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\title{Pascal's Treatise on the Arithmetical Triangle: Mathematical
Induction, Combinations, the Binomial Theorem and Fermat's Theorem\footnote{With
thanks to Joel Lucero-Bryan and Jerry Lodder.}}
\author{David Pengelley\footnote{Mathematical
Sciences; Dept. 3MB, Box 30001; New Mexico State University; Las
Cruces, NM 88003; {\tt davidp@nmsu.edu}.}}
\date{}
\maketitle
\section*{Introduction}
Blaise Pascal (1623--1662) was born in Clermont-Ferrand in central France.
Even as a teenager his father introduced him to meetings for mathematical
discussion in Paris run by Marin Mersenne, who served as a primary conduit for
transmitting mathematical ideas widely at that time, before the existence of
any research journals. He quickly became involved in the development of
projective geometry, the first in a sequence of highly creative mathematical
and scientific episodes in his life, punctuated by periods of religious
fervor. Around age twenty-one he spent several years developing a mechanical
addition and subtraction machine, in part to help his father in tax
computations as a local administrator. It was the first of its kind ever to be
marketed. Then for several years he was at the center of efforts to understand
vacuum, which led to an understanding of barometric pressure. In fact the
scientific unit of pressure is named the \emph{pascal}. He is also known for
Pascal's Law on the behavior of fluid pressure.
Around 1654 Pascal conducted his studies on the Arithmetical Triangle
(\textquotedblleft Pascal's Triangle\textquotedblright) and its relationship
to probabilities. His correspondence with Pierre de Fermat (1601--1665) in
that year marks the beginning of probability theory. Several years later,
Pascal refined his ideas on area problems via the method of indivisibles
already being developed by others, and solved various problems of areas,
volumes, centers of gravity, and lengths of curves. Later in the seventeenth
century, Gottfried Leibniz, one of the two inventors of the infinitesimal
calculus which supplanted the method of indivisibles, explicitly credited
Pascal's approach as stimulating his own ideas on the so-called characteristic
triangle of infinitesimals in his fundamental theorem of calculus. After only
two years of work on the calculus of indivisibles, Pascal fell gravely ill,
abandoned almost all intellectual work to devote himself to prayer and
charitable work, and died three years later at age thirty-nine. In addition to
his work in mathematics and physics, Pascal is prominent for his
\emph{Provincial Letters} defending Christianity, which gave rise to his
posthumously published \emph{Pens\'{e}es} (Thoughts) on religious philosophy
\cite{DPeb,DPGillispie}. Pascal was an extremely complex person, and one of the
outstanding scientists of the mid-seventeenth century, but we will never know
how much more he might have accomplished with more sustained efforts and a
longer life.
Pascal's \emph{Trait\'{e} du Triangle Arithmetique }(in English translation in
\cite[vol. 30]{DPPascal}) makes a systematic study of the numbers in his
triangle. They have simultaneous roles in mathematics as figurate
numbers\footnote{Figurate numbers count the number of equally spaced dots in
geometric figures. Especially important to Pascal were the numbers of dots in
equilateral triangles, triangular pyramids, and so forth in higher
dimensions.}, combination numbers, and binomial coefficients, and he
elaborates on all these. Given their multifaceted nature, it is no wonder that
these ubiquitous numbers had already been in use for over 500 years, in places
ranging from China to the Islamic world \cite{DPKatz}. Pascal, however, was
the first to connect binomial coefficients with combinatorial coefficients in
probability. In fact, a major motivation for Pascal was a question from the
beginnings of probability theory, about the equitable division of stakes in an
interrupted game of chance. The question had been posed to Pascal around 1652
by Antoine Gombaud, the Chevalier de M\'{e}r\'{e}, who wanted to improve his
chances at gambling: Suppose two players are playing a fair game, to continue
until one player wins a certain number of rounds, but the game is interrupted
before either player reaches the winning number. How should the stakes be
divided equitably, based on the number of rounds each player has won \cite[p.
431, 451ff]{DPKatz}? The solution requires the combinatorial properties
inherent in the numbers in the Arithmetical Triangle, as Pascal demonstrated
in his treatise, since they count the number of ways various occurrences can
combine to produce a given result. The Arithmetical Triangle overflows with
fascinating patterns and applications, and we will see several of these in
reading his treatise. We will study parts of Pascal's explanation of the
connections between the numbers in his triangle and combination numbers. The
reader is encouraged to read his entire treatise to see its many other aspects
and connections.
From Pascal's treatise we will also learn the principle of mathematical
induction. Pascal explains this in the specific context of proofs about the
numbers in the triangle. The basic idea of mathematical induction had occurred
in the mathematics of the Islamic world during the Middle Ages, and in
southern Europe in the fourteenth century \cite{DPKatz}, but Pascal's was
perhaps the first text to make a complete explicit statement and justification
of this extremely powerful method of proof in modern mathematics. Mathematical
induction is an astonishingly clever technique that allows us to prove claims
about infinitely many interlinked phenomena all at once, even when proving
just a single one of them in isolation would be very difficult! It will be a
challenging technique to master, but will provide tremendous power for future
mathematical work.
Learning about the connections of the Arithmetical Triangle to the binomial
theorem in algebra will also allow an application to proving a famous and
extremely important theorem on prime numbers discovered by Pascal's
correspondent Pierre de Fermat (1601--1665) of Toulouse, on congruence
remainders and prime numbers. This prepares one to understand the RSA
cryptosystem, which today is at the heart of securing electronic transactions.
We'll see how all these things are interconnected, and along the way we'll
also acquire important mathematical tools, like notations for general
indexing, summations, and products, and learn how to work with recurrence relations.
\section*{Part One: The Arithmetical Triangle and Mathematical Induction}
Let us begin reading Blaise Pascal's
\medskip
\begin{center}
\textsf{{\Large TREATISE ON THE ARITHMETICAL TRIANGLE}}
\medskip
\textsf{\textsc{Definitions}}
\end{center}
\begin{quote}
\textsf{I\ call \emph{arithmetical triangle} a figure constructed as follows:}
\textsf{From any point, G, I draw two lines perpendicular to each other, GV,
G}$\zeta$\textsf{ in each of which I\ take as many equal and contiguous parts
as I please, beginning with G, which I number 1, 2, 3, 4, etc., and these
numbers are the \emph{exponents} of the sections of the lines.}
\textsf{Next I connect the points of the first section in each of the two
lines by another line, which is the base of the resulting triangle.}
\textsf{In the same way I connect the two points of the second section by
another line, making a second triangle of which it is the base.}
\end{quote}
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\begin{quote}
\textsf{And in this way connecting all the points of section with the same
exponent, I\ construct as many triangles and bases as there are exponents.}
\textsf{Through each of the points of section and parallel to the sides
I\ draw lines whose intersections make little squares which I call
\emph{cells}.}
\textsf{Cells between two parallels drawn from left to right are called
\emph{cells of the same parallel row}, as, for example, cells }$G$\textsf{,
}$\sigma,$\textsf{ }$\pi$\textsf{, etc., or }$\varphi,$\textsf{ }$\psi
,$\textsf{ }$\theta,$\textsf{ etc.}
\textsf{Those between two lines are drawn from top to bottom are called
\emph{cells of the same perpendicular row}, as, for example, cells }%
$G,\varphi,A,D,$\textsf{ etc., or }$\sigma,\psi,B,$\textsf{ etc.}
\textsf{Those cut diagonally by the same base are called \emph{cells of the
same base}, as, for example, }$D,B,\theta,\lambda,$\textsf{ or }$A,\psi,\pi
$\textsf{.}
\textsf{Cells of the same base equidistant from its extremities are called
\emph{reciprocals}, as, for example, }$E,R$\textsf{ and }$B,\theta,$\textsf{
because the parallel exponent of one is the same as the perpendicular exponent
of the other, as is apparent in the above example, where }$E$\textsf{ is in
the second perpendicular row and in the fourth parallel row and its
reciprocal, }$R,$\textsf{ is in the second parallel row and in the fourth
perpendicular row, reciprocally. It is very easy to demonstrate that cells
with exponents reciprocally the same are in the same base and are equidistant
from its extremities.}
\textsf{It is also very easy to demonstrate that the perpendicular exponent of
any cell when added to is parallel exponent exceeds by unity the exponent of
its base.}
\textsf{For example, cell }$F$\textsf{ is in the third perpendicular row and
in the fourth parallel row and in the sixth base, and the exponents of rows 3
and 4, added together, exceed by unity the exponent of base 6, a property
which follows from the fact that the two sides of the triangle have the same
number of parts; but this is understood rather than demonstrated.}
\textsf{Of the same kind is the observation that each base has one more cell
than the preceding base, and that each has as many cells as its exponent has
units; thus the second base, }$\varphi\sigma,$\textsf{ has two cells, the third,
}$A\psi\pi,$\textsf{ has three, etc.}
\textsf{Now the numbers assigned to each cell are found by the following
method:}
\textsf{The number of the first cell, which is at the right angle, is
arbitrary; but that number having been assigned, all the rest are determined,
and for this reason it is called the \emph{generator} of the triangle. Each of
the others is specified by a single rule as follows:}
\textsf{The number of each cell is equal to the sum of the numbers of the
perpendicular and parallel cells immediately preceding. Thus cell }%
$F,$\textsf{ that is, the number of cell }$F,$\textsf{ equals the sum of cell
}$C$\textsf{ and cell }$E,$\textsf{ and similarly with the rest.}
\textsf{Whence several consequences are drawn. The most important follow,
wherein I consider triangles generated by unity, but what is said of them will
hold for all others.}
\end{quote}
\bigskip
\begin{center}
\textsc{First Consequence}
\end{center}
\begin{quote}
\textsf{\emph{In every arithmetical triangle all the cells of the first
parallel row and of the first perpendicular row are the same as the generating
cell.} }
\textsf{For by definition each cell of the triangle is equal to the sum of the
immediately preceding perpendicular and parallel cells. But the cells of the
first parallel row have no preceding perpendicular cells, and those of the
first perpendicular row have no preceding parallel cells; therefore they are
all equal to each other and consequently to the generating number.}
\textsf{Thus }%
\begin{tabular}
[t]{ll}%
$\varphi=G+0,$ & \textsf{that is,} $\varphi=G,$\\
$A=\varphi+0,$ & \textsf{that is,} $\varphi,$\\
$\sigma=G+0,$ & $\pi=\sigma+0,$%
\end{tabular}
\textsf{And similarly of the rest.}
\end{quote}
\bigskip
\begin{center}
\textsc{Second Consequence}
\end{center}
\begin{quote}
\textsf{\emph{In every arithmetical triangle each cell is equal to the sum of
all the cells of the preceding parallel row from its own perpendicular row to
the first, inclusive.} }
\textsf{Let any cell, }$\omega,$\textsf{ be taken. I\ say that it is equal to
}$R+\theta+\psi+\varphi,$\textsf{ which are the cells of the next higher
parallel row from the perpendicular row of }$\omega$\textsf{ to the first
perpendicular row.}
\textsf{This is evident if we simply consider a cell as the sum of its
component cells. }
\end{quote}
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$%
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[c]{@{}llllllllll}%
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& \\
& & & \multicolumn{3}{l}{$\underbrace{%
%TCIMACRO{\TeXButton{TeX field}{\hphantom{\theta+B}}}%
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& & & $\theta$ & ${}+{}$ & $B$ & & & & \\
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%TCIMACRO{\TeXButton{TeX field}{\hphantom{\psi+A}}}%
%BeginExpansion
\hphantom{\psi+A}%
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& & & & & & & & $\varphi$, &
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\ \ \ \ $
\noindent\textsf{for }$A$\textsf{ and }$\varphi$\textsf{ are equal to each
other by the preceding consequence.}
\textsf{Therefore }$\omega=R+\theta+\psi+\varphi$\textsf{.}
\bigskip
\end{quote}
\begin{center}
\textsc{Third Consequence}
\end{center}
\begin{quote}
\textsf{\emph{In every arithmetical triangle each cell is equal to the sum of
all the cells of the preceding perpendicular row from its own parallel row to
the first, inclusive.} }
\textsf{Let any cell, }$C,$\textsf{ be taken. I say that it is equal to
}$B+\psi+\sigma,$\textsf{ which are the cells of the preceding perpendicular
row from the parallel row of cell }$C$\textsf{ to the first parallel row.}
\textsf{This is also apparent, as above, simply by the interpretation of
cells.}%
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[c]{llllllll}%
\textsf{For} $C{}={}$ & $B$ & ${}+{}$ & $\theta$ & & & & \\
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\hphantom{\pi\sigma\pi}%
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& & & & & & $\sigma,$ &
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\noindent\textsf{for }$\pi=\sigma$\textsf{ by the first consequence.}
\textsf{Therefore }$C=B+\psi+\sigma.$
\end{quote}
\bigskip
\begin{center}
\textsc{Fourth Consequence}
\end{center}
\begin{quote}
\textsf{\emph{In every arithmetical triangle each cell exceeds by unity the
sum of all the cells within its parallel and perpendicular rows, exclusive.} }
\textsf{Let any cell, }$\xi,$\textsf{ be taken. I say that }$\xi
-G=R+\theta+\psi+\varphi+\lambda+\pi+\sigma+G,$\textsf{ which are all the
numbers between row }$\xi\omega CBA$\textsf{ and row }$\xi S\mu$\textsf{
exclusive.}
\textsf{This is also apparent by interpretation.}
$%
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[c]{llllllllllllllllll}%
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& & & & & & & \\
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& \\
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\hphantom{\sigma+ \psi+ B}%
%EndExpansion
}$} & & & & & & \\
& & & & & & & $\sigma$ & ${}+{}$ & $\psi$ & ${}+{}$ & $B$ & & & & &
& \\
& & & & & & & \multicolumn{1}{c}{} & & & &
\multicolumn{5}{c}{$\underbrace{%
%TCIMACRO{\TeXButton{TeX field}{\hphantom{G+\varphi+A}}}%
%BeginExpansion
\hphantom{G+\varphi+A}%
%EndExpansion
}$} & & \\
& & & & & & & & & & & $G$ & ${}+{}$ & $\varphi$ & ${}+{}$ & $A$ & &
\\
& & & & & & & & & & & & & & & \multicolumn{3}{c}{$\underbrace{%
%TCIMACRO{\TeXButton{TeX field}{\hphantom{AGG}}}%
%BeginExpansion
\hphantom{AGG}%
%EndExpansion
}$}\\
& & & & & & & & & & & & & & & & $G.$ &
\end{tabular}
\ \ \ \ $
\end{quote}
\renewcommand{\arraystretch}{1}
\begin{quote}
\textsf{Therefore }$\xi=\lambda+R+\pi+\theta+\sigma+\psi+G+\varphi+G.$
\textsf{N.B. I\ have written in the enunciation \emph{each cell exceeds by
unity} because the generator is unity. If it were some other number, the
enunciation should read: \emph{each cell exceeds by the generating number}.}
\end{quote}
\bigskip
\begin{enumerate}
\item[1.] Pascal's Triangle and its numbers
\begin{enumerate}
\item Let us use the notation $T_{i,j}$ to denote what Pascal calls the number
assigned to the cell in \emph{parallel row }$i$ (which we today call just
\emph{row }$i$) and \emph{perpendicular row }$j$ (which we today call
\emph{column} $j$). We call the $i$ and $j$ by the name \emph{indices }(plural
of \emph{index}) in our notation. Using this notation, explain exactly what
Pascal's rule is for determining all the numbers in all the cells. Be sure to
give full details. This should include explaining for exactly which values of
the indices he defines the numbers.
\item In terms of our notation $T_{i,j}$, explain his terms $\emph{exponent,}%
$\emph{ base, reciprocal, parallel row, perpendicular row, }and
\emph{generator. }
\item Rewrite Pascal's first two \textquotedblleft
Consequences\textquotedblright\ entirely in the $T_{i,j}$ notation.
\item Rewrite his proofs of these word for word in our notation also.
\item Do you find his proofs entirely satisfactory? Explain why or why not.
\item Improve on his proofs using our notation. In other words, make them
apply for arbitrary prescribed situations, not just the particular examples he
lays out.
\end{enumerate}
\item[2.] Modern mathematical notation
Read in a modern textbook about index, summation, and product notations, and
recurrence relations. Do some exercises.
\end{enumerate}
\bigskip
\begin{center}
\textsc{Fifth Consequence}
\end{center}
\begin{quote}
\textsf{\emph{In every arithmetical triangle each cell is equal to its
reciprocal}. }
\textsf{For in the second base, }$\varphi\sigma,$\textsf{ it is evident that the
two reciprocal cells, }$\varphi,\sigma,$\textsf{ are equal to each other and
to }$G.$
\textsf{In the third base, }$A,\psi,\pi,$\textsf{ it is also obvious that the
reciprocals, }$\pi,A,$\textsf{ are equal to each other and to }$G.$
\textsf{In the fourth base it is obvious that the extremes, }$D,\lambda
,$\textsf{ are again equal to each other and to }$G.$
\textsf{And those between, }$B,\theta,$\textsf{ are obviously equal since
}$B=A+\psi$\textsf{ and }$\theta=\pi+\psi.$\textsf{ But }$\pi+\psi=A+\psi
$\textsf{ by what has just been shown. Therefore, etc.}
\textsf{Similarly it can be shown for all the other bases that reciprocals are
equal, because the extremes are always equal to }$G$\textsf{ and the rest can
always be considered as the sum of cells in the preceding base which are
themselves reciprocals.}
\end{quote}
\bigskip
\begin{enumerate}
\item[3.] Symmetry in the triangle: first contact with mathematical induction
Write the Fifth Consequence using our index notation. Use index notation and
the ideas in Pascal's proof to prove the Consequence in full generality, not
just for the example he gives. Explain the conceptual ideas behind the general proof.
\item[4.] Mathematical induction: gaining more familiarity
\begin{enumerate}
\item Read in a modern textbook about mathematical induction.
\item Prove Pascal's First Consequence by mathematical induction. (Hint: for a
proof by mathematical induction, always first state very clearly exactly what
the $n$-th mathematical statement $P(n)$ says. Then state and prove the base
step. Then state the inductive step very clearly before you prove it.)
\item Write the general form of Pascal's Second Consequence, and give a
general proof using summation notation, but following his approach.
\item Now prove the Second Consequence by mathematical induction, i.e., a
different proof.
\item \textbf{Optional}: More patterns.
\begin{enumerate}
\item Write the Fourth Consequence using summation notation. Hint: You can
write it using a sum of sums. Try writing Pascal's proof in full generality,
using summation notation to help. If you don't complete it his way, explain
why it is difficult.
\item Prove the Fourth Consequence by mathematical induction.
\end{enumerate}
\end{enumerate}
\end{enumerate}
\bigskip
\begin{center}
\textsc{Seventh Consequence}
\end{center}
\begin{quote}
\textsf{\emph{In every arithmetical triangle the sum of the cells of each base
is double that of the preceding base.} }
\textsf{Let any base, }$DB\theta\lambda,$\textsf{ be taken. I say that the sum
of its cells is double the sum of the cells of the preceding base, }$A\psi\pi
$\textsf{.}%
\begin{tabular}
[c]{@{}lcc}%
\textsf{For the extremes}%
%TCIMACRO{\TeXButton{TeX field}{\dotfill}}%
%BeginExpansion
\dotfill
%EndExpansion
& $\underbrace{D,}$ & $\underbrace{\lambda,}$\\
\textsf{are equal to the extremes}%
%TCIMACRO{\TeXButton{TeX field}{\dotfill}}%
%BeginExpansion
\dotfill
%EndExpansion
& $A,$ & $\pi,$\\
\textsf{and each of the rest}%
%TCIMACRO{\TeXButton{TeX field}{\dotfill}}%
%BeginExpansion
\dotfill
%EndExpansion
& $\underbrace{B,}$ & $\underbrace{\theta,}$\\
\textsf{is equal to two cells of the other base} \ldots & $A+\psi,$ &
$\ \ \ \ \psi+\pi.\ \ \ \ $%
\end{tabular}
\textsf{Therefore }$D+\lambda+B+\theta=2A+2\psi+2\pi.$
\textsf{The same thing is demonstrated in the same way of all other bases.}
\end{quote}
\bigskip
\begin{center}
\textsc{Eighth Consequence}
\end{center}
\begin{quote}
\textsf{\emph{In every arithmetical triangle the sum of the cells of each base
is a number of the double progression beginning with unity whose exponent is
the same as that of the base.} }
\textsf{For the first base is unity.}
\textsf{The second is double the first; therefore it is 2.}
\textsf{The third is double the second; therefore it is 4.}
\textsf{And so on to infinity.}
\textsf{N.B. If the generator were not unity but some other number, such as 3,
the same thing would be true. But we should have to take not the numbers of
the double progression beginning with unity, that is, 1, 2, 4, 8, 16, etc.,
but those of the double progression beginning with the generator 3, that is,
3, 6, 12, 24, 48, etc.}
\end{quote}
\bigskip
\begin{enumerate}
\item[5.] Sums of bases in the triangle: a geometric progression
\begin{enumerate}
\item Use our index notation $T_{i,j}$ to explain exactly which are the
numbers in the $n$-th base.
\item In full generality, write the Seventh Consequence and its proof, using
our $T_{i,j}$ notation.
\item Write the statement of the Eighth Consequence in our notation, using
modern exponential notation to describe his double progression. Use summation
notation as needed, and introduce additional new notation if helpful. Then
prove the Eighth Consequence by mathematical induction.
\end{enumerate}
\end{enumerate}
\bigskip
The next consequence is the most important and famous in the whole treatise.
Pascal derives a formula for the ratio of consecutive numbers in a base. From
this he will obtain an elegant and efficient formula for all the numbers in
the triangle.
\begin{center}
\textsc{Twelfth Consequence}
\end{center}
\begin{quote}
\textsf{\emph{In every arithmetical triangle, of two contiguous cells in the
same base the upper is to the lower as the number of cells from the upper to
the top of the base is to the number of cells from the lower to the bottom of
the base, inclusive.} }
\textsf{Let any two contiguous cells of the same base, }$E,$\textsf{ }%
$C,$\textsf{ be taken. I say that}
\noindent%
\begin{tabular}
[c]{@{}ccccccc}%
$E$ & $\ :\ $ & $C$ & $\ ::\ $ & $2$ & $\ :\ $ & $3$\\
\textsf{the } & & \textsf{the } & & \multicolumn{1}{l}{\textsf{because there
are two}} & & \multicolumn{1}{l}{\textsf{because there are three}}\\
\textsf{lower} & & \textsf{upper} & & \multicolumn{1}{l}{\textsf{cells from
}$E$\textsf{ to the }} & & \multicolumn{1}{l}{\textsf{cells from }$C$\textsf{
to the top,}}\\
& & & & \multicolumn{1}{l}{\textsf{bottom, namely }$E,$\textsf{ }$H,$} & &
\multicolumn{1}{l}{\textsf{namely }$C,$\textsf{ }$R,$\textsf{ }$\mu.$}%
\end{tabular}
\textsf{Although this proposition has an infinity of cases, I shall
demonstrate it very briefly by supposing two lemmas:}
\textsf{The first, which is self-evident, that this proportion is found in the
second base, for it is perfectly obvious that }$\varphi:\sigma::1:1$\textsf{;}
\textsf{The second, that if this proportion is found in any base, it will
necessarily be found in the following base.}
\textsf{Whence it is apparent that it is necessarily in all the bases. For it
is in the second base by the first lemma; therefore by the second lemma it is
in the third base, therefore in the fourth, and to infinity.}
\textsf{It is only necessary therefore to demonstrate the second lemma as
follows: If this proportion is found in any base, as, for example, in the
fourth, }$D\lambda,$\textsf{ that is, if }$D:B::1:3$\textsf{, and }%
$B:\theta::2:2$\textsf{, and }$\theta:\lambda::3:1$\textsf{, etc., I say the
same proportion will be found in the following base, }$H\mu$\textsf{, and
that, for example, }$E:C::2:3$\textsf{.}
\textsf{For }$D:B::1:3$\textsf{, by hypothesis.}%
\begin{tabular}
[c]{@{}cccccccc}%
\textsf{Therefore}\ \ & $\underbrace{D+B}$ & $\ :\ $ & $B$ & $\ ::\ $ &
$\underbrace{1+3}$ & $\ :\ $ & $3$\\
& $E$ & $\ :\ $ & $B$ & $\ ::\ $ & $4$ & $\ :$\ & $3$%
\end{tabular}
\textsf{Similarly }$B:\theta::2:2,$\textsf{ by hypothesis}%
\begin{tabular}
[c]{@{}lccccccc}%
\textsf{Therefore}\ \ & $\underbrace{B+\theta}$ & $\ :\ $ & $B$ & $\ ::\ $ &
$\underbrace{2+2}$ & $:$ & $2$\\
\multicolumn{1}{c}{} & $C$ & $\ :\ $ & $B$ & $\ ::\ $ & $4$ & $:$ & $2$\\
\textsf{But}\ \ & $B$ & $\ :\ $ & $E$ & $\ ::\ $ & $3$ & $\ :\ $ & $4$%
\end{tabular}
\textsf{Therefore, by compounding the ratios, }$C:E::3:2.\hspace*{\fill}%
$\textsc{q.e.d}.
\textsf{The proof is the same for all other bases, since it requires only that
the proportion be found in the preceding base,and that each cell be equal to
the cell before it together with the cell above it, which is everywhere the
case.}
\end{quote}
\textsf{\bigskip}
\begin{enumerate}
\item[6.] Pascal's Twelfth Consequence: the key to our modern factorial formula
\begin{enumerate}
\item Rewrite Pascal's Twelfth Consequence as a generalized modern formula,
entirely in our $T_{i,j}$ terminology. Also verify its correctness in a couple
of examples taken from his table in the initial definitions section.
\item Adapt Pascal's proof by example of his Twelfth Consequence into modern
generalized form to prove the formula you obtained above. Use the principle of
mathematical induction to create your proof.
\end{enumerate}
\end{enumerate}
\textsf{\bigskip}
Now Pascal is ready to describe a formula for an arbitrary number in the triangle.
\begin{center}
\textsc{Problem}
\end{center}
\begin{quote}
\textsf{\emph{Given the perpendicular and parallel exponents of a cell, to
find its number without making use of the arithmetical triangle.} }
\textsf{Let it be proposed, for example, to find the number of cell }$\xi
$\textsf{ of the fifth perpendicular and of the third parallel row.}
\textsf{All the numbers which precede the perpendicular exponent, 5, having
been taken, namely 1, 2, 3, 4, let there be taken the same number of natural
numbers, beginning with the parallel exponent, 3, namely 3, 4, 5, 6.}
\textsf{Let the first numbers be multiplied together and let the product be
24. Let the second numbers be multiplied together and let the product be 360,
which, divided by the first product, 24, gives as quotient 15, which is the
number sought.}
\textsf{For }$\xi$\textsf{ is to the first cell of its base, }$V,$\textsf{ in
the ratio compounded of all the ratios of the cells between, that is to say,
}$\xi:V$
\end{quote}
\setlength{\tabcolsep}{6pt}
\begin{quote}%
\begin{tabular}
[c]{@{}lcccc}%
\textsf{in the ratio compounded of} & $\xi:\rho,$ & $\rho:K,$ & $K:Q,$ &
$Q:V$\\
\textsf{or by the twelfth consequence} & $3:4$ & $4:3$ & $5:2$ & $6:1$%
\end{tabular}
\textsf{Therefore} $\xi:V::3\cdot4\cdot5\cdot6:4\cdot3\cdot2\cdot1.$
\textsf{But }$V$\textsf{ is unity; therefore }$\xi$\textsf{ is the quotient of
the division of the product of }$3\cdot4\cdot5\cdot6$\textsf{ by the product
of }$4\cdot3\cdot2\cdot1.$
\textsf{N.B. If the generator were not unity, we should have had to multiply
the quotient by the generator.}
\end{quote}
\medskip
\begin{enumerate}
\item[7.] Pascal's formula for the numbers in the Arithmetical Triangle
\begin{enumerate}
\item Write down the general formula Pascal claims in solving his
\textquotedblleft Problem.\textquotedblright\ Your formula should read
$T_{i,j}=$ \textquotedblleft some formula in terms of $i$ and $j$%
.\textquotedblright\ Also write your formula entirely in terms of factorials.
\item Look at the reason Pascal indicates for his formula for a cell, and use
it to make a general proof for your formula above for an arbitrary $T_{i,j}$.
You may try to make your proof just like Pascal is indicating, or you may
prove it by mathematical induction.
\end{enumerate}
\end{enumerate}
\bigskip
\begin{center}
\textsf{{\large VARIOUS USES OF THE ARITHMETICAL TRIANGLE }}
\textsf{{\large WHOSE GENERATOR IS UNITY}}
\end{center}
\begin{quote}
\textsf{ \emph{Having given the proportions obtaining between the cells and
the rows of arithmetical triangles, I turn in the following treatises to
various uses of those triangles whose generator is unity. But I leave out many
more than I include; it is extraordinary how fertile in properties this
triangle is. Everyone can try his hand. I only call your attention here to the
fact that in everything that follows I am speaking exclusively of arithmetical
triangles whose generator is unity.} }
\end{quote}
\section*{Part Two: Combinations and the Arithmetical Triangle}
We continue reading Pascal's \emph{Treatise on the Arithmetical Triangle}:
\begin{center}
\textsf{{\large USE OF THE ARITHMETICAL TRIANGLE FOR COMBINATIONS}}
\end{center}
\begin{quote}
\textsf{The word \emph{combination} has been used in several different senses,
so that to avoid ambiguity I am obliged to say how I understand it.}
\textsf{When of many things we may choose a certain number, all the ways of
taking as many as we are allowed out of all those offered to our choice are
here called the \emph{different combinations}.}
\textsf{For example, if of four things expressed by the four letters, }%
$A,$\textsf{ }$B,$\textsf{ }$C,$\textsf{ }$D,$\textsf{ we are permitted to
take, say any two, all the different ways of taking two out of the four put
before us are called \emph{combinations}.}
\textsf{Thus we shall find by experience that there are six different ways of
choosing two out of four; for we can take }$A$\textsf{ and }$B,$\textsf{ or
}$A$\textsf{ and }$C,$\textsf{ or }$A$\textsf{ and }$D,$\textsf{ or }%
$B$\textsf{ and }$C,$\textsf{ or }$B$\textsf{ and }$D,$\textsf{ or }%
$C$\textsf{ and }$D.$
\textsf{I do not count }$A$\textsf{ and }$A$\textsf{ as one of the ways of
taking two; for they are not different things, they are only one thing
repeated.}
\textsf{Nor do I\ count }$A$\textsf{ and }$B$\textsf{ and }$B$\textsf{ and
}$A$\textsf{ as two different ways; for in both ways we take only the same two
things but in a different order, and I am not concerned with the order; so
that I could make myself understood at once by those who are used to
considering combinations, simply by saying that I speak only of combinations
made without changing the order.}
\textsf{We shall also find by experience that there are four ways of taking
three things out of four; for we can take }$ABC$\textsf{ or }$ABD$\textsf{ or
}$ACD$\textsf{ or }$BCD.$
\textsf{Finally we shall find that we can take four out of four in one way
only, }$ABCD.$
\textsf{I shall speak therefore in the following terms:}
\end{quote}
\begin{center}%
\begin{tabular}
[c]{l}%
\textsf{1 in 4 can be combined 4 times.}\\
\textsf{2 in 4 can be combined 6 times.}\\
\textsf{3 in 4 can be combined 4 times.}\\
\textsf{4 in 4 can be combined 1 time.}%
\end{tabular}
\end{center}
\begin{quote}
\textsf{Or:}
\end{quote}
\begin{center}%
\begin{tabular}
[c]{l}%
\textsf{the number of combinations of 1 in 4 is 4.}\\
\textsf{the number of combinations of 2 in 4 is 6.}\\
\textsf{the number of combinations of 3 in 4 is 4.}\\
\textsf{the number of combinations of 4 in 4 is 1.}%
\end{tabular}
\end{center}
\begin{quote}
\textsf{But the sum of all the combinations in general that can be made in 4
is 15, because the number of combinations of 1 in 4, of 2 in 4, of 3 in 4, of
4 in 4, when joined together, is 15.}
\textsf{After this explanation I shall give the following consequences in the
form of lemmas:}
\end{quote}
\bigskip
\begin{center}
\textsc{Lemma 1.}
\end{center}
\begin{quote}
\textsf{\emph{There are no combinations of a number in a smaller number; for
example,} }$4$\textsf{ \emph{cannot be combined in}~}$2$\textsf{. }
\medskip
...
\end{quote}
\medskip
\begin{center}
\textsc{Proposition 2}
\end{center}
\begin{quote}
\textsf{\emph{The number of any cell is equal to the number of combinations of
a number less by unity than its parallel exponent in a number less by unity
than the exponent of its base.}}
\textsf{Let any cell be taken, say }$F$\textsf{ in the fourth parallel row and
in the sixth base. I say that is is equal to the number of combinations of 3
in 5, less by unity than 4 and 6, for it is equal to the cells }%
$A+B+C.$\textsf{ Therefore by the preceding proposition, etc.}
\end{quote}
\bigskip
\begin{enumerate}
\item Combinations according to Pascal
\begin{enumerate}
\item Explain in your own words what Pascal says about how many combinations
there are for choosing two things out of four things.
\item Write Pascal's Proposition 2 using our $T_{i,j}$ notation for numbers in
the triangle. In other words, fill in a sentence saying \textquotedblleft%
$T_{i,j}$ is the number of combinations of choosing \underline{\qquad} things
from \underline{\qquad} things.\textquotedblright\ Pascal's justification for
his Proposition 2 is based on his Lemma 4 and Proposition 1, which are not
included in this project. However, the reader is encouraged to study and
understand them, to wit:
\item \textbf{Optional: }From Pascal's treatise \cite[vol. 30]{DPPascal},
rewrite his statements and explanations of his Lemma 4 and Proposition 1 in
your own words. State and prove Lemma 4 in the general case; that is, show
that the number of combinations of $k$ in $n$ is the sum of the combinations
of $k-1$ in $n-1$ and the combinations of $k$ in $n-1$. Also explain why
Proposition 2 follows from Proposition 1.
\end{enumerate}
\item Combinations and Pascal's recursion formula
\begin{enumerate}
\item The modern symbol $\binom{n}{r}$ means the number of ways
(\textquotedblleft combinations\textquotedblright) of choosing $r$ things from
amongst $n$ things. Explain how this is related to what we have been learning
about the Arithmetical Triangle from reading Pascal. In particular, explain
how the numbers $T_{i,j}$ are related to the numbers $\binom{n}{r}$. Do this
by writing an equation expressing $T_{i,j}$ in $\binom{n}{r}$ notation, and
also writing an equation expressing $\binom{n}{r}$ in $T_{i,j}$ notation. Now
use the formula we learned earlier, from Pascal's solution to his
\emph{Problem,}\footnote{\textsf{\emph{Given the perpendicular and parallel
exponents of a cell, to find its number without making use of the arithmetical
triangle.} }} to write a formula for the combination number $\binom{n}{r}$,
and manipulate it to express it entirely in terms of factorials.
\item Now read in a modern textbook about the multiplication rule for counting
possibilities, about permutations, and about combinations. \ Explain how a
combination is different from a permutation.
\item Read in a modern textbook about the algebra of combinations, Pascal's
recursion formula, and how the text presents Pascal's Triangle. How is it
different from Pascal's presentation?
\end{enumerate}
\end{enumerate}
\section*{Part Three: The Binomial Theorem and Fermat's Theorem}
Now we can put together all of what we have learned from Pascal to prove an
extremely important result in number theory, called \emph{Fermat's Little
Theorem}, which is at the heart of today's encryption methods in digital
communications. The ingredients will be the binomial theorem, proof by
mathematical induction as learned from Pascal, Pascal's formula for the
numbers in his triangle (solved in his \emph{Problem}), and uniqueness of
prime factorization.
\begin{enumerate}
\item The\textbf{ }binomial theorem and combinations
Read in a modern textbook about the binomial theorem. Write an explanation of
the proof of the binomial theorem using the idea of counting combinations.
\item Discovering Fermat's \textquotedblleft Little\textquotedblright%
\ Theorem: prime numbers and congruence remainders
\begin{enumerate}
\item Make a table of the remainders of $a^{n}$ upon division by $n$ for
positive integer values of both $a$ and $n$ ranging up to $14$. To do this you
should learn about congruence arithmetic, and figure out how to do these
calculations quickly and easily without a calculator.
\item Based on your table, make a conjecture of the form $a^{p}\equiv?$ (mod
$p$) for $p$ a prime number and $a$ any integer. This is called Fermat's
\textquotedblleft Little\textquotedblright\ Theorem; it is one of the most
important phenomena in number theory. Also make some other interesting
conjectures from patterns in your table, and try to prove them, perhaps using
the binomial theorem.
\item Write up the details of proving Fermat's Theorem by mathematical
induction on $a$, with $p$ held fixed. Use the binomial theorem, our knowledge
of Pascal's \textquotedblleft factorial\textquotedblright\ formula for
binomial coefficients, and the Fundamental Theorem of Arithmetic (uniqueness
of prime factorization) to analyze divisibility of the binomial coefficients
by a prime $p$.
\item \textbf{Optional: }Read about what Fermat was trying to do when he
discovered his Theorem \cite[p. 159ff]{DPLaubenbacher}.
Describe what you find in your own words.
\end{enumerate}
\item \textbf{Optional:} The RSA cryptosystem
Read and study the RSA cryptosystem and its applications to digital security,
including how it works, which follows from Fermat's Theorem. Write up the
details in your own words, with some example calculations.
\end{enumerate}
\section*{Notes to the Instructor}
The project's primary aim is for students of introductory discrete mathematics
to learn the concept of mathematical induction and its application
directly from reading the pioneering
work {\em Treatise on the Arithmetical Triangle} of Blaise Pascal
in the 1650s.
There are three project parts, covering
several standard topics: mathematical
induction, combinations, and the binomial theorem and Fermat's Theorem. The project
even ends with optional application to the RSA cryptosystem.
With the entirety of its applications the project can
constitute as much as 20\% of a semester course.
It works well to have students
complete and submit small pieces of the project as one goes along.
In some places the instructor should give students guidance on reading and
excercises from their textbook recommended by the project.
There are also some places, especially in the third part, where
the project expects either substantial independent learning from the student or
more substantial instructor guidance.
In his treatise, Pascal, after arranging the figurate numbers in a
defining triangular table,
notices several patterns in the table,
which he would like to claim continue indefinitely. Exhibiting unusual
rigor for his day, Pascal offers a condition for the persistence of a
pattern, stated verbally in his Twelfth Consequence, a condition
known today as mathematical induction. This is perhaps the first complete enunciation
and justification in the literature
of the logical principle of mathematical induction, all provided in the context of
a particular application. Moreover,
this Twelfth Consequence immediately results in the modern formula for the combination
numbers or binomial coefficients.
When Pascal's original writing
becomes a student's initial contact for
learning the idea of mathematical induction, a textbook is then merely a supplement.
We have found that students come to
grips with mathematical induction better by first seeing how Pascal eases into the idea
through the proof of several patterns in the triangle, and
then formalizes the principle and applies it further.
Have students first read and work quite a bit with Pascal's verbal description, and then
hold an instructor-moderated class discussion comparing this to the axiomatic formulation
of induction students can read in the textbook. We intentionally wait on having students
read the
textbook approach until they have become comfortable with Pascal's. In fact, many
students become and remain more comfortable with Pascal's more verbal way of handling
mathematical induction than with their textbooks. Students can become so comfortable with
Pascal's treatise that, on a final exam, many will voluntarily choose a question requiring
new analysis of a part of Pascal's treatise using mathematical induction that
they have never seen, over an analogous exercise from their modern textbook.
%An alternative but very fruitful start for the project is to explore the
%beginning of Part Three first, in which students develop experimental
%evidence, based on calculations in congruence arithmetic, which lead them
%to conjecture Fermat's theorem about prime numbers (among other patterns they notice).
%Then the instructor's promise is that by the end of the project,
%they will be able to prove their conjecture, one of the critical
%underpinnings of number theory. This approach has
%the strong advantage of immersing students authentically
%in the process of doing research in mathematics.
%Students will experience firsthand that \emph{experimentation }and \emph{conjecture}
%are the first key steps in the
%creation of new mathematics, followed by attempting to \emph{prove} one's
%conjectures, and then the possibility of \emph{generalization} and interaction
%with other areas for further research.
\begin{thebibliography}{99}
\bibitem{DPeb} \emph{Encyclop\ae dia Britannica}, Chicago, 1986.
\bibitem{DPGillispie} Gillispie, C. C., Holmes, F. L., (editors)
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\end{document}
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