Melis here pointed to the \emph{ex} part of the
word.  The presentation had something of the air of B.S.; but then, in
the absence of examples, so does Vi\`ete's presentation.

Melis had begun by writing \emph{`The Analytic Art'} on the board; I
pointed out that we were reading only the \emph{Introduction} to this,
although I had asked the library to order the whole thing \cite{MR2301766}. 

Finishing with Chapter I, Melis made an obscure reference to Vi\`ete's vague
comment about working not with numbers, but with `species'.  I
indicated the equation~\eqref{eqn:Melis} that she had written, asking
whether, according to Vi\`ete, we were not going to work with such
equations.  Melis didn't have much to say, so I mentioned the footnote
referring to the theory that `species' meant letter, as $A$, $B$, or
$C$. 

Melis proceeded to Chapter II, which lists the `stipulations'; Melis
provided the Turkish translation \emph{\c sart}.  Melis read out the
stipulations, writing out their bracketed symbolic translations in the
text.  I wondered what she thought the value of this was, since she
wasn't really trying to \emph{explain} anything.  She mentioned at the
beginning that the first stipulations were Euclid's Common Notions;
she agreed when I said the later stipulations weren't common notions
for Euclid.  I suggested that the common notion
\begin{quote}\centering
  Equals to the same are equal to each other
\end{quote}
is a lot different from a claim like Vi\`ete's 7th stipulation,
\begin{quote}\centering
  if $a:b::c:d$, then $a:c::b:d$.
\end{quote}
Indeed, the latter doesn't make sense, in Euclid's terms, if, for
example, $a$ and $b$ are triangles in the same parallels, and $c$ and
$d$ are their bases:  In Figure~\ref{fig:tri} we have
\begin{equation*}
  ABD:BCD::AB:BC,
\end{equation*}
but $ABD$ does not even \emph{have} a ratio to $AB$,---much less does
it have the
\emph{same} ratio that $BCD$ has to $BC$.  (I noticed that O\u
guzhan---who always sits in front---was writing this down.)
\begin{figure}[ht]\centering
  \begin{pspicture}(0,-0.5)(4,3)
\psline(1.8,0)(4,0)(2.5,2.5)(1.8,0)(0,0)(2.5,2.5)
\uput[dl](0,0){$A$}
\uput[d](1.8,0){$B$}
\uput[dr](4,0){$C$}
\uput[u](2.5,2.5){$D$}
  \end{pspicture}
  \caption{Ratios in triangles}\label{fig:tri}
\end{figure}

But by Vi\`ete's last stipulations, the proportion 
\begin{equation*}
a:b::c:d
\end{equation*}
is
equivalent to the equation 
\begin{equation*}
ad=bc.
\end{equation*}
I asked what this meant applied
to Euclid's XII.18,
\begin{quote}
  Spheres are to one another in the triplicate ratio of their diameters.
\end{quote}
If the spheres are $S$ and $s$, and the diameters $D$ and $d$, then 
\begin{equation*}
  S:s::D^3:d^3;
\end{equation*}
for Vi\`ete then,
\begin{equation*}
  Sd^3=sD^3;
\end{equation*}
but what does this \emph{mean?}

Ece presented Chapter III, on the `law of homogeneity'.  She wrote a
couple of Vi\`ete's statements as equations:
\begin{quote}\centering
  h + h = homog.\\
  h $\cdot$ h = heterog.
\end{quote}
She admitted in the break that she wasn't sure what the first one
meant, and indeed Vi\`ete's statement,
\begin{quote}
  if a magnitude is added to a magnitude, it is homogeneous with it,
\end{quote}
has pronouns with uncertain antecedents.  But probably Vi\`ete means
that two magnitudes cannot be added unless they are homogeneous.

Before the break, Ece had written down Vi\`ete's `ladder-rungs'---in
Turkish, \emph{merdiven basamaklar\i}.  As Melis had written down
Greek forms, so Ece wrote down the Latin: 
\begin{compactenum}
	\item 
	side (\emph{latus}) or root (\emph{radix}),
	\item
	square (\emph{quadratum}),
	\item
	cube (\emph{cubus}),
	\item
	squared square (\emph{quadrato-quadratum})
\end{compactenum}
---I think Ece stopped there, fortunately, without trying to write
down Vi\`ete's whole list up to the cubed-cubed-cube.
I asked what a squared square was; Ece \emph{didn't} say that my
hypercube was one, so I did, while admitting I had no evidence that
Vi\`ete thought in such terms. 

After the break, Ece continued with the `genera of the compared
magnitudes':
\begin{compactenum}
\item 
length (\emph{longitudo}) or breadth (\emph{latitudo}),
\item
plane (\emph{planum}),
\item
solid (\emph{solidum}),
\item
plane-plane (\emph{plano-planum})
\end{compactenum}
---again she stopped here, without going up to the solid-solid-solid.
I quizzed her about the word \emph{genera,} getting her to admit that
it was the plural of \emph{genus.}  I led her to say that
\emph{homogeneous} meant having the same genus; I'm not sure she had
fully recognized this.  She gave some examples of `conjoined powers'
from the italicized text, which had apparently been added by an
editor.  She quoted the given rule about how many conjoined powers there are
at a given rung; but she gave no sign of having understood it.  (I
don't understand it myself.)

During the break, Salih Kanl\i da\u g---who with MuYaKu would be on
the second team of presenters of Descartes---asked if he could
present, not the coming Thursday, but the one after that, since the
Exam would be next Tuesday, and he had another exam as well. 

`Have you \emph{read} the Descartes?' I asked.

He hadn't; I said I didn't think it would be a problem to prepare for Thursday, so he should at least try.

As it was, Mehmet Do\u gan and G\"ok\c cen finished with their assignment in Descartes. Mehmet made Descartes's argument that lines could be multiplied and divided to produce lines.  He said that for Descartes, 
\begin{equation}\label{eqn:abcd}
a:b::c:d
\end{equation}
meant the same thing as 
\begin{equation*}
ad=bc.  
\end{equation*}
Hadn't Vi\`ete already said that? I
asked.  Mehmet claimed that, for Vi\`ete, the proportion and equation
were merely \emph{equivalent,} not identical.  But he also said that
the notation in~\eqref{eqn:abcd} was merely  the convention of Descartes;
I said I hadn't recalled seeing it in Descartes; in the passage in
question, on the first page, Descartes just wrote out the proportion
in words.

I said that I had recently published a paper \cite{MR2505433} of new
results that had been inspired by Descartes's figure
(Figure~\ref{fig:Descartes}), in which $AB:BD::BC:BE$ because
$DE\parallel AC$. 
\begin{figure}[ht]\centering
        \begin{pspicture}(0,-0.3)(3,2.5)
\psset{unit=5mm}
\psline(0,0)(6,0)(1,4)
\psline(1,0)(1.5,3.6)
\psline(3,0)(3.3,2.16)
\uput[d](1,0){$D$}
\uput[d](3,0){$A$}
\uput[dr](6,0){$B$}
\uput[ur](1.5,3.6){$E$}
\uput[ur](3.3,2.16){$C$}
    \end{pspicture}
\caption{Descartes's geometric arithmetic}\label{fig:Descartes}
\end{figure}

G\"ok\c cen talked about Descartes's formulation of what was in effect
the law of homogeneity.  She had asked me about this during the break:
I don't recall exactly what her question was, but I observed that $1$
cm$^2$ plus $1$ cm was not really $2$ of anything in particular; for
then it should 
also be $100$ mm$^2$ plus $10$ mm, or $110$ of something.  G\"ok\c cen didn't
repeat this example in her presentation, but she said you couldn't
take $a^2+a$ unless you had a designated unit, as $b$; then you could
take $a^2+ab$.

After class, O\u guzhan asked me about Descartes's own example: in
taking the cube root of $a^2b^2-b$, one should think of this as
\begin{equation*}
  \frac{a^2b^2}u-bu^2,
\end{equation*}
where $u$ is the unit.  Why, O\u guzhan wondered, did Descartes
convert everything to a solid?  Why not a plane, say?  I think I
suggested that he could have converted to $a^2b^2-bu^3$, but perhaps
he still had a prejudice against powers higher than $3$.  But now I
see that I missed something:  Descartes wanted to take a \emph{cube}
root, and that's why he wanted the radicand to be a solid.  If the
class were really a \emph{discussion,} and students asked their
questions to the whole class, rather than to me, perhaps somebody
might have pointed this out.

\section{Thursday, March 25}

Today I asked O\u guzhan about his question from last time, and he
said he had later understood the importance of Descartes's taking the
\emph{cube} root.  Somebody wanted me to review for the exam.  I just
mentioned that we had read al-Khw\=arizm\=i, Th\=abit ibn Qurra,
Khayy\=am, and Cardano; students should know how to solve problems in
their styles.  I did make sure that somebody could do the exercise
from \S\ref{ex:u,v}, p.~\pageref{ex:u,v}.  I quickly repeated the
Khayy\=am-style solution of
\begin{equation*}
  x^3+a^2x=a^2b,
\end{equation*}
noting the use of the Law of Homogeneity.
Then I noted that Cardano's method is somewhat neater when applied to
\begin{equation*}
  x^3+3a^2x=2a^2b, 
\end{equation*}
since here if we let $x=u-v$, we get
\begin{gather*}
  u^3=a^2(\sqrt{a^2+b^2}+b),\\
  v^3=a^2(\sqrt{a^2+b^2}-b)
\end{gather*}
(I'm not positive I didn't write $u$ and $v$ instead of $u^3$ and $v^3$).

Salih and MuYaKu made their presentation of pp.~6--11 of Descartes,
though now, five days later, I can't remember just how they divided
this section up.  There wasn't much to say, though Salih did present
the content of footnote 12 on p. 9 (which I hadn't read).

Mehmet Arif \c Sekercio\u glu had stopped by my office, perhaps the
previous day, to ask about his assignment, pp.~17--25, on the locus
problems.  He didn't have a partner; but Mihail, who had joined the
class late, had been absent when I took students for Descartes; so I
told Mehmet that he could work with Mihail.  Mehmet was worried,
though, that his English wasn't good, and Mihail didn't speak
Turkish.  I said I thought Mihail \emph{did} speak Turkish.  (Mihail
and his twin brother are from Turkmenistan, where they attended a
Turkish-language school---I recall learning this when they took a
first-year course with me.)

Ece came later to my office, also trying to figure out the reading
assigned to Mehmet.  Of course everybody should read everything, but
it was good to see Ece taking this seriously.

In class then, I stated a proposition derived from
Taliaferro's appendix to his translation of Apollonius
\cite{MR1660991}.  Proposition III.54 has the result,
\begin{equation*}
  \frac{AF\cdot CG}{AC^2}=\frac{EB^2}{BD^2}\cdot\frac{AD\cdot
    DC}{AE\cdot EC}.
\end{equation*}
If we draw through the arbitrary point $H$ on the conic section the
straight line parallel to $AC$ meeting $AD$ at $Y$ and $DC$ at $Z$,
and the straight line parallel to $DE$ meeting $AC$ at $X$, then, as
an \textbf{exercise,} one can show
\begin{equation*}
  \frac{HX^2}{HZ\cdot HY}=\frac{EB^2}{BD^2}\cdot\frac{DE^2}{AE\cdot
    EC}. 
\end{equation*}
Thus a conic section is a solution to a three-line locus problem.

\section{Tuesday, March 30}

Class was occupied with an exam.  Two hours before the exam, Melis
came to my office to say she had a migraine, but didn't have a medical
report; could she take a make-up?  I said we would work something out.
Aside from Melis the two students who have never come to class---Tolga and
An\i l---only Yasemin didn't come to the exam.  So 19 students took it.

During the exam, three people asked about the first question:
\begin{quote}
  A straight line is cut into equal and unequal segments.  What is the
relationship between the square on the half and the rectangle
contained by the unequal segments?
\end{quote}
They didn't understand what it meant for lines to \emph{contain} a
rectangle.  This was dismaying.  But as last semester, so this
semester, students ended up doing better on the exam than I expected.
This time I think their skills at memorizing formulas helped them.

\section{Thursday, April 1}

\setcounter{equation}{0}

Mehmet Arif \c Sekercio\u glu and Mihail talked about their few pages
of Descartes (17--25).  In stating the three-line locus problem,
Mehmet seemed to think that the angles to each of the three lines
should all be the same, although they need not be right.  I suggested
that the angles could differ, but he didn't agree.  I let it go.

In describing the five-line locus problem, Mehmet suggested that, if
the distances are $a$, $b$, $c$, $d$, and $e$, then the fraction
$abc/de$ should be a given constant.  I said there should be something
else in the denominator, to satisfy the Law of Homogeneity.  I think
\c Sule had already tried to say this, in Turkish, but Mehmet didn't
seem to get the point.  O\u guzhan suggested that, if we had a unit as
Descartes does, then we wouldn't need to worry about the Law of
Homogeneity. 

There were a few minutes left, but Mihail said he needed only a few
minutes.  He mentioned Descartes's claim that, with six to nine lines,
the curve could be found by conic sections.  I emphasized that this
didn't mean the curve \emph{was} a conic section, only that particular
points could be plotted by means of conic sections.  Here I briefly
previewed the first part of Book II, pp.~40--55, which I had asked Ali
and Emir to talk about next week, since no volunteers had been
forthcoming.

Duygu, Yasemin, and Zhala were supposed to present the last part of
Book I in the next class.
After today's class though, Duygu and Yasemin asked to postpone their
presentation, because they had an exam coming up.  Zhala was not
around.  I don't think they had read their section anyway.  They
seemed to think Ali and Emir could skip ahead to Book II, with the
trio then going back to Book I on Thursday.  I pointed out that we had
two hours of class on Tuesday.  Eventually I said \emph{I} would take
their section, if they would be the first to present from Newton. 

The exam had asked:
\begin{quote}
  A cube and nine sides are equal to ten.
Find the side numerically, as the difference of the cube roots of a
\emph{binomium} and an \emph{apotome,} by Cardano's method
(really Tartaglia's 
method); your steps should be clearly justifiable.  
\end{quote}
Many students just used Cardano's formula without justification; I
gave them three out of 5 points.  Ali was one of those students, and
he wasn't too happy about it.  We talked on Friday afternoon (April
2), when Ali seemed to be suggesting that the formula could be
understood as obvious.  He said something about sixth powers that I
didn't understnd; but I had to cut him off in order to go to the
algebra seminar.

Ali then sent me an email, with a new-to-me derivation of the
formula.  We are solving
\begin{equation*}
  x^3+ax=b.
\end{equation*}
Letting $x=u-v$, we get
\begin{equation*}
  x^3=u^3-v^3-3uvx,
\end{equation*}
so
\begin{align*}
  u^3-v^3&=b,& 3uv&=a.
\end{align*}
This is standard.  But then Ali observes:
\begin{gather*}
  \left(\frac{u^3+v^3}2\right)^2
  =\left(\frac{u^3-v^3}2\right)^2+(uv)^3
  =\left(\frac b2\right)^2+\left(\frac a3\right)^3,\\
u^3=\frac{u^3+v^3}2+\frac{u^3-v^3}2
=\sqrt{\left(\frac b2\right)^2+\left(\frac a3\right)^3}+\frac b2,\\
v^3=\frac{u^3+v^3}2-\frac{u^3-v^3}2
=\sqrt{\left(\frac b2\right)^2+\left(\frac a3\right)^3}-\frac b2,
\end{gather*}
so we easily get Cardano's formula,
\begin{equation*}
x=\sqrt[3]{\sqrt{\left(\frac b2\right)^2+\left(\frac
    a3\right)^3}+\frac b2}
-\sqrt[3]{\sqrt{\left(\frac b2\right)^2+\left(\frac
    a3\right)^3}-\frac b2}.
\end{equation*}
I wrote him:
\begin{quote}
  So you are giving an alternative method for solving the simultaneous
  equations 
  \begin{align*}
    uv& = a/3,&
    u^3 - v^3& = b
  \end{align*}
Instead of finding $v = a/3u$ and substituting in the other equation,
you find $u^3 + v^3$ and \emph{then} get $u^3$ and $v^3$ by adding and
subtracting.

I don't know that your alternative is shorter to write down, but it is
more elegant, and by knowing it, one may more easily memorize the
formula for $x$.  Is this your point?  (I'm at home and do not have your
paper here.)

Do you think Cardano found the solution by your method?  Myself, I
don't think I have really understood how Cardano thought about solving
equations.  When he was just working by himself, did he use pen and
paper?  Did he use anything like our modern (Cartesian?) notation?

In Chapter XI, Cardano solves the equation $x^3 + ax = b$ in case $a = 6$
and $b = 20$.  He spends a long time proving what in our notation is
expressed by:
\begin{multline}\label{eqn:Ali-Card}
\text{If $u^3 - v^3 = b$, and $uv = a/3$,}\\
\text{then $(u - v)^3 + a(u - v) = b$.}
\end{multline}
He doesn't appear to say \emph{why} we should let $u$ and $v$ be so.  He
doesn't say, 
\begin{quote}\centering
    If we let $x = u - v$, then $x^3 = u^3 - v^3 - 3uvx$.
\end{quote}
But was he \emph{thinking} of something like this?  If so, \emph{why}
would one 
think to let $x = u - v$?

After Cardano establishes his version of~\eqref{eqn:Ali-Card}, he
immediately says, in 
effect, 
\begin{quote}\sloppy
    $x$ is the difference of the cube roots of the binomial $[ (a/3)^3 +
(b/2)^2 ]^{1/2} + b/2$ and the apotome $[ (a/3)^3 + (b/2)^2 ]^{1/2} -
b/2$.
\end{quote}
How does he know this?  Would we understand this better if we too,
like Cardano aparently, had read Book X of Euclid's Elements, where
the terms `binomial' and `apotome' are defined and used?

I'm not sure that scholars have considered these questions!  The
passage of Cardano appears in both \emph{A Source Book in Mathematics} by
David Eugene Smith and \emph{A Source Book in Mathematics
  [1200--1800]} by D.E. Struik. 
In footnotes, Smith translates Cardano's words into modern notation,
but gives no explanation.  Struik does provide `explanation' in that
he solves $u^3 - v^3 = b$ and $uv = a/3$ by finding $v = a/3u$ and
substituting.  He doesn't address the question of why there is no
indication of such a solution in Cardano.  Do you have any ideas?
\end{quote}
Ali did write back.

\section{Tuesday, March 6}\label{sect:frust}

I signed people up individually for the first ten lemmas in Newton's
\emph{Principia.}  We should talk about the definitions and the axioms
somehow, but I don't know how.

I talked about the last part of Book I of Descartes, on the $n$-line
locus problems, using the notes I had prepared some time before.
Descartes goes to excessive length to prove by example that every line
in the problem involves a new distance of the form $ax+by+c$.  I just
isolated one case to serve for all.  I could see that most people's
minds were elsewhere.

Indeed, it may be hard to get excited about these locus problems.
Anyway, it was break time.

I had also looked up the quadratrix and the conchoid in
\cite{MR13:419a} (the index is in \cite{MR13:419b}), since Descartes
mentions them.  Ali had actually looked it up on Wikipedia, but not
found a picture.  So with the start of the next hour, I described the
curve:

If $ABCD$ is a square as in Figure~\ref{fig:qx}, let $DC$ be moved to
$AB$ at the same time that $AD$ moves to $AB$.
\begin{figure}[ht]\centering
  \begin{pspicture}(-0.5,-0.5)(4.5,4.5)
    \psset{unit=4cm}
\parametricplot{0}{89}{t tan 1 t 90 div sub mul 1 t 90 div sub}
\psarc(0,0){1}{0}{90}
\psline(0,0)(1,0)(1,1)(0,1)(0,0)
\psline(0,0.667)(1,0.667)
\psline(0,0)(0.577,1) 
\uput[dl](0,0){$A$}
\uput[dr](1,0){$B$}
\uput[ur](1,1){$C$}
\uput[ul](0,1){$D$}
\uput[r](0.5,0.866){$E$}
\uput[u](0.385,0.667){$F$}
\uput[d](0.637,0){$G$}
\psdots(0.385,0.667)(0.637,0)
  \end{pspicture}
  \caption{The quadratrix}\label{fig:qx}
\end{figure}
  The intersection of
the two lines in motion, as $E$, traces out the quadratrix, $DEG$.  I
left it as an \textbf{exercise} to show:
\begin{equation*}
  DEB:AD::AD:AF.
\end{equation*}
(Here $DEF$ is a circular arc.)  I said this curve allowed the
squaring of the circle; indeed, the word \emph{quadratrix} translates
the Greek `tetragonizer'---that's how, on the spot, I translated the
Greek \Gk{tetragwn'izousa}; I hadn't looked at the word before.  Ali
asked about this, so I just wrote it out:  Find a straight line $AH$
such that $AH:AD::AD:AF$; then $AH=DEB$, so the triangle with base
$AH$ and height $AD$ is equal to the circular quadrant $ABED$.

Emir had said at the end of the break that he couldn't understand the
Descartes.  I asked if he had talked to Ali, his supposed partner.
He hadn't.  They started talking right there.  But this didn't do much
good.

Actually, Emir started his presentation by writing a table on the
board, listing the three kinds of problems:
\begin{compactenum}[(1)]
\item 
plane,
\item
solid,
\item
linear.
\end{compactenum}
I asked what `linear' meant.  Ali understood that this meant problems
solved by lines in the sense of \emph{curves.}  Under \emph{linear,}
Emir wrote the quadratix and some other things.  But then Emir just
started \emph{reading} Descartes\label{Emir-reading}
out loud.  Eventually I stopped him and asked what the point was.
Then Ali stepped in and said some things.  He didn't understand why
Descartes should exclude the quadratrix but not other curves.  I
suggested we look at some other curves.  Emir agreed to draw
Descartes's funny contraption on p.~46.  He worked out the equations
for the curves drawn by the contraption.

Meanwhile, I noticed that nobody was paying attention.  So when Ali
offered to follow Emir, I suggested instead that we just quit.  But
first I had words with the students.  I said I had hoped to break the
model of education whereby the students face one direction, the
teacher another.  I said I wanted to learn from the students.  Ali
said `asymmetrical education' was the model everywhere in the world.

After everybody else left, Ali, O\u guzhan, and Besmir stayed behind.  I asked
if English was a barrier to classroom participation; they said No.  I
mentioned my general concern that math students took too many math
courses, and that all students entered not just a university, but a
\emph{department,} when they couldn't really have a good idea what
they were getting in for.  O\u guzhan is in electrical engineering,
actually; he said that's what he wanted to do, but he didn't really
know what it meant before he came to METU.

Besmir just wanted to see his exam paper in my office.  We went there
\emph{via} the library, so I could drop off the Newton printouts for
photocopying.  Along the way, Besmir asked about what I had wanted on
Problem 4 of the exam: the solution of a cubic equation.  I said I
wanted a self-justifying solution.  For example, if you solve a
quadratic equation by the quadratic formula, this is not strictly
self-justifying, although I could accept the formula as common
knowledge.  But if one really wanted a self-justifying solution of a
quadratic, one would complete the square.

`What is completing the square?' asked Besmir.

I think I eventually got the point across.  I also talked about Newton.

That night I wondered whether to cancel the attendence requirement,
offering the students the option of basing their grade solely on
exams.

\chapter{Newton}

\section{Thursday, April 8}

I gave up that idea.  Burhan came by Thursday morning (I think it was
then) to see his exam paper, and to make sure that his Newton
assignment (Lemma 6) was really as short as he thought.  I said Yes,
but he should read everything else too.

Ece came by to ask about her assignment, Lemma 1, which didn't make
sense to her.  I said it wouldn't make much sense until we read what
was to be done with the lemma.  But I also handed her Dana Densmore's
book \cite{Densmore}, open to the relevant section, which Ece sat and
read.

In class, I got everybody to sit in a rough
semicircle,\label{desks-again} though there 
had to be empty desks in the middle; they could not all be pushed aside.

 Ali announced that he had found a formula for the
quadratrix, and asked if he could write it down.  Of course, I said.
I recall that his use of letters wasn't so clear, but he did use the
tangent function.  I suggested that, for Descartes, only curves given
by polynomials were `geometrical', though I don't think he has a way
of saying this.

Following Descartes, Ali derived an equation for the device on p.~50
(or p.~320 of the French original).  But Ali didn't know why the
equation defined a hyperbola in particular.  Apparently he hadn't read
the footnote on p.~55 giving Van Schooten's argument.

I gave my own argument, with an additional line: $GM$, parallel to
$DF$, as in Figure~\ref{fig:hyp}.  Then
\begin{figure}[ht]\centering
   \begin{pspicture}(-4,-8)(2.3,0)
 %\psgrid
 \psset{unit=8mm}
     \psline(-5,-8.66)(0,0)(2.588,-10)
     \parametricplot{0.9}{10}{2.07 t div 0.5 t mul sub
 -7.72 t div -0.86602 t mul add}
 \psline(1.03,-3.86)(0.03,-5.59)
 \psline(-3.23,-5.59)(1.50,-5.59)
 \psline(-5,-8.66)(2.32,-8.66)
 \psline(-4.3,-8.66)(0.03,-5.59)(1.22,-4.74)(0.52,-4.74)
 \psline(-4.3,-8.66)(0.22,-0.84)%(0.7,0)
 \uput[u](0,0){$F$}
 \uput[d](0.03,-5.59){$C$}
 \uput[r](1.50,-5.59){$B$}
 \uput[ur](2.32,-8.66){$A$}
 \uput[ul](-5,-8.66){$D$}
 \uput[dl](1.6,-8.66){$E$}
 \uput[dr](-4.3,-8.66){$G$}
 \uput[ul](0.56,-4.7){$N$}
 \uput[ul](-3.2,-5.59){$I$}
 \uput[ur](1,-3.9){$K$}
 \uput[ur](1.2,-4.7){$L$}
 \uput[r](0.2,-0.9){$M$}
   \end{pspicture}
\caption{Descartes's construction of an hyperbola}\label{fig:hyp}
 \end{figure}
 \begin{equation*}
   ML:KL::GL:CL::GA:CB;
 \end{equation*}
 but since $DG=EA=NL$, we have also $GA=DE$ and hence
 \begin{gather*}
   FM:DG::KL:NL::KL:DG,\\
 FM=KL,\\
 ML=FK,\\
 FK:KL::DE:CB,\\
 IC:NL::DE:CB,\\
 IC:EA::DE:CB,\\
 IC\cdot CB=DE\cdot EA,
 \end{gather*}
 which is a condition that $C$ is on a hyperbola with asymptotes $FD$
 and $FA$.  I have just copied the argument from my notes; this is
 what I did in class, though I didn't always stop to follow the
 steps. Ali seemed to follow.  Still, I claimed this argument was more
 faithful to the \emph{picture.}  Descartes gave us a way to just work
 out formulas without really thinking.

 Ece wanted to talk about Newton's Lemma 1, since she would be away the
 following week like Duygu.  (They will be fencing in Bal\i kesir, it
 seems.)  She drew a stream of dots approaching another dot.  I
 suggested that she was just proving that if a limit was approached,
 then the limit was reached (or something like that).

\section{Tuesday, April 13}

Today, Mehmet Do\u gan presented Lemma II; Besmir, III; O\u guzhan,
IV; \c Sule, V; Burhan, VI; Yasemin, VII.

By the way, Mehmet had a facsimile of one of the old printings, not
the Wikipedia transcription that I put in the library.  He didn't seem
to understand well what was going on.  The key to Lemma I, I think, is
the observation that the several rectangles of which the curve forms a
sort of diagonal---these rectangles add up to the tallest of the
circumscribed rectangles.  But when questioned, Mehmet first denied
that this rectangle vanished.  O\u guzhan went to the board to
explain.

Besmir argued out Lemma III orally, but did not make a clear picture;
he just added a line to Mehmet's diagram, as in Newton.  When
questioned by me, Besmir made a picture something in
Figure~\ref{fig:N-III}---mine is not a faithful reproduction,
\begin{figure}\centering
\psset{unit=5mm}
  \begin{pspicture}(10,10)
%\psgrid
\psline(0,10)(0,0)(10,0)
\psline(3,0)(3,10)(0,10)
\psframe(0,0)(1.9,2.4)
\psframe(0,2.4)(2.7,4)
\psframe(0,4)(2.5,7)
\psframe(0,7)(2.2,8.5)
    \psplot{0}{10}{10 x x mul 10 div sub}
  \end{pspicture}
  \caption{Unclear quadrature}\label{fig:N-III}
\end{figure}
but the
point is that Besmir did not make it clear where the stacked-up
rectangles came from.  If he understood, why didn't he make it clear
in the picture?  I got up and drew something like
Figure~\ref{fig:N-III-again}.
\begin{figure}\centering
\psset{unit=5mm}
  \begin{pspicture}(10,10)
%\psgrid
\psline(1.6,0)(1.6,10)(0,10)(0,0)(10,0)
\psline(3.6,0)(3.6,9.744)(0,9.744)
\psline(6.1,0)(6.1,8.704)(1.6,8.704)
\psline(7.9,0)(7.9,6.279)(3.6,6.279)
\psline(10,0)(10,3.759)(6.1,3.759)
\psframe[fillstyle=solid,fillcolor=lightgray](3.6,9.744)(5.2,10)
\psframe[fillstyle=solid,fillcolor=lightgray](3.6,8.704)(5.6,9.744)
\psframe[fillstyle=solid,fillcolor=lightgray](3.6,6.279)(6.1,8.704)
\psframe[fillstyle=solid,fillcolor=lightgray](3.6,3.759)(5.4,6.279)
\psframe[fillstyle=solid,fillcolor=lightgray](3.6,0)(5.8,3.759)
    \psplot{0}{10}{10 x x mul 10 div sub}
  \end{pspicture}
  \caption{Newton's quadrature}\label{fig:N-III-again}
  \end{figure}

O\u guzhan wasn't so clear on Lemma IV either, but then neither is
Newton.  The two figures look the same in the published diagram, when
they probably should be as in Figure~\ref{fig:N-IV}.
\begin{figure}
%\psset{xunit=4mm,yunit=5mm}
\psset{xunit=3.2mm,yunit=4mm}
  \begin{pspicture}(0,-0.5)(10,10)
%\psgrid
    \psline(0,10)(0,0)(10,0)
\psplot{0}{10}{10 x x mul 10 div sub}
\psline(3,0)(3,9.1)(0,9.1)
\psline(5,0)(5,7.5)(3,7.5)
\psline(7.5,0)(7.5,4.375)(5,4.375)
\uput[dr](0,0){$A$}
\uput[d](3,0){$C$}
\uput[d](5,0){$E$}
\uput[ur](3,9.1){$B$}
\uput[ur](5,7.5){$D$}
\uput[ur](7.5,4.375){$F$}
\uput[ur](0,10){$G$}
\uput[d](10,0){$H$}
  \end{pspicture}
\hfill
\psset{xunit=6mm,yunit=2.5mm}
  \begin{pspicture}(0,-1)(10,10)
    \psline(0,10)(0,0)(10,0)
\psplot{0}{10}{10 x x mul 10 div sub}
\psline(3,0)(3,9.1)(0,9.1)
\psline(5,0)(5,7.5)(3,7.5)
\psline(7.5,0)(7.5,4.375)(5,4.375)
\uput[d](0,0){$A'$}
\uput[d](3,0){$C'$}
\uput[d](5,0){$E'$}
\uput[ur](3,9.1){$B'$}
\uput[ur](5,7.5){$D'$}
\uput[ur](7.5,4.375){$F'$}
\uput[u](0,10){$G'$}
\uput[dl](10,0){$H'$}
  \end{pspicture}
  \caption{Proportional areas}\label{fig:N-IV}
\end{figure}
O\u guzhan seemed to say that the bases of the figures were divided
into proportional segments: If the one base is partitioned by $A_0$,
$A_1$, $A_2$, etc., and the other by $B_0$, $B_1$, $B_2$, etc., then
\begin{equation}\label{eqn:O}
  \frac{A_n-A_{n-1}}{B_n-B_{n-1}}=\alpha,
\end{equation}
a constant, he said.  I drew something like Figure~\ref{fig:N-IV} and
wrote that the ratios $AB:A'B'$, $CD:C'D'$, $EF:E'F'$, etc., were
assumed to be
\emph{ultimately} the same, and the conclusion was that $AGH:A'G'H'$
was this ratio.  O\u guzhan said that's what he meant; he
amended~\eqref{eqn:O} to something like
\begin{equation*}
  \frac{r(A_n-A_{n-1})}{r(B_n-B_{n-1})}=\alpha,
\end{equation*}
with $r$ for rectangle.  Anyway, Newton's hypothesis is not that the
ratios $AB:A'B'$ are equal, but that they are \emph{ultimately} equal;
but what can this mean when a given rectangle doesn't obviously
persist through the process of adding more rectangles?

We took a break.  I think Burhan and Yasemin both asked me questions
about their lemmas.  I refrained from scolding them about not having
asked me sooner.  O\u guzhan said there was nothing new in the
corollary to Lemma IV, so we proceeded to \c Sule.  She seemed to
suggest that Lemma V followed directly from IV.  I drew two similar
rectilinear (but non-convex) figures, saying that Euclid had shown
them to be in the duplicate ratio of their sides; now Newton was
saying the same was true even if there were curving sides.  (I didn't
quiz \c Sule on her understanding of \emph{duplicate ratio.}  She had
written things like $\size{AB}=k\size{CD}$ to indicate ratio.)

I don't think Burhan got the point of Lemma VI.  Perhaps he didn't
understand the clause,
\begin{quote}
the arc $ACB$ will contain with the tangent $AD$ an angle equal to a
rectilinear angle.
\end{quote}
I tried to get him to draw this absurd situation, but he couldn't.  So
I did it.  Meanwhile he had quoted something from Math 371,
Differential Geometry, about curvature.  I said we didn't have that knowledge.

Yasemin's\label{Y} Lemma VII gave us something to think about.  She drew a
diagram, but didn't write anything else till I asked her to.  Then she
wrote (for Corollary 2) that $AD$, $DE$, $BF$, $FG$, $AB$, and the arc
$ACB$ were ultimately equal (or `had the ratio of equality' as she
kept saying, just parroting Newton).  That is, she confused $ED$ for
$AD$ (and $FG$ for $GB$). 

I said I didn't believe it.  She said confidently, `It's hard to
believe, but true!'  I asked for a proof, but this was not
forthcoming, just a remark that Newton was smarter than she (Yasemin)
was.  I asked if Yasemin's text had $DE$ or $AD$.  She checked, then
corrected her statement.  (Maybe that's when she said Newton was
smarter than she was.) 

Yasemin had drawn Newton's diagram, but apparently hadn't really seen
the purpose of points $b$ and $d$.  All of the lines with
capital-letter endpoints are vanishing; how can we talk about the
ratios with which they vanish?  I went up to make  the argument.  I
also drew a new line $be$ parallel to $BE$.  I wrote something like: 
\begin{gather*}
	Ab=Ad \text{ ultimately};\\
	Ab:Ad::AB:AD \text{ now; therefore}\\
	AB=AD \text{ ultimately.}\\
	\text{Similarly, } AB=AE.\\
	\text{Thefore } ED:AD \text{ is ultimately zero.}
\end{gather*}
There was some discussion of this.  O\u guzhan didn't believe it at
first.  I had been saying things like `$ED$ vanishes more quickly
than $AD$,' which he apparently thought meant $ED$ got to zero
\emph{first.}  Then he figured it out.  Meanwhile \c Sule seemed to
think Lemma VII followed immediately from Lemma VI. 

It is good that we are getting into propositions that are attractive
like puzzles.  Before Lemma VII, I had wondered if we shouldn't have
just jumped ahead to what Newton labels as Propositions. 

\section{Thursday, April 15}

Salih Kanl\i da\u g presented Lemma VIII.  He was as vague as Newton
about what happens to the `distant points' $b$, $d$, and $r$.  He
seemed to think that, as $B$ approaches $A$, so does $R$; I didn't
think that was necessary.

Makbule had visited my office a few hours before.  She had been absent
on Tuesday, because of an exam in another course.  She was supposed to
present Lemma IX today, but didn't know if she could.  I told her to
work on it, visiting me if she had questions.  She did visit later,
and I discussed the lemma with her.  I sketched some figures.  She
took the paper away with her.  In class, she took that paper to the
board with her.

She didn't draw Newton's figure right though: she didn't make $B$ and
$b$ collinear with $A$.  Salih Acar went to the board to straighten
things out.

Duygu was away, as she had warned; but she was supposed to present
Lemma X.  I figured we could skip it for now, since I was keen to see
how Mihail would do with Lemma XI.  He was fine, but neither he nor
anybody else seemed to know about the osculating circle.  Many
students had taken Math 371, Differential Geometry, and they could
state that the `radius of curvature' is the inverse of the
curvature; but they didn't understand the radius geometrically.  I
pointed out that Newton's $AJ$ is the diameter of the circle of
curvature. 

\section{Tuesday, April 20}

Duygu showed up, but she thought she was supposed to present Lemma
IX (rather than X).  I told her she was wrong.  Otherwise,
the schedule was this:
\begin{compactenum}
\item 
Proposition I: Salih Acar;
\item
Corollary I: Seray;
\item
Corollary II: Zhala;
\item
Corollary III: Melis;
\item
Corollary IV: Mehmet.
\end{compactenum}
However, Salih told me at the beginning of class that Seray had been
in a car accident and would not be coming to class.  It didn't sound
as if Seray was seriously hurt.  Was Salih prepared to take Seray's
part?  No, he had just found out she wasn't coming.  

Otherwise, everybody presented their part; but all I really remember
(writing eight days later) is that Mehmet said his corollary was
immediate, and I accepted this.

I argued that Corollary II should be true by definition of center of
forces, or by the second law of motion.  Indeed, since arcs $AB$ and
$BC$ are traversed in equal times, the chords $AB$ and $BC$ can stand
for the average motions between $A$ and $B$, and $B$ and $C$,
respectively.  The change in the average motions is there represented
by $cC$.  Since this change is effected by forces directed towards
$S$, \emph{ultimately} $cC$ must point towards $S$.

An argument is made in notes by Robert Bart~\cite[pp.~46--47]{Bart}
that I disputed as a
student and still dispute.  I offered it to my own students:
Triangles $SAB$ and $SBC$ are ultimately equal (because the
corresponding sectors of the orbit \emph{are} equal).  Therefore
triangles $SBC$ and $SBc$ are ultimately equal.  `Therefore by Euclid
I, 40' [a supposed interpolation, according to which `equal triangles
  on equal bases and on the same side are in the same parallels']
$Cc$ is ultimately parallel to $SB$.  

But this is bad mathematics:  We could have $B$, $C$, and $c$
collinear, while maintaining the hypothesis that triangles $SBC$ and
$SBc$ are ultimately equal.  I think the attempt here to fit Newton
into the Euclidean mold is wrong-headed.  I told all this to the
students, as a warning not to trust commentators too much.

I also mentioned non-standard analysis \cite{MR1373196}, though I am
not sure how to use it for Newton.  Newton just assumes that
centripetal force can be treated as acting discretely.  In Abraham
Robinson's terms, I suppose this amounts to partitioning time into
intervals of infinitesimal length.

Since Friday was a holiday, and some students were going away
Thursday, and I knew some other teachers cancelled their Thursday
classes, I made my own class that day `optional'.

\section{Thursday, April 22}

Three students came to the optional class: O\u guzhan and Besmir on
purpose, MuYaKu by accident (he hadn't come Tuesday, and didn't know
today's class was optional).  When I asked how they liked Newton, all
I could get was O\u guzhan's exclamation, `Amazing!'  I offered them
a precise definition of \textbf{ultimate equality} of ratios:  $A$ is
to $B$ ultimately as $C$ is to $D$, provided that, for all $k$ and
$m$, we can take the magnitudes far enough (to their ultimate
destination) that, assuming $kA\neq mB$ and $kC\neq mD$,
\begin{equation*}
  kA< mB\iff kC< mD.
\end{equation*}

\section{Tuesday, April 27}

I asked Duygu finally to present Lemma X, and she said she didn't know
she was supposed to present it \emph{today!}  But she said she had
read it a couple of times.  She agreed to try to go through it anyway
at the board, and she did pretty well.  However, where Newton speaks
of the `spaces' described by a body, Duygu thought he was referring
to the `areas described by radii' of Proposition I.  But she seemed
to understand that Newton is obtaining distance by (what we call)
integrating speed with respect to time. 

Duygu later complained about the difficulty of the translation.  I
didn't keep a copy of the Motte translation that I made available to
them, so I've been reading either Donahue~\cite{Densmore} or Cohen and
Whitman~\cite{MR1965137}.

Burhan did Proposition II.  I think he was confused by the language:
\begin{quote}
  And that force by which the body is turned off from its rectilinear
  course, and is made to describe, in equal times, the equal least
  triangles $SAB$, $SBG$, $SCD$, \&c.\ about the immovable point $S$, (by
  prop.\ 40.\ book 1.\ elem.\ and law 2.)\ acts in the place $B$, according
  to the direction of a line parallel to $cC$\dots\footnote{The last
    line was $aC$ in the Wikisource text, but I have now made the correction.} 
\end{quote}
Burhan talked as if $cC$ is \emph{given} as parallel to $AB$, and that
the equality of $SBC$ and $SBc$ follows.  I said that the converse was
what was to be proved, and he claimed to understand, but I had trouble
being sure.

For ambiguity, compare Euclid's \emph{Elements,} Propositions 18 and
19, in Heath's translation:
\begin{quote}
  In any triangle the greater side subtends the greater angle.

\noindent In any triangle the greater angle is subtended by the greater side.
\end{quote}
These are two parts of a biconditional; but which part is which?

O\u guzhan presented Proposition 3, the point of which seems to be
that, in studying the earth--moon system, we can ignore the influence
of the sun.  Indeed, somebody---O\u guzhan or Ali, I think---already
knew that Newton's $L$ could stand for \emph{Luna,} and $T$ for
\emph{Terra.} 

Zhala got started with Proposition 4.  She had earlier visited my
office, so I expected her to be comfortable with proving the main
proposition.  But in class she seemed either to consider the
proposition obvious, or to believe that it could be derived from the
corollaries.  In any case, she stated the main proposition without any
proof that I could recognize, and then she proceeded to the
corollaries.  I objected, but soon we ran out of time.

\section{Thursday, April 29}

We spent the whole time with Zhala's presentation of Proposition 4,
but didn't quite finish.  I think she was better prepared to prove the
main theorem.  Still she was slow, and her notation was confusing.
She would write things like
\begin{equation*}
  \frac{f_1}{f_2}\propto\frac{\ell_1{}^2/r_1}{\ell_2{}^2/r_2}
\end{equation*}
I didn't notice this at first; later I said she should write $=$
instead of $\propto$, or else write
\begin{equation*}
  f\propto\frac{\ell}r.
\end{equation*}
She preferred to continue to work explicitly with ratios.

I recall being at the board for Corollary 3: period is constant if and
only if force varies as the radius, $F\propto R$.  I asked if this
sort of situation actually happened in nature.  I think I managed to
elicit the answer that a spring obeyed such a law of force.  I don't
think I recognized this while reading Newton at St John's College,
by the way; I just thought Newton was having fun finding different
force laws for different orbits.  (See his Proposition 10.)

When I asked who would like a modern translation of the
\emph{Principia,} several people raised hands; so I copied the
relevant pages from the Cohen/Whitman version and put them in the
library.

\section{Tuesday, May 4}

I decided to review what Zhala had proved for Proposition 4.  She
seemed happy enough to 
be relieved of having to say more.  In fact she wanted to cut class
because her father was visiting.  

Given the circle in
Figure~\ref{fig:circles}, 
\begin{figure}[ht]\centering
  \begin{pspicture}(-2,-2.5)(2,2.5)
%\psgrid
    \pscircle(0,0)2
\psline(0,-2)(0,2)
\psline(-1.428,1.4)(0,1.4)
\uput[ul](-1.428,1.4){$B$}
\uput[r](0,1.4){$C$}
\psline(0,2)(-1.428,1.4)(0,-2)
\uput[u](0,2){$A$}
\uput[d](0,-2){$D$}
  \end{pspicture}
  \caption{Uniform circular motion}
  \label{fig:circles}
\end{figure}
we have
\begin{align*}
F&\overset{\ult}{\propto}AC=\frac{AB^2}{AD},&
  AB&\overset{\ult}{=}\arc AB,&
AD=2R,
\end{align*}
and therefore
\begin{equation*}
  F\propto\frac{\arc^2}R
\end{equation*}
---and this is an absolute statement, not an `ultimate' one.
For Cor.~1, since $V\propto\arc$,
\begin{equation*}
  F\propto\frac{V^2}R.
\end{equation*}
For Cor.~2, since $T\propto R/V$, and $F\propto(V/R)^2$,
\begin{equation*}
  F\propto\frac R{T^2}.
\end{equation*}
The next corollaries are a special case of 7:
\begin{equation*}
  F\propto\frac1{R^{2n-1}}\iff T\propto R^n\iff V\propto\frac1{R^{n-1}}.
\end{equation*}
\c Sule then proved Proposition 6 capably, though I complained that
what she wrote on the board did not show the logical connections.

Mehmet Do\u gan was absent, though he was supposed to present Prop.~7
(on a circular orbit with arbitrary center of force); I did it.

Salih Kanl\i da\u g{} presented Proposition 9, on a spiral orbit.  He
presented all of the steps, but admitted to not knowing what it meant
that the figure was `given in shape'.  I went to the board and
distinguished Newton's spiral---our logarithmic spiral---from, say,
the spiral of Archimedes.  Then Salih got the point.

Fuat had discussed his assignment, Lemma 10, with me, and I had argued
that the claim followed from Newton's Lemma 2, since the ellipse was
just a circle dilated in one direction.  But in class he wasn't ready
to make an argument.  After class I photocopied for him the relevant
pages from the appendix of \cite{Densmore}.  Meanwhile, in class, we
went ahead with Mehmet \c Sekercio\u glu's presentation of Proposition
10.  He was sometimes confused, but classmates gave some help.  He
was supposed to present also Corollaries 1 and 2, but I guess we
skipped those.  At the end of class, I made a general comment about
how it could be difficult to think at the board, so people sitting
should give help or corrections, as they had been, and not just wait
for me to do it.  I also told Mehmet not to say `Newton says,' since
we all can read what he says; the point is to tell us the
\emph{truth.}

\section{Thursday, May 6}

Working on Proposition 11, elliptic orbits with center of force at a
focus, \c Sule visited my office a couple of times.  Fortunately I had
notes from working through the proof the night before.  There was a
mistake in the Wikipedia text (now corrected by me); also $Gv\times
vP$ is called $GvP$ there.  \c Sule needed to be reminded of the
relation of points on an ellipse to the foci; we needed to discuss
also how a tangent related to the lines from the focus.

\c Sule caught me again as I was on the way to the classroom.  But
\emph{in} class, first Fuat presented Lemma 12.  I was dismayed when
he went to the board with the photocopy I gave him.  At elaborate
length he reviewed the definitions of \emph{diameter, conjugate
  diameter,} and \emph{ordinate} given there.  Then he worked through
the proof step by step.  This means he repeated the tedium of
Densmore's presentation, which fails to obtain its equation (4)
\emph{immediately} from (2), but just repeats the proof with different
letters.  But the class paid attention (as well as they ever do, at
least).  I hadn't actually read Densmore's proof, thinking it looked
excessively long.  But now I see it's a nice argument.

There were only 15 minutes left, and \c Sule was keen to present
Proposition 11.  We stayed a few minutes late so she could finish.
She never quite proved $PE=AC$, even though I pointed out the gap;
perhaps she was just too excited.

As I was leaving, Ali pointed out that his assignment, Proposition 12,
is almost word for word the same as Prop.~11.  I suggested he give the
alternative argument next time.  He admitted to not having read it.

\section{Tuesday, May 11}

Ali proved Prop.~12, using Newton's main proof; he did \emph{not}
follow my suggestion of presenting the alternative proof.  He did use
his own notes, not the text; but he stood directly in front of his
writing.  When I mentioned this, suggesting that he should explain
better what he was doing, he just asked the class in an ironic tone:
`Does anybody need this explained?'

Ece followed with Lemma 13, but she had missed the whole point: she
had not understood that any point on the parabola could be a vertex.
She proved the lemma for the `principal' vertex only.  I proved it
in general.

Duygu was fine with Lemma 14 and Corollary 1.

Besmir was absent, so he did not present Prop.~13.  Since Mihail was
supposed to do Cor.~1, I thought he might be able to present the main
proposition.  He couldn't, so I did it.  Mihail couldn't say much
about so-called Corollary 1, the converse of Propositions 11--13.
Indeed, I didn't know a proof either, except insofar as Prop.~17 is a
proof.

Fuat, assigned Cor.~2, was absent.  I let Salih Kanl\i da\u g go ahead
with Prop.~14.  When he needed Prop.~13, Cor.~2, I got up to observe
that the result followed from a part of my proof of Prop.~13 itself
that was already on the board.  However, Prop.~13 is about parabolas,
and Cor.~2 is about all conic sections.  I hadn't gone back to check
that the same claim followed for the ellipse and hyperbola.

I think Salih skipped Cor.~1 of Prop.~14, though it was part of his
assignment.

Mehmet \c Sekercio\u glu said he could do Prop.~15 if I wanted, but
there were five minutes left.  I asked what he preferred, and he
refused to give an opinion.  I asked what the class thought, and he
observed that they probably wanted to stop for the day; so we did.

After class, this Mehmet asked me:  Why aren't all orbits in the
universe circular, rather than elliptical?  I tried to argue that a
circle was a limiting case.  I said if we could through a rock fast
enough, and there were no air resistance, we could put the rock into
orbit: but a \emph{circular} orbit would need just the right speed.
But first I drew the wrong picture, the left one of Fig.~\ref{fig:rock};
\begin{figure}[ht]
\mbox{}\hfill
    \begin{pspicture}(-2,-3)(2,1)
\psdots(0,0)
    \pscircle(0,0){1}
\psellipse(0,0)(2,1)
\psellipse(0,0)(0.5,1)
    \end{pspicture}
\hfill
    \begin{pspicture}(-1.8,-3)(1.8,1)
\psdots(0,0)
    \pscircle(0,0){1}
\psellipse(0,-1)(1.732,2)
\psellipse(0,0.333)(0.577,0.667)
    \end{pspicture}
\hfill\mbox{}
  \caption{Various orbits}\label{fig:rock}
  
\end{figure}
I had forgotten that the center of force was at a \emph{focus.}  I
corrected to the right figure.   
O\u guzhan was there; I think he
recognized the problem.

But Mehmet \c S. didn't seem to be satisfied.  It bothered him that
force would \emph{change} as a planet followed its orbit.  I tried to
suggest that the force still obeyed \emph{one law,} the inverse-square
law.

Mehmet said everything happened for a \emph{reason.}  I suggested that
this was only his assumption.  If you assume everything has a reason,
then you can find a reason; but it may not be a good one.

\section{Thursday, May 13}

It's the spring festival, and the few students who showed up were
happy to cancel class and go back to the festival.  I had just been
negociating with Ece in my office.  She is attending Antalya Algebra
Days next week for some reason.  Our second exam is on Tuesday, but
the conference starts Wednesday, and Ece wanted to travel Monday night
with other students.  I said she could travel Tuesday night with Ay\c
se and me, but this was not appealing.  She found out that some other
METU people were taking a 14:00 bus on Tuesday; could she start the
exam earlier than the others on Tuesday, in order to catch that bus?

I said we would finalize our agreement tomorrow; but meanwhile, we
would see each other in class.  Oh, can I go to the Spring Festival
now? Ece asked.  I said I didn't give permission for such things; we
would just have class.  Again, as it happened, we didn't have class,
but I did talk a bit about Newton---about how his work showed that the
earth and the heavens obeyed the same law.  I talked about what might
be on the exam:  Vi\`ete's Law of Homogeneity; Cartesian-style
constructions and their equations; 
proofs as in Newton's first 11 lemmas;

\section{Tuesday, May 25}

The second exam was last Tuesday, and I cancelled Thursday's class to
go to Antalya Algebra Days.  Today, Duygu in particular asked about
the exam: many students were hoping to graduate, but if the exam were
graded by catalogue, perhaps these students couldn't graduate.  I said
they shouldn't worry if they came to class and continued to work.  In
earlier exams, students had often done my problems better then I
expected; this time they did worse.  I said I \emph{liked} the last
exam, and the students \emph{should} be able to do its problems now;
indeed, the analogue of Problem 2 for the ellipse or hyperbola might
appear on the final exam, I said.

Mehmet \c
Sekercio\u glu finally presented Prop.~15 and its corollary.  
I asked \emph{where} on the orbit the distance of a planet from the
center of force was equal to half the major axis.  He didn't know
exactly where, but Ali came to the board to show it.
For the corollary, Mehmet first draw a concentric circle and
ellipse, until he was corrected.

G\"ok\c cen presented Prop.~16 and its corollaries.  She had still
been working with the Motte translation printed out from Wikisource,
and she had visited my office, confused.  There were a number of
mistakes there.  I corrected them on line with her; but it was bad of
me not to have been reading this translation for mistakes all along.
In class, G\"ok\c cen was confused about Corollary 2, I don't know
why.  I drew a diagram for it on the board, as in
Figure~\ref{fig:ell-and-circ}. 
\begin{figure}[ht]\centering
\psset{unit=8mm}
    \begin{pspicture}(-3,-3)(3,3)
\psdots(0,0)
    \pscircle(0,0){1}
\psellipse(1,0)(2,1.732)
\pscircle(0,0){3}
    \end{pspicture}    
  \caption{Tangent orbits}\label{fig:ell-and-circ}
\end{figure}

One proposition was left, 17.  I had assigned to Melis, but she was
not in class.  Well, there was no time left anyway.

\section{Thursday, May 27}


Melis didn't come again today.  I proved Proposition 17.  But the question
of how one determines whether the orbit is an ellipse or an hyperbola
was not fully resolved.  Ali and O\u guzhan in particular were active.

But people asked about the final exam.  Salih Acar asked for sample
problems: I cited the two exams we have had so far.  I admitted I
didn't know how to ask problems about \S\S~2 and 3 of the
\emph{Principia,} except insofar as they concern conic sections.  I
observed that Newton's ideas about tangents and about finding areas
(illustrated in the second exam) continued to be important.

That was the last class.


\appendix

\chapter{Examinations}\label{app:exams}

%\input{history-exams}

These are the examination problems given in the course, along with my
solutions and remarks, which I posted on the web after each exam.
There were only two exams in Math 303; but at least one student who
continued on to Math 304 wanted more exams, so there were three in
Math 304.

\section{Friday, November 6}\label{sect:exam-1-1}

%\emph{Note:}  There are no new diagrams in these solutions, simply
%because creating them electronically is too time-consuming. 


\begin{problem}
  What is wrong with the following proof that all triangles are
  isosceles?  [See Figure~\ref{fig:all-tri-isos}.]
\begin{figure}[ht]\centering
    \setlength{\unitlength}{1.5cm}
  \begin{picture}(0,2.2)(0,-0.0)%(2,0)(-1,1)
  \put(-1,0){\line(1,0)2}
  \put(-1,0){\line(1,2){1}}
   \put(1,0){\line(-1,2){1}}
   \put(0,0){\line(0,1){2}}
   \put(-1,0){\line(2,1){1.6}}
   \put(1,0){\line(-2,1){1.6}}
   \put(0,2.05){\makebox(0,0)[b]{$A$}}
   \put(1,0){\makebox(0,0)[tl]{$C$}}
   \put(-1,0){\makebox(0,0)[tr]{$B$}}
   \put(0,-0.05){\makebox(0,0)[t]{$D$}}
   \put(0,0.6){$E$}
   \put(0.6,0.8){\makebox(0,0)[bl]{$G$}}
   \put(-0.6,0.8){\makebox(0,0)[br]{$F$}}
  \end{picture}
\caption{Are all triangles isosceles?}\label{fig:all-tri-isos}
\end{figure}
  \begin{compactenum}
  \item
  Let a triangle be given, namely $ABC$.
  \item
  Let $BC$ be bisected at $D$.
  \item
  Let a straight line, $DE$, be drawn at right angles to $BC$.
  \item
  Let also the straight line $AE$ bisect the angle $BAC$.
  \item
  Let the straight lines $BE$ and $CE$ be drawn.
  \item
  $BE=CE$.
  \item
  Let the straight line $EF$ be drawn perpendicular to $AB$.
  \item
  Let the straight line $EG$ be drawn perpendicular to $AC$.
   \item
  $AF=AG$ and
  $EF=EG$.
  \item
  $BF=CG$.
  \item
  $AF+FB=AG+GC$.
  \item
  $AF+FB=AB$ and $AG+GC=AC$.
  \item
  $AB=AC$; in particular, $ABC$ is isosceles.
  \end{compactenum}
\end{problem}

\begin{solution}
Step 12 is not justified.  In fact, if $AB>AC$, then $AF+FB=AB$, but $AC+GC=AG$.
\end{solution}

\begin{remark}
\begin{asparaenum}
\item
The diagram is misleading; but (contrary to what some people seemed to
think) the proof never assumes that $AED$ or $BEG$ or $CEF$ is a
straight line.    
\item
Step 4 may \emph{appear} unjustified; however, steps 2, 3, and 4
together say simply that the bisector of angle $BAC$ and the
perpendicular bisector of $BC$ meet at $E$.  This style of writing can
be seen for example in Euclid's Proposition I.44.   
\item
The proof does wrongly assume that $E$ lies within the triangle; but the
proof can easily be adjusted to the case where $E$ lies outside the
triangle.  Euclid usually does not bother to consider all possible
cases: we noted this for example in Proposition I.7.  The real problem
is the assumption that either both $F$ and $G$ lie on the triangle, or
both lie below the triangle. 
\end{asparaenum}
\end{remark}


\begin{problem}\sloppy
Write English translations of the following words:
\begin{inparaenum}[(a)]
\item
\Gk{je'wrhma},
\item
\Gk{pr'oblhma},
\item
\Gk{>an'alusic},
\item
\Gk{sunj'esic},
\item
\Gk{pol'ugwnon},
\item
\Gk{tr'igwnon}.
\end{inparaenum}
\end{problem}

\begin{solution}
Theorem, problem, analysis, synthesis, polygon, triangle.
\end{solution}

\begin{remark}
\begin{asparaenum}\sloppy
\item
A \emph{transliteration} of the words into English (or Latin) letters
would be \emph{theor\^ema, probl\^ema, analysis, synthesis,
  polyg\^onon, trig\^onon,}
but this is not what was asked. 
\item
The first two words on the list have been discussed in class; these
(along with the next two) are also discussed in some notes that I put
on the web.  
\item
The last two words on the list derive from \Gk{gwn'ia} \emph{angle,}
which is apparently related to \Gk{g'onu}; this word shares its
meaning, and an Indo-European ancestor, with the English \emph{knee.}
(Here is a point where English spelling is useful; if \emph{knee} were
spelled phonetically, then its relation with \Gk{g'onu} could not be
seen.)  
\item
As a translation of \Gk{tr'igwnon}, I do find the word \emph{trigon}
in the Oxford English 
Dictionary; but the more usual word is of course \emph{triangle.} 
\end{asparaenum}
\end{remark}

\begin{problem}
Write the letters of the Greek alphabet in the standard order.  Write
only the capital letters \emph{or} only the minuscule letters. 
\end{problem}

\begin{solution}

\Gk{A B G D E Z H J I K L M N X O P R S T U F Q Y W}
or
\Gk{a b g d e z h j i k l m n x o p r s t u f q y w}.
\end{solution}

\begin{problem}
Proclus writes:
\begin{quote}
  Every problem and every theorem that is furnished with all its parts
  should contain the following elements: 
  \begin{compactenum}\renewcommand{\labelenumi}{(\theenumi)}
  \item
  an \emph{enunciation} (\Gk{pr'otasic}), 
  \item
  an \emph{exposition} (or \emph{setting out:} \Gk{>'ekjesic}), 
  \item
  a \emph{specification} (or \emph{definition of goal:} \Gk{diorism'oc}), 
  \item
  a \emph{construction} (\Gk{kataskeu'h}), 
  \item
  a \emph{proof} (\Gk{>ap'o\-deixic}), and 
  \item
  a \emph{conclusion} (\Gk{sump'erasma}). 
  \end{compactenum}
  \end{quote}
  Below is the enunciation (in Heath's translation) of Proposition I.6 of Euclid's \emph{Elements.}  Supply the remaining parts (in your own words, which may or may not be Euclid's).
  \begin{quote}
  \emph{If in a triangle two angles be equal to one another, the sides
    which subtend the equal angles will also be equal to one another.} 
  \end{quote}
 \end{problem}

\begin{solution}
\begin{asparaenum}
\item
(As above, namely:)
If in a triangle two angles be equal to one another, the sides
    which subtend the equal angles will also be equal to one another.
\item
Let $ABC$ be a triangle in which angles $ABC$ and $ACB$ are equal.
\item
We shall show that $AB=AC$.
\item
On $BA$, extended if necessary, let $BD$ be cut off equal to $CA$.
\item
Then triangle $DBC$ is equal to $ACB$, and therefore $D$ must coincide with $A$.  Consequently, $BA=CA$.
\item
Thus we have shown that, if in a triangle two angles be equal to one another, the sides which subtend the equal angles will also be equal to one another.
\end{asparaenum}
\end{solution}

\begin{remark}
Euclid's proof is a \emph{reductio ad absurdum,} that is, a proof by
contradiction.  In particular, Euclid first assumes $AB\neq AC$ and
\emph{then} finds $D$.  In this case, to which of Proclus's six parts
does the hypothesis $AB\neq AC$ belong?  I don't know whether Proclus
considers this question. 
\end{remark}

  \begin{problem}
  \emph{Without} using Euclid's method of `application', prove
  Proposition I.8 of the \emph{Elements,} whose enunciation is, 
  \begin{quote}
  \emph{If two triangles have two sides equal to two sides
    respectively, and have also the base equal to the base, they will
    also have the angles equal whch are contained by the equal
    straight lines.} 
  \end{quote}
  \end{problem}
 
 \begin{solution}
 Suppose $ABC$ and $DEF$ are triangles such that $AB=DE$, $BC=EF$, and
 $AC=DF$.  We shall show that angles $ABC$ and $DEF$ are equal.  To
 this end, let $AG$ be dropped perpendicular to $BC$, extended if
 necessary [by I.12].  On $EF$, extended if necessary, cut off $EH$
 equal to $BG$ [by I.3].  Erect $HK$ perpendicular to $EF$ [by I.11]
 and equal to $AG$ [by I.3 again].  Then $EK= AB$ and angles $KEF$ and
 $ABC$ are equal [by I.4], and similarly, since $HF=GC$, we have
 $FK=CA$.  Hence $EK=ED$ and $FK=FD$.  Therefore $K$ and $D$ coincide
 [by I.7], and in particular, angles $DEF$ and $ABC$ are equal. 
 
Now, we have used two propositions [namely I.11 and 12] that Euclid
proves by means of I.8.  However, alternative proofs are as follows.   

If $A$ does not lie on the straight line $BC$, then by drawing a
circle with center $A$ that cuts the line, we may assume $B$ and $C$
have been chosen so that $AB=AC$.  Draw an equilateral triangle $BCD$
(on the opposite side of $BC$ from $A$) [by I.1].  Draw the straight
line $AD$, which cuts $BC$ at a point $E$.  Then angles $BAD$ and
$CAD$ are equal [by I.5 and 4], and therefore angles $AEB$ and $AEC$
are equal [again by I.4], so the latter angles are right.  Therefore
$AE$ has been dropped perpendicular to $AB$. 

If $A$ does lie on $BC$, we may still assume $AB=AC$.  Draw an equilateral triangle $BCD$ and straight line $AD$.  Then angles $BAD$ and $CAD$ are equal [by I.5 and 4], so they are right.  Thus $AD$ has been erected perpendicular to $BC$.
 \end{solution}
 
 \begin{remark}
 \begin{asparaenum}
 \item
 It is not necessary to name the propositions used.
 \item
 Some people argued by contradiction that if (in the notation above)
 angle $ABC$ is greater than $DEF$, then $BC$ must be greater than
 $EF$.  This is Proposition I.24; but I.24 relies on I.23, which in
 turn relies on I.8.  It is not clear to me that there is a way to
 prove I.24 without first proving I.8. 
 \item
 One person suggested an interesting argument that I understand as
 follows.  If angle $ABC$ is greater than $DEF$, then inside the
 former angle, there must be an angle $ABG$ equal to $DEF$.  We may
 then assume $BG=BC=EF$.  But then $GA=FA$ [by I.4], so we have
 violated I.7, which is absurd; therefore $ABC=DEF$.  Now, if this
 argument is valid, then what is the point of I.3?  If straight line
 $AB$ is greater than straight line $C$, why does Euclid not declare
 that there must be a part of $AB$, namely $AE$, that is equal to $C$?
 Why does Euclid feel the need to \emph{construct} $AE$? 
 \end{asparaenum}
 \end{remark}
 
  \begin{problem}
  In triangle $ABC$, suppose $BC$ is bisected at $D$, and straight
  line $AD$ is drawn.  Assuming $AB$ is greater than $AC$, prove that
  angle $BAD$ is less than $DAC$. 
  \end{problem}
 
 \begin{solution}
 Extend $AD$ to $E$ so that $DE=DA$.  Then angles $DEC$ and $DAB$ are
 equal, and $CE=BA$ [by I.4].  But then angle $CAE$ is greater than
 $CEA$ [by I.18], so $CAD>DAB$. 
 \end{solution}
 
 \begin{remark}
 I think the argument just given is the best of several variants that
 were found by different people.  The argument I had thought
 originally of was more complicated:  Since angle $BDA$ must be
 greater than $ADC$, inside angle $BDA$ we can construct angle $ADE$
 equal to $ADC$, with $DE=DC$.  Then $BE$ is parallel to $AD$ [why?],
 so $E$ lies outside triangle $ABD$.  Therefore angle $BAD$ is less
 than $EAD$; but the latter is equal to $DAC$. 
 \end{remark}

\section{Make-up exam}

\emph{This was given Friday, January 22, to Rashad and Tolga.}

\begin{problem}
  What is wrong with the following proof that angles have no size?---:
\begin{compactenum}
\item
Let an angle $K$ be given [Figure~\ref{fig:angle}].
\begin{figure}[ht]\centering
\begin{pspicture}(-6,-2)(3,0.5)
%\psgrid
\psline(-4,0)(-6,0)(-4.4,-0.8)
\uput[l](-6,0){$K$}
\psline(-2,0)(2,0)(2,-1)(-2,-1)(-2,0)(0,-2)(2,0)(2.4,-0.2)
\psline(-2,0)(2.4,-0.2)(0,-2)(0.2,-0.1)
\psline(-2,-1)(0,-2)(2,-1)(2.4,-0.2)
\psline(0,-2)(0,0)
\uput[u](2,0){$A$}
\uput[ul](-2,0){$B$}
\uput[dl](-2,-1){$C$}
\uput[dr](2,-1){$D$}
\uput[r](2.4,-0.2){$E$}
\uput[u](0,0){$F$}
\uput[dr](0.2,-0.1){$G$}
\uput[d](0,-2){$H$}
\end{pspicture}
\caption{Have angles no size?}\label{fig:angle}
\end{figure}
\item 
Let a rectangle be given, namely $ABCD$.
\item
Let angle $EDA$ be equal to $K$.
\item
Let $DE$ be made equal to $DA$.
\item
Suppose the perpendicular bisectors $FH$ of $AB$ and $GH$ of $BE$ meet at
$H$.
\item
Let the straight lines $HC$, $HB$, $HF$, $HG$, $HA$, $HE$, and $HD$ be
drawn.
\item
$HB=HA$.
\item
$HB=HE$.
\item
$HA=HE$.
\item
Triangles $HAD$ and $HED$ are equal in all respects.
\item
In particular, angle $HDA$ is equal to $HDE$.\footnote{The exam had
  $HDA$ for $HDE$ here.  This went unnoticed till April 29, 2011.}
\item
Angle $EDA$ has no size.
\item
Therefore $K$ has no size.
\end{compactenum}

\end{problem}
 
\begin{problem}
Write English translations of the following words:
\begin{quote}\centering
\begin{inparaenum}[(a)]
\item
\Gk{gramm'h},
\item
\Gk{k'ukloc},
\item
\Gk{k'entron},
\item
\Gk{tr'igwnon},
\item
\Gk{perif'ereia},
\item
\Gk{gewmetr'ia}.
\end{inparaenum}
\end{quote}
\end{problem}
 
\begin{problem}
  What are five postulates of Euclid's \emph{Elements}?
\end{problem}
 
\begin{problem}
Proclus writes:
\begin{quote}
  Every problem and every theorem that is furnished with all its parts
  should contain the following elements: 
  \begin{compactenum}[(1)]
  \item
  an \emph{enunciation} (\Gk{pr'otasic}), 
  \item
  an \emph{exposition} (or \emph{setting out:} \Gk{>'ekjesic}), 
  \item
  a \emph{specification} (or \emph{definition of goal:} \Gk{diorism'oc}), 
  \item
  a \emph{construction} (\Gk{kataskeu'h}), 
  \item
  a \emph{proof} (\Gk{>ap'o\-deixic}), and 
  \item
  a \emph{conclusion} (\Gk{sump'erasma}). 
  \end{compactenum}
  \end{quote}
  Below is the enunciation (in Heath's translation) of Proposition I.7
  of Euclid's \emph{Elements.}  Supply the
  remaining parts (in your own words, which may or may not be
  Euclid's). 
  \begin{quote}
  \emph{Given two straight lines constructed on a straight line [from
      its extremities] and meeting in a point, there cannot be
    constructed on the same straight line [from its extremities], and
    on the same side of it, two other straight lines  meeting in
    another point and equal to the former two respectively, namely
    each to that which has the same extremity with it.}
  \end{quote}
 \end{problem}
   
  \begin{problem}
From I.7, by the method of `application', Euclid proves I.8, whose
enunciation is:
\begin{quote}
  \emph{If two triangles have two sides equal to two sides
    respectively, and have also the base equal to the base, they will
    also have the angles equal which are contained by the equal
    straighte lines.}
\end{quote}
Assuming this proposition, but without using the method of
`application', prove the following (which is part of the enunciation
of I.4):
\begin{quote}
  \emph{If two triangles have two sides equal to two sides
    respectively, and have the angles contained by the equal straight
    lines equal, they will also have the base equal to the base.}
\end{quote}
  \end{problem}
   
  \begin{problem}
In a triangle $ABC$, suppose $D$ is chosen on side $AB$, and $E$ is
chosen on $AC$, so that $DE$ is parallel to $BC$.  Suppose straight
line $DF$ is drawn parallel to $AC$, and $CF$ is drawn parallel to
$AB$, and $DF$ and $CF$ meet at $F$.  Similarly, suppose $BG$ is drawn
parallel to $AC$, and $EG$ is drawn parallel to $AB$, and $BG$ and
$EG$ meet at $G$.  Let straight line $GF$ be drawn.

Prove that $GF$ is parallel to $BC$.  You may use only propositions
from Book I of the \emph{Elements.}
  \end{problem}
   


 
\section{Tuesday, January 12}

\begin{problem}
In the 8th century \textsc{b.c.e.,} the colony of Cumae (\Gk{K'umh})
was founded, near what is now Naples, by settlers from Euboea (E\u
griboz), and also from Cyme (\Gk{K'umh}) in western Anatolia near what
is now Alia\u ga \cite{Harvey, Umar}. From the Greek
alphabet as used in Cumae, the Latin alphabet was ultimately derived;
this came to have 23 letters:  
\begin{quote}\centering
\textup{A B C D E F G H I K L M N O P Q R S T V X Y Z}.
\end{quote}
In the year 863 \textsc{c.e.}, a monk from Salonica named Cyril invented
the so-called Glagolitic alphabet in order to translate holy scripture
from Greek into Old Bulgarian.  Soon after that, the simpler Cyrillic
alphabet was invented \cite{MR0351689}.\footnote{Many
  alphabets can be seen in \cite{Faulmann}.}  After
some changes (such as the abolition of a few letters by the Soviet
government in 1918), the Cyrillic alphabet became the 33-letter
Russian alphabet of today:
\newcommand{\gap}{\hfill} 
\begin{quote}\centering
\Ru{A\gap B\gap V\gap G\gap D\gap E\gap \"E\gap Zh\gap Z\gap I\gap
  \CYRISHRT\gap K\gap L\gap M\gap N\gap O\gap P\\
R\gap S\gap T\gap
  U\gap F\gap H\gap C\gap Q\gap X\gap W\gap \CYRHRDSN\gap Y\gap
  \CYRSFTSN\gap \CYREREV\gap Yu\gap Ya.} 
\end{quote}
This alphabet retains 19 of the 24 letters of the Greek alphabet, in
their original order, though not always in the original form.  What
\emph{are} the 24 letters of the Greek alphabet? 
\end{problem}


\begin{solution}
\Gk{A B G D E Z H J I K L M N X O P R S T U F Q Y W},
or
\Gk{a b g d e z h j i k l m n x o p r s t u f q y w}.
\end{solution}

\begin{remark}
Most people seem to have learned the alphabet for this exam.  If this had
been so on the first exam, I may not have asked for the
alphabet on \emph{this} exam.\footnote{Note added, September 18, 2013:
  A horrible possibility that I did not consider is that some students
  were able to cheat.}  
\end{remark}

%\newpage
\begin{problem}
Does a square have a ratio to its side?  Explain.
  \end{problem}

\begin{solution}
No, since no multiple of the side can exceed the square.
\end{solution}

\begin{remark}
This problem alludes to Definition 4 of Book V of the \emph{Elements:}
\begin{quote}
Magnitudes are said to \emph{have a ratio} to one another which are
capable, when multiplied, of exceeding one another. 
\end{quote}
Euclid does not seem to \emph{refer} to this definition later; but (as we
discussed in class) he \emph{uses} the definition implicitly, in
Proposition V.16 for example, where there is an unstated assumption
that $A$ and $C$ have a ratio, and (therefore) $B$ and $D$ have a
ratio.  In his `quadrature of the parabola,' discussed on the last day
of class, Archimedes assumes that, if two areas are unequal, then their
difference \emph{has a ratio} (in the sense of Euclid) to either of the areas.  
\end{remark}

  \begin{problem}
  Suppose a magnitude $A$ has a ratio to a magnitude $B$, and a
  magnitude $C$ has a ratio to a magnitude $D$.  What does it mean to
  say that $A$ has the \emph{same} ratio to $B$ that $C$ has to $D$
  (according to Definition 5 of Book V of Euclid's \emph{Elements})? 
  \end{problem}
 
 \begin{solution}
 If equimultiples $mA$ and $mC$ of $A$ and $C$ be taken, and other
 equimultiples $nB$ and $nD$ of $B$ and $D$ be taken, then 
 \begin{gather*}
 mA>nB \text{ if and only if } mC>nD,\\
 mA=nB \text{ if and only if } mC=nD,\\
 mA<nB \text{ if and only if } mC<nD.
 \end{gather*}
 \end{solution}
 
 \begin{remark}
 The definition of ratio is perhaps the most important sentence in Euclid.
 Euclid of course does not use special notation for a multiple of a magnitude.
 \end{remark}
 
  
  \begin{problem}
  Suppose a straight line $AB$ is bisected at $C$, and another point,
  $D$, is chosen on $AB$.  What is the relation between the squares on
  $AC$ and $CD$ and the rectangle contained by $AD$ and $DB$? 
  \end{problem}
  
  \begin{solution}
  $AC^2=CD^2+AD.DB$ [by Euclid's II.5].
  \end{solution}
\begin{problem}
  In the diagram [Figure~\ref{fig:circ}], $BAC$ is the diameter of a
  circle, $A$ is the 
  center, and $AD$ is at right angles to $BC$.  
\begin{figure}[ht]\centering
    \begin{pspicture}(-2,-2)(2.1,1.6)
%\psgrid
\psset{unit=8mm}
  \pscircle(0,0)2
  \psline(-2,0)(2,0)
  \psline(0,-2)(0,0)(-1.6,-1.2)(0,-2)(2,0)
  \psline(-1.6,-1.2)(0,-1.2)
  \uput[l](-2,0){$B$}
  \uput[u](0,0){$A$}
  \uput[r](2,0){$C$}
  \uput[d](0,-2){$D$}
  \uput[dl](-1.6,-1.2){$E$}
  \uput[ur](0,-1.2){$F$}
  \end{pspicture}
\caption{The swing of a pendulum}\label{fig:circ}
\end{figure}
Straight line $DC$ is
  drawn.  From a point $E$ on the circumference between $B$ and $D$,
  the straight line $EF$ is drawn at right angles to $AD$, and $EA$
  and $ED$ are drawn.    
  Show that the square on $DE$ has the same ratio to the square on
  $DC$ that the straight line $DF$ has to $DA$.  (\emph{Suggestion:}
  express $DE^2$ and $DC^2$ in terms of $DF$, $FA$, and $DA$.)
  \end{problem}

\begin{solution}
Just compute: $DC^2=2DA^2$, while
\begin{align*}
DE^2=DF^2+FE^2&=DF^2+EA^2-FA^2\\
&=DF^2+DA^2-FA^2\\
&=2DF^2+2DF.FA=2DF.DA,
\end{align*}
so
$DE^2:DC^2::2DF.DA:2DA.DA::DF:DA$.
\end{solution}


\begin{remark}
The equation
$DA^2+DF^2=2DF.DA+FA^2$
happens to be the symbolic expression of Euclid's Proposition II.7.  I
obtained this problem from Isaac Newton, who writes in the
\emph{Principia,} in the scholium after the Laws of Motion:   
\begin{quote}
It is a proposition very well known to geometers that the velocity of
a pendulum at the lowest point is as the chord of the arc which it
describes in falling. 
\end{quote}
\end{remark}

\begin{problem}
In the diagram [Figure~\ref{fig:parab}], $ABC$ is an axial triangle of a cone
whose base is the circle $CDEBFG$, and $DKG$ and $EMF$ are at right
angles to $BC$.  
\begin{figure}[ht]\centering
       \begin{pspicture}(-1.5,-6.2)(4,0.5)
%\psgrid
\psset{unit=5mm}
\psset{plotpoints=200}
 \parametricplot{72.54}{287.46}
{3.5 1 t cos sub div t 60 add cos 60 sin 15 sin mul 2 mul add mul
 3.5 1 t cos sub div t 60 add sin 60 cos 15 sin mul 2 mul sub 30 sin
 mul 30 cos 2 mul sub mul 
 }
 \parametricplot{95.74}{264.26}
{5.5 1 t cos sub div t 60 add cos 60 sin 15 sin mul 2 mul add mul
 5.5 1 t cos sub div t 60 add sin 60 cos 15 sin mul 2 mul sub 30 sin
 mul 30 cos 2 mul sub mul 
 }
 \psline(0,0)(4.741,-7.142)(-0.259,-11.472)(0,0)
 \parametricplot{0}{360}
 {2.241 5 t cos mul add
 -9.307 5 t sin mul 30 sin mul add}
 \psline(-2.32,-8.28)(6.34,-10.76)
 \psline(-1.14,-7.46)(7.12,-9.84)
 \psline(3,-8.65)(-0.1,-4)
 \psline(2,-9.52)(-0.14,-6.3)
\uput[u](0,0){$A$}
\uput[dl](-0.26,-11.47){$B$}
\uput[ur](4.74,-7.14){$C$}
\uput[d](-1.14,-7.46){$D$}
\uput[dr](7.12,-9.84){$G$}
\uput[d](-2.32,-8.28){$E$}
\uput[dr](6.34,-10.76){$F$}
\uput[d](3,-8.65){$K$}
\uput[ul](-0.1,-4){$H$}
\uput[ul](-0.14,-6.3){$L$}
\uput[d](2,-9.52){$M$}
\end{pspicture}
\caption{Two parabolas in a cone}\label{fig:parab}
\end{figure}
Planes through $DKG$ and $EMF$ cut the cone, making
sections $DHG$ and $ELF$, with diameters $HK$ and $LM$, respectively;
and these diameters are parallel to $AC$.  The \textbf{parameters} (the
`upright sides' or \emph{latera recta}) of the sections are not shown;
but let them be $HN$ and 
$LP$.  What is the ratio of $HN$ to
$LP$ (in terms of straight lines that \emph{are} shown in the diagram)? 
\end{problem}

\begin{solution}
Since $HN:HA::BC^2:BA.AC$ and $LP:LA::BC^2:BA.AC$ [by I.11 of
  Apollonius], we have $HN:HA::LP:LA$, and alternately 
\begin{equation*}
HN:LP::HA:LA.
\end{equation*}
\end{solution}

\begin{remark}
One may alternatively observe that $DK^2=HN.HK$, but also
$DK^2=BK.KC$, and similarly for $EM$.  Hence 
\begin{equation}\label{eqn:P}
DK^2:EM^2::HN:LP \comprat HK:LM,
\end{equation}
but also
\begin{align*}
DK^2:EM^2&::BK:BM\comprat KC:MC\\
&::HK:LM\comprat HA:LA,
\end{align*}
and therefore $HN:LP::HA:LA$.  Now, from~\eqref{eqn:P}, one might write
\begin{align*}
HN:LP&::DK^2:EM^2\comprat LM:HK\\
&::DK^2.LM:EM^2.HK;
\end{align*}
but this isn't the best answer.  A better answer is
$HN:LP::CK:CM$,
but this still refers to the particular choice of base for the cone,
when the parabolas themselves do not depend on this choice. 
\end{remark}

\begin{problem}
We know that an ellipse or an hyperbola has two `conjugate' diameters,
each diameter being situated ordinatewise with respect to the other.
A parabola cannot have conjugate diameters in this sense.
Nonetheless, suppose, in the diagram [Figure~\ref{fig:cub}], $AB$ is the
diameter of a parabola, and $AC$ is drawn ordinatewise, and $AC$ is
also the diameter of another parabola, and  $AB$ is situated
ordinatewise with respect to $AC$.  
\begin{figure}[ht]\centering
  \begin{pspicture}(-0.5,-0.6)(2,2)
%\psgrid
\psset{unit=7mm}
\parametricplot{-1}{2}{t t mul 2 div t 2 div add t}
\parametricplot{-1.73}{1.73}{t t t mul 2 div add t t mul}
\psline(0,0)(1.5,3)
\psline(0,0)(3,0)
\psline(1.26,0)(2.05,1.59)(0.79,1.59)
\uput[dr](0,0){$A$}
\uput[dr](1.26,0){$B$}
\uput[ul](2.05,1.59){$D$}
\uput[ul](0.79,1.59){$C$}
\end{pspicture}
  \caption{Two intersecting parabolas}\label{fig:cub}  
\end{figure}
Suppose the two parabolas meet at
$D$ (as well as at $A$).  Let the respective ordinates $DB$ and $DC$
be dropped.  Finally, suppose the parabola with diameter $AB$ has
parameter $E$ (not shown), and the parabola with diameter $AC$ has
parameter $F$. 

Show that 
\begin{align*}
E:AC&::AC:AB,&AC:AB&::AB:F.
\end{align*}
(\emph{Remark.}  It follows then that $E$ is to $F$ as the \emph{cube}
on $AC$ is to 
the cube on $AB$.  In particular, if $E$ is twice $F$, then the cube
on $AC$ is double the cube on $AB$.
According to Eutocius in his
\emph{Commentary on Archimedes's Sphere and Cylinder,} Menaechmus
discovered this method of `duplicating' the cube, along with another
method involving a parabola and a hyperbola.  This work is the earliest
known use of conic sections.
For Menaechmus however, the angle $BAC$ would have been
right.) 
\end{problem}

 \begin{solution}
Since $AB.E=BD^2=AC^2$, we have $E:AC::AC:AB$; the other proportion is
similar. 
\end{solution} 

\begin{problem}
In the triangle $ABC$ shown [Figure~\ref{fig:tri-par}], $FG$ is
parallel to $DC$, and $DE$ is 
parallel to $AG$.  
\begin{figure}[ht]\centering
    \begin{pspicture}(-0.5,0)(4,4)
%\psgrid
\psset{unit=7mm}
\psline(0,0)(2,5)(5,0)(0.4,1)(3.2,3)(1.2,3)
\psline(1.2,3)(3.5,2.5)(0,0)
\uput[dl](0,0){$A$}
\uput[u](2,5){$B$}
\uput[dr](5,0){$C$}
\uput[l](0.4,1){$D$}
\uput[r](3.2,3){$E$}
\uput[l](1.2,3){$F$}
\uput[r](3.5,2.5){$G$}
%\psline[linestyle=dashed](0,0)(5,0)
\psline(0,0)(5,0)
\end{pspicture}
  \caption{Parallels in a triangle}\label{fig:tri-par}
\end{figure}
Show that $AC$ is parallel to $FE$.  (You may use
the theory of proportion developed in Books V and VI of the
\emph{Elements.}  In that case, you will probably want to use
\emph{alternation:} if $A:B::C:D$, then $A:C::B:D$.  You may use also
that if $A:B::E:F$ and $B:C::D:E$, then $A:C::D:F$.  Alternatively, it
is possible to
avoid the theory of proportion by showing, as a lemma, that, in the
diagram, $FE$ is
parallel to $AC$ if and only if the parallelogram bounded by $BF$ and
$BC$, in the angle $B$, is equal to the parallelogram bounded by $BE$
and $BA$.  Or maybe you can find another method.
In modern
terms, this problem can be set in a two-dimensional
vector-space; but if the scalar field of that space is non-commutative,
then the claim is false.) 
\end{problem}

\begin{solution}
Because of the parallels, we have
\begin{align*}
BF:BD&::BG:BC,&
BD:BA&::BE:BG;
\end{align*}
therefore [by the suggested result, which is V.23 of Euclid]
$BF:BA::BE:BC$,
which yields the parallelism of $FE$ and $AC$.
\end{solution}

\begin{remark}
I learned this short proof from some students' papers.  I had
previously found a longer argument, which \emph{did} use alternation.

Really, Euclid's VI.2 gives us only (for example) $DF:FB::CG:GB$; this
is equivalent to $DB:FB::CB:GB$ by V.17 and 18. 

As noted, we don't really need to use proportions, just that, in the
diagram here [Figure~\ref{fig:assoc}], the parallelograms $ABEG$ and
$BCKF$ are equal (by cutting
and pasting) if and only if $FE$ is parallel to $AC$.  
\begin{figure}[ht]
\centering
    \begin{pspicture}(-0.5,-1.4)(3,3)
%\psgrid
\psset{unit=5mm}
\psline(0,0)(2,5)(5,0)(0,0)
\psline(0,0)(1.2,-2)(3.2,3)(1.2,3)(4.2,-2)(5,0)
\uput[dl](0,0){$A$}
\uput[u](2,5){$B$}
\uput[dr](5,0){$C$}
\uput[r](3.2,3){$E$}
\uput[l](1.2,3){$F$}
\uput[d](1.2,-2){$G$}
\uput[d](4.2,-2){$K$}
\end{pspicture}
  \caption{Parallelograms}\label{fig:assoc}
\end{figure}
Let's use
$BA.BE$ and $BF.BC$ to denote these parallelograms respectively.  In
the problem then, we have $BF.BC=BG.BD=BA.BE$, so $AE\parallel BE$. 
This problem is inspired by Descartes, who, in his \emph{Geometry,}
observes that, if (in the original diagram) $BF$ is a unit length, and
$BG=a$, while $BD=b$,
then we can define the product $ba$ as (the length of) $BC$.
Descartes does not show that the multiplication so defined is
commutative.  But it \emph{is} commutative, by this problem.  Indeed,
if $BE=BF$, then $BA=ab$, but also $BA=BC$, so $ab=ba$. 

However, if you know about the skew-field $\mathbb H$ of
\emph{quaternions,} then suppose the diagram sits in the vector-space
$\mathbb H^2$ as shown below [Figure~\ref{fig:quat}].
\begin{figure}[ht]
\centering
      \begin{pspicture}(-1,-0.5)(4.5,4)
%\psgrid
\psset{unit=7mm}
\psline(0,0)(2,5)(5,0)(0.4,1)(3.2,3)(1.2,3)
\psline(1.2,3)(3.5,2.5)(0,0)
\uput[dl](0,0){$(\mathrm{ji},0)$}
\uput[u](2,5){$(0,0)$}
\uput[dr](5,0){$(0,\mathrm{ij})$}
\uput[l](0.4,1){$(\mathrm i,0)$}
\uput[r](3.2,3){$(0,1)$}
\uput[l](1.2,3){$(1,0)$}
\uput[r](3.5,2.5){$(0,\mathrm j)$}
%\psline[linestyle=dashed](0,0)(5,0)
\psline(0,0)(5,0)
\end{pspicture}
  \caption{Parallels in a triangle again}\label{fig:quat}
\end{figure}
Then the assumptions of parallelism in the problem hold here, since
for example $(0,\mathrm{ij})-(\mathrm i,0)$ is a scalar multiple of
$(0,\mathrm j)-(1,0)$.  However,
$(0,\mathrm{ij})-(\mathrm{ji},0)=\mathrm{ij}(-1,1)$, which is not a
scalar multiple of $(0,1)-(1,0)$. 
\end{remark}


\begin{bonus}
How can this exam and this course be improved?  (Responses may be
submitted also by 
email in the next few days: \url{dpierce@metu.edu.tr}.
Meanwhile, \emph{iyi \c cal\i\c smalar; ondan sonra, iyi tatiller!}) 
\end{bonus}



\section{Tuesday, March 30}

You may use modern notation in your work; but
  Problems~\ref{prob:quad} and~\ref{prob:OK} should involve diagrams.

\begin{problem}
A straight line is cut into equal and unequal segments.  What is the
relationship between the square on the half and the rectangle
contained by the unequal segments?
  \end{problem}

\begin{solution}
The square exceeds the rectangle by the square on the straight line
between the points of section. 
\end{solution}

\begin{remark}
This problem is based on Proposition II.5 of Euclid's \emph{Elements.}
The language follows the style of Heath's translation of Euclid (on
the course webpage).
\end{remark}

  \begin{problem}\label{prob:quad}\sloppy
    A square is equal to three roots and twenty-eight dirhams.  
What is the root?
Give a geometrical justification of your answer (as Mu\d
hammad ibn M\=us\=a al-Khw\=arizm\=\i{} or Th\=abit ibn Qurra did).
  \end{problem}

\begin{solution}
In Figure~\ref{fig:304-1}, the root is $AB$; $AC=3$; and $D$ bisects
$AC$.  Then 
\begin{gather*}
		DB^2=28+DC^2
        =28+\left(\frac32\right)^2
        =\frac{121}4,\\
      DB=\frac{11}2,\\
      AB=AD+DB=\frac32+\frac{11}2=7;
\end{gather*}
so the root is $7$.
\begin{figure}[ht]
\centering
\psset{unit=4mm}
\begin{pspicture}(7,7)
%\psgrid
%\psset{unit=5mm}
\psline(3,0)(7,0)(7,7)(3,7)(3,0)(0,0)(0,7)(3,7)
\psline[linestyle=dashed](7,1.5)(1.5,1.5)(1.5,7)
\psline[linestyle=dashed](1.5,5.5)(3,5.5)
\rput(5,3.5){$28$}
\uput[u](0,7){$A$}
\uput[u](7,7){$B$}
\uput[u](3,7){$C$}
\uput[u](1.5,7){$D$}
\end{pspicture}
\caption{Analysis of a square}\label{fig:304-1}
\end{figure}
\end{solution}

\begin{remark}
Euclid's Proposition II.6 is behind this problem.
\end{remark}

\begin{problem}\label{prob:OK}\sloppy
  Suppose a cube and nine sides are equal to ten.
Find the side by taking the
intersection of two conic sections (as Omar Khayy\=am did).  It is
preferable if one of those sections is a circle.
\end{problem}

\begin{solution}
\mbox{}[\emph{Analysis:}]\vspace{-1.5\baselineskip}
\begin{gather}\notag
	x^3+9x=10,\\\notag
	x^3=10-9x,\\\label{eqn:Kh1}
	\frac{x^2}9=\frac{10/9-x}x,\\\label{eqn:Kh2}
	\frac x3=\frac yx=\frac{10/9-x}y,\\\notag
x^2=3y\quad\&\quad y^2=x\left(\frac{10}9-x\right).
\end{gather}
[\emph{Synthesis:}] As in Figure~\ref{fig:2}, 
\begin{figure}[ht]
\centering
\psset{unit=3cm}
\begin{pspicture}(0,-0.1)(1,0.7)
%\psgrid
%\parametricplot{0}{180}{20 9 div t cos mul 20 9 div add t 20 9 div t sin mul}
\parametricplot{0}{180}{t cos 1 add 5 9 div mul t sin 5 9 div mul}
\parametricplot{-0.5}{1.5}{t t t mul 3 div}
\psline(0,0.75)(0,0)(1.11,0)
\psline(1,0)(1,0.33)(0,0.33)
\uput[d](0,0){$A$}
\uput[dr](1.11,0){$B$}
\uput[r](1,0.33){$C$}
\uput[l](0,0.33){$D$}
\uput[d](1,0){$E$}
\end{pspicture}
\caption{Circle and parabola}\label{fig:2}
\end{figure}
let $ABC$ be a
semicircle with diameter $10/9$, and let $AD$, perpendicular to $AB$,
be the axis of a parabola with parameter $3$.  The semicircle and
parabola intersect at a point $C$ (as well as at $A$).  Let $CE$ be
dropped perpendicular 
to $AB$; and $CD$, to $AD$.  Then $AE=CD$; either of these is the
desired ``side''.  Indeed, 
\begin{gather*}
  CD^2=3AD,\\
    CD:3::AD:CD
        ::EC:AE::EB:EC,\\
AE^2:9::CD^2:9::EB:AE
              ::\left(\frac{10}9-AE\right):AE,\\
AE^3=10-9AE,\\
AE^3+9AE=10.
\end{gather*}
\end{solution}

\begin{remark}
\begin{asparaenum}[(i).]
\item
  In the solution, \emph{analysis} and \emph{synthesis} are used in
  the sense attributed to Theon (presumably Theon of Smyrna, that is,
  \.Izmir) by Vi\`ete at the beginning of Chapter 1 of 
  the \emph{Introduction to the Analytic Art.}  In his solutions of
  cubic equations, Omar Khayy\=am gives only the synthesis; we can
  only speculate whether he had some sort of analysis like ours.  
\item
In
  our analysis, equations~\eqref{eqn:Kh1} and~\eqref{eqn:Kh2} could
  have been
  \begin{gather*}
	x^2=\frac{10-9x}x,\\
	x=\frac yx=\frac{10-9x}y,
  \end{gather*}
yielding the parabola given by $y=x^2$ and the ellipse given by
$y^2=x(10-9x)$.  This is why the problem says, ``It is
preferable if one of those sections is a circle.''
\item
I think it is better to understand the circle through the equation
$y^2=x(10/9-x)$ than to convert this equation to the more usual
modern form,
\begin{equation*}
  y^2+\left(x-\frac59\right)^2=\left(\frac59\right)^2.
\end{equation*}
\end{asparaenum}
\end{remark}

\begin{problem}
Again, a cube and nine sides are equal to ten.
  \begin{compactenum}
  \item
Find the side numerically, as the difference of the cube roots of a
\emph{binomium} and an \emph{apotome,} by Cardano's method
(really Tartaglia's 
method); your steps should be clearly justifiable.  
\item
The side is in fact a whole
number; which one? 
\end{compactenum}
\end{problem}

\begin{solution}
\begin{compactenum}
\item
We have to solve $x^3+9x=10$.  We let $x=u-v$, so
\begin{equation*}
x^3=u^3-v^3-3uv(u-v)=u^3-v^3-3uvx.
\end{equation*}
So we let
\begin{align*}
u^3-v^3&=10,&uv&=3,
\end{align*}
which we can solve:
\begin{gather*}
	u^6-u^3v^3=10u^3,\\
u^6-27=10u^3,\\
u^3=\sqrt{5^2+27}+5=2\sqrt{13}+5,\\
v^3=\frac{3^3}{2\sqrt{13}+5}=2\sqrt{13}-5.
\end{gather*}
Therefore
\begin{equation*}
x=\sqrt[3]{2\sqrt{13}+5}-\sqrt[3]{2\sqrt{13}-5}.
\end{equation*}
\item $x=1$.
\end{compactenum}
\end{solution}

\begin{remark}
  \begin{asparaenum}[(i).]\sloppy
    \item
Cardano does give a formula for finding $x$, without clear
explanation.  However, this problem said ``steps should be clearly
justifiable''; so for full credit, the answer should be
\emph{derived,} as above, not just obtained from a memorized formula.
Some people who tried to memorize, remembered wrongly.
\item
Of course, the solution above did rely on the (memorized) quadratic
formula.  Memory does have its uses.
\item
Note here that $u^3$ could have been $-2\sqrt{13}+5$; but
$x$ in the end would have been the same.  Two other values of $x$ can
be obtained by considering \emph{complex} cube roots; but Cardano does
not know about these.
  \end{asparaenum}
\end{remark}

\begin{problem}
  A square-square, twelve squares, and thirty-six are equal to
  seventy-two sides.  In finding the side by Cardano's method (really
  Ferrari's method), you first solve a cubic equation.
  \begin{compactenum}
  \item\label{part:1}
Obtain that cubic equation. %in the form ``cube and squares equal to
                           %number''. 
\item
Convert that cubic equation to an equation of the form ``cube equal to roots and
number''. 
\item
The cubic equation in~\eqref{part:1} should have $6$ as a root.  Use this to
find the side in the original fourth-degree equation.
  \end{compactenum}
\end{problem}

\begin{solution}
\begin{asparaenum}
\item
$\begin{gathered}[t]
	x^4+12x^2+36=72x,\\
	(x^2+6)^2=72x,\\
	(x^2+6+t)^2=2tx^2+72x+t^2+12t,\\
	2t(t^2+12t)=36^2=2^43^4,\\
	t^3+12t^2=2^33^4=648.
\end{gathered}$
\item
Let $t=s-4$; then
\begin{gather*}
	s^3-48s+12\cdot16-64=2^33^4,\\
	s^3-48s=2^33^4+2^6-2^63=2^3(3^4-2^4)=8\cdot65=520.
\end{gather*}
\item
$\begin{gathered}[t]
\begin{aligned}[t]
	(x^2+12)^2&=12x^2+72x+108,\\
	&=12(x^2+6x+9)\\
	&=12(x+3)^2,
	\end{aligned}\\
	x^2+12=2\sqrt3(x+3),\\
	x^2=2\sqrt3-6(2-\sqrt3),\\
	x=\sqrt3+\sqrt{3-6(2-\sqrt3)}=\sqrt3+\sqrt{6\sqrt3-9}.
\end{gathered}$
\end{asparaenum}
\end{solution}

\begin{remark}
If we believe in negative numbers, then from $(x^2+12)^2=12(x+3)^2$ we
should obtain $x^2+12=\pm2\sqrt3(x+3)$; but the negative sign here
leads to a negative value of $x$.  The problem asks for the ``side'',
which is implicitly positive. 
\end{remark}

\section{Tuesday, May 18}\label{sect:exam-I-1}

\begin{problem}\mbox{}
The ellipse $AEB$ [Figure~\ref{fig:exam-ell}] is
determined as follows. 
\begin{figure}[ht]
\centering
\psset{unit=6mm}
  \begin{pspicture}(-4,-4)(4,4)
%\psgrid
\pscircle[linestyle=dashed](0,0)4
    \psellipse(0,0)(4,3)
\psline(-4,-4.5)(-4,0)(4,0)(-4,-4.5)
%\psline(1,2.905)(1,-2.905)(-4,-2.905)
\psline(2.1,3.404)(2.1,-1.069)
\psline(2.1,-1.069)(-4,-1.069)
\uput[l](-4,0){$A$}
\uput[r](4,0){$B$}
\uput[l](-4,-4.5){$C$}
\uput[ul](2.1,0){$D$}
\uput[dl](2.1,2.553){$E$}
\uput[dr](2.1,-1.069){$F$}
\uput[l](-4,-1.069){$G$}
\uput[ur](2.1,3.404){$H$}
  \end{pspicture}
  \caption{Concentric circle and ellipse}\label{fig:exam-ell}
\end{figure}
Triangle $ABC$ is given, the angle at $A$ being right.  If a point $D$
is chosen at random on $AB$, and $DE$ is erected at right angles to
$AB$, then $E$ lies on the ellipse if (and only if) the square on $DE$
is equal to the rectangle $ADFG$ (which is formed by letting $ED$,
extended as necessary, meet $BC$ at $F$).  Let also the circle $AHB$
with diameter $AB$ be given.

Find $h$ (in terms of the given straight lines) such that $h$ is
to $AB$ as the ellipse is to the circle.
Prove that your answer is correct, using Newton's lemmas as needed.
\end{problem}

\begin{remark}
  The ellipse appears to result from contracting the circle in one
  direction.  If this is so, then by Newton's Lemma 4, the ratio of
  ellipse to circle is the factor of contraction, which should be
  $DE/DH$.  So one should find this ratio and check that it is indeed
  independent of the choice of $D$.

Two students solved this problem perfectly.  Five others used without
proof a rule for the area of an ellipse; but we do not officially have
such a rule, and in fact the point of this problem is to establish
this rule.
\end{remark}

  \begin{solution}
    By construction and the similarity of the triangles $BDF$ and $BAC$,
    \begin{equation*}
      DE^2=ADFG=AD\times DF=AD\times DB\times\frac{AC}{AB}.
    \end{equation*}
In the circle,
\begin{equation*}
  DH^2=AD\times DB.
\end{equation*}
Let $h$ be a mean proportional of $AB$ and $AC$, so
\begin{align*}
  h^2&=AB\times AC,&
\frac{AC}{AB}=\frac{h^2}{AB^2}.
\end{align*}
Then
\begin{align*}
  \frac{DE^2}{DH^2}&=\frac{AC}{AB},&
\frac{DE}{DH}=\frac h{AB}.
\end{align*}
If we inscribe series of parallelograms in the ellipse and circle, all
of the same breadth, then corresponding parallelograms will be to each
other as $DE$ to $DH$, that is, $h$ to $AB$.  Therefore this is the
ratio of the ellipse to the circle [by Newton's Lemma 4].
  \end{solution}


  \begin{problem}
We have used without proof Propositions I.33 and 49 of the
\emph{Conics} of Apollonius.  This problem is an opportunity to prove
those propositions, using the techniques of Descartes and Newton as
appropriate.

    A straight line $\ell$ (not shown), a curved line
    $ABE$ [Figure~\ref{fig:DA}],
    \begin{figure}[ht]
\centering
%\psset{yunit=0.8cm}
\psset{xunit=8mm,yunit=6.4mm}
      \begin{pspicture}(-1,-1)(6,7)
%\psgrid
    \psplot{-1}{5.5}{x x mul 4 div}
\psline(0,1)(2,1)(2,6.25)
\psline(0,-1)(5,4)(5,6.25)
\psline(0,-1)(0,6.25)(5,6.25)(2,3.25)
\uput[dl](0,0){$A$}
\uput[dr](2,1){$B$}
\uput[l](0,1){$C$}
\uput[d](0,-1){$D$}
\uput[dr](5,6.25){$E$}
\uput[l](2,3.25){$F$}
\uput[u](2,6.25){$G$}
\uput[ul](0,6.25){$H$}
\uput[dr](5,4){$K$}
  \end{pspicture}
      \caption{Parabola and tangent}\label{fig:DA}
    \end{figure}
and a straight line
    $AC$ are given such that, whenever a point $B$ is chosen at random
    on $ABE$, and straight line $BC$ is dropped
    perpendicular to $AC$, then the square on $BC$ is equal to the
    rectangle bounded by $\ell$ and $AC$.  So $ABE$ is a parabola with
    axis $AC$.

Let $B$ now be fixed; so we may write $BC=a$ and $AC=b$.  Extend $CA$ to $D$
so that $AD=AC$.  Draw straight line $DBK$, and let $c=BD$.

Let a point $E$ be chosen at random on the parabola $ABE$.  Draw
straight lines $BF$ parallel to $AC$, and $EF$ parallel to $BD$.
\begin{compactenum}
\item
Show that the parabola
$ABE$ must indeed lie all on one side of $DBK$.
\item 
Show that the square on $EF$ varies as $BF$, and find $m$ (in
terms of $a$, $b$, and $c$ only) such that $m\times BF$ is equal to the
square on $EF$.  For your computations, let $x=EF$ and $y=BF$.
\item
Explain why $BD$ is tangent to the parabola at $B$.
\end{compactenum}
  \end{problem}


  \begin{remark}
    One approach to (a) is showing that $E$ lies above $K$.  The
    height of $E$ above $D$ is the length of $DH$; by similarity of
    triangles, the height of $K$ above $D$ is $2b/a$ times $EH$.  The
    point of using $DH$ and $EH$ is that we know how their lengths are
    related.  Two students solved this problem perfectly; one other
    was partially successful.

In (b), we want to find $x^2/y$ in terms of fixed magnitudes.  We have
one equation, $EH^2=\ell\times AH$, and we can write this in terms of
$x$ and $y$ (and fixed magnitudes) by using the similar triangles
$BCD$ and $EGF$.  Three students solved this problem completely; two
others got halfway there.

For (c), one student showed that $DB$ is the only straight line
passing through $B$ and meeting $AD$ that meets the parabola exactly
once.  A number of students observed that $DB$ does meet the parabola
just once; but this is not enough to establish that $DB$ is a
tangent.  Note also that $BG$ also meets the parabola exactly once,
but is not a tangent.
  \end{remark}

  \begin{solution}
    \begin{compactenum}
    \item 
Assuming $KE$ is parallel to $AC$, drop a perpendicular $KL$ to $AC$.
We want to show $DH\geq DL$ or $AH\geq AL$.  We have
\begin{align*}
  AH&=\frac{EH^2}{\ell},& DL&=LK\times\frac{2b}a=EH\times\frac{2b}a,
\end{align*}
so
\begin{align*}
  \ell\times(DH-DL)
&=EH^2+b\ell-EH\times\frac{2b\ell}a\\
&=EH^2+a^2-EH\times2a\\
&=(EH-a)^2,
\end{align*}
which is positive when $E$ is not $B$; so $DH>DL$.
\item
We have $EH^2=\ell\times AH$.
Since $EG=ax/c$ and $GF=2bx/c$, this means
\begin{gather*}
  \Bigl(a+\frac{ax}c\Bigr)^2=\ell\Bigl(y+\frac{2bx}c+b\Bigr),\\
a^2+\frac{2a^2x}c+\frac{a^2x^2}{c^2}=\ell y+\frac{2b\ell x}c+b\ell,
\end{gather*}
and since $a^2=\ell b$, we have
\begin{align*}
  \frac{a^2x^2}{c^2}&=\ell y,& x^2&=\frac{c^2}{a^2}\ell y,&
m&=\frac{c^2}b.
\end{align*}
\item
In the figure, as $E$ approaches $B$, $EK$ varies as $BK^2$.
Therefore $EK/BK$ varies as $BK$, so the angle $EBK$ ultimately vanishes.
    \end{compactenum}
  \end{solution}

\section{Saturday, June 12}

\renewcommand{\theenumi}{\textbf{\Alph{enumi}}}
\renewcommand{\labelenumi}{\textbf{\theenumi.}}


\begin{problem}
This problem is about the cubic equations
\begin{gather}\label{eqn:x}
  x^3+3x^2=6x+17,\\\label{eqn:t}
  t^3=9t+9.
\end{gather}
  \begin{asparaenum}
\item
Explain the relation between the solutions of~\eqref{eqn:x} and 
%the solutions of
\eqref{eqn:t}. 
\item
For one of~\eqref{eqn:x} and~\eqref{eqn:t}, find a solution geometrically,
by intersecting conic sections (as Omar Khayyam does).
\item
Find \emph{three} solutions in the same way (some might be negative).
\item
Find a solution of~\eqref{eqn:x} or~\eqref{eqn:t} numerically (in the
manner suggested by Cardano); your steps should be justifiable.  Your
answer will involve square roots of negative numbers. 
  \end{asparaenum}
\end{problem}

\begin{solution}
  \begin{compactenum}
  \item 
The substitution $x=t-1$ converts \eqref{eqn:x} into \eqref{eqn:t}; so $x$ is a
solution to \eqref{eqn:x} if and only if $x+1$ is a solution to
\eqref{eqn:t}).   
\item
From \eqref{eqn:t} we have
\begin{align*}
  \frac{t^2}9&=\frac{t+1}t,&
\frac t3&=\frac yt=\frac{t+1}y
\end{align*}
for some $y$; that is, we can solve \eqref{eqn:t} by simultaneously
solving $t/3=y/t$ and $y/t=(t+1)/y$, that is,
\begin{align*}
  t^2&=3y,&y^2&=t(t+1).
\end{align*}
These equations define a parabola and a hyperbola, respectively, as
below [Figure~\ref{fig:parab-hyp}].  Then $AB$ is a solution to
\eqref{eqn:t}.
\begin{figure}[ht]
\centering
\psset{unit=8mm}
  \begin{pspicture}(-3,-1)(4,4)
%\psgrid
    \psplot{-3}{3.6}{x x mul 3 div}
    \psplot{0}{4}{x x 1 add mul sqrt}
    \psplot{0}{0.6}{x x 1 add mul sqrt neg}
    \psplot{-3}{-1}{x x 1 add mul sqrt}
    \psplot{-1.6}{-1}{x x 1 add mul sqrt neg}
\psline(-3,0)(5,0)
\psline(0,-1)(0,4.5)
\psline(0,3.88)(3.42,3.88)
\psline(0,1.64)(-2.20,1.64)
\psline(0,0.48)(-1.20,0.48)
\uput[l](0,3.88){$A$}
\uput[dr](3.42,3.88){$B$}
\uput[r](0,1.64){$C$}
\uput[dl](-2.20,1.64){$D$}
\uput[ul](0,0.48){$E$}
\uput[dl](-1.20,0.48){$F$}
\psdots(0,3.88)(3.42,3.88)(0,1.64)(-2.20,1.64)(0,0.48)(-1.20,0.48)(0,0)
\uput[dl](0,0){$G$}
\end{pspicture}
\caption{Intersecting parabola and hyperbola}\label{fig:parab-hyp}
\end{figure}
\item
The negative solutions of \eqref{eqn:t} are $CD$ and $EF$.  (The parabola and
hyperbola intersect also at $G$, but no solution to \eqref{eqn:t}
corresponds to this, since the corresponding value of $y$ is $0$.)
\item
Let $t=u+v$; then
\begin{equation*}
  t^3=3uvt+u^3+v^3.
\end{equation*}
Then \eqref{eqn:t} holds, provided
$uv=3$ and $u^3+v^3=9$.
Solving these, we have
\begin{gather*}
  u^6+u^3v^3=9u^3,\\
u^6+27=9u^3,\\
u^3=\frac92\pm\sqrt{\frac{81}4-27}=\frac{9\pm3\sqrt{-3}}2.
\end{gather*}
So if $u$ is a cube root of $(9+3\sqrt{-3})/2$, then
 one solution to \eqref{eqn:t} is $u+3/u$.
  \end{compactenum}
\end{solution}

\begin{remark}
  Cardano could not give a meaning to the solution we found in the
  last part; today we can, and the three choices of the cube root give
  the three solutions found geometrically earlier.
\end{remark}


\begin{problem}
This problem shows that every line through the
center of an ellipse is a diameter with certain properties.  The
method is based on Apollonius; but the algebraic geometry of Descartes
makes some simplifications possible. 
\begin{figure}[ht]
\mbox{}\hfill
%\centering
%\psset{xunit=2.5cm,yunit=3.5cm,plotpoints=360}
\psset{xunit=2cm,yunit=2.8cm,plotpoints=360}
\begin{pspicture}(-1,-1.2)(1,1.4)
%\psgrid
  \parametricplot{0}{360}{t cos t sin}
\psline(-0.6,-0.8)(0.75,1)(0,1)
\psline(0.6,0.8)(0,1.25)(0,-1)
\psline(0.6,0.8)(0,0.8)
\psline[linestyle=dashed](0.15,0.2)(-0.98,0.2)(0,-0.53)
\uput[ul](0,1){$A$}
\uput[d](0,-1){$B$}
\uput[ul](0,0.2){$C$}
\uput[l](-0.98,0.2){$D$}
%\uput[r](0.98,0.2){$D'$}
\uput[r](0.6,0.8){$E$}
\uput[l](0,0.8){$F$}
\uput[u](0,1.25){$G$}
\uput[dr](0,0){$H$}
\uput[u](0.75,1){$K$}
\uput[r](0,-0.53){$L$}
\uput[dr](0.15,0.2){$M$}
\uput[dl](-0.6,-0.8){$N$}
\uput[l](-0.26,-0.34){$P$}
\end{pspicture}
\hfill
\begin{pspicture}(-1,-1.2)(1,1.4)
%\psgrid
  \parametricplot{0}{360}{t cos t sin}
\psline(-0.68,-0.9)(0.75,1)(0,1)
\psline(0.6,0.8)(0,1.25)(0,-1)
\psline(0.6,0.8)(0,0.8)
\psline[linestyle=dashed](-0.28,-0.37)(0.44,-0.9)(-0.68,-0.9)
\uput[ul](0,1){$A$}
\uput[d](0,-1){$B$}
\uput[ul](0,-0.9){$C$}
\uput[dr](0.44,-0.9){$D$}
\uput[r](0.6,0.8){$E$}
\uput[l](0,0.8){$F$}
\uput[u](0,1.25){$G$}
\uput[dr](0,0){$H$}
\uput[u](0.75,1){$K$}
\uput[ur](0,-0.57){$L$}
\uput[d](-0.68,-0.9){$M$}
\uput[l](-0.6,-0.8){$N$}
\uput[ul](-0.28,-0.37){$P$}
\end{pspicture}
\hfill\mbox{}
\caption{Diameters of ellipses}\label{fig:2-ell}
\end{figure}
Straight line $AB$ is given, and angle $BAK$ is given.  The point $C$
moves along $AB$, and as it moves, straight line $CD$ remains parallel
to $AK$.  But $D$ moves along $DC$ as $C$ moves, so that $D$ traces
out a curvilinear figure $ADB$, as shown [Figure~\ref{fig:2-ell}] with
two possible positions of $DC$.   

Recall that the curvilinear figure $ADB$ is an \textbf{ellipse} with
\textbf{diameter} $AB$ and \textbf{ordinates} parallel to $AK$ if and
only if 
\begin{equation}\label{eqn:p}
  CD^2\propto AC\times CB
\end{equation}
(that is, the square on $CD$ varies as the rectangle formed by $AC$
and $CB$).

Let $E$ be chosen at random on $ADB$, and let straight line $EF$ be
drawn parallel to $KA$, meeting $AB$ at $F$.  Let straight line $EG$
be drawn, meeting $BA$ extended at $G$ 
so that
\begin{equation}
  \frac{AG}{GB}=\frac{AF}{FB}.
\end{equation}
Let $H$ be the midpoint of $AB$, and let straight line $HE$ be drawn and extended to meet $AK$ at $K$.  Let $L$ be taken on $AB$ (extended if necessary) so that
straight line $DL$ is parallel to $GE$.  Finally, let $M$ be the point
of intersection of $DC$ and $HK$ (both extended if necessary).

For computations, let
\begin{align*}
  AH&=b,&
EF&=c,&
HF&=d,&
CD&=x,&
CH&=y.
\end{align*}
Also, let $a$ be such that
\begin{equation}
\frac{a^2}{b^2}=\frac{EF^2}{AF\times FB}=\frac{c^2}{b^2-d^2}.
\end{equation}


\begin{asparaenum}
\item\label{item:a}
Show that~\eqref{eqn:p} holds if and only if
\begin{equation}
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.
\end{equation}
\item
Find  $HG$ in terms of $b$ and $d$.
\item\label{item:tri}
Show that~\eqref{eqn:p} holds if and only if
\begin{equation}\label{eqn:ell}
  \triangle CDL=\triangle AHK-\triangle CHM.
\end{equation}
(Angle $BAK$ is not assumed to be a right
angle; but the computations can be performed as if it were.)
\item\label{item:b}
Assuming~\eqref{eqn:p} holds (and hence~\eqref{eqn:ell} holds, for \emph{all} possibilities for $C$), show
\begin{equation*}
\triangle AHK=\triangle GHE.
\end{equation*}
\item
Assume~\eqref{eqn:p} holds.  Let
$EH$ be extended to meet the ellipse again at $N$, and let $EN$ meet $DL$ (extended as necessary) at $P$.  Show that the curvilinear figure $ADB$
is an ellipse with diameter $EN$ whose ordinates are parallel to
$EG$.  (You will probably want to use part~\ref{item:tri}, translated
appropriately.) 
%(As far as I know, the best way to do this is by means of
       %part~\eqref{item:b}.) 
\end{asparaenum}
\end{problem}


\begin{solution}
  \begin{compactenum}
  \item 
If ($\ddag$) holds, then in particular it holds when $C$ is $F$.
Therefore ($\ddag$) is equivalent to
\begin{gather*}
  \frac{CD^2}{AC\times CB}=\frac{EF^2}{AF\times FB}=\frac{a^2}{b^2},\\
\frac{x^2}{b^2-y^2}=\frac{a^2}{b^2},\\
b^2x^2=a^2b^2-a^2y^2,
\end{gather*}
which is equivalent to ($\parallel$).
\item
Let $HG=e$.  Then ($\mathsection$) becomes
\begin{equation*}
  \frac{e-b}{e+b}=\frac{b-d}{b+d},
\end{equation*}
which yields $e=b^2/d$.
\item
Since $CDL\sim FEG$, and
\begin{equation*}
  FEG=\frac12\Bigl(\frac{b^2}d-d\Bigr)c,
\end{equation*}
we have
\begin{equation*}
  CDL=\frac{x^2}{c^2}FEG=\frac{x^2}{2c}\Bigl(\frac{b^2}d-d\Bigr).
\end{equation*}
We assume angle $BAK$ is right; otherwise, we can just
multiply throughout by its sine.)
Also $AHK$ and $CHM$ are both similar to $FHE$, which is $cd/2$; so
\begin{equation*}
  AKH-CHM=\frac{cd}2\Bigl(\frac{b^2}{d^2}-\frac{y^2}{d^2}\Bigr).
\end{equation*}
So ($**$) holds if and only if
\begin{gather*}
  \frac{x^2}c\Bigl(\frac{b^2}d-d\Bigr)
=c\frac{b^2-y^2}d,\\
x^2(b^2-d^2)=c^2(b^2-y^2),\\
b^2x^2=a^2(b^2-y^2),
\end{gather*}
which is equivalent to ($\parallel$).
\item
In ($**$), let $C$ be $F$; then the equation becomes
\begin{equation*}
  FEG=AHK-FHE,
\end{equation*}
so $AHK=FEG+FHE=GHE$.
\item
By part \textbf C, it is enough to show
\begin{equation*}
  PDM=EHG-PHL.
\end{equation*}
We have
\begin{align*}
  PDM
&=CDL+CHM-PHL&&\\
&=AHK-PHL&&\text{ [by ($**$)]}\\
&=EHG-PHL&&\text{ [by \textbf D]}.
\end{align*}
  \end{compactenum}
\end{solution}


\begin{bonus}
  What are your suggestions for improving the course?
\end{bonus}

\emph{Geldi\u giniz i\c cin te\c sekk\"urler.  \.Iyi tatiller!}



\chapter{Student comments}\label{app:comments}

%\input{student-comments}

Here are the comments invited by the `bonus' questions on the final exams.


\section{Fall}

\subsection*{Original comments}

\emph{Some of these came by email; others are transcribed from exam
  papers.  In either case, I do not make any stylistic corrections;
  mistakes can serve as a reminder that the students are not
  native English speakers.  Nothing is left out; dots of ellipsis [\dots] are
  by the students.}

\minisec{Yunus}
Lesson was generally nice.  Although presentations of students are
necessary, I think that you should talk more because you know the
connections of propositions with other things.  And learning these
connections is very exciting.

For the exam, I think greek alphabet part is not necessary.  I just
memorized it and unfortunately I am sure that I will forget it.

%\mbox{}\hfill ---Yunus

\minisec{Elif}

First of all thank you that this course is opened. History and philosophy of
mathematics are interesting and I am glad to take such kind of course from my
own department this time. I like course material (conics is a bit difficult
compared to elements but it is also good choice) and the connection with
language add a variation. Presentations shows us our deficiencies, so they were
very useful for us. Maybe it were possible that not to choose before the lesson
the person who presents the proposition in order to make everyone prepared and
have higher interaction level. Homeworks or quizzes about alternative proofs
may force us to consider them more frequently. Exams were parallel to lessons
and measure what we have learned. In short, I am pleased to have attended to
this course.

%\mbox{}\hfill ---Elif [by email]

\minisec{Tolga}

Firstly, I didn't suppose that this course might be exciting for me. I  
added this course after the add-drop week, so I couldn't attend the  
lesson until this week. We have studied `The Elementa', Euclid, so I  
had a chance to study on this book. After all, this book helped me to  
learn how to examine on mathematics. secondly, this final exam is  
extraordinary. because of that I coudn't well done that I supposed but  
it is certain that this exam gave a chance to apply the propositions  
we learned. Moreover, the lessons were interactive so that it  
increases my attandences. Shortly, I think it is a good and exciting  
course for me. I am happy to take this course. Thanks for everything.
Good holiday.

%\mbox{}\hfill ---Tolga [by email]

\minisec{Besmir}

I wish the course was a little more math history, that would be very
nice I think.  The exam maybe should be more alike the things we are
usually used to in the texts we had.  Considering the risk I now find
my self in, maybe a second midterm would be good.

%\mbox{}\hfill ---Besmir

\minisec{Ali}

due to excessive amount of sidenotes, there is not much place left for
us to write.  it'd be proper to give those problems on a 5-page exam
paper.  also, size de iyi tatiller\footnote{Happy holidays to you too.}

%\mbox{}\hfill ---Ali

\minisec{Tu\u gba}

I had trouble in understanding the propositions of the last book we
covered in class, Apollonius.  It would be better if the books which
are easier to understand and follow in lessons are covered

%\mbox{}\hfill ---Tu\u gba

\minisec{Rashad}

I think exam and course were nice.  \emph{Maybe} a \emph{little bit}
more \emph{history} will be good.  I wish you to have a nice holiday,
too.

%\mbox{}\hfill ---Rashad

\minisec{Melis}

The course can be improved by making it `more' history included and
`less' mathematics included.  Maybe other than learning history of
mathematical concepts, we also need to learn the lives of the
mathematicians who discovered those mathematical concepts.

%\mbox{}\hfill ---Melis

\minisec{Se\c cil}

   Firstly it is good to learn about Euclid an Apollonius. We think  
(as a mathematics student) we know some geometry but we doesn't even  
know Euclid's Elements.During this semester if we read some articles  
or some books about Euclid and Apollonius lecture would be more  
educational.Sometimes classes were like a geometry class.I think we  
still don't understand why Euclid and Apollonius are so important in  
the history of mathematics.Moreover I think we should talk about  
science in Ancient Greek .

     Secondly, final exam was a good exam .I just blame myself since I  
didn'T read definitions  carefully.
      It was a good lecture thank you for being so nice to us.

      Have a nice holiday.

%\mbox{}\hfill ---Se\c cil [by email]

\minisec{M\"ursel}

It is a good chance learning history of mathematics.  In our country
these terms are not investigated at high school.  At this stage I
think this will be better if we learn than before.  I like this course
but it can be improved by more modern terms.  At class when you
compare those think with later work, we enjoyed to much.  Thank you!

%\mbox{}\hfill ---M\"ursel

\minisec{Nur}

The book Appollionus was really hard to understand, it can be reduced
from the consept of lecture.  Giving some exercise sheets for exams
can help the students to guide how to work for exam.  For example I
try to memorize everything covered but still can not get a high grade
:(

%\mbox{}\hfill ---Nur

\minisec{\"Ozge}

Before taking this course, I've heard something about it from my
friends who took in the previous years.  They were presenting the
biographies of the ancient mathemecians and their famous theorems
u.s.w.  We had partially done the same by presenting Euclid's Elements
and Apollonius' Conics but it would be nice if we learnt something
about also their lives, how did they become who they are\dots

%\mbox{}\hfill ---\"Ozge

\minisec{Tolgay}

I wanted to write my opinions about the lecture. Sorry about late-sending this
email. But you know\dots finals.

First of all, I really enjoyed the course. The thing is I am not really
motivated for a big part of our courses. I like geometry, I like history so the
idea of this lecture was good for me. But the `history' part was not that
emphasized. I mean when we check our website of department
\begin{quote}
MATH 303 History of Mathematical Concepts I (3-0)3 Mathematics in Egypt and
Mesopotamia, Ionia and Pythagoreans, paradoxes
of Zeno and the heroic age. Mathematical works of Plato, Aristotle, Euclid
of Alexandria, Archimedes, Appolonius and Diophantus. Mathematics in China
and India. \\ 
Prerequisite: Consent of the instructor.
\end{quote}
It seems its more historical. I think I prefer what we did but still\dots

Maybe there could be a part where the other matematicians' works are presented. 
I think the students of mathematics should know some basic history of maths but
I am don't know if it's here to learn it. They can, if they're interested,
easily read and make research about them.

And lastly, maybe a course website can be constructed by students as well\dots 

Thank you.

%\mbox{}\hfill ---Tolgay [by email]

\minisec{Taner}

The lecture is especially not boring as the other courses that I have
taken from our department.You made us think about the proofs and formalize
the mathematical stuff by using geometry(g\"orsel kan\i tlama y\"ontemleri in
Turkish).
But grading is more important for the students at METU so you can name the
cost(howmany points we are supposed to take from a presentation{proof}) of
the classes.I accept that this is not nice but you will see that we
students will attempt the course more than we do now.
Also if we learn the greek alphabet better this may benefit us.To give an
example, I sometimes couldn't understand the proofs from the books that
are in your site.Or may be some more books can be suggested for us to have
a look for the proofs of the propositions.
This is all that I can find for now.Thanks a lot for your help during the
classes, and your understanding about being a teacher =)
See you in the next semester hocam

%\mbox{}\hfill ---Taner [by email]

\subsection*{My responses}

\emph{Here I summarized some responses to the bonus problem on the
  last exam, and I added my comments.  I did not finish or distribute
  this work; but students' comments did influence my writing of the
  course webpage for Math 304; see the beginning of Part~\ref{part:two}.}

\begin{asparaenum}
\item
\emph{There was not enough space on the exam paper.}
Sorry! 
\item
\emph{There were no problems on ellipses or hyperbolas on the exam.
  Exercise sheets would have been useful in preparing for exams.} 
In a course like calculus, the objective is to be able to solve
problems, and the purpose of the textbook is to help you meet this
objective.  Such is probably not the purpose of Euclid or Apollonius.

Briefly, I see the objective of this course as to gain some insight
into what mathematics \emph{is.}  Two millenia and a few centuries
ago, Euclid, Archimedes, and Apollonius were doing something that we
can recognize as mathematics; but is it really the same as what we
call mathematics today?  The only way I know to answer this is to read
these mathematicians and try to understand what they were doing.

This course has exams because exams are a standard means today for
assessing student progress.  But my hope is that, if one does the
readings for the course with sufficient\dots

\item
\emph{Learning the Greek alphabet was not necessary.}  Indeed, I'm
sure that most mathematicians outside of Greece cannot recite the
Greek alphabet.  Unfortunately most mathematicians have not read
Euclid or Apollonius either.  But mathematicians \emph{use} every
letter of the Greek alphabet (except perhaps \Gk o), and they may have
opinions about ancient Greek mathematics (such as `Euclid discovered
the parallel postulate', or `Euclid invented the axiomatic
method').  Such opinions really ought
to be based on reading the original 
works.  It would be best to read these works in the original Greek,
since translations can introduce distortions.  Indeed, as we discussed
in class:
\begin{compactenum}
\item 
Euclid says `Let a straight line have been drawn,' but the translator
might say `Let a straight line be drawn';
\item
Euclid says `Let the given straight line be $AB$,' but the translator
might say `Let $AB$ be the given straight line.'
\end{compactenum}
These distinctions are subtle and are perhaps not mathematically
important.  Also, we cannot all learn ancient Greek.  However, as
mathematicians, we ought to be able to recognize words like
\Gk{k'ukloc} and \Gk{parabol'h}. 
\item
We should have done more \emph{history.}
\item
\emph{Apollonius is difficult; students have trouble presenting some
propositions, and this causes difficulties for others in the class.}
\end{asparaenum}



\section{Spring}\label{sect:comments-spring}

\minisec{Salih A.}

It will be more helpful for us, if you explain (teach) the course
instead of the students.  When we make presentations all of us know
only their subject well, because we cannot concentrate on other
students subjects.  You can give homeworks or some other projects
instead of teaching.  Because listening subject from a lecturer or a
student is very different.  Thanks for everything\dots

\minisec{Ece}\label{B2Ece}

In this course, if we want to solve questions we need to think and
work on more about them.  To make presentations is a good idea.  At
least some of the students get prepared the course and know the
propositions or corrollaries\dots

I think, take home exams or, only homework questions without exams
will be better idea.  Because I think if the exams would be take home
style, the students (we) meet all together and think together.  In
this way we all need to learn all the corollaries or etc., because in
the questions I can use some of the properties and the other student
can use other ones.  So none of us can solve questions, but we all can
solve some part of them.  But if we do them together, we may solve the
questions.  As a old people say `Bir elin nesi var, iki elin sesi var.'\footnote{`What has one hand got?  Two hands have a sound.'}

\minisec{Duygu}

Don't get angry but I totally think student presentations is not a
good idea.  Personally I like old-fashioned classroom style, the
teacher lectures and kids listen and takes notes.  If you try the
classical method, I think both you and students will be happier.  (and
it would be better for exams too, it's good to have a proper notebook
for the exams.)  Also thank you for recommendation letters \smiley

\minisec{Burhan}

expecially, A book which has a fluently english and `g\"uncel'\footnote{Current.} english
words.  Also at the class, you can be more active about teaching
lesson because when students try to teach it can be difficult to
understand.  Also, we are not recognize ancient terms about
mathematics so I think before these course, Math 303 need to be a
prerequisite lesson.

\minisec{G\"ok\c cen}

This course wasn't that clear because of the language of the texts. It was a
little bit strange and challenging to understand the content of them. Maybe due
to this, I couldn't enjoy than I expected at the beginning of this
semester\dots Because for me mathematics is getting much more enjoyable and
attractive when I can understand and can do something about it. Only these two
points bothered me during whole this semester\dots
(Again thanks for your understanding\dots)             

\minisec{Seray}

This semester I started working and this course was the only one
requiring attendence among 3 courses I've taken.  I could only attend
this course at the beginning of the semester.  Therefore I don't have
a lot to say about improving the course.  But I have some
observations.  First, the presentations weren't effective enough to
get us to the level of being successful at your exams.  Also written
group homeworks would be more motivating than making presentations.

\minisec{Besmir}

I think the conic section questions are too confusing and I find it
hard to see what is going on.  Thank you.

\minisec{Ali}\label{B2Ali}

Due to lots of mathematicians being studied in the semester, students
may grow tired, because each new mathematician requires a new, more or
less, mindset than the previous one.  So reducing the number of people
studied may be a good idea.  Also, if you're lecture this course next
year, do put previous exams on the net so that students may see what
your style is, what kind \&\ type of questions you ask.

\minisec{O\u guzhan}

Well I believe that understanding this stuff is not the main trouble
for us; however when it comes to applying to questions in the exam, we
are a bit confused (at least me!!).  So it may be quite useful if some
applications of these propositions / lemmas are distributed to the
class (similar to recitation hour!)  I enjoyed attending this class.
Thank you.

\minisec{Yasemin}

First of all, I thank you for your kindness and help during this
semester.  In order to improve the course, in my opinion it is better
for the lectures to be guided by you more, instead of the students.
For participation in class, student's proving the statements is good;
however, it would be better for anyone to prove some important part
from the lecture notes which is choosen by the student at the
beginning of the semester, and then you may expect a more qualified
presentation and proof, digged into the topic by the student itself,
and this presentation might worth more credits for this course, such
as \% 25.

Also you may give some bonus tasks, since this course is not a simple
one, then the catalogue grading would fit everyone.  Have a nice
holiday.  Best regards.

\minisec{Melis}

This course is all about the history of mathematical concepts, but I
would wish to learn more about the `people' who discovered those
concepts.  Surely, it's totally up to me to learn about them by
myself, but I would prefer to be asked about the people rather than
what they discovered.  We, as the upcoming mathematicians, are
supposed to know about the history of mathematics in all aspects.  For
instance, last semester, in Math 303 course, I was very glad to be
taught the Greek alphabet although I can already speak Greek.  It was
a completely different perspective for both me and the rest of the
class.

My answer to this question has become `my feedback to the course' more
than `my suggestions for improving the course', but I'm finding it
useful to transfer my ideas about the course.  For one thing, I
enjoyed attending the classes of Math 303 more.  I could concentrate
on it more.  I don't think that this is about the easiness of Math
303.  It's about that I liked the content of the course more.  The
reason why I took Math 304 without hesitate is exactly this.  As you
must have realized, I couldn't focus on Math 304 during the semester.

I hope my feedback gives you some idea about how this course made the
class feel.  Thank you.

\minisec{Salih K.}

I can not speak English very well.  So I can not explain my sentences
to teacher.  This is my problem, I know.  In my opinion, Cem TEZER
must be the instructor.  David Hoca is a good teacher except for me.

\minisec{Zhala}

My suggestions for improving the course are that:

Firstly, I think the content of the course is good but the system of
learning is not.  I mean, presentations handed by students should be
more well-prepared.  Visuals should be used.  Moreover, Publishes of
T\"UB\.ITAK can be used in order to improve our analytic thinking on
certain problems coming from history of mathematics.  Also, grp
presentations can be held and can make more understandable the
content.

Thanks a lot.

\minisec{\c Sule }

(I think) Giving take home quizzes may help students to think on
lecture materials.  (To be familiar with problems etc.)

\minisec{Mehmet \c S.}

I think this course needs a book which is understood easier than the
current textbook.

\minisec{Makbule}

The content of the course is becoming harder at the end of the term.
Lemma's and theorems that we covered after midterm seem as high
levelled mathematical concepts.  Therefore, the course may be clearer
when you tell the idea of the propositions, theorems etc.\ to us
first.  It is really convinient and efficient to go over the notes
which we take during the class.  I mean, it is easier to understand
your own notes than the ones in the book.

The presentations make the course more interactive.  I'm happy with
the idea of presentations.  But it may be more beneficial if you repeat
the main points and the idea of the propositions after our
presentations.  I talked to almost everyone in the class about this
and the general idea about presentations is parallel to mine.

The last thing I want to mention is that the lackness of the questions
related to the topics we covered in the class.  It is really difficult
to handle with the questions for the first time during the exam.  It
can be better if you give some exercises before exams.


\chapter{Collingwood on history}\label{app:Col}

%\input{collingwood}

From the \emph{Autobiography}
\cite{Collingwood-Auto}:%---a book that gives little information about
                        %the author's non-academic life.
\begin{quote}
I expressed this new
   conception of history in the phrase: `all history is the history
   of thought.' You are thinking historically, I meant, when you say
   about anything, `I see what the person who made this (wrote this,
   used this, designed this, \&c.)\ was thinking.' Until you can say
   that, you may be trying to think historically, but you are not
   succeeding. And there is nothing except thought that can be the
   object of historical knowledge. Political history is the history of
   political thought: not `political theory', but the thought which
   occupies the mind of a man engaged in political work: the formation
   of a policy, the planning of means to execute it, the attempt to
   carry it into effect, the discovery that others are hostile to it,
   the devising of ways to overcome their hostility, and so forth\dots
   Military history, again, is not a description of weary marches in
   heat or cold, or the thrills and chills of battle or the long agony
   of wounded men. It is a description of plans and counter-plans: of
   thinking about strategy and thinking about tactics, and in the last
   resort of what men in the ranks thought about the battle. 

    On what conditions was it possible to know the history of a
    thought? First, the thought must be expressed: either in what we
    call language, or in one of the many other forms of expressive
    activity\dots Secondly, the historian must be able to think over
    again for himself the thought whose expression he is trying to
    interpret\dots If some one, hereinafter called the mathematician,
    has written that twice two is four, and if some one else,
    hereinafter called the historian, wants to know what he was
    thinking when he made those marks on paper, the historian will
    never be able to answer this question unless he is mathematician
    enough to think exactly what the mathematician thought, and
    expressed by writing that twice two are four. When he interprets
    the marks on paper, and says, `by these marks the mathematician
    meant that twice two are four', he is thinking simultaneously:
    (a) that twice two are four, (b) that the mathematician thought
    this, too; and (c) that he expressed this thought by making these
    marks on paper\dots

    This gave me a second proposition: `historical knowledge is the
    re-enactment in the historian's mind of the thought whose history
    he is studying.'\enlargethispage{\baselineskip}
\end{quote}

\chapter{Departmental correspondence}\label{app:cor}

%\input{correspondence}


Here are some emails about the course that were shared within the METU
mathematics department.

\section{Wednesday, April 28, at 13:01}%{Wed, Apr 28, 2010 at 1:01 PM}

\emph{I wrote to \url{odtu-math}:}

%Dear colleagues,

Since it is time to make our course requests for next fall, I thought I
would talk about what I have been doing this year with Math 303 and 304,
`History of Mathematical Concepts'.  I would be happy to teach this course
again next year; but I would also be happy if somebody else was interested
in teaching the course as I have been.

I have three principles for the course:
\begin{compactenum}[1.]
\item
Our only textbook is original sources (in translation as necessary):
Euclid, Apollonius, al-Khwarizmi, Descartes, \dots
\item
The `teacher' does not lecture; the students present at the blackboard
what they have read.
\item
Attendence is required.
\end{compactenum}
There are details on the web pages
\begin{quote}\centering
\url{http://www.metu.edu.tr/~dpierce/Courses/303/}

\url{http://www.metu.edu.tr/~dpierce/Courses/304/}
\end{quote}
In addition, I keep a journal of what goes on in class.  My record of the
first semester is 42 pages.  If you want to see it, let me know.

We spent most of last semester reading Euclid.  Many of the propositions
were familiar to the students; but the students had not proved these
propositions before.  It seemed a shame that the students had had to wait
till their third or fourth year at university to prove these propositions.
Euclid's propositions were part of the basic education of most of the
great mathematicians whose theorems we try to teach.  Indeed, it might be
good if Math 111 consisted (in part or whole) of reading and presenting
Euclid.  For example, \textbf{proportion} as Euclid defines it is an excellent
example of an equivalence relation.

Meanwhile, Math 303 is a place where our students can read Euclid (and
Apollonius, or Archimedes, or \dots).  As I said, I am happy if either I or
somebody else does this reading with them next year.

%Respectfully,

%David Pierce

\section{Wednesday, April 28, at 18:55}%{Wed, Apr 28, 2010 at 6:55 PM}

\emph{Sergey responded on \url{odtu-math}:}

%Dear David,

In my opinion, a few first lectures should be really devoted to mathematics
of antique times and Middle Ages, but the most importand and most
interesting events in mathematics happened later, and one should spare
enough time to discuss the works of Newton, Euler, Lagrange, Galois, Abel,
Gauss, Riemann, Klein, Hilbert, Poincare, and many other giants.
 One should discuss the history of geometry (famous old problems,
non-Euclidean geometry, Italian Algebro-geometric school), of algebra
(evolution of the concepts, like numbers, groups), of Analysis
(the Newton-Leibnitz dispute, the problem of foundations, notorious
mistaken `theorems', fake references like `L'Hospital rule', etc.).
 It is good to say about the history of the first Math journals,
Academies of Science, about their Competitions and Awards.
 One should certainly discuss Hilbert's problems, the history of Fields
medals, solution of the most outstanding problems (of Fermat, of
Poincare, Four-colour, etc.), some new theories and trends in Math in the
last century, and may be stop with the Millenium problems.

I can give such a course myself, or welcome anybody else who would do it!

%Sergey Finashin

\section{Thursday, April 29, at 11:44}%{Thu, Apr 29, 2010 at 11:44 AM}

\emph{I responded:}

Sergey wrote:
\begin{quote}
In my opinion, a few first lectures should be really devoted to
mathematics of antique times and Middle Ages, but the most importand and
most interesting events in mathematics happened later, and one should
spare enough time to discuss the works of Newton, Euler, Lagrange,
Galois, Abel, Gauss, Riemann, Klein, Hilbert, Poincare, and many other
giants.
\end{quote}
%Dear Sergey,

Thanks for writing.  However, I don't know what your point is.  You refer
to `a first few lectures'.  A few first lectures of \textbf{what?}  You are
replying to my email about Math 303/4, so maybe you are referring to
lectures in this course.  However, I wrote:

\begin{quote}
I have three principles for the course:\dots
\begin{compactenum}[1.]\setcounter{enumi}1
\item
The `teacher' does not lecture; the students present at the
blackboard what they have read.
\end{compactenum}
\end{quote}
So there are no lectures in my course.  Or rather, everybody in the
classroom is a lecturer.  You don't seem to address this point.  But my
undergraduate education consisted entirely of classes like this.  I was
very happy with the arrangement, and I decided to see if it would work at
METU.  I believe I have had some success.

However, we are reading Newton's \emph{Principia} now.  I don't know if we shall
have time for anything else.  Recently a young woman whom I'll call
`Yolanda' [Yasemin] was presenting Lemma VII [see p.~\pageref{Y}
  above], which you can see at:
\begin{quote}\centering
\url{http://en.wikisource.org/wiki/The_Mathematical_Principles_of_Natural_Philosophy_(1729)/Book_1/Section_1#Lem7}
\end{quote}
When `Yolanda' got to Corollary 2, she said and wrote that $AD$, $DE$,
$BF$, $FG$, $AB$, and the arc ACB had [ultimately] the ratio of equality.

I said I didn't believe it.  In fact she had miscopied Newton.  I hoped
she would try to work out a proof and see her mistake.  But she only
beamed at me and said, `It's hard to believe, but true!'

I finally asked `Yolanda' to check her text.  She saw that she should have
had $AE$ for $DE$, and $BG$ for $FG$.  But she couldn't give a proof of the
correct statement.  She just muttered something about how Newton was
smarter than she was.

I went to the board and suggested a proof.  One of the most interested and
active students in the class, `Oscar' [O\u guzhan], was skeptical; but
when you are
talking (for the first time in history, perhaps) about the ratios with
which quantities vanish, skepticism is to be expected.  `Sara' [\c
  Sule] seemed to think at first that Lemma VII followed immediately
from Lemma VI. 

And so the discussion continued.  Thus a number of students became
collectively engaged in puzzling out what Newton was talking about.
Unfortunately it doesn't happen much in my class.  Students come to class
and present the propositions assigned to them, but often they haven't
really understood the point of the propositions, or their proof.  In their
presentations, they may say, `he says this, then he says that', rather
than saying \textbf{we} have this, and \textbf{therefore} we have that.

How can they do anything else?  One difficulty is that students are taking
several other courses, in particular math courses, which also demand their
attention.  Of greater concern is that students are trained to believe
that books and teachers are unquestionable authorities.  I hope to
encourage them to see things differently.  This is the main point.

By the way, reading Cardano's \emph{Ars Magna} in my course was perhaps useful
for this purpose.  I hadn't read Cardano before, but I thought that, in
Math 304, we might read his solution of cubic equations.  Then I found
more and more of Cardano that seemed worth reading in class.

After reading more carefully with the students, I had to conclude that
either Cardano was a bad writer, or else he really didn't understand what
he was doing.  He also makes computational mistakes, which students
discovered.

Unfortunately the only available English translation of Cardano is
unsatisfactory, because it uses modern algebraic notation.  Therefore I
also gave the students the original Latin to look at.  I think there is no
point to studying pre-Cartesian mathematics unless one tries to forget
about our modern symbolic tools.  Descartes thought the Ancients really
had such tools too; but this is not at all clear.

Sergey, let me repeat what you said:
\begin{quote}
In my opinion, a few first lectures should be really devoted to
mathematics of antique times and Middle Ages, but the most importand and
most interesting events in mathematics happened later, and one should
spare enough time to discuss the works of Newton, Euler, Lagrange,
Galois, Abel, Gauss, Riemann, Klein, Hilbert, Poincare, and many other
giants.
\end{quote}
The most important and most interesting events happened later?  This makes
as much sense as saying that \emph{War and Peace} is more important and more
interesting than the \emph{Iliad} and the \emph{Odyssey.}  But have
you \textbf{read}
Newton, Euler, Lagrange, and the others you mention?  Do you propose to
read them in class \textbf{with students?}

For Math 304, I wondered if we could use something like Struik's \emph{Source
Book in Mathematics, 1200--1800.}  I decided against it.  There is not
much point in reading the short passages provided by Struik, just so that
one can say, `I've read Leibniz' or `I've read Bernoulli'.  A writer worth
reading is worth spending time with, over the course of many pages.

Struik gives a passage from Cardan with a solution of what we write as
\begin{equation}\label{eqn:SC}
x^3 + 6x = 20.
\end{equation}
A third part of each of Struik's pages is filled with footnotes explaining
what Cardan is doing.  Only Cardan is \textbf{not} doing what is in the
footnotes. As Struik shows in his notes, \textbf{we} can solve the
cubic equation 
\begin{equation*}
       x^3 + px = q
\end{equation*}
by substituting
\begin{equation*}
       x = u - v
\end{equation*}
and solving first for $u$ and $v$.  But either Cardan doesn't really see this
himself, or else he is hiding it.  Cardan gives a formula for $x$, and he
can prove it is correct by substitution; but he shows no interest in
\textbf{deriving} the formula.  Struik does not address this point.

Neither does Boyer, whose text has (I believe) been traditionally used for
Math 303/4.  In his section on Cardan's solution, Boyer just writes,
\begin{quote}
       The solution of this equation covers a couple of pages of
       rhetoric that we should now put in symbols as follows:
       Substitute $u - v = x$\dots
\end{quote}
In that `rhetoric', Cardan shows in effect that, \textbf{if} we have the
simultaneous equations
\begin{align*}
       u^3 - v^3& = 20,&
       uv& = 2,
\end{align*}
\textbf{then} $x = u - v$  in~\eqref{eqn:SC} above.  He doesn't say
\textbf{why} we 
should start with 
those simultaneous equations.  Neither does he explain how to solve those
simultaneous equations: he just tells you the formula for the solution.
Actually this seems to be just what mathematics is in the minds of (some
of) our students, who love formulas, no matter where they come from, as
long as they can be used on the exam.  But I blame the university entrance
exam system for this (and perhaps teachers who are earlier products of
this system).
\begin{quote}
 One should discuss the history of geometry (famous old problems,
non-Euclidean geometry, Italian Algebro-geometric school), of algebra
(evolution of the concepts, like numbers, groups), of Analysis (the
Newton-Leibnitz dispute, the problem of foundations, notorious mistaken
`theorems', fake references like `L'Hospital rule', etc.).
 It is good to say about the history of the first Math journals,
Academies of Science, about their Competitions and Awards.
 One should certainly discuss Hilbert's problems, the history of Fields
medals, solution of the most outstanding problems (of Fermat, of
Poincare, Four-colour, etc.), some new theories and trends in Math in the
last century, and may be stop with the Millenium problems.

I can give such a course myself, or welcome anybody else who would do it!
\end{quote}
Well Sergey, there is a procedure for opening new courses.  Or if you mean
to be describing how \textbf{Math 303/4} should be taught, then please say so.
You seem to be describing a lecture course; if so, it is not a course that
I would consider myself competent to teach.

Lecturing \textbf{mathematics} is fine, since the listeners can check the
lecturer's claims by using the critical powers of their own reason.
Again though, I am sorry that even some students in Math 303/4 don't use
these critical powers very much.  In any case, lecturing about what
happened in the past is a different matter.  For example, perhaps we all
grew up with the idea that there was a crisis in ancient mathematics owing
to the discovery of incommensurable magnitudes.  We may tell students
about this if we happen to prove to them the irrationality of the square
root of $2$.  However, it seems there is simply no evidence of an ancient
crisis.

Of course events of more recent centuries may be better documented.

%Respectfully,

%David

\section{Friday, April 30, at 12:27}%{Fri, Apr 30, 2010 at 12:27 PM}

\emph{Sergey wrote again to \url{odtu-math}:}

%Dear David,

 It would be interesting to discuss your ideas: I would prefer to do it
privately, not involving people not interested in this subject.
 I will just try now to state clearly my opinion which differs from yours:
a course `History of Math concepts' is really needed for our Department
just because it helps to understand better mathematics.
There are impotant topics to be covered, and they should not be missed.
 For undergraduate students, an idea to replace a lecture course by
a seminar course does not seem good: stdents may really have more
fun\label{fun} 
(like in a course of singing, or dancing), and even study a few
selected topics better, but overall they will be far behind the syllabus.
 A kind of a seminar that you proposed instead of lectures would be
perfect for graduate students in History Department, who really need to
learn how to work with the original sources.



\chapter{Notes on Greek mathematics}\label{app:background}

%\input{background}

[I put these notes on the Math 303 webpage at the beginning of the semester.]

\section{Introduction}\label{sect:intro}

Some time in the 3rd century \textsc{b.c.e.,} Apollonius of Perga wrote eight
books on \textbf{conic sections.}  We have the first four books
\cite{MR1660991,Apollon4} in the original
Greek; the next three books survive in Arabic translation
\cite{MR1053158}; the eighth book is lost.
As Apollonius tells us in an introductory letter, his first four books
are part of an elementary course on the conic sections.
%The present notes are for use in such a course.

Before Apollonius, around 300 \textsc{b.c.e.,} Euclid published the
thirteen books of the \emph{Elements}
\cite{MR17:814b,MR1932864,euclid-fitzpatrick}, a work of mathematics
of which some parts could well be used as a textbook today.  The
\emph{Elements} provide a good example of mathematical exposition and
of what it means to prove something.

In 2008, getting ready to teach a course on the conic sections,%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%
\footnote{At the
Nesin Mathematics Village, \c Sirince, Sel\c cuk, \.Izmir, Turkey.} 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
I wrote some notes on ancient mathematics.  Using those notes, I have
prepared the present notes, for use in a course called `History of
Mathematical Concepts I' at METU---a course in which participants will
read Euclid and Apollonius.

In the latter sections of these notes, I look at
some general features of ancient mathematics as I understand it.
Meanwhile, in
\S\ref{sect:syn-anal}, I jump forward in history to Descartes, to see
the sorts of improvements that he thought he was making to the
mathematical practice of mathematicians like Euclid and Apollonius.

Because I shall occasionally refer to some Greek words, I review the Greek
alphabet in [Appendix~\ref{app:Gk}.]

\section{Why read the Ancients?}

As an undergraduate, I attended a college\footnote{St John's
  College, with campuses in Annapolis, Maryland, and Santa Fe,
  New Mexico, USA [see p.~\pageref{SJC}].} where Euclid and Apollonius were used as
  textbooks.  They were so used, I think, not because they were considered to
  be the \emph{best} textbooks, but because they \emph{had
  been} textbooks for countless generations of mathematicians:
  therefore (the idea was), one might gain some understanding of
  humanity and oneself by reading these books.  (The same is true for
  Homer, Aeschylus, Plato, and the other great books read at the college.)

Now, having become a professional mathematician, I ask what Euclid and
Apollonius have to offer the mathematician of today.  It is in pursuit
of an answer to this question that I prepare these notes---which
therefore are part of an ongoing project.

I prepare these notes also for the sake of honesty about what students
are asked to learn.  The curves called  \textbf{conic
sections} are a standard part of an elementary course of mathematics.
The origin of such curves is in the name: they are obtained by slicing
a cone.  Apollonius treated the curves in this way.
But in math courses today, the conic sections are usually given as the curves
defined by certain equations, such as  
\begin{align*}
ay&=x^2&
&\text{or}&
\frac{x^2}{a^2}\pm\frac{y^2}{b^2}&=1.  
\end{align*}
Or perhaps the curves are given in terms of foci and directrices.  A
textbook may \emph{assert}
that the curves so defined can indeed by obtained as sections of cones;
but it is rare that this assertion is justified.

One calculus textbook\footnote{James Stewart, \emph{Calculus,} fifth
  edition, p.~720.  This text is currently in use at METU.}
  writes:
\begin{quote}
  In this section we give geometric definitions of parabolas,
  ellipses, and hyperbolas and derive their standard equations.  They
  are called \textbf{conic sections,} or \textbf{conics,} because they
  result from intersecting a cone with a plane as shown in Figure 1.
\end{quote}
(I omit the author's figure.)
The conic sections result from intersecting a cone with a plane: this
can be understood as a \emph{definition} of the conic sections.  Let
us call it Definition I.
%I omit the book's Figure 1; but it can serve as one definition of the
%conic sections.  
More precisely, this definition distinguishes three kinds of conic
sections, depending on the angle of the plane with respect to the cone.
One kind of conic section is called the
\emph{parabola,} and the text continues under the heading
\emph{Parabolas:}
\begin{quote}
  A \textbf{parabola} is the set of points in a plane that are
  equidistant from a fixed point $F$ (called the \textbf{focus}) and a
  fixed line (called the \textbf{directrix})\dots In the 16th century
  Galileo showed that the path of a projectile that is shot into the
  air at an angle to the ground is a parabola.  Since then, parabolic
  shapes have been used in designing automobile headlights, reflecting
  telescopes, and suspension bridges\dots We obtain a
  particularly simple equation for a parabola if we place its vertex
  at the origin $O$\dots
\end{quote}
Here then is another definition of the parabola; call it Definition
II.  Definitions I and II are equivalent in that they define the same
objects; but the author does not clearly say so, much less prove it.
I don't think he \emph{needs} to prove the 
equivalence; but at least he ought to state that he is not going to
prove it.   

Perhaps the author expects the reader to \emph{infer} the equivalence
of Definitions I and II.  But this is not his style.  He
is usually eager to give his readers every assistance.  Note for
example that he apparently does not
trust readers to infer for themselves that parabolas are worth 
studying.  Before concluding anything from his definition of
parabolas, the author feels the need to tell the reader how
\emph{useful} parabolas are. 

Another textbook\footnote{Robert A. Adams, \emph{Calculus: a complete
    course,} fourth edition, p.~476.  This text was formerly used
 at METU.} follows a similar procedure, first defining the conic
    sections \emph{as such,} then defining them in terms of foci and
    directrices.  Between the two definitions, the writer observes that the
    intersection of a cone and a plane will be given by a
    second-degree equation.  This suggests that the quadratic
    equations to be derived presently in the book may indeed define conic
    sections.  However, no attempt is made to prove
    that every curve defined by a quadratic equation can be obtained
    as the section of a cone.  The author observes:
\begin{quote}
  After straight lines the conic sections are the simplest of plane
  curves.  They have many properties that make them useful in
  applications of mathematics; that is why we include a discussion of
  them here.  Much of this material is optional from the point of view
  of a calculus course, but familiarity with the properties of conics
  can be very important in some applications.  Most of the properties
  of conics were discovered by the Greek geometer, Apollonius of
  Perga, about 200 BC.  It is remarkable that he was able to obtain
  these properties using only the techniques of classical Euclidean
  geometry; today most of these properties are expressed more
  conveniently using analytic geometry and specific coordinate
  systems. 
\end{quote}
Again, the justification offered for the study of the conic sections
is their usefulness.  But as for `expressing' the properties of conic
sections, which of the following expresses better what a conic section
\emph{is?} 
\begin{compactenum}[1.]
  \item
It is the intersection of a cone and a plane.
\item
It is the intersection of the surfaces defined by the equations
\begin{align*}
ax+by+cz+d&=0,&
(x-ez)^2+y^2&=fz^2.
\end{align*}
\end{compactenum}
What the author means, I think, is that it is convenient to define
certain curves `ana\-lytically'---that is, in a coordinate system such
as Descartes introduced; properties of the curves can then be obtained
by further
analysis.  But showing that those curves are conic sections is a whole
other problem, not addressed in the book.

By the way, despite what the last quotation suggests, I am not
sure that obtaining nice results with limited 
mathematical tools is remarkable in itself.  The tools of an artisan
depend on what is available in the physical environment; but the tools
of a mathematician depend only on imagination.  A
mathematician without the imagination to come up with the best tool
for the job would seem to be an unremarkable mathematician.


The first chapter of Hilbert and Cohn-Vossen's \emph{Geometry and the
  Imagination}~\cite{MR0046650} contains a
beautiful account of how various properties of the conic sections
arise from consideration of the cones from which the sections are
obtained.  However, the cones considered by the authors
are all \emph{right} cones.  Apollonius does not make this
restriction.  Hilbert and Cohn-Vossen give an etymology for the names
of the ellipse, the hyperbola, and the parabola: it involves
eccentricity.  The etymology is plausible, but it appears to be
literally incorrect, as a reading of Book I of Apollonius would show.

Mathematics reveals underlying correspondences between seemingly
dissimilar things.  Sometimes we treat these correspondences as
identities.  This can be a mistake.  There is a correspondence between
conic sections and quadratic equations.  But are the sections
\emph{really} the equations?  One cannot answer the question without
considering \emph{conic sections as such,} as Apollonius considered
them. 


\section{Synthesis and analysis}\label{sect:syn-anal}

It may be said that, in reading Euclid and Apollonius, we are
going to do \textbf{pre-Cartesian} mathematics: mathematics as done
before (well before) the time of Ren\'e Descartes (1596--1650).

The geometry pioneered by
Ren\'e Descartes is called \textbf{analytic
  geometry;} by contrast, the geometry of ancient mathematicians like
Euclid and Apollonius is sometimes called 
\textbf{synthetic geometry.}  But what does this \emph{mean?}
  The word \Engl{synthetic} comes from the Greek \Gk{sunjetik'oc,} meaning
  \Engl{skilled in putting together} or \Engl{constructive.}  This Greek
  adjective derives from the verb \Gk{sunt'ijhmi} \Engl{put together,
  construct} (from \Gk{sun} \Engl{together} and \Gk{t'ijhmi}
  \Engl{put}).  The word \Engl{analytic} is the English form of
  \Gk{>analutik'oc}, which derives from the verb \Gk{>anal'uw}
  \Engl{undo, set free, dissolve} (from \Gk{>ana} \Engl{up,} \Gk{l'uw}
  \Engl{loose}).    
Although we refer to ancient geometry as synthetic, the Ancients
evidently recognize both analytic and synthetic
methods.  Around 320 \textsc{c.e.,} Pappus of Alexandria
writes \cite[p.~597]{MR13:419b}:
\begin{quote}
  Now \textbf{analysis} (\Gk{>an'alusic}) is a method of taking that
  which is sought as
  though it were admitted and passing from it through its consequences
  in order to something which is admitted as a result of
  synthesis; for in analysis we suppose that which is sought to be
  already done, and we inquire what it is from which this comes about,
  and again what is the antecedent cause of the latter, and so on
  until, by retracing our steps,  we light upon something already
  known or ranking as a first principle; and such a method we call
  analysis, as being a \emph{reverse solution}
  (\Gk{>an'apalin l'usic}).

But in \textbf{synthesis} (\Gk{sunj'esic}), proceeding in the opposite
  way, we suppose to be already done that which was last reached in
  the analysis, and arranging in their natural order as consequents
  what were formerly antecedents and linking them one with another, we
  finally arrive at the construction of what was sought; and this we
  call synthesis. 

Now analysis is of two kinds, one, whose object is to seek the truth,
being called \textbf{theoretical} (\Gk{jewrhtik'oc}), and the other,
whose object is to find something set for finding, being called
\textbf{problematical} (\Gk{problhmatik'oc}). 
\end{quote}
This passage is not very useful without examples: I shall propose one
presently.  Meanwhile, I note that Pappus elsewhere
\cite[pp.~564--567]{MR13:419b} says more
about the distinction between theorems and problems:
  \begin{quote}\sloppy
    Those who favor a more technical terminology in geometrical
    research use \textbf{problem} (\Gk{pr'oblhma}) to
    mean a [proposition\footnote{Ivor Thomas \cite[p.~567]{MR13:419b}
    uses \Engl{inquiry} here in his translation;
    but there is \emph{no} word in the Greek original corresponding to
    this or to \Engl{proposition.}}] in which it is proposed to do or construct
    [something]; and \textbf{theorem} (\Gk{je'wrhma}), a [proposition]
    in which the
    consequences and necessary implications of certain hypotheses are
    investigated; but among the ancients some described them all as
    problems, some as theorems.
  \end{quote}

What really distinguishes Cartesian geometry from what came before is
perhaps suggested by the first sentence of Descartes's \emph{Geometry}
\cite[p.~2]{Descartes-Geometry}:
\begin{quote}
  Any problem in geometry can easily be reduced to such terms that a
  knowledge of the lengths of certain straight lines is sufficient for
  its construction.
\end{quote}
From a straight line, Descartes abstracts something called \emph{length.}  A
length is
something that we might today call a positive real number.  

Descartes
takes the edifice of geometry that has been built up or
`synthesized' over the centuries, and reduces or `analyzes' its study
into the manipulation of numbers.  To be more precise, he `takes that
which is sought as though it were admitted' in the following way.  In
Figure~\ref{fig:Descartes2},
\begin{figure}[ht!]
\centering
\psset{unit=0.75cm}
%\psset{unit=0.1cm}
    \begin{pspicture}(-2,-3.5)(5,3.8)
%\psgrid
      \psline(-1.364,-1.364)(3.333,3.333)
      \psline(0,0)(4,0)
      \psline(1.5,-3)(3.714,1.429)
      \psset{linestyle=dotted}
      \psline(4.388,-1.94)(1.5,-3)
      \psline(4.388,-1.94)(3.333,3.333)
      \psline(-1.364,-1.364)(4.388,-1.94)
      \uput[ul](3,0){$A$}
      \uput[ur](4,0){$B$}
      \uput[dr](4.388,-1.94){$C$}
      \uput[dl](1.5,-3){$D$}
      \uput[ul](0,0){$E$}
      \uput[ul](-1.364,-1.364){$F$}
      \uput[l](3.714,1.429){$R$}
      \uput[ul](3.333,3.333){$S$}
    \end{pspicture}
\caption{Descartes's diagram}\label{fig:Descartes2}
\end{figure}
 straight lines $BE$, $DR$, and $FS$ are
given in position (meaning their endpoints themselves are not fixed);
and the sizes of angles $ABC$, $ADC$, and $CFE$ are given.  It is required to
find the point $C$ so that the
rectangle with sides $BC$ and $CD$ has a given ratio to the square on
$CF$.  (This is a simplified version of the problem that Descartes
takes up in the \emph{Geometry.})

In his analytic approach, Descartes assumes that $C$ has already been
found, as in
the figure.  We denote $AB$ by $x$, and $BC$ by $y$.  The ratio
$AB:BR$ is given; call it $z:b$.  Then
\begin{align*}
  RB&=\frac{bx}z,&CR&=y+\frac{bx}z=\frac{zy+bx}z.
\end{align*}
But $CR:CD$ is given; call it $z:c$.  Then
\begin{equation*}
  CD=\frac{czy+bcx}{z^2}.
\end{equation*}
Also $AE$ is given; call it $k$.  And let $BE:BS=z:d$.  Then
\begin{align*}
  BE&=k+x,& BS&=\frac{dk+dx}z,& CS&=\frac{zy+dk+dx}z.
\end{align*}
Finally, if $CS:CF=z:e$, then
\begin{equation*}
  CF=\frac{ezy+dek+dex}{z^2}.
\end{equation*}
So it is given that the ratio
\begin{equation*}
y\cdot\frac{czy+bcx}{z^2}:\left(\frac{ezy+dek+dex}{z^2}\right)^2
\end{equation*}
is constant.  This gives us a quadratic equation in the unknowns $x$ and
$y$.  

Descartes's method does not use explicitly drawn
\emph{axes} with respect to which $x$ and $y$ are measured. Also, the
straight lines called $x$ and $y$ are not 
required to be perpendicular: they are merely not parallel.

Through analysis, we have found an equation
that determines the point $C$.  Since the equation is quadratic, the
point $C$ lies on (a curve that turns out to be) a conic section.  When there
are more straight lines in the problem, then the resulting equation
may have a higher degree.

We do not get any sense here for what the curve of $C$ \emph{looks like.}
We might get some sense by analyzing the equation for $C$.
Apollonius will give us a sense for what conic sections look like by
showing \emph{how they are related to the cones that they come from.}

\section{Theorems and problems}\label{sect:Proclus}

The text of Apollonius as we have it consists almost entirely of
theorems and problems (in the sense of the last section).  There are
some introductory remarks, some
definitions, but nothing else.  The theorems and problems can be
analyzed in a way described by Proclus,\footnote{Proclus was born in
  Byzantium (that is, Constantinople, now \.Istanbul), but his parents were from
  Lycia (Likya), and he was educated first in Xanthus.  He moved to
  Alexandria, then Athens, to study philosophy
  \cite[p.~xxxix]{MR1200456}.} in the
fifth century \textsc{c.e.,} in his 
commentaries on Euclid \cite[p.~159]{MR1200456}:
\begin{quote}\sloppy
  Every problem and every theorem that is furnished with all its parts
  should contain the following elements: an \emph{enunciation}
  (\Gk{pr'otasic}), an \emph{exposition} (\Gk{>'ekjesic}), a
  \emph{specification}
  (\Gk{diorism'oc}), a \emph{construction} (\Gk{kataskeu'h}), a \emph{proof}
  (\Gk{>ap'o\-deixic}), and a \emph{conclusion} (\Gk{sump'erasma}).  Of these,
  the enunciation states what is given and what is being sought from
  it, for a perfect enunciation consists of both these parts.  The
  exposition takes separately what is given and prepares it in advance
  for use in the investigation.  The specification takes separately
  the thing that is sought and makes clear precisely what it is.  The
  construction adds what is lacking in the given for finding what is
  sought.  The proof draws the proposed inference by reasoning
  scientifically from the propositions that have been admitted.  The
  conclusion reverts to the enunciation, confirming what has been
  proved. 

So many are the parts of a problem or a theorem.  The most essential
ones, and those which are always present, are enunciation, proof, and
conclusion.
\end{quote}
Alternative translations are: for
\Gk{>'ekjesic}, \emph{setting out,} and 
for \Gk{diorism'oc}, \emph{definition of
  goal} \cite[p.~10]{MR1683176}.

For an illustration, we may analyze Proposition 1 of Book I of
Euclid's \emph{Elements} (in Fitzpatrick's translation
\cite{euclid-fitzpatrick}).  The 
proposition is a \emph{problem:} 

\begin{enunciation}\sloppy
To construct an equilateral triangle on a given finite
straight-line.
\end{enunciation}

\begin{exposition}
    Let AB be the given finite straight-line.
\end{exposition}

\begin{specification}
    So it is required to construct an equilateral triangle on
the straight-line AB.
\end{specification}

\begin{construction}
    Let the circle BCD with center A and radius AB have
been drawn, and again let the circle ACE with
center B and radius BA have been drawn. And
let the straight-lines CA and CB have been joined from
the point C, where the circles cut one another, to the
points A and B (respectively).
\end{construction}

\begin{proof}
    And since the point A is the center of the circle CDB,
AC is equal to AB. Again, since the point
B is the center of the circle CAE, BC is equal to BA. But CA was also
shown (to be) equal to AB. 
Thus, CA and CB are each equal to AB. But things equal
to the same thing are also equal to one another.
Thus, CA is also equal to CB. Thus, the three (straight-lines) CA, AB, and BC are equal to one another.
\end{proof}

\begin{conclusion}
    Thus, the triangle ABC is equilateral, and has been
constructed on the given finite straight-line AB. (Which
is) the very thing it was required to do.
\end{conclusion}

\section{Conversational implicature}\label{sect:c-imp}

One apparent difference between the ancient and modern approaches to
 mathematics may result from a modern habit that is exemplified
in a Russian textbook of the Soviet period \cite[pp.~9 f.]{DPR}:\label{Soviet}
\begin{quote}
The student of mathematics must at all times have a clear-cut
  understanding of all fundamental mathematical concepts\dots
  The student will also recall the signs of weak inequalities: $\leq$
  (less than or equal to) and~$\geq$ (greater than or equal to).  The
  student usually finds no difficulty when using them in formal
  transformations, but examinations have shown that many students do
  not fully comprehend their meaning.

To illustrate, a frequent answer to: ``\emph{Is the inequality $2\leq
  3$ true?}'' is ``No, since the number $2$ is less than $3$.''  Or,
  say, ``\emph{Is the inequality $3\leq3$ true?}'' the answer is often
  ``No, since $3$ is equal to $3$.''  Nevertheless, students who
  answer in this fashion are often found to write the result of a
  problem as $x\leq3$.  Yet their understanding of the sign $\leq$
  between concrete numbers signifies that not a single specific number
  can be substituted in place of $x$ in the inequality $x\leq3$, which
  is to say that the sign $\leq$ cannot be used to relate any numbers
  whatsoever.
\end{quote}
The students referred to, who will not allow that $2\leq3$, are following
a habit of ordinary language, whereby the \emph{whole} truth must be
told.  According to this habit, one does not say $2\leq3$, because one
can make a stronger, more informative statement, namely $2<3$.  This
habit would appear to be an instance of \emph{conversational
  implicature:} this is the ability of people to convey or \emph{implicate}
statements that are not logically \emph{implied} by their words
\cite[ch.~1, \S5, pp.~36--40]{CGEL}.  In saying \Engl{$A$ or $B$ [is true],} one
usually `implicates' that one does not know \emph{which} is true.

This habit of implicature may be reflected in the ancient
understanding,
according to which \emph{one} (\Gk{<'en}) is not a \emph{number}
(\Gk{>arijm'oc}).
In Book VII of the \emph{Elements,} Euclid somewhat
obscurely defines a
\textbf{unit} (\Gk{mon'ac}) as that by virtue of which each being is
called `one'. (This English version of the definition is based on the
Greek text supplied in
\cite[Vol.~2, p.279]{MR17:814b}.)  Then a \textbf{number} is defined
as a \emph{multitude}
(\Gk{pl~hjoc}) composed of units.  In particular, a unit is not a
number, because it is not a multitude: it is one.  Euclid does not
bother to state explicitly this distinction between units and numbers,
but it can be inferred, for example,
from his presentation of what we now call the Euclidean algorithm.
Proposition~VII.1 of the \emph{Elements} involves 
a pair of numbers such that the algorithm, when applied to
them, yields a unit (\Gk{mon'ac}).  Then
this unit is \emph{not} considered as a greatest common divisor of the
numbers; the numbers do not \emph{have} a greatest common divisor; the
numbers are simply relatively prime.  If the numbers are
\emph{not} relatively prime, then the \emph{same} algorithm yields
their greatest common divisor.  This observation appears to be the
contrapositive of the first, but Euclid distinguishes it as
Proposition VII.2 of the \emph{Elements.}

Conversational implicature may be seen in Apollonius's treating of the
circle as different from an ellipse.

\section{Lines}\label{sect:lines}

In the old understanding, a \textbf{line} need not be
straight.  A line may have endpoints, or it may be, for
example, the circumference of a \emph{circle.}  Indeed, according to
the definition in Euclid's \emph{Elements,}
%\cite{MR17:814b,bones,MR1932864}
\begin{quote}
A \textbf{circle} (\Gk{k'ukloc}) is a plane figure contained by one
line (\Gk{gramm'h}) such that all
the straight lines falling upon it from one point among those lying
within the figure are equal to one another.
\end{quote}
A \emph{straight} line (\Gk{e>uje~ia gramm'h}) does have endpoints;
but the straight line may
be \emph{produced} (extended) beyond these endpoints, as far as desired.

\chapter{The Greek Alphabet}\label{app:Gk}

I have heard a rumor (see p.~\pageref{DHH}) that students can
improve their mathematics simply by learning this alphabet.
\begin{figure}[ht]
\centering
  \renewcommand{\arraystretch}{1}
      \begin{tabular}{| c  l | c  l | c l |} \hline
\Gk{A a}&\textbf alpha  &\Gk{I i} & \textbf iota   &\Gk{R r}   &\textbf{rh}o\\ 
\Gk{B b}&\textbf beta   &\Gk{K k} & \textbf kappa  &\Gk{S sv,c}&\textbf sigma\\
\Gk{G g}&\textbf gamma  &\Gk{L l} & \textbf lambda &\Gk{T t}   &\textbf tau\\
\Gk{D d}&\textbf delta  &\Gk{M m} & \textbf mu     &\Gk{U u}&\textbf upsilon\\ 
\Gk{E e}&\textbf epsilon&\Gk{N n} & \textbf nu     &\Gk{F f}   &\textbf{ph}i\\ 
\Gk{Z z}&\textbf zeta   &\Gk{X x} & \textbf xi     &\Gk{Q q}   &\textbf{ch}i\\ 
\Gk{H h}&\textbf{\=e}ta &\Gk{O o} & \textbf omicron&\Gk{Y y}   &\textbf{ps}i\\
\Gk{J j}&\textbf{th}eta &\Gk{P p} & \textbf pi     &\Gk{W
  w}&\textbf{\=o}mega\\\hline 
  \end{tabular}
\caption[The Greek alphabet]{The Greek alphabet}\label{fig:Greek}
\end{figure}

The first letter or two of
  the (Latin) name for a Greek 
letter provides a transliteration
for that letter.  However, upsilon is also transliterated by \Engl{y.}
The diphthong \Gk{ai} often comes into English (\emph{via} Latin)
  as \Engl{ae,} while \Gk{oi} may come as \Engl{oe.}
The second form of the small sigma is used at the ends of words.
  In texts, the rough-breathing mark (\Gk{<}) over 
an initial
vowel (or \Gk r) is transcribed as a preceeding (or following) \letter
h (as in \Gk{<o <r'omboc} \emph{ho rhombos} `the rhombus').  The
  smooth-breathing mark~(\Gk{>}) and the three tonal accents
%(\Gk{'{\ }}, \Gk{~}, \Gk{`{\ }}) 
(\Gk{'a}, \Gk{~a}, \Gk{`a}) 
can be ignored.  Especially in the dative case (the Turkish
\Tur{-e~hali}), some long vowels may be given the iota  
subscript (\Gk{a|}, \Gk{h|}, \Gk{w|}), representing what was once a
following iota (\Gk{ai}, \Gk{hi}, \Gk{wi}).

%\smaller
\relscale{.90}

% \bibliographystyle{amsplain}
% \bibliography{../../references}

\def\cprime{$'$} \def\cprime{$'$} \def\cprime{$'$}
\providecommand{\bysame}{\leavevmode\hbox to3em{\hrulefill}\thinspace}
\providecommand{\MR}{\relax\ifhmode\unskip\space\fi MR }
% \MRhref is called by the amsart/book/proc definition of \MR.
\providecommand{\MRhref}[2]{%
  \href{http://www.ams.org/mathscinet-getitem?mr=#1}{#2}
}
\providecommand{\href}[2]{#2}
\begin{thebibliography}{10}

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\end{thebibliography}


\end{document}
