43 files changed, 4471 insertions, 0 deletions
diff --git a/.gitattributes b/.gitattributes
new file mode 100644
index 0000000..6833f05
--- /dev/null
+++ b/.gitattributes
@@ -0,0 +1,3 @@
+* text=auto
+*.txt text
+*.md text
diff --git a/9934-pdf.pdf b/9934-pdf.pdf
new file mode 100644
index 0000000..6c12a1a
--- /dev/null
+++ b/9934-pdf.pdf
diff --git a/9934-pdf.zip b/9934-pdf.zip
new file mode 100644
index 0000000..a2c06df
--- /dev/null
+++ b/9934-pdf.zip
diff --git a/9934-t.zip b/9934-t.zip
new file mode 100644
index 0000000..e255e50
--- /dev/null
+++ b/9934-t.zip
diff --git a/9934-t/9934-t.tex b/9934-t/9934-t.tex
new file mode 100644
index 0000000..982129f
--- /dev/null
+++ b/9934-t/9934-t.tex
@@ -0,0 +1,4455 @@
+% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %
+%                                                                         %
+% The Project Gutenberg EBook of A Primer of Quaternions, by Arthur S. Hathaway
+%                                                                         %
+% This eBook is for the use of anyone anywhere in the United States and most
+% other parts of the world at no cost and with almost no restrictions     %
+% whatsoever.  You may copy it, give it away or re-use it under the terms of
+% the Project Gutenberg License included with this eBook or online at     %
+% www.gutenberg.org.  If you are not located in the United States, you'll have
+% to check the laws of the country where you are located before using this ebook.
+%                                                                         %
+%                                                                         %
+%                                                                         %
+% Title: A Primer of Quaternions                                          %
+%                                                                         %
+% Author: Arthur S. Hathaway                                              %
+%                                                                         %
+% Release Date: April 25, 2015 [EBook #9934]                              %
+%                                                                         %
+% Language: English                                                       %
+%                                                                         %
+% Character set encoding: ASCII                                           %
+%                                                                         %
+% *** START OF THIS PROJECT GUTENBERG EBOOK A PRIMER OF QUATERNIONS ***   %
+%                                                                         %
+% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %
+
+\def\ebook{9934}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%                                                                  %%
+%% Packages and substitutions:                                      %%
+%%                                                                  %%
+%% book:  Required.                                                 %%
+%%                                                                  %%
+%% amsmath:  AMS mathematics enhancements. Required.                %%
+%% amssymb:  AMS extra symbols. Required.                           %%
+%% amsthm:  AMS theorem environments. Required.                     %%
+%% makeidx:  Indexing. Required.                                    %%
+%%                                                                  %%
+%% alltt:    Fixed-width font environment. Required.                %%
+%%                                                                  %%
+%% graphicx: Graphics. Required.                                    %%
+%%                                                                  %%
+%% Producer's Comments:                                             %%
+%%                                                                  %%
+%%   This ebook was originally produced in 2003; boilerplate for    %%
+%%   auto-compiling at Project Gutenberg added April 2015.          %%
+%%                                                                  %%
+%% PDF pages: 85                                                    %%
+%% PDF page size: US Letter (8.5 x 11in)                            %%
+%%                                                                  %%
+%% Images: 34 png diagrams                                          %%
+%%                                                                  %%
+%% Summary of log file:                                             %%
+%% * Three overfull hboxes (10.75pt too wide).                      %%
+%% * Two underfull hboxes.                                          %%
+%%                                                                  %%
+%% Command block:                                                   %%
+%%                                                                  %%
+%%     pdflatex x2                                                  %%
+%%                                                                  %%
+%%                                                                  %%
+%% April 2015: pglatex.                                             %%
+%%   Compile this project with:                                     %%
+%%   pdflatex 9934-t.tex ..... TWO times                            %%
+%%                                                                  %%
+%%   pdfTeX, Version 3.1415926-2.5-1.40.14 (TeX Live 2013/Debian)   %%
+%%                                                                  %%
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\listfiles
+\documentclass[oneside,12pt]{book}[2005/09/16]
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%% PACKAGES %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\usepackage[leqno]{amsmath}[2000/07/18] %% Displayed equations
+\usepackage{amssymb}[2009/06/22]
+\usepackage{amsthm}[2009/07/02]
+
+\usepackage{alltt}[1997/06/16]   %% boilerplate, credits, license
+
+\usepackage{graphicx}[1999/02/16]
+
+\providecommand{\ebook}{00000}    % Overridden during white-washing
+
+%%%% Fixed-width environment to format PG boilerplate %%%%
+\newenvironment{PGtext}{%
+\begin{alltt}
+\fontsize{9.2}{10.5}\ttfamily\selectfont}%
+{\end{alltt}}
+
+%%%% Global style parameters %%%%
+% Loosen horizontal spacing
+\allowdisplaybreaks[1]
+\setlength{\emergencystretch}{1.5em}
+\newcommand{\loosen}{\spaceskip 0.5em plus 1em minus 0.25em}
+
+%%%% Major document divisions %%%%
+\newcommand{\PGBoilerPlate}{%
+  \frontmatter
+  \pagenumbering{Alph}
+  \pagestyle{empty}
+}
+\newcommand{\PGLicense}{%
+  \backmatter
+  \pagenumbering{Roman}
+}
+
+\newcommand{\TranscribersNote}{%
+  \begin{minipage}{0.85\textwidth}
+    \small
+    \subsection*{\centering\normalfont\scshape\normalsize Transcriber's Note}
+      Minor typographical corrections and presentational changes have been
+  made without comment. The \LaTeX\ source file may be downloaded from
+  \begin{center}
+    \texttt{www.gutenberg.org/ebooks/\ebook}.
+  \end{center}
+  \end{minipage}
+}
+
+\newcommand{\MainMatter}
+{
+  \mainmatter
+  \pagenumbering{arabic}
+  \pagestyle{plain}
+}
+
+%%%%%%%%%%%%%%%%%%%%%%%% START OF DOCUMENT %%%%%%%%%%%%%%%%%%%%%%%%%%
+\begin{document}
+%%%% PG BOILERPLATE %%%%
+\PGBoilerPlate
+\begin{center}
+\begin{minipage}{\textwidth}
+\small
+\begin{PGtext}
+The Project Gutenberg EBook of A Primer of Quaternions, by Arthur S. Hathaway
+
+This eBook is for the use of anyone anywhere in the United States and most
+other parts of the world at no cost and with almost no restrictions
+whatsoever.  You may copy it, give it away or re-use it under the terms of
+the Project Gutenberg License included with this eBook or online at
+www.gutenberg.org.  If you are not located in the United States, you'll have
+to check the laws of the country where you are located before using this ebook.
+
+
+
+Title: A Primer of Quaternions
+
+Author: Arthur S. Hathaway
+
+Release Date: April 25, 2015 [EBook #9934]
+
+Language: English
+
+Character set encoding: ASCII
+
+*** START OF THIS PROJECT GUTENBERG EBOOK A PRIMER OF QUATERNIONS ***
+\end{PGtext}
+\end{minipage}
+\end{center}
+\clearpage
+
+%%%% Credits and transcriber's note %%%%
+\begin{center}
+\begin{minipage}{\textwidth}
+\begin{PGtext}
+Produced by Cornell University, Joshua Hutchinson, John
+Hagerson, and the Online Distributed Proofreading Team
+\end{PGtext}
+\end{minipage}
+\vfill
+\TranscribersNote
+\end{center}
+%%%%%%%%%%%%%%%%%%%%%%%%%%% FRONT MATTER %%%%%%%%%%%%%%%%%%%%%%%%%%
+\cleardoublepage
+
+\iffalse %%%%% Start of original header %%%%
+\documentclass[oneside]{book}
+\usepackage[leqno]{amsmath}
+\usepackage{amssymb,amsthm,graphicx}
+\begin{document}
+\frontmatter
+
+\thispagestyle{empty}
+\small
+\begin{verbatim}
+
+\end{verbatim}
+\normalsize
+\fi
+%%%%% End of original header %%%%
+\newpage
+
+\begin{center}
+\bigskip \huge
+A PRIMER OF QUATERNIONS
+
+\bigskip\bigskip
+\footnotesize BY
+
+\bigskip
+\large ARTHUR S. HATHAWAY
+
+\bigskip
+\footnotesize PROFESSOR OF MATHEMATICS IN THE ROSE POLYTECHNIC \\
+INSTITUTE, TERRE HAUTE, IND.
+
+\bigskip\bigskip
+\normalsize 1896
+\end{center}
+
+\newpage
+
+\pagenumbering{roman}
+\pagestyle{plain}
+
+\section*{Preface}
+
+The Theory of Quaternions is due to Sir William Rowan Hamilton,
+Royal Astronomer of Ireland, who presented his first paper on the
+subject to the Royal Irish Academy in 1843. His Lectures on
+Quaternions were published in 1853, and his Elements, in 1866,
+shortly after his death. The Elements of Quaternions by Tait is
+the accepted text-book for advanced students.
+
+The following development of the theory is prepared for average
+students with a thorough knowledge of the elements of algebra and
+geometry, and is believed to be a simple and elementary treatment
+founded directly upon the fundamental ideas of the subject. This
+theory is applied in the more advanced examples to develop the
+principal formulas of trigonometry and solid analytical geometry,
+and the general properties and classification of surfaces of
+second order.
+
+In the endeavour to bring out the \textit{number} idea of
+Quaternions, and at the same time retain the established
+nomenclature of the analysis, I have found it necessary to abandon
+the term ``\textit{vector}'' for a directed length. I adopt
+instead Clifford's suggestive name of ``\textit{step},'' leaving
+to ``\textit{vector}'' the sole meaning of ``\textit{right
+quaternion}.'' This brings out clearly the relations of this
+number and line, and emphasizes the fact that Quaternions is a
+natural extension of our fundamental ideas of number, that is
+subject to ordinary principles of geometric representation, rather
+than an artificial species of geometrical algebra.
+
+The physical conceptions and the breadth of idea that the subject
+of Quaternions will develop are, of themselves, sufficient reward
+for its study. At the same time, the power, directness, and
+simplicity of its analysis cannot fail to prove useful in all
+physical and geometrical investigations, to those who have
+thoroughly grasped its principles.
+
+On account of the universal use of analytical geometry, many
+examples have been given to show that Quaternions in its
+semi-cartesian form is a direct development of that subject. In
+fact, the present work is the outcome of lectures that I have
+given to my classes for a number of years past as the equivalent
+of the usual instruction in the analytical geometry of space. The
+main features of this primer were therefore developed in the
+laboratory of the class-room, and I desire to express my thanks to
+the members of my classes, wherever they may be, for the interest
+that they have shown, and the readiness with which they have
+expressed their difficulties, as it has been a constant source of
+encouragement and assistance in my work.
+
+I am also otherwise indebted to two of my students,---to Mr.\
+H.~B.\ Stilz for the accurate construction of the diagrams, and to
+Mr.\ G.\ Willius for the plan (upon the cover) of the plagiograph
+or mechanical quaternion multiplier which was made by him while
+taking this subject. The theory of this instrument is contained in
+the step proportions that are given with the diagram.\footnote{See
+Example 19, Chapter I.}
+
+\begin{flushright}
+ARTHUR S.\ HATHAWAY.
+\end{flushright}
+
+\tableofcontents
+
+\MainMatter
+\chapter{Steps}
+
+\addcontentsline{toc}{section}{Definitions and Theorems}
+
+\begin{enumerate}
+
+\item \textsc{Definition.} \textit{A step is a given length
+measured in a given direction.}
+
+\textit{E.g., 3 feet east, 3 feet north, 3 feet up, 3 feet
+north-east, 3 feet north-east-up,} are steps.
+
+\item \textsc{Definition.} \textit{Two steps are equal when, and
+only when, they have the same lengths and the same directions.}
+
+\textit{E.g., 3 feet east}, and \textit{3 feet north}, are not
+equal steps, because they differ in direction, although their
+lengths are the same; and \textit{3 feet east, 5 feet east}, are
+not equal steps, because their lengths differ, although their
+directions are the same; but all steps of \textit{3 feet east} are
+equal steps, whatever the points of departure.
+
+\item We shall use bold-faced $\mathbf{AB}$ to denote the step
+whose length is $AB$, and whose direction is from $A$ towards $B$.
+
+\begin{center}
+\includegraphics[width=80mm]{images/Image1.png}
+\end{center}
+
+Two steps $\mathbf{AB}$, $\mathbf{CD}$, are obviously equal when,
+and only when, $ABDC$ is a parallelogram.
+
+\item \textsc{Definition.} \textit{If several steps be taken in
+succession, so that each step begins where the preceding step
+ends, the step from the beginning of the first to the end of the
+last step is the sum of those steps.}
+
+\begin{center}
+\includegraphics[width=80mm]{images/Image2.png}
+\end{center}
+
+\textit{E.g., 3 feet east + 3 feet north = $3\sqrt{2}$ feet
+north-east = 3 feet north + 3 feet east}. Also $\mathbf{AB + BC =
+AC}$, whatever points $A$, $B$, $C$, may be. Observe that this
+equality between \textit{steps} is not a length equality, and
+therefore does not contradict the inequality $AB + BC > AC$, just
+as 5 \textit{dollars credit} + 2 \textit{dollars debit} = 3
+\textit{dollars credit} does not contradict the inequality
+\textit{5 dollars + 2 dollars $>$ 3 dollars}.
+
+\begin{center}
+\includegraphics[width=80mm]{images/Image3.png}
+\end{center}
+
+\item \textit{If equal steps be added to equal steps, the sums are
+equal steps.}
+
+Thus if $\mathbf{AB = A'B'}$, and $\mathbf{BC=B'C'}$, then
+$\mathbf{AC = A'C'}$, since the triangles $ABC$, $A'B'C'$ must be
+equal triangles with the corresponding sides in the same
+direction.
+
+\item \textit{A sum of steps is commutative} (\textit{i.e.}, the
+components of the sum may be added in any order without changing
+the value of the sum).
+
+\begin{center}
+\includegraphics[width=80mm]{images/Image4.png}
+\end{center}
+
+For, in the sum $\mathbf{AB + BC + CD + DE + \cdots}$, let
+$\mathbf{BC' = CD}$; then since $BCDC'$ is a parallelogram,
+therefore $\mathbf{C'D = BC}$, and the sum with $\mathbf{BC}$,
+$\mathbf{CD}$, interchanged is $\mathbf{AB + BC' + C'D + DE +
+\cdots}$, which has the same value as before. By such
+interchanges, the sum can be brought to any order of adding.
+
+\item \textit{A sum of steps is associative} (\textit{i.e}., any
+number of consecutive terms of the sum may be replaced by their
+sum without changing the value of the whole sum).
+
+For, in the sum $\mathbf{AB + BC + CD + DE + \cdots}$, let
+$\mathbf{BC}$, $\mathbf{CD}$, be replaced by their sum
+$\mathbf{BD}$; then the new sum is $\mathbf{AB + BD + DE +
+\cdots}$, whose value is the same as before; and similarly for
+other consecutive terms.
+
+\item \textit{The product of a step by a positive number is that
+step lengthened by the multiplier without change of direction.}
+
+\textit{E.g.}, $\mathbf{2AB = AB + AB}$, which is $\mathbf{AB}$
+doubled in length without change of direction; similarly $\frac
+{1}{2}\mathbf{AB} = $(step that doubled gives $\mathbf{AB}$) $=$
+($\mathbf{AB}$ halved in length without change of direction). In
+general, $m\mathbf{AB} = m$ lengths $AB$ measured in the direction
+$\mathbf{AB}$; $\frac{1}{n} \mathbf{AB} = \frac {1}{n}$th of
+length $AB$ measured in the direction $\mathbf{AB}$; etc.
+
+\item \textit{The negative of a step is that step reversed in
+direction without change of length}.
+
+For the negative of a quantity is that quantity which added to it
+gives zero; and since $\mathbf{AB + BA = AA} = 0$, therefore
+$\mathbf{BA}$ is the negative of $\mathbf{AB}$, or
+$\mathbf{BA=-AB}$.
+
+\begin{itemize}
+\item \textsc{Cor.\ 1.} \textit{The product of a step by a negative
+number is that step lengthened by the number and reversed in
+direction.}
+
+For $-n\mathbf{AB}$ is the negative of $n\mathbf{AB}$.
+
+\item \textsc{Cor.\ 2.} \textit{A step is subtracted by reversing
+its direction and adding it.}
+
+For the result of subtracting is the result of adding the negative
+quantity. \textit{E.g.}, $\mathbf{AB-CB = AB+BC = AC}$.
+\end{itemize}
+
+\item \textit{A sum of steps is multiplied by a given number by
+multiplying the components of the sum by the number and adding the
+products.}
+
+\begin{center}
+\includegraphics[width=80mm]{images/Image5.png}
+\end{center}
+
+Let $n \cdot \mathbf{AB = A'B'}, n \cdot \mathbf{BC=BC'}$; then
+$ABC, A'B'C'$ are similar triangles, since the sides about $B$,
+$B'$ are proportional, and in the same or opposite directions,
+according as $n$ is positive or negative; therefore $AC$, $A'C'$
+are in the same or opposite directions and in the same ratio;
+\textit{i.e.}, $n\mathbf{AC = A'C'}$, which is the same as
+$n(\mathbf{AB+BC}) = n\mathbf{AB}+n\mathbf{BC}$.
+
+This result may also be stated in the form: \textit{a multiplier
+is distributive over a sum}.
+
+\item \textit{Any step may be resolved into a multiple of a given
+step parallel to it; and into a sum of multiples of two given
+steps in the same plane with it that are not parallel; and into a
+sum of multiples of three given steps that are not parallel to one
+plane.}
+
+\begin{center}
+\includegraphics[width=80mm]{images/Image6.png}
+\end{center}
+
+\item It is obvious that if the sum of two finite steps is zero,
+then the two steps must be parallel; in fact, if one step is
+$\mathbf{AB}$, then the other must be equal to $\mathbf{BA}$.
+Also, if the sum of three finite steps is zero, then the three
+steps must be parallel to one plane; in fact, if the first is
+$\mathbf{AB}$, and the second is $\mathbf{BC}$, then the third
+must be equal to $\mathbf{CA}$. Hence, \textit{if a sum of steps
+on two lines that are not parallel (or on three lines that are not
+parallel to one plane) is zero, then the sum of the steps on each
+line is zero,} since, as just shown, the sum of the steps on each
+line cannot be finite and satisfy the condition that their sum is
+zero. We thus see that an equation between steps of one plane can
+be separated into two equations by resolving each step parallel to
+two intersecting lines of that plane, and that an equation between
+steps in space can be separated into three equations by resolving
+each step parallel to three lines of space that are not parallel
+to one plane. We proceed to give some applications of this and
+other principles of step analysis in locating a point or a locus
+of points with respect to given data (Arts.\ 13-20).
+
+\addcontentsline{toc}{section}{Centre of Gravity}
+\section*{Centre of Gravity}
+
+\item \textit{The point $P$ that satisfies the condition
+$l\mathbf{AP} + m\mathbf{BP} = 0$ lies upon the line $AB$ and
+divides $AB$ in the inverse ratio of $l:m$ (i.e., $P$ is the
+centre of gravity of a mass $l$ at $A$ and a mass $m$ at $B$).}
+
+The equation gives $l\mathbf{AP} = m\mathbf{PB}$; hence:
+
+$\mathbf{AP}$, $\mathbf{PB}$ are parallel; $P$ lies on the line
+$AB$; and $\mathbf{AP}:\mathbf{PB} = m:l = $ \textit{inverse of}
+$l:m$.
+
+If $l:m$ is positive, then $\mathbf{AP}$, $\mathbf{PB}$ are in the
+same direction, so that $P$ must lie between $A$ and $B$; and if
+$l:m$ is negative, then $P$ must lie on the line $AB$ produced. If
+$l = m$, then $P$ is the middle point of $AB$; if $l=-m$, then
+there is no finite point $P$ that satisfies the condition, but $P$
+satisfies it more nearly, the farther away it lies upon $AB$
+produced, and this fact is expressed by saying that \textit{``$P$
+is the point at infinity on the line $AB$.''}
+
+\item By substituting $\mathbf{AO + OP}$ for $\mathbf{AP}$ and
+$\mathbf{BO + OP}$ for $\mathbf{BP}$ in $l\mathbf{AP} +
+m\mathbf{BP} = 0$, and transposing known steps to the second
+member, we find the point $P$ with respect to any given origin
+$O$, viz.,
+
+\begin{enumerate}
+\item $(l+m)\mathbf{OP} = l\mathbf{OA} + m\mathbf{OB}$, where $P$
+divides $AB$ inversely as $l:m$.
+\end{enumerate}
+
+\begin{center}
+\includegraphics[width=80mm]{images/Image7.png}
+\end{center}
+
+\begin{itemize}
+\item \textsc{Cor.} \textit{If $\mathbf{OC} = l\mathbf{OA} +
+m\mathbf{OB}$, then $OC$, produced if necessary, cuts $AB$ in the
+inverse ratio of $l:m$, and $\mathbf{OC}$ is $(l + m)$ times the
+step from $O$ to the point of division.}
+
+For, if $P$ divide $AB$ inversely as $l:m$, then by \textit{(a)}
+and the given equation, we have
+\begin{equation*}
+\mathbf{OC}=(l+m)\mathbf{OP}.
+\end{equation*}
+\end{itemize}
+
+\item \textit{The point $P$ that satisfies the condition
+$l\mathbf{AP} + m\mathbf{BP} + n\mathbf{CP} = 0$ lies in the plane
+of the triangle $ABC$; $AP$ (produced) cuts $BC$ at a point $D$
+that divides $BC$ inversely as $m:n$, and $P$ divides $AD$
+inversely as $l:m+n$ (i.e., $P$ is the center of gravity of a mass
+$l$ at $A$, a mass $m$ at $B$, and a mass $n$ at $C$). Also the
+triangles $PBC$, $PCA$, $PAB$, $ABC$, are proportional to $l$,
+$m$, $n$, $l+m+n$.}
+
+\begin{center}
+\includegraphics[width=80mm]{images/Image8.png}
+\end{center}
+
+The three steps $l\mathbf{AP}$, $m\mathbf{BP}$, $n\mathbf{CP}$
+must be parallel to one plane, since their sum is zero, and hence
+$P$ must lie in the plane of $ABC$. Since $\mathbf{BP=BD + DP}$,
+$\mathbf{CP = CD + DP}$, the equation becomes, by making these
+substitutions, $l\mathbf{AP} + (m+n)\mathbf{DP} + m\mathbf{BD} +
+n\mathbf{CD} = 0$. This is an equation between steps on the two
+intersecting lines, $AD$, $BC$, and hence the resultant step along
+each line is zero; i.e., $m\mathbf{BD} + n\mathbf{CD} = 0$ (or $D$
+divides $BC$ inversely as $m:n$), and
+
+\begin{equation}
+\tag{a} l\mathbf{AP} + (m+n)\mathbf{DP} = 0
+\end{equation}
+
+(or $P$ divides $AD$ inversely as $l:m+n$). Also, we have, by
+adding $l\mathbf{PD} + l\mathbf{DP} = 0$ to (\textit{a}),
+\begin{equation*}
+l\mathbf{AD}+(l+m+n)\mathbf{DP} = 0.
+\end{equation*}
+Hence
+\begin{equation*}
+l:l+m+n = \mathbf{PD:AD} = PBC:ABC,
+\end{equation*}
+since the triangles $PBC$, $ABC$ have a common base $BC$, (We must
+take the ratio of these triangles as positive or negative
+according as the vertices $P$, $A$ lie on the same or opposite
+sides of the base $BC$, since the ratio $\mathbf{PD:AD}$ is
+positive or negative under those circumstances.) Similarly,
+\begin{gather*}
+PCA:ABC = m:l+m+n,
+\intertext{and}
+PAB:ABC = n:l+m+n.
+\intertext{Hence, we have,}
+PBC:PCA:PAB:ABC = l:m:n:l+m+n.
+\end{gather*}
+
+\item By introducing in $l\mathbf{AP} + m\mathbf{BP} +
+n\mathbf{CP} = 0$ an origin $O$, as in Art.\ 14, we find
+
+(\textit{a}) $(l+m+n) \mathbf{OP} = l\mathbf{OA} + m\mathbf{OB} +
+n\mathbf{OC}$, \textit{where $P$ divides $ABC$ in the ratio $l:m:n$.}
+
+\small \textsc{Note.} As an exercise, extend this formula for the center
+of gravity $P$, of masses $l$, $m$, $n$, at $A$, $B$, $C$, to four
+or more masses. \normalsize
+
+\addcontentsline{toc}{section}{Curve Tracing, Tangents}
+\section*{Curve Tracing. Tangents.}
+
+\item \textit{To draw the locus of a point $P$ that varies
+according to the law $\mathbf{OP} = t\mathbf{OA} +
+\frac{1}{2}t^2\mathbf{OB}$, where $t$ is a variable number.}
+(\textit{E.g.}, $t=$ number of seconds from a given epoch.)
+
+Take $t = -2$, and $P$ is at $D'$, where
+\begin{equation*}
+\mathbf{OD'} = -2\mathbf{OA} + 2\mathbf{OB}.
+\end{equation*}
+
+Take $t = -1$, and $P$ is at $C'$, where
+\begin{equation*}
+\mathbf{OC'} = -\mathbf{OA} + \frac{1}{2} \mathbf{OB}
+\end{equation*}
+
+Take $t = 0$, and $P$ is at $O$. Take $t = 1$, and $P$ is at $C$,
+where $\mathbf{OC} = \mathbf{OA} + \frac{1}{2}\mathbf{OB}$. Take
+$t = 2$, and $P$ is at $D$, where $\mathbf{OD} = 2\mathbf{OA} +
+2\mathbf{OB}$. It is thus seen that when $t$ varies from -2 to 2,
+then $P$ traces a curve $D'C'OCD$. To draw the curve as accurately
+as possible, we find the tangents at the points already found. The
+method that we employ is perfectly general and applicable to any
+locus.
+
+\begin{center}
+\includegraphics[width=80mm]{images/Image9.png}
+\end{center}
+
+(\textit{a}) To find the direction of the tangent to the locus at
+the point $P$ corresponding to any value of $t$.
+
+Let $P$, $Q$ be two points of the locus that correspond to the
+values $t$, $t + h$ of the variable number. We have
+\begin{gather*}
+\mathbf{OP} = t\mathbf{OA} + \frac{1}{2} t^2 \mathbf{OB},\\
+\mathbf{OQ} = (t + h) \mathbf{OA} + \frac{1}{2} (t + h)^2
+\mathbf{OB},
+\intertext{and therefore}
+\mathbf{PQ} = \mathbf{OQ-OP} = h\left[\mathbf{OA} + (t + \frac{1}{2}
+h)\mathbf{OB}\right].
+\end{gather*}
+
+Hence (dropping the factor $h$) we see that $\mathbf{OA} + (t +
+\frac{1}{2} h) \mathbf{OB}$ is always \textit{parallel} to the
+chord $PQ$. Make $h$ approach 0, and then $Q$ approaches $P$, and
+the (indefinitely extended) chord $PQ$ approaches coincidence with
+the tangent at $P$. Hence making $h = 0$, in the step that is
+parallel to the chord, we find that $\mathbf{OA} + t\mathbf{OB}$
+is parallel to the tangent at $P$.
+
+Apply this result to the special positions of $P$ already found,
+and we have: $\mathbf{D'A'} = \mathbf{OA - 2OB}=$
+tangent at $D'$; $\mathbf{C'S = OA-OB} =$ tangent at
+$C'$; $\mathbf{OA = OA + 0 \cdot OB =}$ tangent at $O$;
+$\mathbf{SO = OA + OB =}$ tangent at $C$; $\mathbf{AD = OA +2 OB
+=}$ tangent at $D$.
+
+This is the curve described by a heavy particle thrown from $O$
+with velocity represented by $\mathbf{OA}$ on the same scale in
+which $\mathbf{OB}$ represents an acceleration of $32$
+\textit{feet per second per second downwards}. For, after $t$
+seconds the particle will be displaced a step $t \cdot
+\mathbf{OA}$ due to its initial velocity, and a step
+$\frac{1}{2}t^2\cdot \mathbf{OB}$ due to the acceleration
+downwards, so that $P$ is actually the step $\mathbf{OP} =
+t\mathbf{OA} + \frac{1}{2}t^2 \cdot \mathbf{OB}$ from $O$ at time
+$t$. Similarly, since the velocity of $P$ is increased by a
+velocity represented by $\mathbf{OB}$ in every second of time,
+therefore $P$ is moving at time $t$ with velocity represented by
+$\mathbf{OA} + t\mathbf{OB}$, so that this step must be parallel
+to the tangent at $P$.
+
+\item \textit{To draw the locus of a point $P$ that varies
+according to the law}
+\begin{equation*}
+\mathbf{OP} = \cos (nt + e) \cdot \mathbf{OA} + \sin (nt + e)
+\cdot \mathbf{OB},
+\end{equation*}
+\textit{where $\mathbf{OA,OB}$ are steps of equal length and
+perpendicular to each other, and $t$ is any variable number.}
+
+With centre $O$ and radius $OA$ draw the circle
+$ABA'B'$. Take arc $AE=e$ radians in the direction
+of the quadrant $AB$ (\textit{i.e.} an arc of $e$ radii of the
+circle in length in the direction of $AB$ or $AB'$
+according as $e$ is positive or negative). Corresponding to any
+value of $t$, lay off arc $EP=nt$ radians in the direction of the
+quadrant $AB$. Then arc $AP=nt+e$ radians. Draw $LP$ perpendicular
+to $OA$ at $L$. Then according to the definitions of the
+trigonometric functions of an angle we have,
+\[
+\cos(nt+e)=\overline{OL}/OP,\quad \sin(nt+e)=\overline{LP}/OP.
+\footnote{Observe the distinctions: $\mathbf{OL}$, a step;
+$\overline{OL}$, a positive or negative length of a directed axis;
+$OL$, a length.}
+\]
+Hence we have for all values of $t$,
+\begin{gather*}
+\mathbf{OL}=\cos(nt+e) \mathbf{OA}, \quad \mathbf{LP}=\sin(nt+e)
+\mathbf{OB},
+\intertext{and adding these equations, we find that}
+\mathbf{OP}=\cos(nt+e)\mathbf{OA}+\sin(nt+e)\mathbf{OB}.
+\end{gather*}
+Hence, \textit{the locus of the required point $P$ is the circle
+on $\mathbf{OA, OB}$ as radii}.
+
+\begin{center}
+\includegraphics[width=80mm]{images/Image10.png}
+\end{center}
+
+Let $t$ be the number of seconds that have elapsed since epoch.
+Then, at epoch, $t = 0$, and $P$ is at $E$; and since in $t$
+seconds $P$ has moved through an arc $EP$ of $nt$ radians,
+therefore $P$ moves uniformly round the circle at the rate of $n$
+radians per second. Its velocity at time $t$ is therefore
+represented by $n$ times that radius of the circle which is
+perpendicular to $OP$ in the direction of its motion, or by
+$\mathbf{OP'} = n\mathbf{OQ}$, where arc ${PQ} = \frac{\pi}{2}$
+radians. Hence, since arc ${AQ}=(nt+e + \frac{\pi}{2})$ radians,
+therefore $\mathbf{OP'} =n\left[\cos{\left(nt + e +
+\frac{\pi}{2}\right)} \cdot \mathbf{OA} + \sin{\left(nt + e +
+\frac{\pi}{2}\right)} \cdot \mathbf{OB}\right]$. The point $P\prime$ also
+moves uniformly in a circle, and this circle is the hodograph of
+the motion. The velocity in the hodograph (or the acceleration of
+$P$) is similarly $\mathbf{OP''}= n^2\mathbf{PO}$.
+
+\addcontentsline{toc}{section}{Parallel Projection}
+\section*{Parallel Projection}
+
+\item \textit{If $\mathbf{OP} = x\mathbf{OA} + y\mathbf{OB}$,
+$\mathbf{OP'}=x\mathbf{OA} + y\mathbf{OB'}$, where $x$, $y$ vary
+with the arbitrary number $t$ according to any given law so that
+$P$, $P'$ describe definite loci (and have definite motions when $t$
+denotes time), then the two loci (and motions) are parallel
+projections of each other by rays that are parallel to $BB'$},
+
+For, by subtracting the two equations we find $\mathbf{PP'} =
+y\mathbf{BB'}$, so that $PP'$ is always parallel to $BB'$; and as $P$
+moves in the plane $AOB$ and $P'$ moves in the plane $AOB'$, therefore
+their loci (and motions) are parallel projections of each other by
+rays parallel to $BB'$. The parallel projection is definite when
+the two planes coincide, and may be regarded as a projection
+between two planes $AOB$, $AOB'$, that make an indefinitely small
+angle with each other.
+
+\begin{center}
+\includegraphics[width=80mm]{images/Image11.png}
+\end{center}
+
+\item \textit{The motion of $P$ that is determined by}
+\begin{equation*}
+\mathbf{OP} = \cos(nt + e)\mathbf{OA} + \sin(nt + e)\mathbf{OB}
+\end{equation*}
+\textit{is the parallel projection of uniform circular motion.}
+
+For, draw a step $\mathbf{OB'}$ perpendicular to $\mathbf{OA}$ and
+equal to it in length. Then, by Art.\ 18, the motion of $P'$
+determined by
+\begin{equation*}
+\mathbf{OP'} = \cos(nt + e)\mathbf{OA} + \sin(nt + e)\mathbf{OB'}
+\end{equation*}
+is a uniform motion in a circle on $\mathbf{OA}$, $\mathbf{OB'}$
+as radii; and by Art.\ 19 this is in parallel perspective with the
+motion of $P$.
+
+\addcontentsline{toc}{section}{Step Proportion}
+\section*{Step Proportion}
+
+\item \textsc{Definition.} \textit{Four steps} $\mathbf{AC}$,
+$\mathbf{AB}$, $\mathbf{A'C'}$, $\mathbf{A'B'}$ \textit{are in
+proportion when the first is to the second in respect to both
+relative length and relative direction as the third is to the
+fourth in the same respects.}
+
+\begin{center}
+\includegraphics[width=80mm]{images/Image12.png}
+\end{center}
+
+This requires, first, that the lengths of the steps are in
+proportion or
+\begin{equation*}
+AC: AB = A'C': A'B';
+\end{equation*}
+and secondly, that $\mathbf{AC}$ deviates from $\mathbf{AB}$ by
+the same plane angle in direction and magnitude that
+$\mathbf{A'C'}$ deviates from $\mathbf{A'B'}$.
+
+Hence, first, the triangles $ABC$, $A'B'C'$ are similar, since the
+angles $A$, $A'$ are equal and the sides about those angles are
+proportional; and secondly, one triangle may be turned in its
+plane into a position in which its sides lie in the same
+directions as the corresponding sides of the other triangle. Two
+such triangles will be called \textit{similar and congruent
+triangles}, and corresponding angles will be called
+\textit{congruent angles}.
+
+\item We give the final propositions of Euclid, Book V., as
+exercises in step proportion.
+
+\begin{itemize}
+\item (xi.) \textit{If four steps are proportionals, they are also
+proportionals when taken alternately.}
+
+\item (xii.) \textit{If any number of steps are proportionals,
+then as one of the antecedents is to its consequent, so is the sum
+of the antecedents to the sum of the consequents.}
+
+\item (xiii.) \textit{If four steps are proportionals, the sum (or
+difference) of the first and second is to the second as the sum
+(or difference) of the third and fourth is to the fourth.}
+
+\item (xiv.) If $\mathbf{OA:OB = OP:OQ}$ and $\mathbf{OB:OC =
+OQ:OR}$,\\
+then $\mathbf{OA:OC = OP:OR}$.
+
+\item (xv.) If $\mathbf{OA:OB = OC:OD}$ and $\mathbf{OE:OB =
+OF:OD}$,\\
+then $\mathbf{OA+OE:OB = OC+OF:OD}$.
+
+\item (xvi.) If $\mathbf{OA:OB = OB:OX = OC:OD =
+OD:OY}$,\\
+then $\mathbf{OA:OX = OC:IO}$.
+\end{itemize}
+\end{enumerate}
+
+\addcontentsline{toc}{subsection}{Examples}
+\subsection*{Examples}
+
+\small We shall use $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$, as
+symbols for \textit{unit length east}, \textit{unit length north},
+and \textit{unit length up}, respectively.
+
+\begin{enumerate}
+\item Mark the points whose steps from a given point are
+$\mathbf{i}+2\mathbf{j}$, $-3\mathbf{i}-\mathbf{j}$. Show that the
+step from the first point to the second is
+$-4\mathbf{i}-3\mathbf{j}$, and that the length is 5.
+
+\item Show that the four points whose steps from a given point are
+$2\mathbf{i}+\mathbf{j}$, $5\mathbf{i}+4\mathbf{j}$,
+$4\mathbf{i}+7\mathbf{j}$, $\mathbf{i}+4\mathbf{j}$ are the
+angular points of a parallelogram. Also determine their centre of
+gravity, with weights $1, 1, 1, 1$; also with weights $1, 2, 3, 4$;
+also with weights $1, -2, 3, -4$.
+
+\item If $\mathbf{OA} = \mathbf{i}+2\mathbf{j}$, $\mathbf{OB} =
+4\mathbf{i}+3\mathbf{j}$, $\mathbf{OC} = 2\mathbf{i}+3\mathbf{j}$,
+$\mathbf{OD}= 4\mathbf{i}+\mathbf{j}$, find $CD$ as sums of
+multiples of $\mathbf{CA}$, $\mathbf{CB}$, and show that
+$\mathbf{CD}$ bisects $\mathbf{AB}$.
+
+\item If $\mathbf{OP} = x\mathbf{i}+y\mathbf{j}$, $\mathbf{OP'} =
+x'\mathbf{i}+y'\mathbf{j}$, then $\mathbf{PP'} = (x'-x)\mathbf{i}
++ (y'-y)\mathbf{j}$ and ${\overline{PP'}}^2 = (x'-x)^2+(y'-y)^2$.
+
+\item Show that $\mathbf{AB}$ is bisected by $\mathbf{OC =
+OA+OB}$, and trisected by \linebreak $\mathbf{OD = 2OA+OB}$,
+$\mathbf{OE = OA+2OB}$, and divided inversely as $2:3$ by
+\linebreak $\mathbf{OF = 2OA+3OB}$.
+
+\item Show that $\mathbf{AA'+BB' = 2MM'}$, where $MM'$ are the
+middle points of $AB$, $A'B'$, respectively.
+
+\item Show that $\mathbf{2AA'+3BB' = (2+3)CC'}$, where $C$, $C'$
+are the points that divide $AB$, $A'B'$, inversely as $2:3$.
+Similarly, when 2, 3 are replaced by $l$, $m$.
+
+\item Show that the point that divides a triangle into three equal
+triangles is the intersection of the medial lines of the triangle.
+
+\item Show that the points which divide a triangle into triangles
+of equal magnitude, one of which is negative (the given triangle
+being positive), are the vertices of the circumscribing triangle
+with sides parallel to the given triangle.
+
+\item If $a$, $b$, $c$ are the lengths of the sides $BC$, $CA$,
+$AB$ of a triangle, show that $\frac{1}{b}\mathbf{AC} \pm
+\frac{1}{c}\mathbf{AB}$ (drawn from $A$) are interior and exterior
+bisectors of the angle $A$; and that when produced they cut the
+opposite side $BC$ in the ratio of the adjacent sides.
+
+\item The $\left\lbrace
+\begin{matrix}
+  \text{lines} \\
+  \text{points}
+\end{matrix}\right.$ that join the $\left\lbrace
+\begin{matrix}
+  \text{vertices} \\
+  \text{sides}
+\end{matrix}\right.$ of a triangle $ABC$ to any $\left\lbrace
+\begin{matrix}
+  \text{point } P \\
+  \text{line } p
+\end{matrix}\right.$ in its plane divide the sides $BC$, $CA$,
+$AB$ in ratios whose product is $\left\lbrace
+\begin{matrix}
+  +1 \\
+  -1
+\end{matrix}\right.$; and conversely $\left\lbrace
+\begin{matrix}
+  \text{lines from} \\
+  \text{points on}\end{matrix}\right.$ the $\left\lbrace
+\begin{matrix}
+  \text{vertices} \\
+  \text{sides}
+\end{matrix}\right.$ that so divide the sides $\left\lbrace
+\begin{matrix}
+  \text{meet in a point.} \\
+  \text{lie in a line.}\end{matrix}\right.$
+
+\item Prove by Exs.\ 10, 11, that the three interior bisectors of
+the angles of a triangle (also an interior and two exterior
+bisectors) meet in a point; and that the three exterior bisectors
+(also an exterior and two interior bisectors) meet the sides in
+colinear points.
+
+\item Determine the locus (and motion) of $P$, given by
+$\mathbf{OP = OA}+t\mathbf{OB}$; also of $\mathbf{OP} =
+(1+2t)\mathbf{i}+(3t-2)\mathbf{j}$.
+
+\item Compare the loci of $P$ determined by the following
+\textit{pairs} of \textbf{step} and \textbf{length} equations:
+\begin{gather*}
+\mathbf{AP} = 2\text{ east},\quad AP = 2; \qquad
+  \mathbf{AP} = 2\mathbf{BP}, \quad AP = 2BP; \\
+\mathbf{AP+BP = CD}, \quad  AP+BP = CD
+\end{gather*}
+
+\item Draw, by points and tangents, the locus of $P$ determined by
+each of the following values of $\mathbf{OP}$, in which $x$ is any
+number:
+\begin{gather*}
+x\mathbf{i}+\frac{1}{2}x^2\mathbf{j}; \quad
+x\mathbf{i}+\frac{2}{x}\mathbf{j}; \quad
+x\mathbf{i}+\frac{1}{3}x^3\mathbf{j}; \quad
+x\mathbf{i}+(\frac{1}{3}x^3-x^2+2)\mathbf{j}; \\
+x\mathbf{i}+\frac{8}{x^2+4}\mathbf{j}; \quad
+x\mathbf{i}+\sqrt{4-x^2}\mathbf{j}; \quad
+x\mathbf{i}+\frac{1}{2}\sqrt{4-x^2}\mathbf{j}.
+\end{gather*}
+
+\item Take three equal lengths making angles $120^{\circ}$ with each
+other as projections of $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$,
+and construct by points the projection of the locus of $P$, where
+$\mathbf{OP} = 2(\cos{x}{\cdot}\mathbf{i} +
+\sin{x}{\cdot}\mathbf{j}) + x{\cdot}\mathbf{k}$, $x$ varying from
+0 to $2\pi$. Show that this curve is one turn of a helix round a
+vertical cylinder of altitude $2\pi$, the base being a horizontal
+circle of radius 2 round $O$ as centre.
+
+\item A circle rolls inside a fixed circle of twice its diameter;
+show that any point of the plane of the rolling circle traces a
+parallel projection of a circle.
+
+\item A plane carries two pins that slide in two fixed rectangular
+grooves; show that any point of the sliding plane traces a
+parallel projection of a circle.
+
+\item $OACB$ is a parallelogram whose sides are rigid and jointed
+so as to turn round the vertices of the parallelogram; $APC$,
+$BCQ$ are rigid similar and congruent triangles. Show that
+$\mathbf{AC:AP = BQ:BC = OQ:OP}$, and that therefore $P$, $Q$
+trace similar congruent figures when $O$ remains stationary (21,
+22, xii.). [See cover of book.]
+
+\item If the plane pencil $OA$, $OB$, $OC$, $OD$ is cut by any
+straight line in the points $P$, $Q$, $B$, $S$, show that the
+\textit{cross-ratio}
+$(\overline{PR}:\overline{RQ}) : (\overline{PS}:\overline{SQ})$ is
+constant for all positions of the line.
+\begin{equation*}
+[\mathbf{OC} = l\mathbf{OA}+m\mathbf{OB} = lx\mathbf{OP}+ my\mathbf{OQ}
+  \text{ gives } \overline{PR}:\overline{RQ} = my:lx].
+\end{equation*}
+
+\item Two roads run north, and east, intersecting at $O$. $A$ is
+60 \textit{feet south} of $O$, walking 3 \textit{feet per second
+north}, $B$ is 60 \textit{feet west} of $O$, walking 4
+\textit{feet per second east}. When are $A$, $B$ nearest together,
+and what is $B$'s apparent motion as seen by $A$?
+
+\item What is $B$'s motion relative to $A$ in Ex.\ 21 if $B$ is
+accelerating his walk at the rate of 3 \textit{inches per second
+per second}?
+
+\item In Ex.\ 21, let the east road be 20 feet above the level of
+the north road; and similarly in Ex.\ 22.
+
+\item A massless ring $P$ is attached to several elastic strings
+that pass respectively through smooth rings at $A$, $B$, $C$,
+$\cdots$ and are attached to fixed points $A'$, $B'$, $C'$,
+$\cdots$ such that $A'A$, $B'B$, $C'C$, $\cdots$ are the natural
+lengths of the strings. The first string has a tension $l$ per
+unit of length that it is stretched (Hooke's law), the second a
+tension $m$, the third a tension $n$, etc. Find the resultant
+force on $P$ and its position of equilibrium.
+
+\item The same as Ex.\ 24, except that the ring has a mass $w$.
+\end{enumerate} \normalsize
+
+\newpage
+\chapter{Rotations. Turns. Arc Steps}
+
+\addcontentsline{toc}{section}{Definitions and Theorems of
+Rotation}
+
+\begin{enumerate}
+\setcounter{enumi}{22}
+\item \textsc{Definitions of Rotation}
+
+A step is \textbf{rotated} when it is revolved about an axis through its
+initial point as a rigid length rigidly attached to the axis. The
+step describes a conical angle about the axis except when it is
+perpendicular to the axis.
+
+\begin{center}
+\includegraphics[width=80mm]{images/Image13.png}
+\end{center}
+
+If a rotation through a diedral angle of given magnitude and
+direction in space be applied to the radii of a sphere of unit
+radius and centre $O$, the sphere is rotated as a rigid body about
+a certain diameter $PP'$ as axis, and a plane through $O$
+perpendicular to the axis intersects the sphere in the
+\textbf{equator} of the rotation.
+
+Either of the two directed arcs of the equator from the initial
+position $A$ to the final position $A'$ of a point of the rotated
+sphere that lies on the equator is the \textbf{arc} of the
+rotation. If these two arcs he bisected at $L$, $M$ respectively,
+then the two arcs are $2\overset\frown{AL}$, $2\overset\frown{AM}$
+respectively, and $\overset\frown{AL}$, $\overset\frown{AM}$ are
+supplementary arcs in opposite directions, each less than a
+semicircle. When these half-arcs are $0^{\circ}$ and $180^{\circ}$
+respectively, they represent a rotation of the sphere into its
+original position, whose axis and equator are indeterminate, so
+that such arcs may be measured on any great circle of the sphere
+without altering the corresponding rotation.
+
+\item \textit{A rotation is determined by the position into which
+it rotates two given non-parallel steps.}
+
+For let the radii $\mathbf{OB}$, $\mathbf{OC}$ rotate into the
+radii $\mathbf{OB'}$, $\mathbf{OC'}$. Any axis round which
+$\mathbf{OB}$ rotates into $\mathbf{OB'}$ must be equally inclined
+to these radii; \textit{i.e.}, it is a diameter of the great
+circle $PKL$ that bisects the great arc $\overset\frown{BB'}$ at
+right angles.
+
+\begin{center}
+\includegraphics[width=80mm]{images/Image14.png}
+\end{center}
+
+\textit{E.g.}, $OK$, $OL$, $OP$, $\cdots$ are such axes.
+Similarly, the axis that rotates $\mathbf{OC}$ into $\mathbf{OC'}$
+must be a diameter of the great circle $PN$ that bisects the great
+arc $\overset\frown{CC'}$ at right angles. Hence there is but
+one axis round which $\mathbf{OB}$, $\mathbf{OC}$ rotate into
+$\mathbf{OB'}$, $\mathbf{OC'}$; viz., the intersection $OP$ of the
+planes of these two bisecting great circles: the equator is the
+great circle whose plane is perpendicular to this axis, and the
+arcs of the rotation are the intercepts on the equator by the
+planes through the axis and either $B$, $B'$ or $C$, $C'$. [When
+the two bisecting great circles coincide (as when $C$, $C'$ lie on
+$BP$, $B'P$), then their plane bisects the diedral angle
+$BC-O-B'C'$, whose edge $OP$ is the only axis of rotation.]
+
+\small \textsc{Note.} Since $\overset\frown{BC}$,
+$\overset\frown{B'C'}$ may be any two positions of a marked arc on
+the surface of the sphere, we see that any two positions of the
+sphere with centre fixed determine a definite rotation of the
+sphere from one position to the other. \normalsize
+
+\item \textit{A marked arc of a great circle of a rotating sphere
+makes a constant angle with the equator of the rotation.}
+
+For the plane of the great arc makes a constant angle both with
+the axis and with the equator of the rotation.
+
+\item \textit{If the sphere $O$ be given a rotation
+$2\overset\frown{A_0C}$ followed by a rotation
+$2\overset\frown{CB_0}$, the resultant rotation of the sphere is
+$2\overset\frown{A_0B_0}$.}
+
+\begin{center}
+\includegraphics[width=80mm]{images/Image15.png}
+\end{center}
+
+For produce the arcs $\overset\frown{A_0C}$,
+$\overset\frown{B_0C}$ to $A_1$, $B'$ respectively, making
+$\overset\frown{CA_1} = \overset\frown{A_0C}$,
+$\overset\frown{B'C} = \overset\frown{CB_0}$. Then the spherical
+triangles $A_0B_0C$, $A_1B'C$ are equal, since the corresponding
+sides about the equal vertical angles at $C$ are by construction
+equal. Therefore the sides $\overset\frown{A_0B_0}$,
+$\overset\frown{B'A_1}$ are equal in length, and the corresponding
+angles $A_0$, $A_1$ and $B_0$, $B'$ are equal. Therefore, by Art.\
+25, if a marked arc $\overset\frown{AB}$ of the sphere coincide
+initially with $\overset\frown{A_0B_0}$, the first rotation
+$2\overset\frown{A_0C} = \overset\frown{A_0A_1}$ will bring
+$\overset\frown{AB}$ into the position $\overset\frown{A_1B_1}$ on
+$\overset\frown{B'A_1}$ produced, and the second rotation
+$2\overset\frown{CB_0} = \overset\frown{B'B_1}$ will bring
+$\overset\frown{AB}$ into the position $\overset\frown{A_2B_2}$ on
+$\overset\frown{A_0B_0}$ produced, where $\overset\frown{B_0A_2}$
+= $\overset\frown{A_0B_0}$. Hence the resultant rotation of the
+sphere is $2\overset\frown{A_0B_0}$ = $\overset\frown{A_0A_2}$.
+
+\small \textsc{Note.} This theorem enables one to find the
+resultant of any number of successive rotations, by replacing any
+two successive rotations by their resultant, and so on until a
+single resultant is found. \normalsize
+
+\addcontentsline{toc}{section}{Definitions of Turn and Arc Steps}
+\item \textsc{Definitions of Turn}
+
+A step is turned when it is made to describe a \textit{plane
+angle} round its initial point as centre.
+
+\begin{center}
+\includegraphics[width=80mm]{images/Image16.png}
+\end{center}
+
+If a turn through a plane angle of given magnitude and direction
+in space be applied to the radii of the sphere $O$, it turns the
+great circle that is parallel to the given plane angle as a rigid
+circle, and does not affect the other radii of the sphere.
+\textit{E.g.}, only horizontal radii can be turned through a
+horizontal plane angle. The circle that is so turned is the great
+circle of the turn.
+
+A directed arc of the great circle of a turn from the initial
+position $A$ to the final position $B$ of a point on the great
+circle, and less than a semi-circumference, is the arc of the
+turn. When this arc is $0^{\circ}$ or $180^{\circ}$, it represents
+a turn that brings a step back to its original position or that
+reverses it; and since such turns may take place in any plane with
+the same results, therefore such arcs may be measured on any great
+circle of the sphere without altering their corresponding turns.
+
+The \textbf{axis} of a turn is that radius of the sphere $O$ which
+is perpendicular to its great circle and lies on that side of the
+great circle from which the arc of the turn appears
+counter-clockwise.
+
+\item \textit{A turn is determined by the position into which it
+displaces any given step.}
+
+For, let the radius $\mathbf{OA}$ turn into the radius
+$\mathbf{OB}$. Then, the great circle $O-AB$ must be the great
+circle of the turn, and $\overset\frown{AB}$, the arc of the turn.
+
+\item \textsc{Definitions.} The resultant of two successive turns
+$\overset\frown{AB}$, $\overset\frown{BC}$ is the turn
+$\overset\frown{AC}$.
+
+\begin{center}
+\includegraphics[width=80mm]{images/Image17.png}
+\end{center}
+
+When the arc of the turns are not given with the first ending
+where the second begins, each arc may be moved as a rigid arc
+round its great circle until they do so end and begin, without
+altering their turning value. When the two great circles are not
+the same, then the common point of the two arcs must be one or the
+other point of intersection $(B, B')$ of the two great circles.
+The figure shows that the same resultant is found from either of
+these points.
+
+\subsubsection{ARC STEPS}
+
+We may call the great arc $\overset\frown{AB}$ the \textbf{arc
+step} from $A$ to $B$ on the surface of the sphere; and call two
+arc steps \textbf{equal} when they are arcs of the same great
+circle of the same length and direction; and call
+$\overset\frown{AC}$ the \textbf{sum} of $\overset\frown{AB}$,
+$\overset\frown{BC}$ or the sum of any arc steps equal to these.
+The half-arc of a resultant rotation is thus the sum of the
+half-arcs of its components, and the arc of a resultant turn is
+the sum of the arcs of the components. The sum of several arcs is
+found by replacing any two successive arcs of the sum by their
+sum, and so on, until a single sum is found. An arc of $0^{\circ}$
+or $180^{\circ}$ may be measured on any great circle without
+altering its value as the representative of a half-rotation, a
+turn, or an arc step.
+
+\item \textit{The resultant of two successive rotations or turns
+(i.e., the sum of two arc steps) is commutative only when the arcs
+are cocircular.}
+
+For let the half-arcs of the rotations, or the arcs of the turns,
+be $\overset\frown{AB} = \overset\frown{BA'}$, and
+$\overset\frown{C'B} = \overset\frown{BC}$; then the sums
+$\overset\frown{AB}+\overset\frown{BC}$,
+$\overset\frown{C'B}+\overset\frown{BA'}$ in opposite orders are
+respectively $\overset\frown{AC}$, $\overset\frown{C'A'}$; and
+from the figure those arcs are equal when, and only when, the
+given arcs are cocircular.
+
+\begin{center}
+\includegraphics[width=80mm]{images/Image18.png}
+\end{center}
+
+\begin{itemize}
+\item \textsc{Cor.\ 1.} \textit{An arc of $0^{\circ}$ or
+$180^{\circ}$ is commutative with any other arc.}
+
+For it may be taken cocircular with the other arc.
+
+\item \textsc{Cor.\ 2.} \textit{The magnitudes of the sums of two
+arcs in opposite orders are equal.}
+
+For $ABC$, $A'BC'$ are equal spherical triangles by construction,
+and therefore $\overset\frown{AC}$, $\overset\frown{C'A'}$ are
+equal in length.
+\end{itemize}
+
+\item \textit{A sum of successive arc steps is associative.}
+
+\begin{center}
+\includegraphics[width=80mm]{images/Image19.png}
+\end{center}
+
+For, consider first three arcs upon the great circles $LQ'$,
+$Q'R$, $RL$. If the arcs are such as to begin and end
+successively, the proof is the same as for step addition,
+\textit{e.g.}, in the sum $\overset\frown{AQ'} +
+\overset\frown{Q'R} + \overset\frown{RB} = \overset\frown{AB}$,
+the first two may be replaced by their sum $\overset\frown{AR}$,
+or the second and third by their sum $\overset\frown{Q'B}$ without
+altering the whole sum. In the more general case when the three
+arcs are
+\begin{equation*}
+\overset\frown{AQ'} = \overset\frown{S'P'}, \quad
+\overset\frown{Q'Q} = \overset\frown{R'R}, \quad
+\overset\frown{RB} = \overset\frown{PT},
+\end{equation*}
+the sum of the first two is $\overset\frown{AQ} =
+\overset\frown{SP}$, whose sum with the third is
+$\overset\frown{ST}$; and the sum of the second and third is
+$\overset\frown{R'B} = \overset\frown{P'T'}$, whose sum with the
+first is $\overset\frown{S'T'}$; and we must prove that
+$\overset\frown{ST}$, $\overset\frown{S'T'}$ are equal arcs of the
+same great circle in the same direction.
+
+\smallskip
+[Observe that in the construction $P$ is determined as the
+intersection of $QA$ and $RB$, and $P'$ as the intersection of
+$Q'A$ and $R'B$.]
+
+\smallskip
+Let the three given arcs be the half-arcs of successive rotations
+of the sphere $O$. Then by Art.\ 26, the rotation
+$2\overset\frown{AQ} = 2\overset\frown{SP}$ gives the sphere the
+same displacement as the first and second rotations, so that
+$2\overset\frown{ST}$ gives the sphere the same displacement as
+the three rotations. Similarly, the rotation $2\overset\frown{R'B}
+= 2\overset\frown{P'T}$ gives the sphere the same displacement as
+the second and third rotations, so that $2\overset\frown{S'T'}$
+gives the sphere the same displacement as the three rotations.
+Hence $\overset\frown{ST}$, $\overset\frown{S'T'}$ are arcs of the
+same great circle, and either equal (and in the same direction) or
+supplementary (and in opposite directions), since they are
+half-arcs of the same rotation. This is true wherever $Q$ may be.
+Suppose that $Q$ is slightly displaced towards $R$; then
+$\overset\frown{ST}$, $\overset\frown{S'T'}$ are slightly
+displaced, and if equal at first, they must remain equal, since a
+slight change in each of two equal arcs could not change them to
+supplementary arcs in opposite directions.\footnote{When both arcs
+are nearly $90^{\circ}$, a slight change in each could change them
+from equals to supplements in the same direction.} Hence by moving
+$Q$ continuously towards $R$ and finding how the arcs
+$\overset\frown{ST}$, $\overset\frown{S'T'}$ are related when $Q$
+reaches $R$, we find how they are related for any position of $Q$,
+since there is no change in the relation when $Q$ is moved
+continuously. But when $Q$ is at $R$, it was shown above that both
+arcs were equal; therefore $\overset\frown{ST}$,
+$\overset\frown{S'T'}$ are always equal.
+
+\smallskip
+So, in general, for a sum of any number of successive arcs, any
+way of forming the sum by replacing any two successive terms by
+their sum and so on, must give a half-arc of the resultant of the
+rotations through double each of the given arcs. Hence any two
+such sums are either equal or opposite supplementary arcs of the
+same great circle; and since by continuous changes of the
+component arcs, they may be brought so that each begins where the
+preceding arc ends, in which position the two sums are equal,
+therefore they are always equal.
+
+\begin{itemize}
+\item \textsc{Cor.\ 1.} \textit{An arc of $0^{\circ}$ or
+$180^{\circ}$ may have any position in a sum.} [Art.\ 30, Cor.\ 1.]
+
+\item \textsc{Cor.\ 2.} \textit{The magnitude of a sum of arcs is
+not changed by a cyclic change in the order of its terms.}
+
+For $(\overset\frown{AB} + \overset\frown{CD} + \cdots) +
+\overset\frown{HK}$ and $\overset\frown{HK} + (\overset\frown{AB}
++ \overset\frown{CD} + \cdots)$ have equal magnitudes. [Art.\ 30,
+Cor.\ 2.]
+\end{itemize}
+\end{enumerate}
+
+\addcontentsline{toc}{subsection}{Examples}
+\subsection*{EXAMPLES}
+
+\small \begin{enumerate}
+
+\item Show that $2(\overset\frown{AB} + \overset\frown{BC})$ and
+$2\overset\frown{AB} + 2\overset\frown{BC}$ are in general
+unequal.
+
+\item If (2, $30^{\circ}$) denote a turn of $30^{\circ}$
+counter-clockwise in the plane of the paper and a doubling, and
+(3, $-60^{\circ}$) denote a turn of $60^{\circ}$ clockwise in the
+plane of the paper and a trebling, express the resultant of these
+two compound operations (\textit{versi-tensors}) in the same
+notation.
+
+\item Find the resultant of (2, $30^{\circ}$), (3, $60^{\circ}$),
+(4, $-120^{\circ}$), (1, $180^{\circ}$).
+
+\item Show that either (2, $-60^{\circ}$) or (2, $120^{\circ}$)
+taken twice have the resultant (4, $-120^{\circ}$).
+
+\item Would you consider the resultants of \textit{versi-tensors}
+as their sums or their products, and why?
+
+\item Let the base $QR$ of a spherical triangle $PQR$ slide as a
+rigid arc round its fixed great circle, and let the great circles
+$QP$, $RP$, always pass through fixed points $A$, $B$
+respectively. Show that if points $S$, $T$ lie on the great
+circles $QP$, $RP$ so as always to keep $\overset\frown{PS} =
+\overset\frown{QA}$ and $\overset\frown{PT} = \overset\frown{RB}$,
+then the arc $\overset\frown{ST}$ is an arc of fixed length and
+direction that slides around a fixed great circle as
+$\overset\frown{QR}$ slides round its fixed great circle. [Let
+$P'$, $Q'$, $R'$, $S'$, $T'$, be given positions of $P$, $Q$, $R$,
+$S$, $T$, and use Art.\ 31 and figure.]
+
+\item Show that the locus of the radius $OP$ in Ex.\ 6 is an
+oblique circular cone of which $OA$, $OB$ are two elements, and
+that the fixed great circles $QR$, $ST$ are parallel to its
+circular sections. [Draw a fixed plane parallel to $OQR$ and
+cutting the radii $OA$, $OB$, in the fixed points $A'$, $B'$, and
+cutting $OP$ in the variable point $P'$, and show that $P'$
+describes a circle in this plane through the fixed points $A'$,
+$B'$; similarly, for a fixed plane parallel to $OST$.]
+
+\textsc{Note.}---The locus of $P$ on the surface of the sphere is
+called a \textbf{spherical conic} (the intersection of a sphere
+about the vertex of a circular cone as centre with the surface of
+the cone); and the great circles $QR$, $ST$ (parallel to the
+circular sections of the cone) are the \textit{cyclic} great
+circles of the spherical conic. The above properties of a
+spherical conic and its cyclic great circles become properties of
+a plane conic and its \textit{asymptotes} when the centre $O$ of
+the sphere is taken at an indefinitely great distance.
+
+\item State and prove Ex.\ 6 for a plane, and construct the locus
+of $P$.
+\end{enumerate} \normalsize
+
+\newpage
+\chapter{Quaternions}
+
+\addcontentsline{toc}{section}{Definitions and Theorem}
+
+\begin{enumerate}
+\setcounter{enumi}{31}
+\item \textsc{Definitions.} A \textbf{quaternion} is a number that
+alters a step in length and direction by a given ratio of
+extension and a given turn. \textit{E.g.}, in the notation of Ex.\
+2, II, (2, $30^{\circ}$), (2, $-60^{\circ}$) are quaternions.
+
+Two quaternions are \textbf{equal} when, and only when, their
+ratios of extension are equal and their turns are equal.
+
+A \textbf{tensor} is a quaternion that extends only;
+\textit{i.e.}, a tensor is an ordinary positive number. Its turn
+is $0^{\circ}$ in any plane.
+
+A \textbf{versor} or \textbf{unit} is a quaternion that turns
+only. \textit{E.g.}, $1$, $-1 = (1, 180^{\circ})$, $(1,
+90^{\circ})$, $(1, 30^{\circ})$, are versors.
+
+A \textbf{scalar} is a quaternion whose product lies on the same
+line or ``scale'' as the multiplicand; \textit{i.e.}, a scalar is an
+ordinary positive or negative number. Its turn is $0^{\circ}$ or
+$180^{\circ}$ in any plane.
+
+A \textbf{vector} is a quaternion that turns $90^{\circ}$.
+\textit{E.g.}, $(2, 90^{\circ})$, $(1,-90^{\circ})$, are vectors.
+
+\item \textsc{Functions of a Quaternion} $q$. The \textbf{tensor}
+of $q$, or briefly $Tq$, is its ratio of extension. \textit{E.g.},
+$T2 = 2 = T(-2) = T(2, 30^{\circ})$.
+
+The \textbf{versor} of $q$ ($Uq$) is the versor with the same arc
+of turn as $q$. \textit{E.g.},
+\begin{equation*}
+U2 = 1, \quad U(-2) = -1, \quad U(2, 30^{\circ}) = (1, 30^{\circ}).
+\end{equation*}
+
+The \textbf{arc, angle, axis, great circle,} and \textbf{plane} of
+$q$, are respectively the \textit{arc, angular magnitude, axis,
+great circle,} and \textit{plane} of its turn. \textit{E.g.},
+$\mathrm{arc } (2, 30^{\circ})$ is a counter-clockwise arc of
+$30^{\circ}$ of unit radius in the plane of the paper, and
+$\mathrm{arc } (2, -30^{\circ})$ is the same arc oppositely
+directed; $\angle(2, 30^{\circ}) = \angle(2,-30^{\circ}) =
+30^{\circ} = \frac{\pi}{6} \mathrm{ radians}$; $\mathrm{axis }(2,
+30^\circ)$ is a unit length perpendicular to the plane of the
+paper directed towards the reader, and $\mathrm{axis }(2,
+-30^{\circ})$ is the same length oppositely directed; etc.
+
+\begin{center}
+\includegraphics[width=80mm]{images/Image20.png}
+\end{center}
+
+If $q\mathbf{OA} = \mathbf{OB}$, and if $L$ be the foot of the
+perpendicular from $B$ upon the line $OA$, then $\mathbf{OL}$,
+$\mathbf{LB}$ are called \textit{the components of $q$'s product}
+\textit{respectively parallel and perpendicular to the
+multiplicand;} also, \textit{the projections of $\mathbf{OB}$
+parallel and perpendicular to $\mathbf{OA}$.}
+
+The \textbf{scalar} of $q$ ($Sq$) is the scalar whose product
+equals the component of $q$'s product parallel to the
+multiplicand; \textit{viz.}, $Sq\cdot\mathbf{OA} = \mathbf{OL}$.
+
+\textit{E.g.}, $S(2, 30^{\circ}) = \sqrt{3}, \quad
+S(2, 150^{\circ}) = -\sqrt{3}$.
+
+The \textbf{vector} of $q$ ($Vq$) is the vector whose product
+equals the component of $q$'s product perpendicular to the
+multiplicand; \textit{viz.}, $Vq\cdot\mathbf{OA} = \mathbf{LB}$.
+
+\textit{E.g.}, $V(2, 30^{\circ}) = (1, 90^{\circ})= V(2, 150^{\circ}),
+\quad V(2, -60^{\circ}) = (\sqrt{3}, -90^{\circ})$.
+
+The \textbf{reciprocal} of $q$ ($1/q$ or $q^{-1}$) is the
+quaternion with reciprocal tensor and reversed turn.
+\textit{E.g.}, $(2, 30^{\circ})^{-1}=(\frac{1}{2},-30^{\circ})$.
+
+The \textbf{conjugate} of $q(Kq)$ is the quaternion with the same
+tensor and reversed turn. \textit{E.g.,}
+\begin{equation*}
+K(2,30^{\circ}) = (2,-30^{\circ}).
+\end{equation*}
+
+\item From the above diagram and the definitions of the cosine and
+sine of an angle, we have
+\begin{gather}
+\tag{a} Sq = \frac{\mathbf{OL}}{\mathbf{OA}} = \frac{\overline{OL}}{OA} =
+  \frac{OB}{OA} \cdot \frac{\overline{OL}}{OB} = Tq \cdot \cos\angle{q} \\
+\tag{b} TVq = \frac{LB}{OA} = \frac{OB}{OA} \cdot \frac{LB}{OB} =
+  Tq \cdot \sin\angle{q}
+\end{gather}
+
+\small \textsc{Note.} \textit{Arc $Vq$ is a quadrant on the great
+circle of $q$ in the direction of arc $q$.} \normalsize
+\end{enumerate}
+
+\addcontentsline{toc}{subsection}{Examples}
+\subsection*{Examples}
+
+\small \begin{enumerate}
+\item If equal numbers multiply equal steps, the products are
+equal; and if they multiply unequal steps, the products are
+unequal.
+
+\item If the products of two steps by equal numbers are equal,
+then the two steps are equal; and if the products of two equal
+steps by two numbers are equal, then the numbers are equal.
+
+\item If several steps be multiplied by equal numbers, then any
+product is to its multiplicand as any other product is to its
+multiplicand.
+
+\item If two steps be multiplied by reciprocal numbers, then
+corresponding products and multiplicands are reciprocally
+proportional.
+
+\item Construct the following products, where $\mathbf{OA}$ is a
+unit step to the right in the plane of the paper, and determine
+the functions of each multiplier that are defined in Art.\ 33.
+
+\begin{enumerate}
+\item $2\cdot\mathbf{OA = OL}$, \quad
+$(4, 60^{\circ})\cdot\mathbf{OA = OB}$, \quad
+$(4,-60^{\circ})\cdot\mathbf{OA = OB'}$, \\
+$(2\sqrt{3},90^{\circ})\cdot\mathbf{OA = OM}$, \quad
+$(2\sqrt{3},-90^{\circ})\cdot\mathbf{OA = OM'}$, \\
+$(1, 60^{\circ})\cdot\mathbf{OA = OB_1}$, \quad
+$(1,-60^{\circ})\cdot\mathbf{OA = OB_1'}$, \\
+$(1, 90^{\circ})\cdot\mathbf{OA = OM_1}$, \quad
+$(1,-90^{\circ})\cdot\mathbf{OA = OM_1'}$.
+
+\item The same as (a) with $120^{\circ}$ in the place of
+$60^{\circ}$.
+\end{enumerate}
+
+\item Show that $SSq = Sq$, \quad $SVq = 0$, \quad $VSq = 0$,
+\quad $VVq = Vq$, \\ $SKq = KSq = Sq$, \quad $VKq = KVq$, \quad
+$USq = \pm 1$, \quad $UTq = 1 = TUq$.
+\end{enumerate} \normalsize
+
+\addcontentsline{toc}{section}{Multiplication}
+\section*{Multiplication}
+
+\begin{enumerate}
+\setcounter{enumi}{34}
+\item \textsc{Definition.} The \textbf{product} of two or more
+numbers is that number whose extension and turn are the resultants
+of the successive extensions and turns of the factors (beginning
+with the right-hand factor).
+
+\textit{E.g.}, if $r\mathbf{OA = OB}$, $q\mathbf{OB = OC}$,
+$p\mathbf{OC = OD}$, then we have $pqr\cdot\mathbf{OA} =
+pq\mathbf{OB} = p\mathbf{OC} = \mathbf{OD}$.
+
+\item The product is, however, independent of whether a step
+$\mathbf{OA}$ can be found or not, such that each factor operates
+upon the product of the preceding factor; \textit{i.e.}, we have
+by definition,
+
+(\textit{a}) $T(\cdots pqr) = \cdots Tp \cdot Tq \cdot Tr$.
+
+(\textit{b}) $\text{arc }(\cdots pqr) = \text{arc }r + \text{arc
+}q + \text{arc }p + \cdots$.
+
+\item \textit{The product of a tensor and a versor is a number
+with that tensor and versor; and conversely, a number is the
+product of its tensor and its versor.}
+
+For if $n$ be a tensor, and $q'$ a versor, then $nq'$ turns by the
+factor $q'$ and extends by the factor $n$, and \textit{vice versa}
+for $q'n$; hence either of the products, $nq'$, $q'n$, is a
+quaternion with tensor $n$ and versor $q\prime$. Similarly,
+\begin{equation*}
+q = Tq \cdot Uq = Uq \cdot Tq.
+\end{equation*}
+
+\item \textit{Any successive factors of a product may be replaced
+by their product without altering the value of the whole product;
+but in general such factors can be changed in order without
+altering the value of the product only when those factors are
+cocircular.}
+
+For replacing successive factors by their product does not alter
+the tensor of the whole product by Art.\ 36(a), nor the arc of the
+product by Art.\ 31, 36(b); but by Art.\ 30 the arc of the product
+is altered if two factors be interchanged except when those
+factors are cocircular.
+
+\begin{itemize}
+\item \textsc{Cor.\ 1.} \textit{A scalar factor may have any
+position in the product without altering the value of the
+product.} [Art.\ 31, Cor.\ 1.]
+
+\item \textsc{Cor.\ 2.} \textit{The angle of a product is not
+altered by a cyclic change in the order of the factors.} [Art.\
+31, Cor.\ 2.]
+
+\item \textsc{Cor.\ 3.} \textit{The scalar, and the tensor of the
+vector, of a product are not altered by a cyclic change in the
+order of the factors.} [Art.\ 34, \textit{a, b}.]
+\end{itemize}
+
+\item \textit{The product of two numbers with opposite turns
+equals the product of the tensors of the numbers; and conversely
+if the product of two numbers is a tensor, then the turns of the
+factors are opposites.} [36 \textit{a, b}.]
+
+\begin{itemize}
+\item \textsc{Cor.\ 1.} \textit{The product of two conjugate
+numbers equals the square of their tensor; and if the product of
+two numbers with equal tensors is a tensor, then the two numbers
+are conjugates.}
+
+\item \textsc{Cor.\ 2.} \textit{The conjugate of a product equals
+the product of the conjugates of the factors in reverse order.}
+
+For $(pqr)(Kr\cdot Kq\cdot Kp) = (Tp)^2 \cdot (Tq)^2 \cdot (Tr)^2$
+since $rKr = (Tr)^2$, may have any place in the product, and may
+be put first; and then $(qKq) = (Tq)^2$, may be put second, and
+then $(pKp) = (Tp)^2$. [Cor.\ 1, 38 Cor.\ 1.]
+
+Hence, $K(pqr) = Kr \cdot Kq \cdot Kp$. [Cor.\ 1.]
+
+\item \textsc{Cor.\ 3.} \textit{The product of two reciprocal
+numbers is unity; and conversely, if the product of two numbers
+with reciprocal tensors is unity, then the numbers are
+reciprocals.}
+
+\item \textsc{Cor. 4.} \textit{The reciprocal of a product equals
+the product of the reciprocals of the factors in reverse order.}
+
+For $(pqr)(r^{-1}q^{-1}p^{-1}) = 1$.
+\end{itemize}
+
+\item \textit{The square of a vector is $-1$ times the square of
+its tensor; and conversely, if the square of a number is a
+negative scalar, then the number is a vector.} [36, \textit{a, b}.]
+
+\begin{itemize}
+\item \textsc{Cor.\ 1.} \textit{The conjugate of a vector is the
+negative vector.} [39 Cor.\ 1.]
+
+\item \textsc{Cor.\ 2.} \textit{The conjugate of a product of two
+vectors is the product of the same vectors in reverse order.}
+[Art.\ 39, Cor.\ 2.]
+
+\item \textsc{Cor.\ 3.} \textit{The conjugate of a product of
+three vectors is the negative of the product of the same vectors
+in reverse order.} [Art.\ 39, Cor.\ 2.]
+\end{itemize}
+
+\addcontentsline{toc}{section}{The Rotator $q()q^{-1}$}
+\section*{The Rotator $q()q^{-1}$}
+
+\item We may consider the ratio of two steps as determining a
+number, the antecedent being the product and the consequent the
+multiplicand of the number; \textit{viz.}, ${\mathbf OB/OA}$
+determines the number $r$ such that $r{\mathbf OA = OB}$. By Art.\
+21, equal step ratios determine equal numbers.
+
+If the several pairs of steps that are in a given ratio $r$ be
+given a rotation whose equatorial arc is $2\text{ arc }q$, they
+are still equal ratios in their new positions and determine a new
+number $r'$ that is called \textit{the number $r$ rotated through
+$2\text{ arc }q$.} In other words, the rotation of $r$ produces a
+number with the same tensor as $r$, and whose great circle and arc
+are the rotated great circle and arc of $r$.
+
+\item \textit{The number $r$ rotated through $2\text{ arc }q$ is
+the number $qrq^{-1}$.}
+
+For, 1st, $Tqrq^{-1} = Tq \cdot Tr(Tq)^{-1} = Tr$.
+
+\begin{center}
+\includegraphics[width=80mm]{images/Image21.png}
+\end{center}
+
+2d, let $A$ be an intersection of the great circle of $r$ with the
+great circle of $q$ and construct
+\begin{gather*}
+\overset\frown{AB} = \overset\frown{BA^{\prime}} = \text{arc }q, \quad
+\overset\frown{AC} = \text{arc }r,
+\intertext{and}
+\overset\frown{C'B} = \overset\frown{BC} = \text{arc }rq^{-1};
+\intertext{then}
+\overset\frown{C'A'} = \overset\frown{A'C''} = \text{arc }qrq^{-1}.
+\end{gather*}
+But by construction, the spherical triangles $ABC$, $A'BC'$ are
+equal, and therefore $\overset\frown{AC}$ and
+$\overset\frown{C'A'} ( = \overset\frown{A'C''})$ are arcs of
+equal length, and the corresponding angles at $A$, $A'$ are equal.
+Hence, when arc $r ( = \overset\frown{AC})$ is rotated through
+$2\text{ arc }q( = \overset\frown{AA'})$, it becomes arc
+$qrq^{-1}( = \overset\frown{A'C''})$.
+
+\addcontentsline{toc}{section}{Powers and Roots}
+\section*{Powers and Roots}
+
+\item An integral power, $q^n = q \cdot q \cdot q \cdots$ \textit{to
+n factors}, is determined by the equations,
+
+\begin{enumerate}
+\item $T \cdot q^n = Tq \cdot Tq \cdot Tq \cdots  = (Tq)^n$.
+
+\item $\text{arc }q^n = \text{arc }q + \text{arc }q + \text{arc }q
+\cdots = n \text{ arc }q \pm (\text{whole circumferences})$.
+
+To find $q^{\frac{1}{n}}$, \textit{the number whose $n$th power is
+$q$}, we have, by replacing $q$ by $q^{\frac{1}{n}}$ in
+(\textit{a}), (\textit{b}),
+
+\item $Tq = (T \cdot q^{\frac{1}{n}})^n$ or
+$T \cdot q^{\frac{1}{n}} = (Tq)^{\frac{1}{n}}$
+
+\item $\text{Arc }q = n\text{ arc }q^{\frac{1}{n}} \pm$ whole
+circumferences, or, arc $q^{\frac{1}{n}} = {\frac{1}{n}}$
+(arc $q \pm$ whole circumferences) $= {\frac{1}{n}}$ arc $q
++ {\frac{m}{n}}$ circumferences $\pm$ whole circumferences),
+where $m = 0,\,1,\,2,\,3,\,\cdots\, n-1$, successively.
+\end{enumerate}
+
+There are therefore $n$ $n$th roots of $q$ whose tensors are all
+equal and whose arcs lie on the great circle of $q$.
+
+When the base is a scalar, its great circle may be any great
+circle, so that there are an infinite number of quaternion $n$th
+roots of a scalar. On this account, the roots as well as the
+powers of a scalar are \textbf{limited to scalars.} By ordinary
+algebra, there are $n$ such $n$th roots, real and imaginary. There
+are also imaginary $n$th roots of $q$ besides the $n$ real roots
+found above; \textit{i.e.}, roots of the form $a+b\sqrt{-1}$,
+where $a$, $b$ are real quaternions.
+
+\addcontentsline{toc}{section}{Representation of Vectors}
+\section*{Representation of Vectors}
+
+\item Bold-face letters will be used as symbols of vectors only.
+In particular, $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$ will
+denote \textit{unit} vectors whose axes are respectively a unit
+length east, a unit length north, and a unit length up. More
+generally we shall use the step $\mathbf{AB}$ to denote the vector
+whose axis is a unit length in the direction of $\mathbf{AB}$, and
+whose tensor is the numerical length of $\mathbf{AB}$ ($= AB$:
+{\textit unit length}).
+
+This use of a step $\mathbf{AB}$ as the symbol of a vector is
+analogous to the use of $AB$ to represent a tensor ($AB$: {\textit
+unit length}), or of $\overline{AB}$ to represent a positive or
+negative scalar, according as it is measured in or against the
+direction of its axis of measurement. In none of these cases is
+the concrete quantity an absolute number; {\textit i.e.}, the
+value of the number that it represents varies with the assumed
+unit of length. When desirable, we distinguish between the vector
+$\mathbf{OA}$ and the step $\mathbf{OA}$ by enclosing the vector
+in a parenthesis.
+
+\item {\it If $q(\mathbf{OA}) = (\mathbf{OB})$, then
+$q \cdot \mathbf{OA} = \mathbf{OB}$, and conversely.}
+
+The tensor of $q$ in either equation is $OB:OA$. It is therefore
+only necessary to show that the arc of $q$ in one equation equals
+the arc of $q$ in the other equation in order to identify the two
+numbers that are determined by these two equations as one and the
+same number.
+
+\begin{center}
+\includegraphics[width=80mm]{images/Image22.png}
+\end{center}
+
+Draw the sphere of unit radius and centre $O$, cutting
+$\mathbf{OA}$, $\mathbf{OB}$ in $A'$, $B'$; then
+$\overset\frown{A'B'}$ is the arc of $q$ in the second equation.
+Draw the radius $OL$ perpendicular to the plane $OA'B'$ on the
+counter-clockwise side of $\overset\frown{A'B'}$, and draw
+counter-clockwise round $OA'$, $OB'$ as axes the quadrants
+$\overset\frown{LM}$, $\overset\frown{LN}$ respectively; then
+these are the arcs of $(\mathbf{OA})$, $(\mathbf{OB})$
+respectively, and since $\overset\frown{LM} + \overset\frown{MN} =
+\overset\frown{LN}$, therefore $\overset\frown{MN}$ is the arc of
+$q$ in the first equation. But since $\overset\frown{LM}$,
+$\overset\frown{LN}$ are quadrants, therefore the plane $OMN$ is
+perpendicular to $OL$, and must therefore coincide with the plane
+$OA'B'$, which is by construction also perpendicular to $OL$.
+Hence $\overset\frown{MN}$ lies on the great circle of
+$\overset\frown{A'B'}$, and by the construction of the figure, it
+must, when advanced $90^{\circ}$ on that great circle, coincide
+with $\overset\frown{A'B'}$. Hence the theorem.
+
+\small \textsc{Note}. This theorem shows that a number extends and
+turns vectors into vectors in the same way that it extends and
+turns steps into steps. Moreover, when the vector is not
+perpendicular to the axis of the multiplier, there is no resulting
+vector, since in the case of the corresponding step there is no
+resulting step. In the case of a vector multiplicand, that is
+oblique to the axis of $q$, the product is an actual quaternion
+that is not a vector, while in the case of the corresponding step
+multiplicand the product belongs to that class of products in
+which the multiplicand does not admit of the operation of the
+multiplier, as in \textit{$\sqrt{2}$ universities, -2 countries,
+etc}. \normalsize
+
+\begin{itemize}
+\item \textsc{Cor.\ 1.} \textit{The product of two vectors is a
+vector when, and only when, the factors are perpendicular to each
+other; the product is perpendicular to both factors; and its
+length (its tensor) is equal to the area of the rectangle on the
+lengths of the factors.}
+
+\small \textsc{Note}. The direction of the product $\mathbf{OA}
+\cdot \mathbf{OB = OC}$ is obtained by turning $\mathbf{OB}$ about
+$\mathbf{OA}$ as axis through a counter-clockwise right angle;
+thus $\mathbf{OC}$ lies on that side of the plane $OAB$ from which
+the right angle $AOB$ appears counter-clockwise. \normalsize
+
+\item \textsc{Cor.\ 2.} \textit{The product of two perpendicular
+vectors changes sign when the factors are interchanged.}
+($\mathbf{OB}\cdot \mathbf{OA = OC' = -OC}$.)
+
+\begin{center}
+\includegraphics[width=80mm]{images/Image23.png}
+\end{center}
+
+\item \textsc{Cor.\ 3.} \textit{The condition that $\alpha$ is
+perpendicular to $\beta$ is that $\alpha\beta = $ vector, or
+$S\alpha\beta = 0$}.
+\end{itemize}
+
+\item \textit{If $\mathbf{AB}$, $\mathbf{CD}$ are parallel, then
+$\mathbf{AB \cdot CD = CD \cdot AB} = -\overline{AB} \cdot
+\overline{CD}$, a scalar; and conversely, the product of two
+vectors is a scalar only when they are parallel.}
+
+Since the axes of the vectors $\mathbf{AB}$, $\mathbf{CD}$ are
+parallel, therefore their product is commutative. When the vectors
+are in the same direction, then each turns $90^{\circ}$ in the same
+direction, the resultant turn is $180^{\circ}$, and the product is
+negative; and when the vectors are in opposite direction, their
+turns are in opposite directions, the resultant turn is $0^{\circ}$,
+and the product is positive. This is just the opposite of the
+product of the corresponding scalars $\overline{AB}$,
+$\overline{CD}$, which is positive when the scalars are in the
+same direction (or both of the same sign), and negative when the
+scalars are in opposite directions; \textit{i.e.},
+$\mathbf{AB{\cdot}CD} = -\overline{AB}\cdot\overline{CD}$.
+
+Conversely, the product $\mathbf{AB}$, $\mathbf{CD}$ can be a
+scalar only when the resultant of their two turns of $90^{\circ}$
+each is a turn of $0^{\circ}$ or $180^{\circ}$; \textit{i.e.},
+only when the turns are cocircular, and therefore their axes
+parallel.
+
+\begin{itemize}
+\item \textsc{Cor.} \textit{The condition that $\alpha$ is parallel to
+$\beta$ is $\alpha\beta =$ scalar, or $V\alpha\beta = 0$}.
+\end{itemize}
+\end{enumerate}
+
+\addcontentsline{toc}{subsection}{Examples}
+\subsection*{Examples}
+
+\small \begin{enumerate}
+\item Prove by diagram that $(pq)^2$ and $p^2q^2$ are in general
+unequal.
+
+\item Find the 2d, 3d, 4th, 5th, 6th powers of $(2, 90^{\circ})$,
+$(2, -60^{\circ})$.
+
+\item Find the square roots and cube roots of $(4, 30^{\circ}),
+(8, -120^{\circ})$.
+
+\begin{enumerate}
+\item Find the values of $[(2, 50^{\circ})^6]^\frac{1}{3}$, $[(2,
+50^{\circ})^\frac{1}{3}]^{6}$, and $(2, 50^{\circ})^\frac{6}{3}$.
+\end{enumerate}
+
+\item What numbers are represented by 2 \textit{feet}, 2
+\textit{feet east}, the unit of length being a foot, a yard, an
+inch?
+
+\item Show that $\mathbf{i^2 = j^2 = k^2 = ijk = -1}$; $\mathbf{jk
+= i = -kj}$; $\mathbf{ki = j = -ik}$; $\mathbf{ij = k = -ji}$.
+
+\item Let $e^\mathbf{(AB)}$ denote the versor that turns
+counter-clockwise round the axis $AB$ through an arc that is
+formed by bending the length AB into an arc of unit radius. Show
+that if facing the west, and holding the paper in a north and
+south vertical plane, then $e^\mathbf{i}$, $e^\mathbf{2i}$,
+${\cdots}e^\mathbf{-i}$, $e^\mathbf{-2i}$, turn respectively 1, 2,
+$\cdots$ radians counter-clockwise, and 1, 2, $\cdots$ radians
+clockwise in the plane of the paper. Also show that
+$e^{\pm\frac{\pi}{2}\mathbf{i}} = \pm\mathbf{i}$,
+$e^{\pm\pi\mathbf{i}} = -1$ $e^{2n\pi\mathbf{i}} = 1$, where $n$
+is any integer.
+
+\item Show by diagram that $Se^{\theta\mathbf{i}} = \cos\theta$,
+$Ve^{\theta\mathbf{i}} = i\sin\theta$, where $\theta$ is any
+positive or negative number and the unit of angle is a radian.
+
+\item Show that if $\mathbf{OA}$ rotate into $\mathbf{OB}$ through
+$2\text{ arc }q$, then $(\mathbf{OB}) = q(\mathbf{OA})q^{-1}$.
+
+\item Show that if $\alpha$ be a vector in the plane of $q$, then
+$Kq = \alpha{q}\alpha^{-1} = \alpha^{-1}q\alpha$.
+
+\item Show that $pq$ rotates into $qp$, and determine two such
+rotations.
+
+\item Show that $SKq = Sq$, $VKq = -Vq$.
+
+\item Show that $K\alpha\beta = \beta\alpha$, $S\alpha\beta =
+S\beta\alpha$, $V\alpha\beta = -V\beta\alpha$.
+
+\item Show that $K\alpha\beta\gamma = -\gamma\beta\alpha$;
+$V\alpha\beta\gamma = V\gamma\beta\alpha$; $S\alpha\beta\gamma =
+S\beta\gamma\alpha = S\gamma\alpha\beta = -S\gamma\beta\alpha =
+-S\beta\alpha\gamma = -S\alpha\gamma\beta$. (\textit{a}) Determine the
+conjugate of a product of $n$ vectors.
+
+\item Prove by diagram that $Kpq = Kq \cdot Kp$.
+\end{enumerate} \normalsize
+
+\addcontentsline{toc}{section}{Addition}
+\section*{Addition}
+
+\begin{enumerate}
+\setcounter{enumi}{46}
+
+\item \textsc{Definition}. The sum $(p+q)$ is the number
+determined by the condition that its product is the sum of the
+products of $p$ and $q$.
+
+\begin{center}
+\includegraphics[width=80mm]{images/Image24.png}
+\end{center}
+
+Thus let $\mathbf{OA}$ be any step that is multiplied by both $p$
+and $q$, and let $p\mathbf{OA = OB}$, $q\mathbf{OA = OC}$, and
+$\mathbf{OB + OC = OD}$, then $(p+q)\mathbf{OA = OD}$. It is
+obvious that any change in $\mathbf{OA}$ alters $\mathbf{OB}$,
+$\mathbf{OC}$, $\mathbf{OD}$, proportionally, so that the value of
+the sum $p+q(= \mathbf{OD:OA})$ is the same for all possible
+values of $\mathbf{OA}$.
+
+Similarly, any quaternion, $r$, may be added to the sum $p+q$,
+giving the sum $(p+q)+r$; and we may form other sums such as
+$p+(q+r)$, $(q+r)+p$, etc. It will be shown later that all such
+sums of the same numbers are equal, or that quaternion addition is
+\textit{associative} and \textit{commutative}.
+
+\item \textit{The sum of a scalar and a vector is a quaternion
+with that scalar and that vector, and conversely, a quaternion is
+the sum of its scalar and its vector.}
+
+\begin{center}
+\includegraphics[width=80mm]{images/Image25.png}
+\end{center}
+
+For let $w$ be any scalar, and $\rho$ any vector, and let
+$w\mathbf{OA} = \mathbf{OL}$, \linebreak[4] $\rho\mathbf{OA =
+OM}$, then completing the rectangle $\mathbf{OLBM}$, we have
+\newline $(w + \rho)\mathbf{OA = OB}$, and the scalar of $w+\rho$
+is $w$, and its vector is $\rho$, since $\mathbf{OL}$,
+$\mathbf{OM}$ are the components of $\mathbf{OB}$ parallel and
+perpendicular to $\mathbf{OA}$. Similarly,
+\begin{equation*}
+q = Sq+Vq.
+\end{equation*}
+
+\item \textit{The scalar, vector, and conjugate, of any sum equals
+the like sum of the scalars, vectors, and conjugates of the terms
+of the sum.} [\textit{I.e., $S$, $V$, $K$, are distributive over a
+sum.}]
+
+\begin{center}
+\includegraphics[width=80mm]{images/Image26.png}
+\end{center}
+
+For let
+\begin{gather*}
+p\mathbf{OA} = \mathbf{OB}, \quad q\mathbf{OA = OC}, \\
+(p+q) \mathbf{OA} = \mathbf{OB + OC} = \mathbf{OD}.
+\end{gather*}
+Then the components of $\mathbf{OD}$ parallel and perpendicular to
+$\mathbf{OA}$ are, by the figure, the sums of the like components
+of $\mathbf{OB}$, $\mathbf{OC}$; \textit{i.e.}, $S(p+q) \cdot
+\mathbf{OA} = Sp \cdot \mathbf{OA} + Sq \cdot \mathbf{OA}$, or
+$S(p+q) = Sp+Sq$; and $V(p+q) \cdot \mathbf{OA} = Vp \cdot
+\mathbf{OA} + Vq \cdot \mathbf{OA}$, or $V(p+q) = Vp+Vq$.
+
+Also, if $OB'D'C'$ be the parallelogram that is symmetric to the
+parallelogram $OBDC$ with reference to $OA$ as axis of symmetry,
+then \linebreak $Kp \cdot \mathbf{OA = OB'}$, $Kq \cdot
+\mathbf{OA = OC'}$, and $K(p+q) \cdot \mathbf{OA = OD'}$, and
+since \linebreak $\mathbf{OB'+OC' = OD'}$, therefore $K(p+q) =
+Kp + Kq$.
+
+These results extend to any given sum; \textit{e.g.},
+$V[(p+q)+r] = V(p+q)+Vr = (Vp+Vq)+Vr$, etc.
+
+\item \textit{If $\mathbf{(OA)+(OB) = (OC)}$, then $\mathbf{OA +
+OB = OC}$, and conversely.}
+
+\begin{center}
+\includegraphics[width=80mm]{images/Image27.png}
+\end{center}
+
+For erect a pin $OD$ of unit length perpendicular to the plane of
+the angle $AOB$ on its counter-clockwise side; and turn $AOB$
+round $OD$ as axis through a clockwise right angle as seen from
+$D$ into the position $A'OB'$. Then since $({\mathbf OA})$ is the
+vector that turns through a counter-clockwise right angle round
+$OA$ as axis, and extends unit length into $OA = OA'$, therefore
+${\mathbf (OA)OD = OA'}$, and similarly ${\mathbf (OB)OD = OB'}$,
+and therefore ${\mathbf (OC)OD = OA'+OB' = OC'}$, where $OA'C'B'$
+is a parallelogram. Hence the step ${\mathbf OC}$ of proper length
+and direction to give the tensor and axis of the vector ${\mathbf
+(OC)}$ must be the diagonal of the parallelogram on ${\mathbf
+OA}$, ${\mathbf OB}$ as sides; and therefore ${\mathbf OA+OB =
+OC}$. Conversely, if ${\mathbf OA+OB = OC}$, then turning the
+parallelogram $OACB$ into the position $OA'C'B'$, we have, since
+${\mathbf OA'+OB' = OC'}$, that ${\mathbf (OA)+(OB) = (OC)}$.
+
+\begin{itemize}
+\item \textsc{Cor.\ 1.} \textit{Vectors add in the same way as
+their corresponding steps, and all the laws of addition and
+resolution of steps extend at once to vectors.}
+
+\item \textsc{Cor.\ 2.} \textit{A sum of quaternions is
+associative and commutative.}
+\end{itemize}
+
+For since by Cor.\ 1 a sum of vectors is independent of the way in
+which its terms are added, and since we know that a sum of scalars
+(i.e., ordinary numbers) is independent of the way in which its
+terms are added, therefore by Art.\ 49 the scalar and the vector of
+a sum are independent of the way in which the sum is added. Hence
+the sum is independent of the way in which it is added, since it
+is equal to the sum of its scalar and its vector.
+
+\item \textsc{Lemma.} \textit{If $p$, $q$ be any quaternions, then
+$(1+p)q = q+pq$.}
+
+\begin{center}
+\includegraphics[width=80mm]{images/Image28.png}
+\end{center}
+
+For take $\mathbf{OB}$ in the intersection of the planes of $p$,
+$q$ and draw $\mathbf{OA}$, $\mathbf{OC}$ such that
+$q \cdot \mathbf{OA = OB}$, $p\mathbf{OB = OC}$; then
+$(1+p)q \cdot \mathbf{OA} = (1+p)\mathbf{OB} = \mathbf{OB+OC} =
+q\mathbf{OA} + pq\mathbf{OA}$. Hence,
+\begin{equation*}
+(1+p)q=q+pq.
+\end{equation*}
+
+\item \textit{If $p$, $q$, $r$ be any quaternions, then $(p+q)r =
+pr+qr$.}
+
+For we have, $(1+qp^{-1})p \cdot r = (1+qp^{-1}) \cdot pr$, and
+expanding each member by the preceding lemma, we have, $(p+q)r =
+pr+qr$.
+
+This result extends to any sum; \textit{e.g.},
+\begin{equation*}
+(p+q+r+s)t = [(p+q)+(r+s)]t = (p+q)t+(r+s)t = pt+qt+rt+st.
+\end{equation*}
+
+\begin{itemize}
+\item \textsc{Cor.\ 1.} $r(p+q) = rp+rq$.
+
+For let $p'$, $q'$, $r'$ be the conjugates of $p$, $q$, $r$. Then
+from $(p'+q')r' = p'r'+q'r'$, we have, by taking the conjugates of
+each member, $r(p+q) = rp+rq$. [Art.\ 39, Cor.\ 2; Art.\ 49.]
+
+\item \textsc{Cor.\ 2.} \textit{A product of sums equals the sum
+of all partial products that may be formed from the given product
+by multiplying together, in the order in which they stand, a term
+from each factor of the product.}
+
+\textit{E.g., $(p+q)(r+s) = pr+ps+qr+qs$.}
+\end{itemize}
+
+\small \textsc{Note}.---This rule should be used even when the
+factors are commutative, as it prevents all danger of taking out
+the same partial product twice; \textit{e.g.}, from taking both
+$pr$ and $rp$ from the above product. To be sure that all the
+partial products are found, some system of arrangement should be
+adopted; also the total number of partial products should be
+determined. \normalsize
+
+\textit{E.g.}, $(p+q)(p+q)(p+q)$ may be arranged according to the
+degrees of the terms in $p$, and there are $2\times2\times2 = 8$
+terms. This product is then easily seen to be
+\begin{gather*}
+p^3+(p^2q+pqp+qp^2)+(pq^2+qpq+q^2p)+q^3,
+\intertext{when $p$, $q$ are not commutative, and}
+p^3+3p^2q+3pq^2+q^3,
+\end{gather*}
+when $p$, $q$ are commutative.
+
+\addcontentsline{toc}{section}{Formulas}
+\section*{Formulas. For Exercise and Reference}
+
+\item \begin{enumerate} \item $q = Tq \cdot Uq = Uq \cdot Tq$.
+
+\item $q = Sq+Vq$; $Kq = Sq-Vq$.
+
+\item $Sq = Tq\cos\angle{q}, = r\cos\theta$, say;
+$TVq = Tq\sin\angle{q} = r\sin\theta.$
+
+\item $Vq = TVq \cdot UVq = r\sin\theta\epsilon$ where $\epsilon =
+UVq$.
+
+\item $q = r(\cos\theta + \epsilon \cdot \sin\theta) =
+re^{\theta\epsilon}$, $Kq = r (\cos\theta -
+\epsilon\sin\sin\theta) = re^{-\theta\epsilon}$
+
+\item $e^{\theta\epsilon} \cdot e^{\theta'\epsilon} =
+e^{(\theta+\theta')\epsilon}$.
+
+\item $Sq = \frac{1}{2}(q+Kq)$; $Vq = \frac{1}{2}(q-Kq)$.
+
+\item $Tq^2 = qKq = Kq \cdot q = (Sq)^2-(Vq)^2 = (Sq)^2+(TVq)^2$.
+
+\item $q^{-1} = Kq/Tq^2$.
+\end{enumerate}
+
+\textit{As a further exercise find the $T$, $U$, $S$, $V$, $K$ of
+the $T$, $U$, $S$, $V$, $K$ of $q$, in terms of $r$, $\theta$,
+$\epsilon$.}
+
+\item \begin{enumerate} \item $T(\cdots pqr) = \cdots Tp \cdot Tq
+\cdot Tr$.
+
+\item $U(\cdots pqr) = \cdots Up \cdot Uq \cdot Ur$.
+
+\item $\angle(\cdots pqr) = \angle(r \cdots pq) =
+\angle(qr \cdots p)$, etc.
+
+\item $S(\cdots pqr) = S(r \cdots pq) = S(qr \cdots p)$, etc.
+
+\item $TV(\cdots pqr)= TV(r \cdots pq) = TV(qr \cdots p)$, etc.
+
+\item $\text{arc }(\cdots pqr) = \text{arc }r + \text{arc }q +
+\text{arc }p + \cdots$.
+
+\item $(\cdots pqr)^{-1} = r^{-1}q^{-1}p^{-1}\cdots$.
+
+\item $K(\cdots pqr) = Kr \cdot Kq \cdot Kp \cdots$.
+
+\item $S(xp+yq+zr) = xSp+ySq+zSr$, [$x$, $y$, $z$,
+\textit{scalars}] \textit{and similarly for $V$ or $K$ instead of
+$S$}.
+\end{enumerate}
+
+\item \begin{enumerate} \item $K\alpha = -\alpha$; $T\alpha^2 =
+-\alpha^2$; $S\alpha = 0$; $V\alpha = \alpha$.
+
+\item $K\alpha\beta = \beta\alpha$; $S\alpha\beta = S\beta\alpha$;
+$V\alpha\beta = -V\beta\alpha$.
+
+\item $\alpha\beta + \beta\alpha = 2S\alpha\beta$,
+$\alpha\beta - \beta\alpha = 2V\alpha\beta$.
+
+\item $(\alpha \pm \beta)^2 = \alpha^2 \pm 2S\alpha\beta +
+\beta^2$.
+
+\item $V(x\alpha + y\beta)(x'\alpha + y'\beta) = (xy' - x'y)V\alpha\beta
+= \begin{vmatrix}
+x & y \\
+x'& y'\\
+\end{vmatrix} Va\beta$, say. [$x$, $y$, $x'$, $y'$, scalars.]
+
+\item $V(x\alpha + y\beta + z\gamma)(x'\alpha + y'\beta + z'\gamma)=
+\begin{vmatrix} y&z \\ y'&z' \end{vmatrix}V\beta\gamma +
+\begin{vmatrix} z&x \\ z'&x' \end{vmatrix}V\gamma\alpha +
+\begin{vmatrix} x&y \\ x'&y' \end{vmatrix}V\alpha\beta$. [$x$, $y$,
+$z$, $x'$, $y'$, $z'$, scalars.]
+\end{enumerate}
+
+\item \begin{enumerate} \item $K\alpha\beta\gamma =
+-\gamma\beta\alpha$.
+
+\item $\alpha\beta\gamma - \gamma\beta\alpha = 2S\alpha\beta\gamma
+= -2S\gamma\beta\alpha$.
+
+Hence the scalars of the six products of $\alpha$, $\beta$,
+$\gamma$ are equal to one of two negative numbers according to the
+cyclic order of the product; and an interchange in two factors
+(which changes the cyclic order) changes the sign of the scalar of
+the product. When two of the three factors are equal, the scalar
+of their product must therefore be zero, since an interchange of
+the equal factors changes the sign without changing the value.
+
+\item $S \cdot (x\alpha + y\beta + z\gamma)(x'\alpha + y'\beta + z'\gamma)
+(x''\alpha + y''\beta + z''\gamma) \\
+= \Bigl\{ x \begin{vmatrix} y'&z' \\ y''&z'' \end{vmatrix}
++ y \begin{vmatrix} z'&x' \\ z''&x'' \end{vmatrix}
++ z \begin{vmatrix} x'&y' \\ x''&y'' \end{vmatrix} \Bigr\}
+S\alpha\beta\gamma \\
+= \begin{vmatrix} x&y&z \\ x'&y'&z' \\ x''&y''&z'' \end{vmatrix}
+S\alpha\beta\gamma$, say. [$x$, $y$, $z$, etc., scalars.]
+
+\item $S\alpha\beta\gamma = S\alpha V\beta\gamma = S\beta
+V\gamma\alpha = S\gamma V\alpha\beta.$
+
+[Replace $\beta\gamma$ by $S\beta\gamma + V\beta\gamma$, expand by
+54 (\textit{i}), and note that \linebreak[3] $S \cdot \alpha
+S\beta\gamma = 0.$]
+
+\item $\alpha\beta\gamma + \gamma\beta\alpha = 2V\alpha\beta\gamma
+= 2V\gamma\beta\alpha.$
+
+\small \textsc{Note}.---Insert between the two terms of the first
+member of (\textit{e}), the null term $(\alpha\gamma\beta -
+\alpha\gamma\beta - \gamma\alpha\beta + \gamma\alpha\beta)$, and
+it becomes $\alpha(\beta\gamma + \gamma\beta)-(\alpha\gamma
++\gamma\alpha)\beta + \gamma(\alpha\beta + \beta\alpha)$. Hence,
+using (55 \textit{c}), we have (\textit{f}). \normalsize
+
+\item $V\alpha\beta\gamma = \alpha S\beta\gamma - \beta
+S\gamma\alpha + \gamma S\alpha\beta.$
+
+Transpose the first term of the second member of (\textit{f}) to
+the first member, noting that $\alpha S\beta\gamma = V\cdot\alpha
+S\beta\gamma$, and $\beta\gamma - S\beta\gamma = V\beta\gamma$,
+and we have
+
+\item $V\alpha V\beta\gamma = -\beta S\gamma\alpha + \gamma
+S\alpha\beta$;
+
+($g'$) $V\cdot (V\beta\gamma)\alpha = \beta S\gamma\alpha - \gamma
+S\alpha\beta.$
+
+\item \begin{align}
+V\cdot(V\alpha\beta)V\gamma\delta &= -\gamma S\alpha\beta\delta
+    + \delta S\alpha\beta\gamma \tag*{[(\textit{g}), (\textit{d})]} \\
+ &= \alpha S\beta\gamma\delta - \beta S\alpha\gamma\delta.
+   \tag*{[($g'$), (\textit{d})]}
+\end{align}
+
+\item \begin{equation}
+\delta S\alpha\beta\gamma = \alpha S\beta\gamma\delta +
+\beta S\gamma\alpha\delta + \gamma S\alpha\beta\delta. \tag*{[(\textit{h})]}
+\end{equation}
+
+Replace $\alpha, \beta, \gamma,$ by $V\beta\gamma, V\gamma\alpha,
+V\gamma\beta,$ noting that $V \cdot (V\gamma\alpha \cdot
+V\alpha\beta) = -\alpha S\alpha\beta\gamma,$ etc., and that
+$S(V\beta\gamma \cdot V\gamma\alpha \cdot V\alpha\beta)
+= -(S\alpha\beta\gamma)^2$, and we have
+
+\item $\delta S\alpha\beta\gamma = V\beta\gamma S\alpha\delta +
+V\gamma\alpha S\beta\delta + V\alpha\beta S\gamma\delta$.
+
+\small \textsc{Note}.---(\textit{i}), (\textit{j}) may be obtained
+directly by putting $\delta = x\alpha + y\beta + z\gamma$ or
+$xV\beta\gamma + yV\gamma\alpha + zV\alpha\beta$, and finding $x$,
+$y$, $z$, by multiplying in the first case by $\beta\gamma$,
+$\gamma\alpha$, $\alpha\beta$, and in the second case by $\alpha$,
+$\beta$, $\gamma$, and taking the scalars of the several products.
+\normalsize
+\end{enumerate}
+
+\item \begin{enumerate} \item $\mathbf{i}^2 = \mathbf{j}^2 =
+\mathbf{k}^2 = \mathbf{ijk} = -1$; $\mathbf{jk} = \mathbf{i} =
+-\mathbf{kj}$; $\mathbf{ki} = \mathbf{j} = -\mathbf{ik}$,
+$\mathbf{ij} = \mathbf{k} = -\mathbf{ji}$.
+
+\item $\rho = \mathbf{i}S\mathbf{i}\rho -
+\mathbf{j}S\mathbf{j}\rho - \mathbf{k}S\mathbf{k}\rho$. [56
+(\textit{i}) or (\textit{j}) or directly as in note.]
+
+Let $\rho = x\mathbf{i} + y\mathbf{j} + \mathbf{z}k$, $\rho' =
+x'\mathbf{i} + y'\mathbf{j} + z'\mathbf{k}$, etc. [$x$, $y$, $z$,
+etc. scalars.]
+
+Then, prove by direct multiplication,
+
+\item $-\rho^2 = x^2 + y^2 + z^2 = T\rho^2$.
+
+\item $-S\rho\rho' = xx' + yy' + zz' = -s\rho'\rho$.
+
+\item $v\rho\rho'%
+    = \left|\begin{matrix}y & z \\ y' & z' \end{matrix}\right|\mathbf{i}%
+    + \left|\begin{matrix}z & x \\ z' & x' \end{matrix}\right|\mathbf{j}%
+    + \left|\begin{matrix}x & y \\ x' & y' \end{matrix}\right|\mathbf{k}%
+    = -V\rho'\rho$.
+
+\item $-S\rho\rho'\rho''
+   = \left|\begin{matrix} x   & y   & z   \\%
+                          x'  & y'  & z'  \\%
+                          x'' & y'' & z'' \end{matrix}\right|%
+    = -S\rho V\rho'\rho''$.
+\end{enumerate}
+
+\addcontentsline{toc}{section}{Geometric Theorems}
+\section*{Geometric Theorems}
+
+\item \textit{The angle of $\alpha\beta$ equals the supplement of
+the angle $\theta$ between $\alpha$, $\beta$.}
+
+\begin{center}
+\includegraphics[width=80mm]{images/Image29.png}
+\end{center}
+
+For, since $\alpha\beta \cdot \beta^{-1} = \alpha$, therefore
+$\alpha\beta$ turns through the angle from $\beta^{-1}$ to
+$\alpha$, which is the supplement of the angle $\theta$ from
+$\alpha$ to $\beta$.
+
+\begin{itemize}
+\item \textsc{Cor.} $S\alpha\beta = -T\alpha\beta\cos\theta$,
+$TV\alpha\beta = T\alpha\beta\sin\theta$. [$Sq = Tq\cos\angle{q}$,
+etc.]
+\end{itemize}
+
+\item \textit{The scalar of $\alpha\beta$ equals the product of
+$\alpha$ and the projection of $\beta$ upon it; the vector of
+$\alpha\beta$ equals the product of $\alpha$ and the projection of
+$\beta$ perpendicular to it, and $V\alpha\beta$ is a vector
+perpendicular to $\alpha$, $\beta$ on their counter-clockwise side
+whose length equals the area of the parallelogram on $\alpha$,
+$\beta$ as sides.}
+
+\begin{center}
+\includegraphics[width=80mm]{images/Image30.png}
+\end{center}
+
+Let $\beta_1$, $\beta_2$ be the components of $\beta$ parallel and
+perpendicular to $\alpha$, then $\beta = \beta_1 + \beta_2$ and
+$\alpha\beta = \alpha\beta_1 + \alpha\beta_2 =
+\mathit{scalar+vector}$. Hence
+\begin{equation*}
+S\alpha\beta = \alpha\beta_1, \text{ as stated; and } V\alpha\beta =
+\alpha\beta_2,
+\end{equation*}
+which is $\beta_2$ turned a counter-clockwise right angle round
+$\alpha$ and lengthened by $T\alpha$. Hence $V\alpha\beta$ is
+perpendicular to $\alpha$, $\beta$ on their counterclockwise side
+(towards the reader in the figure), and its length is
+$T\alpha \cdot T\beta_2 = $ area parallelogram on $\alpha$,
+$\beta$, as sides.\footnote{The parallelogram on $\alpha$, $\beta$
+may be considered as bounded by the path of a point that receives
+the displacement $\alpha$, then the displacement $\beta$, then the
+displacement $-\alpha$, then the displacement $-\beta$. This area
+is therefore bounded \textit{counter-clockwise} round
+$V\alpha\beta$ as axis; and $V\alpha\beta$ may therefore be called
+the vector measure of the area of this directed parallelogram or
+of any parallel plane area of the same magnitude and direction of
+boundary.}
+
+\begin{itemize}
+\item \textsc{Cor.\ 1.} \textit{The projections of $\beta$
+parallel and perpendicular to $\alpha$ equal
+$\alpha^{-1}S\alpha\beta$ and $\alpha^{-1}V\alpha\beta$.}
+
+\item \textsc{Cor.\ 2.} \textit{The scalar measure of the
+projection of $\beta$ upon $\alpha$ is
+$\frac{-S\alpha\beta}{T\alpha}$, and the tensor measure of the
+projection of $\beta$ perpendicular to $\alpha$ is
+$\frac{TV\alpha\beta}{T\alpha}$.} [Also from 58, Cor.]
+
+\item \textsc{Cor.\ 3.} \textit{If $\theta$ be the angle between
+$\alpha$, $\beta$, then $\cos\theta =
+-\frac{S\alpha\beta}{T\alpha\beta}$, $\sin\theta =
+\frac{TV\alpha\beta}{T\alpha\beta}$.} [Also from 58, Cor.]
+\end{itemize}
+
+\item \textit{The volume of a parallelepiped on $\alpha$, $\beta$,
+$\gamma$ as edges is $-S\alpha\beta\gamma$ (the volume being
+positive or negative according as $\alpha$ lies on the
+counter-clockwise or clockwise side of $\beta$, $\gamma$).}
+
+\begin{center}
+\includegraphics[width=80mm]{images/Image31.png}
+\end{center}
+
+For let $\alpha_1$ be the projection of $\alpha$ upon
+$V\beta\gamma$; then taking the face $\beta$, $\gamma$ of the
+parallelepiped as the base, we have by Art.\ 59 that
+$TV\beta\gamma$ is the area of the base; also $T\alpha_1$ is the
+altitude. Hence \textit{numerical volume}
+\begin{align}
+&= T\alpha_1 \cdot TV\beta\gamma = \mp\alpha_1 V\beta\gamma \tag*{[Art.\ 46.]}\\
+&= \mp S\alpha{V}\beta\gamma = \mp{S}\alpha\beta\gamma. \tag*{[59,
+56, \textit{d}.]}
+\end{align}
+The upper or lower sign must be taken according as $\alpha_1$,
+$V\beta\gamma$ are in the same or opposite directions. This
+numerical result must be multiplied by -1 when $\alpha$ lies on
+the clockwise side of $\beta$, $\gamma$; i.e., when $\alpha_1$,
+$V\beta\gamma$ are opposites (since $V\beta\gamma$ lies on the
+counter-clockwise side of $\beta$, $\gamma$).
+Hence---$S\alpha\beta\gamma$ is the required algebraic volume.
+
+\begin{itemize}
+\item \textsc{Cor.} \textit{The condition that $\alpha$, $\beta$,
+$\gamma$ are coplanar vectors is that $S\alpha\beta\gamma = 0$ (or
+$\alpha\beta\gamma =$ a vector)}.
+\end{itemize}
+\end{enumerate}
+
+\addcontentsline{toc}{subsection}{Examples}
+\subsection*{Examples}
+
+\small \begin{enumerate}
+\item Expand $(p+q+r)^2$, $(\alpha+\beta+\gamma)^2$, $(p+q)(p-q)$,
+$(p-q)(p+q)$, $(p+q)K(p+q)$. Show that $T(p+q)^2 = Tp^2 + 2SpKq +
+Tq^2$.
+
+\item Solve $q^2+ 4\mathbf{k}q-8 = 0$ for $q$. [$q$ must be
+cocircular with $\mathbf{k}$. Hence $q = -2\mathbf{k}\pm 2$ are
+the real solutions.]
+
+\item Find the tensor, versor, scalar, vector, and angle of each
+of the numbers: $2$, $-3$, $3\mathbf{i}$, $2 + 3\mathbf{i}$,
+$\mathbf{i+j}$, $3\mathbf{i}+4\mathbf{j}$,
+$5e^{\frac{\pi}{3}\mathbf{i}}$,
+$(2\mathbf{i}+3\mathbf{j}+6\mathbf{k})^2$.
+
+\item Show that the three quaternion cube roots of $-1$, with
+horizontal great circle, are $-1$,
+$\frac{1}{2}\pm\frac{1}{2}\sqrt{3}\mathbf{k}$.
+
+\item Show geometrically that
+$e^{\theta\epsilon} + e^{-\theta\epsilon} = 2\cos\theta$,
+$\epsilon^{\theta\epsilon} - e^{-\theta\epsilon} =
+2\epsilon\sin\theta$.
+
+\item The numbers $e^\alpha$ and
+\begin{equation*}
+1+\alpha+\frac{\alpha^2}{2!}+\frac{\alpha^3}{3!}+\frac{\alpha^4}{4!}+\cdots
+\end{equation*}
+are equal. Verify this approximately by geometric construction
+when $T\alpha = 1$, and when $T\alpha = 2$. [For the series,
+construct $\mathbf{OA}$, $\mathbf{AB} = \alpha\mathbf{OA}$,
+$\mathbf{BC} = \frac{1}{2}\alpha\mathbf{AB}$, $\mathbf{CD} =
+\frac{1}{3}\alpha\mathbf{BC}$, $\mathbf{DE} =
+\frac{1}{4}\alpha\mathbf{CD}$, etc.]
+
+\item In the plane triangle $ABC$, whose sides opposite $A$, $B$,
+$C$, are $a$, $b$, $c$, show by squaring $\mathbf{BC = AC-AB}$,
+that as $a^2 = b^2 + c^2 - 2bc\cos{A}$; also from
+\begin{equation*}
+V\mathbf{BCCA} = V\mathbf{CAAB} = V\mathbf{ABBC}
+\end{equation*}
+show that $a:b:c = \sin{A}:\sin{B}:\sin{C}$.
+
+\item From $e^{\theta\epsilon} = \cos\theta + \epsilon\sin\theta$,
+$e^{\theta\epsilon} \cdot e^{\theta'\epsilon} =
+e^{(\theta+\theta')\epsilon}$, show that
+\begin{align*}
+\cos(\theta+\theta') &= \cos\theta\cos\theta' - \sin\theta\sin\theta', \\
+\sin(\theta+\theta') &= \sin\theta\cos\theta' + \cos\theta\sin\theta'.
+\end{align*}
+
+\item Show that $(\cos\theta + \epsilon\sin\theta)^n =
+\cos{n\theta} + \epsilon\sin{n\theta}$.
+
+\item Show that $Spq = SpSq + S(Vp \cdot Vq)$, and hence that
+$\cos\angle{pq} = \cos\angle{p} \cdot \cos\angle{q} + \sin\angle{p}
+\cdot \sin\angle{q} \cdot \cos\angle{(Vp \cdot Vq)}$.
+
+\item If $\text{arc } q = \overset\frown{BA}$, $\text{arc } p
+= \overset\frown{AC}$, show that the last equation of Ex.\ 10 is
+the property
+\begin{equation*}
+\cos{a} = \cos{b}\cos{c}+\sin{b}\sin{c}\cos{A}
+\end{equation*}
+of the spherical triangle $ABC$. [Draw $\overset\frown{B'A} =
+\text{arc } Vq$, $\overset\frown{AC'} = \text{arc } Vp$.]
+
+\item If $\alpha$, $\beta$, $\gamma$, $\alpha'$, $\beta'$,
+$\gamma'$ are vectors from $O$ to the vertices $A$, $B$, $C$,
+$A'$, $B'$, $C'$ of two spherical triangles on the unit sphere
+$O$, where $\alpha' = UV\beta\gamma$, $\beta' = UV\gamma\alpha$,
+$\gamma' = UV(\alpha\beta)$; then $\alpha = UV\beta'\gamma'$,
+$\beta = UV\gamma'\alpha'$, $\gamma = UV\alpha'\beta'$, and the
+two triangles are polar triangles.
+
+\item In Ex.\ 12 show that $\cos\alpha = -S\beta\gamma$,
+$\sin\alpha = TV\beta\gamma$, etc.; $\cos{A} = S(UV\gamma\alpha
+\cdot UV\alpha\beta) = S\beta'\gamma' = -\cos\alpha'$, etc. Hence
+$\angle{A}$, $\angle\alpha'$ are supplements, etc.
+
+\item Show that the equation of Ex.\ 11 follows from the identity,
+$-S\beta\gamma = S(\beta\gamma \cdot \gamma\alpha) = S\beta\gamma
+\cdot S\gamma\alpha + S(V\gamma\alpha \cdot V\alpha\beta)$.
+
+\item From $V(V\gamma\alpha \cdot V\alpha\beta) =
+-\alpha{S}\alpha\beta\gamma$, and the similar equations found by
+advancing the cyclic order $\alpha$, $\beta$, $\gamma$, show that
+we have in the spherical triangle $ABC$,
+\begin{equation*}
+\sin{a}:\sin{b}:\sin{c} = \sin{A}:\sin{B}:\sin{C}.
+\end{equation*}
+
+\item Show that if $\alpha$, $\beta$, $\gamma$ are coplanar unit
+vectors, then $\alpha\beta\gamma = -\alpha\beta^{-1} \cdot \gamma$
+$=$ ($\gamma$ turned through the angle from $\beta$ to $\alpha$
+and reversed) $=$ ($\beta$ rotated $180^{\circ}$ about the exterior
+bisector of the angle between $\alpha$, $\gamma$) $=$
+$(\alpha-\gamma)\beta(\alpha-\gamma)^{-1}$.
+
+\item Show that $(V\mathbf{ABCD})^{-1}S\mathbf{AC}V\mathbf{ABCD}$
+is the shortest vector from the line $AB$ to the line $CD$.
+[Project $\mathbf{AC}$ upon the common perpendicular to $AB$,
+$CD$.]
+
+\item If $\alpha$, $\beta$, $\gamma$ be the vector edges about a
+vertex of an equilateral pyramid (whose edges are unit lengths),
+then $\beta-\gamma$, $\gamma-\alpha$, $\alpha-\beta$, are the
+remaining vector edges. Hence show that $S\beta\gamma =
+S\gamma\alpha = S\alpha\beta = -\frac{1}{2}$, and $V\alpha
+V\beta\gamma = (-\beta)S\gamma\alpha + \gamma{S}\alpha\beta =
+\frac{1}{2}(\beta-\gamma)$. Also show that:
+
+\begin{enumerate}
+\item The face angles are $60^{\circ}$, the area of a face is
+$\frac{1}{4}\sqrt{3}$, and its altitude is $\frac{1}{2}\sqrt{3}$.
+
+\item Opposite edges are perpendicular, and their shortest
+distance is $\frac{1}{2}\sqrt{2}$.
+
+\item The angle between a face and an edge is
+$\cos^{-1}\frac{1}{3}\sqrt{3}$.
+
+\item The angle between two adjacent faces is
+$\sin^{-1}\frac{2}{3}\sqrt{2}$.
+
+\item The volume and altitude of the pyramid are
+$\frac{1}{12}\sqrt{2}$, $\frac{1}{3}\sqrt{3}$.
+\end{enumerate}
+
+\item The cosines of the angles that a vector makes with
+$\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$, are called its
+\textit{direction cosines}. Find the lengths and direction cosines
+of
+\begin{equation*}
+\mathbf{2i-3j+6k}, \mathbf{i+2j-2k},
+x\mathbf{i}+y\mathbf{j}+z\mathbf{k}.
+\end{equation*}
+
+\item Show that the sum of the squares of the direction cosines of
+a line equals 1.
+
+\item If $(l, m, n)$, $(l', m', n')$ are the direction cosines of
+two lines, show that $l\mathbf{i}+m\mathbf{j}+n\mathbf{k}$,
+$l'\mathbf{i}+m'\mathbf{j}+n'\mathbf{k}$, are unit vectors in the
+directions of the lines, and that if $\theta$ be the angle between
+the lines, then $\cos\theta = ll'+mm'+nn'$; also that $\sin^2
+\theta =
+\begin{vmatrix}
+m & n   \\
+m' & n' \\
+\end{vmatrix}^2
++
+\begin{vmatrix}
+n & l   \\
+n' & l' \\
+\end{vmatrix}^2
++
+\begin{vmatrix}
+l & m   \\
+l' & m' \\
+\end{vmatrix}^2$,
+and that the three terms of the second member, respectively
+divided by $\sin^2\theta$, are the squares of the direction
+cosines of a line that is perpendicular to the given lines. [Art.\
+57.]
+
+\item If $O$ be a given origin, then the vector $\mathbf{OP} =
+\rho = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$ say, is called the
+\textit{vector of $P$ with reference to the given origin}. If
+$OX$, $OY$, $OZ$ be axes in the directions of $\mathbf{i}$,
+$\mathbf{j}$, $\mathbf{k}$, the scalar values of the projections
+of $\mathbf{OP}$ upon these axes, \textit{i.e.}, $(x,y,z)$, are
+called the \textit{co\"{o}rdinates of $P$ with reference to the given
+axes}. Let the co\"{o}rdinates of the vertices of the pyramid $ABCD$
+be, respectively, (8,2,7), (10,6,3), (1,6,3), (9,10,11). Draw
+this pyramid with reference to a perspective of $\mathbf{i}$,
+$\mathbf{j}$, $\mathbf{k}$, showing co\"{o}rdinates and vectors. Also:
+
+\begin{enumerate}
+\item Find the vectors and co\"{o}rdinates of the middle points of the
+edges. \linebreak[4] [$\mathbf{OM} = \frac{1}{2}\mathbf{(OA + OB)}$, etc.]
+
+\item Find the lengths and direction cosines of the edges.
+[$\mathbf{-AB^2} = AB^2$, etc.]
+
+\item Find vectors that bisect the face angles. [$U\mathbf{AC}
+{\pm}U\mathbf{AD}$ bisects $\angle CAD$.]
+
+\item Find altitudes of the faces and the vectors of their feet.
+[If $L$ be the foot of the perpendicular from $B$ on $AC$, then
+$\mathbf{AL = AB^{-1}}S\mathbf{ABAC}$, etc.]
+
+\item Find the areas of the faces.
+
+\item Find the volume and altitudes of the pyramid.
+
+\item Find the angles between opposite edges, and their (shortest)
+distance apart. [Ex.\ 17.]
+
+\item Find the angle between two adjacent faces.
+\end{enumerate}
+\end{enumerate} \normalsize
+
+\newpage
+\chapter{Equations of First Degree}
+
+\addcontentsline{toc}{section}{Scalar Equations, Plane and
+Straight Line}
+
+\begin{enumerate}
+\setcounter{enumi}{60}
+
+\item The general equation of first degree in an unknown vector
+$\rho$ is of the form,
+
+\begin{enumerate}
+\item $q_1{\rho}r_1+q_2{\rho}r_2+\cdots = q$,
+\end{enumerate}
+
+where $q$, $q_1$, $r_1$, $q_2$, $r_2$, $\cdots$ are known numbers.
+
+This equation may be resolved into two equations by taking the
+scalar and the vector of each member; and we shall consider these
+equations separately.
+
+\item Taking the scalar of (\textit{a}), Art.\ 61, the term
+$Sq_1{\rho}r_1$ becomes, by a cyclic change in the factors,
+$S \cdot r_1q_1\rho$, and this becomes [by dropping the vector
+$(Sr_1q_1)\rho$, since its scalar is zero] $S(Vr_1q_1 \cdot \rho)$;
+and similarly for the other terms. Hence if we put
+$Vr_1q_1 + Vr_2q_2 + \cdots = \delta$, and $Sq = d$, the general
+scalar equation of first degree in $\rho$ becomes,
+
+\begin{enumerate}
+\item $S\delta\rho = d$ or $S\delta(\rho - d\delta^{-1}) = 0$.
+\end{enumerate}
+
+One solution of this equation is obviously $\rho = d\delta^{-1}$.
+This is not the only solution, since by Art.\ 45, Cor.\ 3, the
+second factor may be any vector that is perpendicular to $\delta$.
+Hence the general solution is $\rho = d\delta^{-1} + V\sigma\delta$,
+where $\sigma$ is an arbitrary vector.
+
+\item Hence, draw $\mathbf{OD} = \delta$, take $N$ on the line
+$OD$ so that $\mathbf{ON} = d\delta^{-1}$, and draw any vector
+$\mathbf{NP} = V\sigma\delta$ that is perpendicular to the line
+$OD$; then $\rho = \mathbf{OP}$ is a solution of the equation
+$S\delta\rho = d$.
+
+\begin{center}
+\includegraphics[width=80mm]{images/Image32.png}
+\end{center}
+
+The locus of $P$ is therefore a plane perpendicular to $OD$ at the
+point $N$; and this plane is called \textit{the locus of the
+equation $S\delta\rho = d$, with respect to the origin O}. [The
+locus is the assemblage of all points that satisfy the equation.]
+
+\item \textit{The vector perpendicular distance from the plane
+$S\delta\rho - d = 0$ to the point $P'$ (whose vector is $\rho'$) is
+$\delta^{-1}(S\delta\rho' - d)$, and the corresponding scalar
+distance measured upon $\delta$ is}
+\begin{equation*}
+\frac{-(S\delta\rho'-d)}{T\delta}.
+\end{equation*}
+
+For the perpendicular distance of $P'$ is the projection of
+$\mathbf{NP'} = (\rho' - d\delta^{-1})$, upon $\mathbf{OD} =
+\delta$.
+
+\item \textit{The locus of the simultaneous equations $Sa\rho =
+a$, $S\beta\rho = b$ is a straight line, viz., the intersection of
+the two plane loci of these equations taken separately.}
+
+For in order that $\rho = \mathbf{OP}$ may satisfy both equations,
+$P$ must lie in both planes, and its locus is therefore the
+intersection, of those planes.
+
+\item The equation $V\delta\rho = \delta'$, or
+\begin{equation*}
+V\delta(\rho - \delta^{-1}\delta') = 0,
+\end{equation*}
+is a consistent equation only when $\delta'$ is perpendicular to
+$\delta$, since $V\delta\rho$ is always perpendicular to $\delta$.
+When $\delta'$ is perpendicular to $\delta$, then
+$\delta^{-1}\delta'$ is a vector (Art.\ 45, Cor.\ 1), and the
+general solution of this equation is $\rho =
+\delta^{-1}\delta' + x\delta$, where $x$ is an arbitrary scalar
+(Art.\ 46, Cor.). Hence draw $\mathbf{ON} = \delta^{-1}\delta'$,
+and $\mathbf{NP} = x\delta$ (any vector parallel to $\delta$), and
+then $\rho = \mathbf{OP}$ is a solution of the given equation. The
+locus of $P$ is therefore the straight line through $N$ parallel
+to $\delta$, and $\mathbf{ON}$ is the perpendicular from the
+origin upon the line. The equations of Art.\ 65 take this form by
+multiplying the first by $\beta$, the second by $\alpha$, and
+subtracting, remembering that
+\begin{equation*}
+V(V\alpha\beta \cdot \rho) = \alpha{S}\beta\rho - \beta{S}\alpha\rho.
+\end{equation*}
+
+\begin{center}
+\includegraphics[width=80mm]{images/Image33.png}
+\end{center}
+
+\item \textit{The vector perpendicular distance from the line
+$V\delta\rho-\delta' = 0$ to the point $\mathbf{P'}$ is
+$\delta^{-1}(V\delta\rho'-\delta')$, where $\rho' =
+\mathbf{OP'}$.}
+
+For the required perpendicular distance of $P'$ is the projection
+of $\mathbf{NP'}$, $=(\rho' - \delta^{-1}\delta')$, perpendicular
+to $\delta$.
+
+\item \textit{The point of intersection of the three planes}
+$S\alpha\rho = a$, $S\beta\rho = b$, $S\gamma\rho = c$ \textit{is}
+\begin{equation}
+\rho = \frac{(aV\beta\gamma + bV\gamma\alpha +
+cV\alpha\beta)}{S\alpha\beta\gamma}. \tag*{[Art. 56, (\textit{j}).]}
+\end{equation}
+\end{enumerate}
+
+\addcontentsline{toc}{subsection}{Examples}
+\subsection*{Examples}
+
+\small \begin{enumerate}
+\item Find the equation of the locus of a point that moves
+so that its numerical distances from two fixed points are equal.
+
+\item A point moves so that its scalar distances from two fixed
+planes are equal; show that its locus is a plane bisector of the
+diedral angle of the given planes.
+
+\item A point moves so that the sum or difference of its scalar
+distances from two fixed planes is constant; show that its locus
+is a plane parallel to the interior or exterior bisector of the
+diedral angle of the given planes.
+
+\item A point moves so that the ratio of its scalar distances from
+two fixed planes is constant; show that its locus is a plane.
+
+\item A point moves so that its numerical distances from two
+intersecting lines are equal; find its locus. [Take the point of
+intersection as origin.]
+
+\item A point moves so that its numerical distances from three
+fixed points are equal; find its locus.
+
+\begin{enumerate}
+\item The same with coplanar lines instead of points. [Four
+straight lines perpendicular to the plane of the lines.]
+\end{enumerate}
+
+\item Find the vector of the centre of the sphere whose surface
+passes through four given points.
+
+\item A point moves so that its tangential distances from two
+given spheres are numerically equal; find its locus.
+
+\item On the chord $\mathbf{OQ}$ of a given sphere a point $P$ is
+taken so that $\mathbf{OP \cdot OQ} = -a^2$; when $Q$ moves round
+the sphere find the locus of $P$. [A plane perpendicular to the
+diameter $OD$.]
+
+\item The locus of the point $P$ whose co\"{o}rdinates $(x, y, z)$
+satisfy $lx + my + nz + d = 0$ is a plane perpendicular to the
+vector $l\mathbf{i} + m\mathbf{j} + n\mathbf{k}$, at a distance
+from the origin of $-d/\sqrt{l^2 + m^2 + n^2}$, measured in the
+direction of this vector. (\textit{a}) Show that the equation of
+this plane may be put in the form $x\cos\alpha + y\cos\beta +
+z\cos\gamma - p = 0$, where $p$ is the perpendicular distance from
+$O$ to the plane and the cosines are the direction cosines of this
+perpendicular.
+
+\item Find the perpendicular distance of $P' = (x', y', z')$, from
+the plane of Ex.\ 10,
+\begin{equation*}
+[x'\cos\alpha + y'\cos\beta + z'\cos\gamma - p].
+\end{equation*}
+
+\item The locus of the point $P$ whose co\"{o}rdinates $(x, y, z)$
+satisfy $\frac{(x - a)}{l} = \frac{(y - b)}{m} = \frac{(z -
+c)}{n}$ is a line parallel to the vector $l\mathbf{i} +
+m\mathbf{j} + n\mathbf{k}$ through the point $(a, b, c)$. If $P$
+satisfy the first two of these three equations, its locus is a
+plane through the line, perpendicular to the plane of $XOY$. [If
+$t$ be the common value of the three ratios, then $\rho =
+a\mathbf{i} + b\mathbf{j} + c\mathbf{k} + t(l\mathbf{i} +
+m\mathbf{j} + n\mathbf{k})$.]
+
+\item Find the perpendicular distance of $P' = (x', y', z')$ from
+the line of Ex.\ 12.
+
+\smallskip
+In the following examples $A$, $B$, $C$, $D$, $P$ are points whose
+co\"{o}rdinates are $(8, 2, 7)$, $(10, 6, 3)$, $(1, 6, 3)$, $(9,
+10,11)$, $(x, y, z)$.
+
+\smallskip
+\item The equation of the plane through $A$ perpendicular to $OD$
+is $S\mathbf{ODAP} = 0$, or $9x + 10y + 11z = 169$.
+
+\item The equation of the plane through $AB$ parallel to $CD$ is
+$S \cdot \mathbf{AP}V\mathbf{ABCD} = 0$, or $2x - 2y - z = 5$.
+
+\item The equation of the plane $ABC$ is $S \cdot
+\mathbf{AP}V\mathbf{ABAC} = 0$ or $y + z = 9$.
+
+\item Find the perpendicular distance of $D$ from the planes in
+Exs.\ 14, 15, 16.
+
+\item The equation of the plane through $AB$ that contains the
+common perpendicular to $AB$, $CD$ is
+\begin{equation*}
+S \cdot \mathbf{AP}V(\mathbf{AB}V\mathbf{ABCD}) = 0, \text{ or }
+2x + y + 2z = 32.
+\end{equation*}
+
+\item The equation of the line through $A$ parallel to $OD$ is
+$V\mathbf{ODAP} = 0$ or $\mathbf{AP} = t\mathbf{OD}$, or $\frac{(x -
+8)}{9} = \frac{(y - 2)}{10} = \frac{(z - 7)}{11}$.
+
+\item The equation of the line $AB$ is $V\mathbf{ABAP} = 0$, or
+$\mathbf{AP} = t\mathbf{AB}$ or $\frac{(x - 8)}{2} = \frac{(y -
+2)}{4} = \frac{(z - 7)}{-4}$.
+
+\item The equation of the common perpendicular to $AB$, $CD$ is
+the equation of Ex.\ 18 and $x + 2y - 2z = 7$.
+
+\item Find the distance of $D$ from the lines in Exs.\ 19, 20, 21.
+
+\item Find $\mathbf{OD}$ in the form $l\mathbf{OA} + m\mathbf{OB}
++ n\mathbf{OC}$, and find the ratios in which $OD$ cuts the
+triangle $ABC$.
+\end{enumerate} \normalsize
+
+\addcontentsline{toc}{section}{\textbf{Nonions}}
+\section*{Nonions}
+
+\addcontentsline{toc}{section}{Vector Equations, the Operator
+$\phi$}
+
+\begin{enumerate}
+\setcounter{enumi}{68}
+
+\item The vector equation of first degree is
+\begin{equation}
+Vq_1\rho r_1 + Vq_2\rho r_2 + \cdots = Vq. \tag*{(\textit{a})}
+\end{equation}
+
+To solve this equation we resolve it along $\mathbf{i, j, k}$, by
+multiplying it by these vectors and taking the scalars of the
+products. We thus find three scalar equations of first degree from
+which $\rho$ may be immediately found as in Art.\ 68. Hence
+(\textit{a}) has in general one, and only one, solution which
+corresponds to the intersection of three given planes. [See
+further Art.\ 81.]
+
+\item The first member of Art.\ 69 (\textit{a}) is a
+\textit{linear, homogeneous vector function of} $\rho$;
+\textit{i.e.}, it is of first degree in $\rho$, every term is of
+the same degree in $\rho$, and it is a vector.
+
+We may denote the operator
+\begin{equation*}
+Vq_1()r_1 + Vq_2()r_2 + \cdots
+\end{equation*}
+by a single letter, $\phi$, so that $\phi\rho, \phi\sigma, \cdots$
+denote the vectors that result from putting $\rho, \sigma, \cdots$
+in the places occupied by the parenthesis.
+
+\item \textit{The operator} $\phi$ \textit{is distributive over a
+sum and commutative with scalars; i.e.},
+\begin{equation*}
+\phi(x\rho + y\sigma) = x\phi\rho + y\phi\sigma.
+\end{equation*}
+
+This is immediately verified by putting $x\rho + y\sigma$ in the
+places occupied by the parentheses of $\phi$ and expanding the
+several terms.
+
+\item We have $\rho = x\alpha + y\beta + z\gamma$, where $\alpha,
+\beta, \gamma$ are given non-coplanar vectors, and $x, y, z$, are
+scalars, each of first degree in $\rho$, as shown in
+56(\textit{i}) with $\rho$ in the place of $\delta$; hence,
+\begin{equation}
+\phi\rho = x\phi\alpha + y\phi\beta + z\phi\gamma. \tag*{(\textit{a})}
+\end{equation}
+
+The complete operation of $\phi$ is therefore determined when the
+three vectors $\phi\alpha, \phi\beta, \phi\gamma$ are known. Since
+each of these vectors involves three scalar constants
+(\textit{e.g.}, the multiples of the given non-coplanar vectors
+$\alpha$, $\beta$, $\gamma$, that express it), therefore the value
+of $\phi$ depends upon nine scalar constants. The operator $\phi$
+may therefore be called a \textit{nonion}. Scalars and rotators
+are particular forms of nonions.
+
+\small \textsc{Note.}---It is readily shown that nonions have the
+same laws of addition and multiplication among themselves as
+quaternions. Products are not in general commutative. A product
+\begin{equation*}
+(\phi - g_1) (\phi - g_2) (\phi - g_3),
+\end{equation*}
+where $g_1$, $g_2$, $g_3$ are scalars, is commutative, since
+$\phi$ is commutative with scalars by Art.\ 71. Hence this product
+multiplies out as if $\phi$ were a scalar, and is
+\begin{equation*}
+\phi^3 - (g_1 + g_2 + g_3)\phi^2 + (g_2 g_3 + g_3 g_1 + g_1
+g_2)\phi - g_1 g_2 g_3.
+\end{equation*} \normalsize
+
+\addcontentsline{toc}{section}{Linear Homogeneous Strain}
+\section*{Linear Homogeneous Strain}
+
+\item An elastic solid is \textit{subjected to the strain $\phi$
+with respect to an origin $O$}, when all its particles, $A$, $B$,
+$C$, etc., are displaced to positions $A'$, $B'$, $C'$, etc., that
+are determined by $\mathbf{OA'} = \phi \mathbf{OA}$, $\mathbf{OB'}
+= \phi \mathbf{OB}$, $\mathbf{OC'} = \phi \mathbf{OC}$, etc. In
+general, any particle $P$ whose vector is $\mathbf{OP} = \rho$
+occupies after the strain the position $P'$, whose vector is
+$\mathbf{OP'} = \phi\rho$. The particle at $O$ is not moved, since
+its vector after strain is $\phi\mathbf{OO} = \phi 0 = 0$.
+
+\begin{center}
+\includegraphics[width=80mm]{images/Image34.png}
+\end{center}
+
+\begin{enumerate}
+\item We have, also, $\phi\mathbf{AP} = \mathbf{A'P'}$, etc.
+\end{enumerate}
+
+For,
+\begin{align*}
+\mathbf{A'P'} &= \mathbf{OP'} - \mathbf{OA'} = \phi\mathbf{OP} - \phi\mathbf{OA} \\
+              &= \phi(\mathbf{OP} - \mathbf{OA}) = \phi\mathbf{AP}, \text{ etc.}
+\end{align*}
+
+\item \textit{A straight line of particles parallel to $\alpha$ is
+homogeneously stretched and turned by the strain $\phi$ into a
+straight line of particles parallel to $\phi\alpha$, and the ratio
+of extension and turning is $\phi\alpha / \alpha$.}
+
+For let $AP$ be a line parallel to $\alpha$, and let $A$, $P$
+strain into $A'$, $P'$. Then, since $\mathbf{AP} = x\alpha$,
+therefore, by 73\textit{a}, $\mathbf{A'P'} = x\phi\alpha$, and the
+ratio of extension and turning is $\mathbf{A'P'/AP} = \phi
+\alpha/\alpha$.
+
+\small \textsc{Note.}---This property that parallel lengths of
+the substance strain into parallel lengths and are stretched
+proportionally, is the physical definition of \textit{linear
+homogeneous strain}. \normalsize
+
+\item \textit{A plane of particles parallel to $\alpha$, $\beta$
+is homogeneously spread and turned by the strain $\phi$ into a
+plane of particles parallel to $\phi\alpha$, $\phi\beta$, and the
+ratio of extension and turning is
+$V\phi\alpha\phi\beta / V\alpha\beta$.}
+
+For let $APQ$ be a plane parallel to $\alpha$, $\beta$, and let
+$A$, $P$, $Q$ strain into $A'$, $P'$, $Q'$. Then, since
+$\mathbf{AP} = x\alpha + y\beta$, $\mathbf{AQ} = x'\alpha +
+y'\beta$, therefore
+\begin{equation*}
+\mathbf{A'P'} = x\phi\alpha + y\phi\beta, \quad
+\mathbf{A'Q'} = x'\phi\alpha + y'\phi\beta.
+\end{equation*}
+
+By Arts.\ 59, 55, (\textit{e}), the directed area of the triangle
+$APQ$ is $\frac{1}{2} V \cdot \mathbf{AP} \cdot \mathbf{AQ} =
+\frac{1}{2}(xy' - x'y)V\alpha\beta$, and the directed area of the
+triangle $A'P'Q'$ is the same multiple of $V\phi\alpha\phi\beta$.
+Hence the ratio of the extension and turning of directed area is
+$V\phi\alpha\phi\beta / V\alpha\beta$.
+
+\item \textit{A volume of particles is homogeneously dilated by
+the strain $\phi$ in the ratio
+\begin{equation*}
+S\phi\alpha\phi\beta\phi\gamma / S\alpha\beta\gamma,
+\end{equation*}
+where $\alpha$, $\beta$, $\gamma$ are any given non-coplanar
+vectors.}
+
+For let the pyramid $APQR$ strain into the pyramid $A'P'Q'R'$.
+Then since
+\begin{equation*}
+\mathbf{AP} = x\alpha + y\beta + z\gamma, \quad
+\mathbf{AQ} = x'\alpha + y'\beta + z'\gamma,
+\end{equation*}
+$\mathbf{AR} = x''\alpha + y''\beta + z''\gamma$, therefore
+$\mathbf{A'P'}$, $\mathbf{A'Q'}$, $\mathbf{A'R'}$ have these
+values with $\phi\alpha$, $\phi\beta$, $\phi\gamma$ instead of
+$\alpha$, $\beta$, $\gamma$. The volume of the pyramid $APQR$
+relative to the order $AP$, $AQ$, $AR$ of its edges is, by Arts.\
+60, 56, (\textit{c}),
+\begin{equation*}
+-\frac{1}{6}S\mathbf{APAQAR} =
+-\frac{1}{6}\left|\begin{array}{lll}
+                            x   & y   & z   \\
+                            x'  & y'  & z'  \\
+                            x'' & y'' & z'' \end{array} \right|
+  S\alpha\beta\gamma,
+\end{equation*}
+while that of the strained pyramid is the same multiple of
+$S\phi\alpha\phi\beta\phi\gamma$. Hence the ratio of dilation of
+volume is $S\phi\alpha\phi\beta\phi\gamma / S\alpha\beta\gamma$.
+
+\item The ratio of dilation of $\phi$ is called its
+\textit{modulus}.
+
+\begin{enumerate}
+\item It is obvious from the signification of the modulus that
+\textit{the modulus of a product of nonions equals the product of
+the moduli of the factors; e.g.}, $\textit{mod } \phi\psi =
+\textit{mod } \phi \cdot \textit{mod } \psi$.
+\end{enumerate}
+
+When $\text{mod }\phi$ is positive, the parts of the volume are
+in the same order before and after strain. When $\text{mod }\phi$
+is negative, the order of the parts is reversed by the strain;
+\textit{i.e.}, if $AP$ lie on the counter-clockwise side of the
+plane $AQR$, then $A'P'$ lies on the clockwise side of $A'Q'R'$,
+so that the particles along $AP$ have been strained through the
+particles of the plane $AQR$. Such a strain is obviously not a
+physical possibility.
+
+\addcontentsline{toc}{section}{Finite and Null Strains}
+\section*{Finite and Null Strains}
+
+\item \textit{If an elastic solid which fills all space be
+subjected to a strain $\phi$, the strained solid fills all space
+if $\text{mod }\phi$ be finite, and it fills only an indefinite
+plane or line through the origin or reduces to the origin if
+$\text{mod }\phi$ be zero.}
+
+For if $S\phi\alpha\phi\beta\phi\gamma$ be finite, then
+$\phi\alpha$, $\phi\beta$, $\phi\gamma$ are non-coplanar vectors,
+so that
+\begin{equation*}
+\phi\rho( = x\phi\alpha + y\phi\beta + z\phi\gamma)
+\end{equation*}
+may be made any vector by properly choosing
+$\rho(= x\alpha + y\beta + z\gamma)$. But if
+$S\phi\alpha\phi\beta\phi\gamma = 0$, then $\phi\alpha$,
+$\phi\beta$, $\phi\gamma$ are coplanar vectors or colinear vectors
+or each zero, so that $\phi\rho$ will be a vector in a given plane
+or line through $O$ or the vector of $O$, whatever value be given
+to $\rho$.
+
+When $\text{mod }\phi$ is zero, $\phi$ is called a \textit{null}
+nonion; and it is called \textit{singly} or \textit{doubly} or
+\textit{triply} null, according as it strains a solid into a
+\textit{plane} or a \textit{line} or a \textit{point}. If
+$\phi\alpha = 0$, then $\alpha$ is called a \textit{null
+direction} of $\phi$.
+
+\item \textit{Null strains, and only null strains, can have null
+directions; a singly null strain has only one null direction; a
+doubly null strain has a plane of null directions only; a triply
+null strain has all directions null.}
+
+For when $\text{mod }\phi = 0$, then $\phi\alpha$, $\phi\beta$,
+$\phi\gamma$ are coplanar or colinear vectors, and we have a
+relation $l\phi\alpha + m\phi\beta + n\phi\gamma = 0$, \textit{i.e.},
+$l\alpha + m\beta + n\gamma$ is a null direction of $\phi$.
+Conversely, if $\phi$ have a null direction, take one of the three
+non-coplanar vectors $\alpha$, $\beta$, $\gamma$, in that
+direction, say $\alpha$, and we have
+$S\phi\alpha\phi\beta\phi\gamma = 0$, since $\phi\alpha = 0$, and
+therefore $\mathrm{mod}\phi = 0$.
+
+Also, if $\phi$ have only one null direction, $\alpha$, then
+$\phi\beta$, $\phi\gamma$, are not parallel, since $\phi\beta =
+l\phi\gamma$ makes $\beta - l\gamma$ a second null direction. Since
+$\rho = x\alpha + y\beta + z\gamma$, therefore $\phi\rho =
+y\phi\beta + z\phi\gamma$, which is any vector in the plane through
+$O$ parallel to $\phi\beta$, $\phi\gamma$; hence $\phi$ is singly
+null.
+
+But if $\phi$ have two null directions, $\alpha$, $\beta$, then
+$\phi\rho = z\phi\gamma$, which is any vector in the line through
+$O$ parallel to $\phi\gamma$, and therefore $\phi$ is doubly null.
+Also, since $\phi(x\alpha + y\beta) = 0$, therefore any direction in
+the plane of $\alpha$, $\beta$ is a null direction of $\phi$.
+
+If $\phi$ have three non-coplanar null directions $\alpha$,
+$\beta$, $\gamma$, then $\phi\rho = 0$ for all values of $\rho$;
+\textit{i.e.}, a \textit{triply} null nonion is identically zero.
+
+\item \textit{A singly null nonion strains each line in its null
+direction into a definite point of its plane; and a doubly null
+nonion strains each plane that is parallel to its null plane into
+a definite point of its line.}
+
+For when $\phi$ is singly null, say $\phi\alpha = 0$, then
+$x\phi\beta + y\phi\gamma$ is the vector of any point in the plane
+of $\phi$, and all particles that strain into this point have the
+vectors $\rho = x\alpha + y\beta + z\gamma$, where $x$ is
+arbitrary, since $\phi\alpha = 0$; \textit{i.e.}, they are
+particles of a line parallel to $\alpha$. So, if $\phi$ is doubly
+null, say $\phi\alpha = 0$, $\phi\beta = 0$, then any point of the
+line of $\phi$ is $z\phi\gamma$, and the particles that strain
+into this point have the vectors
+\begin{equation*}
+\rho = x\alpha + y\beta + z\gamma,
+\end{equation*}
+in which $x$, $y$ are arbitrary; \textit{i.e.}, they are particles
+of a plane parallel to $\alpha$, $\beta$.
+
+\small \textsc{Note.}---It follows similarly that the strain
+$\phi$ alters the dimensions of a line, plane, or volume by as
+many dimensions as the substance strained contains independent
+null directions of $\phi$, and no more. Hence, a product
+$\phi\psi$ has the null directions of the first factor, and the
+null directions of the second factor that lie in the figure into
+which the first factor strains, and so on; the order of nullity of
+a product cannot exceed the sum of the orders of its factors, and
+may be less; etc. \normalsize
+
+\addcontentsline{toc}{section}{Solution of $\phi\rho = \delta$}
+\section*{Solution of $\phi\rho = \delta$}
+
+\item The solutions of $\phi\rho = \delta$ are, by definition of
+the strain $\phi$, the vectors of the particles that strain into
+the position whose vector is $\delta$. Hence:
+
+\begin{enumerate}
+\item When $\phi$ is finite, there is one, and only one, solution.
+
+\item When $\phi$ is singly null, and $\delta$ does not lie in the
+plane of $\phi$, there is no finite solution. Divide the equation
+by $T\rho$, and make $T\rho$ infinite, and we find $\phi U \rho =
+0$; \textit{i.e.}, the vector of the point at infinity in the null
+direction of $\phi$ is a solution.
+
+\item When $\phi$ is singly null, and $\delta$ lies in the plane
+of $\phi$, there are an infinite number of solutions,
+\textit{viz.}, the vectors of the particles of a line that is
+parallel to the null direction of $\phi$.
+
+\item When $\phi$ is doubly null, and $\delta$ does not lie in the
+line of $\phi$, there is no finite solution. As in (2) the vectors
+of the points of the line at infinity in the null plane of $\phi$
+are solutions.
+
+\item When $\phi$ is doubly null, and $\delta$ lies in the line of
+$\phi$, there are an infinite number of solutions, \textit{viz.},
+the vectors of the particles of a plane that is parallel to the
+null plane of $\phi$.
+\end{enumerate}
+
+These results correspond to the intersections of three planes,
+viz.:
+\begin{enumerate}
+\item The three planes meet in a point.
+\item The three planes parallel to a line.
+\item The three planes meet in a common line.
+\item The three planes parallel.
+\item The three planes coincide.
+\end{enumerate}
+
+\addcontentsline{toc}{section}{Derived Moduli. Latent Roots}
+\section*{Derived Moduli of $\phi$}
+
+\item The ratio in which the nonion $\phi+g$ dilates volume is,
+\begin{equation*}
+\text{mod }(\phi+g) = S(\phi\alpha + g\alpha)(\phi\beta +
+g\beta)(\phi\gamma + g\gamma) / S\alpha\beta\gamma.
+\end{equation*}
+This is independent of the values of the non-coplanar vectors
+$\alpha$, $\beta$, $\gamma$ in terms of which it is expressed. If
+$g$ is a scalar, this modulus is an ordinary cubic in $g$, whose
+coefficients will therefore depend only upon $\phi$. The constant
+term is $\text{mod }\phi$, and the coefficients of $g$, $g^2$,
+are called $\text{mod}_1 \phi$, $\text{mod}_2 \phi$, so that,
+\begin{equation}
+\text{mod }(\phi + g) = g^3 + g^2 \text{mod}_2 \phi
++ g \text{mod}_1 \phi + \text{mod } \phi. \tag*{(\textit{a})}
+\end{equation}
+\begin{align*}
+[\mathrm{mod}_1 \phi &= S(\alpha\phi\beta\phi\gamma
+     + \beta\phi\gamma\phi\alpha + \gamma\phi\alpha\phi\beta)
+   / S\alpha\beta\gamma; \\
+\mathrm{mod}_2 \phi &= S(\beta\gamma\phi\alpha
+     + \gamma\alpha\phi\beta + \alpha\beta\phi\gamma)
+   / S\alpha\beta\gamma].
+\end{align*}
+
+\item The roots $g_1$, $g_2$, $g_3$, of the cubic
+\begin{equation*}
+\text{mod }(\phi-g) = 0
+\end{equation*}
+are called \textit{the latent roots of} $\phi$. We have from
+82 (\textit{a}) with $-g$ in the place of $g$, and the
+theory of equations,
+\begin{gather*}
+\text{mod }\phi = g_1 g_2 g_3, \quad
+\text{mod}_1 \phi = g_2 g_3 + g_3 g_1 + g_1 g_2, \\
+\text{mod}_2 \phi = g_1 + g_2 + g_3.
+\end{gather*}
+
+\begin{enumerate}
+\item \textit{The latent roots of $\phi-g_1$ are those of $\phi$
+diminished by $g_1$.}
+
+For the roots of $\text{mod }(\phi - g_1 - g) = 0$, are $g=0$,
+$g_2 - g_1$, $g_3 - g_1$. \textit{E.g.}, $g = g_2 - g_1$ gives
+\begin{equation*}
+\text{mod }[\phi - g_1 -(g_2 - g_1)] = \text{mod }(\phi - g_2) = 0.
+\end{equation*}
+
+\item \textit{The order of nullity of $\phi$ cannot exceed the
+number of its zero latent roots.}
+
+For if $\phi$ has one null direction $\alpha$, then $\phi\alpha =
+0$ makes $\text{mod }\phi = 0$, so that at least one of the
+latent roots is zero, say $g_1$; and if $\phi$ has a second null
+direction $\beta$, then $\phi\alpha = 0$, $\phi\beta = 0$, makes
+$\mathrm{mod}_1\phi = 0$ or $g_2g_3 = 0$, so that another latent
+root is zero, etc.
+
+\item \textit{The order of nullity of $\phi - g_1$ cannot exceed
+the number of latent roots of $\phi$ that equal $g_1$.}
+[(\textit{a}), (\textit{b})]
+\end{enumerate}
+
+\addcontentsline{toc}{section}{Latent Lines and Planes}
+\section*{Latent Lines and Planes of $\phi$}
+
+\item Those lines and planes that remain unaltered in geometrical
+position by the strain $\phi$ are called \textit{latent lines and
+planes of $\phi$}.
+
+\begin{enumerate}
+\item \textit{The latent directions of $\phi$ are the null
+directions of $\phi - g_1$, $\phi - g_2$, $\phi - g_3$, and $g_1$,
+$g_2$, $g_3$ are the corresponding ratios of extension in those
+directions.}
+\end{enumerate}
+
+For if $\rho$ is any latent direction, and $g$ is the ratio of
+extension in that direction, then we have $\phi\rho = g\rho$ or
+$(\phi - g)\rho = 0$. Hence $\phi - g$ is a null nonion, or
+$\text{mod }(\phi - g) = 0$, so that $g$ is a latent root of
+$\phi$; also $\rho$ is a null direction of $\phi - g$.
+
+\small \textsc{Note.}---Since a cubic with real coefficients has
+at least one real root, therefore a real nonion has at least one
+latent direction. Also if two roots are imaginary, they are
+conjugate imaginaries, and the corresponding latent directions
+must also be conjugate imaginaries. \normalsize
+
+\item \textit{If $\alpha$, $\beta$, $\gamma$ be the latent
+directions corresponding to $g_1$, $g_2$, $g_3$, then $(\beta,
+\gamma)$, $(\gamma, \alpha)$, $(\alpha, \beta)$ determine latent
+planes of $\phi$ in which the ratios of spreading are $g_2 g_3$,
+$g_3 g_1$, $g_1 g_2$. E.g.,}
+\begin{equation*}
+V \phi \beta \phi \gamma = V (g_2 \beta \cdot g_3 \gamma) = g_2
+g_3 V \beta \gamma .
+\end{equation*}
+
+Hence, in the general case when the latent roots are all unequal,
+the latent vectors $\alpha$, $\beta$, $\gamma$ must form a
+non-coplanar system, since any two of the latent lines or planes
+determined by them have unequal ratios of extension, and cannot,
+therefore, coincide.
+
+\begin{enumerate}
+\item \textit{The plane of $\phi - g_1$ is the latent plane
+corresponding to $g_2$, $g_3$.}
+\end{enumerate}
+
+For $(\phi - g_1) \rho = y (g_2 - g_1) \beta + z (g_3 - g_1)
+\gamma$, ($=$ plane of $\beta$, $\gamma$). [The plane and null
+line of $\phi - g_1$ may be called \textit{corresponding} latents
+of $\phi$.]
+
+\addcontentsline{toc}{section}{The Characteristic Equation}
+\section*{The Characteristic Equation of $\phi$}
+
+\item We have also,
+
+\begin{enumerate}
+\item $(\phi - g_1)(\phi - g_2)(\phi - g_3) = 0$. For the first
+member has the three non-coplanar null directions $\alpha$,
+$\beta$, $\gamma$. [See 80 note, 72 note.]
+\end{enumerate}
+
+\addcontentsline{toc}{section}{Conjugate Nonions}
+\section*{Conjugate Nonions}
+
+\item Two nonions $\phi$, $\phi'$ are conjugate when
+
+\begin{enumerate}
+\item $S\rho\phi\sigma = S\sigma\phi'\rho$ for all values of
+$\rho$, $\sigma$.
+\end{enumerate}
+When $\phi$ is known, this determines $\phi'$ without ambiguity.
+Thus, put $\sigma = \mathbf{i, j, k}$, in turn, and we have by
+Art.\ 57 (\textit{b}),
+\begin{equation*}
+\phi'\rho = -\mathbf{i}S\rho\phi \mathbf{i} - \mathbf{j}S\rho\phi
+\mathbf{j} - \mathbf{k}S\rho\phi \mathbf{k}.
+\end{equation*}
+
+Conversely, this function satisfies (\textit{a}), for we have
+$S\sigma\phi'\rho = S\rho\phi(-\mathbf{i}S\mathbf{i}\sigma -
+\mathbf{j}S\mathbf{j}\sigma - \mathbf{k}S\mathbf{k}\sigma) =
+S\rho\phi\sigma$.
+
+\item From this definition of conjugate strains we have
+
+\begin{enumerate}
+\item $(a\phi + b\psi)' = a\phi' + b\psi'$; $(\phi\psi)' = \psi'\phi'$.
+\item $(Vq()q)' = Vq()p$, $[\alpha S\beta()]' = \beta S\alpha()$.
+\end{enumerate}
+\begin{align*}
+\textit{E.g.}, S\sigma(\phi\psi)'\rho &= S\rho\phi\psi\sigma
+           = S\rho\phi(\psi\sigma) \\
+    &= S\psi\sigma\phi'\rho = S\sigma\psi'\phi'\rho,
+\end{align*}
+and therefore $(\phi\psi)' = \psi'\phi'$. [If $S\sigma(\alpha -
+\beta) = 0$ for all values of $\sigma$, then $\alpha - \beta = 0$,
+since no vector is perpendicular to every vector $\sigma$. Hence,
+comparing the first and last member of the above equation, we have
+$(\phi\psi)'\rho = \psi'\phi'\rho$.]
+
+\item \textit{Two conjugate strains have the same latent roots and
+moduli, and a latent plane of one is perpendicular to the
+corresponding latent line of the other.}
+
+For since $(\phi - g_1)\alpha = 0$, therefore
+\begin{equation*}
+0 = S\rho(\phi - g_1)\alpha = S\alpha(\phi' - g_1)\rho,
+\end{equation*}
+and therefore $\phi' - g_1$ is a null nonion whose plane is
+perpendicular to $\alpha$. Hence $g_1$ is a latent root of
+$\phi'$, and the latent plane of $\phi'$ corresponding to $\phi' -
+g_1$ is perpendicular to the latent line of $\phi$ corresponding
+to $\phi - g_1$. [Art.\ 85, (\textit{a}).]
+
+\addcontentsline{toc}{section}{Self-conjugate Nonions}
+\section*{Self-conjugate Nonions}
+
+\item A nonion $\phi$ is \textit{self-conjugate} when $\phi' =
+\phi$ or when $S\rho\phi\sigma = S\sigma\phi\rho$ for all values
+of $\rho$, $\sigma$. In consequence of this relation a
+self-conjugate strain has only six scalar constants, three of the
+nine being equal to three others, \textit{viz.},
+\begin{equation*}
+S\mathbf{i}\phi \mathbf{j} = S\mathbf{j}\phi \mathbf{i}, \quad
+S\mathbf{i}\phi \mathbf{k} = S\mathbf{k}\phi \mathbf{i}, \quad
+S\mathbf{j}\phi \mathbf{k} = S\mathbf{k}\phi \mathbf{j}.
+\end{equation*}
+
+\item A self-conjugate strain has by Art.\ 88 three mutually
+perpendicular latent directions, and conversely, if $\phi$ have
+three mutually perpendicular latent directions, $\mathbf{i}$,
+$\mathbf{j}$, $\mathbf{k}$, corresponding to latent roots $a$,
+$b$, $c$, then
+\begin{equation*}
+\phi\rho = -a\mathbf{i}S\mathbf{i}\rho -
+b\mathbf{j}S\mathbf{j}\rho - c\mathbf{k}S\mathbf{k}\rho,
+\end{equation*}
+which is self-conjugate. [68 \textit{b}.]
+
+\item \textit{A real self-conjugate strain has real latent roots.}
+
+For let $\alpha' = \alpha + \beta\sqrt{-1}$, $\beta' = \alpha -
+\beta\sqrt{-1}$ be latent directions corresponding to conjugate
+imaginary roots $a$, $b$ of a real nonion $\phi$; then, if $\phi$
+is self-conjugate, we have
+\begin{equation*}
+S\alpha'\phi\beta' = S\beta'\phi\alpha' = bS\alpha'\beta' =
+aS\alpha'\beta',
+\end{equation*}
+or, since $a$, $b$ are unequal, therefore $S\alpha'\beta' = 0$;
+but this is impossible, since $S\alpha'\beta' = \alpha^2 +
+\beta^2$, a negative quantity. Therefore $\phi$ is not
+self-conjugate if it has imaginary latent roots.
+
+\item A nonion $\phi$ is \textit{negatively self-conjugate} when
+$\phi' = -\phi$, or when $S\sigma\phi\rho = -S\rho\phi\sigma$.
+Such a nonion has therefore only three scalar constants, since
+$S\mathbf{i}\phi\mathbf{i} = -S\mathbf{i}\phi\mathbf{i}$ shows
+that $S\mathbf{i}\phi\mathbf{i} = 0$, and similarly,
+$S\mathbf{j}\phi\mathbf{j} = 0$, $S\mathbf{k}\phi\mathbf{k} = 0$,
+while the other six constants occur in negative pairs
+\begin{equation*}
+S\mathbf{i}\phi\mathbf{j} = -S\mathbf{j}\phi\mathbf{i}, \text{ etc.}
+\end{equation*}
+
+\begin{enumerate}
+\item The identity $S\rho\phi\rho = 0$ gives (by putting $\rho =
+x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$ where $x$, $y$, $z$ are
+arbitrary) all the above relations between the constants of
+$\phi$, and is therefore the sufficient condition that $\phi$ is
+negatively self-conjugate. It shows that $\phi\rho$ is
+perpendicular to $\rho$ or that $\phi\rho = V\epsilon\rho$, where
+$\epsilon$ must be independent of $\rho$ since $\phi\rho$ is
+linear in $\rho$.
+\end{enumerate}
+
+\item \textit{Any nonion $\phi$ may be resolved into a sum of a
+conjugate and a negatively self-conjugate nonion in only one way.}
+
+For if $\phi = \overline{\phi}+\psi$, where $\overline{\phi'} =
+\overline{\phi}$, $\psi' = -\psi$, then $\phi' =
+\overline{\phi}-\psi$, and adding and subtracting, we have
+$\overline{\phi} = \frac{1}{2}(\phi+\phi')\psi =
+\frac{1}{2}(\phi-\phi')$, and
+\begin{equation}
+\phi\rho = \frac{1}{2}(\phi + \phi')\rho + \frac{1}{2}(\phi
+- \phi')\rho = \overline{\phi}\rho + V\epsilon\rho. \tag*{(\textit{a})}
+\end{equation}
+
+To find $\epsilon$ in terms of the constants of $\phi$, we have
+$\rho = -\mathbf{i}S\mathbf{i}\rho - \mathbf{j}S\mathbf{j}\rho -
+\mathbf{k}S\mathbf{k}\rho$, and therefore
+\begin{align*}
+\phi\rho  &= -\phi\mathbf{i}S\mathbf{i}\rho - \text{ etc.} \\
+\phi'\rho &= -\mathbf{i}S\rho\phi\mathbf{i} - \text{ etc.} \tag*{[88
+\textit{b.}]}
+\end{align*}
+Hence
+\begin{align*}
+\frac{1}{2}(\phi - \phi')\rho &=
+   \frac{1}{2}(\mathbf{i}S\rho\phi\mathbf{i}
+             - \phi\mathbf{i}S\mathbf{i}\rho) + \text{ etc.} \\
+  &= \frac{1}{2}V \cdot (V\mathbf{i}\phi\mathbf{i})\rho +
+\textit{ etc.} = V\epsilon\rho,
+\end{align*}
+and therefore
+\begin{equation}
+\epsilon = \frac{1}{2}V(\mathbf{i}\phi\mathbf{i} +
+\mathbf{j}\phi\mathbf{j} + \mathbf{k}\phi\mathbf{k}). \tag*{(\textit{a})}
+\end{equation}
+\end{enumerate}
+
+\addcontentsline{toc}{subsection}{Examples}
+\subsection*{Examples}
+
+\small \begin{enumerate}
+\item Find the equation of a sphere whose centre is $A(\mathbf{OA}
+= \alpha)$ and radius $\alpha$.
+
+\item Show that the square of the vector tangent from the sphere
+of Ex.\ 1 to $P'$ is $(\rho' - \alpha)^2 + a^2$.
+
+\item Find the locus of the point $P$ such that $PP'$ is cut in
+opposite ratios by the sphere of Ex.\ 1; show that it is the plane
+of contact of the tangent cone from $P'$ to the sphere and is
+perpendicular to $AP'$.
+
+\item Let $P'$ be any point on the sphere $A$ of Ex.\ 1, and take
+$P$ on $OP'$ so that $\mathbf{OP} \cdot \mathbf{OP'} + c^2 = 0$;
+find the locus of $P$. [$P$, $P'$ are called \textit{inverse}
+points with respect to $O$, and the locus of $P$ is the
+\textit{inverse} of the given sphere $A$. It is a sphere with
+centre $A'$ on $OA$, or a plane perpendicular to $OA$ if the given
+sphere $A$ pass through $O$.]
+
+\item Show that the inverse of a plane is a sphere through $O$.
+
+\item Show that the general scalar equation of second degree is
+$S\rho\phi\rho + 2S\delta\rho + d = 0$, where $\phi$ is a
+self-conjugate nonion.
+
+\item Show that $S\rho\phi\rho = 0$ is the equation of a cone
+with vertex at $O$.
+
+\item Show that the line $\rho = \alpha + x \beta$ cuts the
+quadric surface of Ex.\ 6 in two points; apply the theory of
+equations to determine the condition that this line is a tangent
+to the surface, or an element of the surface, or that it meets the
+surface in one finite point and one point at infinity, or that the
+point whose vector is $\alpha$ lies midway between the points of
+intersection.
+
+\item Show that the solution of $\phi\rho + \delta = 0$ is the
+vector of a centre of symmetry of the quadric surface of Ex.\ 6.
+Hence classify quadric surfaces as \textit{central},
+\textit{non-central}, \textit{axial}, \textit{non-axial},
+\textit{centro-planar}.
+
+\item Show that the locus of the middle points of chords parallel
+to $\beta$ is a diametric plane perpendicular to $\phi\beta$.
+
+\item Show that an axial quadric is a cylinder with elements
+parallel to the null direction of its nonion $\phi$.
+
+\item Show that a non-axial quadric is a cylinder with elements
+parallel to the null plane of its nonion $\phi$ and perpendicular
+to its vector $\delta$.
+
+\item Show that a centro-planar quadric consists of two planes
+parallel to the null plane of its nonion $\phi$.
+
+\item Show that the equation of a central quadric referred to its
+centre as origin is $S\rho\phi\rho + 1 = 0$. Show that the latent
+lines and planes of $\phi$ are axes and planes of symmetry of the
+quadric; also that $\phi\rho$ is perpendicular to the tangent
+plane at the point whose vector is $\rho$. (\textit{a}) Show that
+the axes and planes of symmetry of the general quadric are
+parallel to the latent lines and planes of $\phi$.
+
+\item Show that if $\psi^2 = \phi$, then the equation of the
+central quadric is $(\psi\rho)^2 + 1 = 0$; and that therefore the
+quadric surface when strained by $\psi$ becomes a spherical
+surface of unit radius.
+
+\item Show that if $g$, $\alpha$ are corresponding latent root and
+direction of $\phi$, then $g^n$, $\alpha$ are the same for
+$\phi^n$. Find the latent lines and planes, the latent roots and
+moduli of the following nonions and their powers:
+
+\begin{enumerate}
+\item $(a\alpha S\beta\gamma\rho + b\beta S\gamma\alpha\rho +
+c\gamma S\alpha\beta\rho) / S\alpha\beta\gamma$.
+
+\item $[a\alpha S\beta\gamma\rho + (a\beta + b\alpha)S\gamma\alpha\rho
+  + (c\gamma S\alpha\beta\rho] / S\alpha\beta\gamma$.
+
+\item $[a\alpha S\beta\gamma\rho + (a\beta + b\alpha)S\gamma\alpha\rho
+  + (a\gamma + c\beta)S\alpha\beta\rho] / S\alpha\beta\gamma$.
+
+\item $V\epsilon\rho$, $q{\rho}q^{-1}$.
+\end{enumerate}
+
+\item Show that the latent roots of $e\rho - fV\alpha\rho\beta$
+($f>0$, $T\alpha = T\beta = 1$) are $e + f$, $e + fS\alpha\beta$,
+$e - f$, corresponding to latent directions $\alpha + \beta$,
+$V\alpha\beta$, $\alpha - \beta$; and that this is therefore a
+general form for self-conjugate nonions. Determine the latent
+directions and roots in the limiting case when $\alpha = \beta$,
+or $-\beta$ or $f=0$.
+
+\item Show that the nonion of Ex.\ 16 takes the form $b\rho -
+f(\alpha S\beta\rho + \beta S\alpha\rho)$, where $b$ is the mean
+latent root.
+
+\item Substitute the nonion of Ex.\ 18 for $\phi$ in Ex.\ 6 and show
+that the quadric surface is cut in circles by planes perpendicular
+to $\alpha$ or $\beta$. When is the surface one of revolution?
+
+\item If the conjugate of a nonion is its reciprocal, and the
+modulus is positive, then the nonion is a rotation; and conversely
+every rotation satisfies this condition. [If $R$, $R^{-1}$ are
+conjugate nonions, then $\rho^2 = S\rho R^{-1} R \rho = SR\rho
+R\rho = (R\rho)^2$; \textit{i.e.}, $TR\rho / \rho = 1$. Also
+$S\rho\sigma = S\rho R^{-1} R\sigma = SR\rho R\sigma$ and
+therefore the angle between $\rho, \sigma = \angle$ between
+$R\rho, R\sigma$. Therefore $R$ strains a sphere with centre $O$
+into another sphere with centre $O$ in which the angles between
+corresponding radii are equal and their order in space is the
+same, since $\text{mod } R$ is positive. Hence the strain is a
+rotation.]
+
+\begin{enumerate}
+\item Show that $(R\phi R^{-1})^n = R\phi^n R^{-1}$.
+\end{enumerate}
+
+\item Show that $\phi'\phi$ is self-conjugate, and that its latent
+roots are positive, and that therefore there are four real values
+of $\psi$ that satisfy $\psi^2 = \phi'\phi$, $\text{mod } \psi =
+\text{mod } \phi$. [Let $\phi'\phi \mathbf{i} = a\mathbf{i}$; then
+$a = -S\mathbf{i}\phi'\phi\mathbf{i} = (T\phi\mathbf{i})^2$.]
+
+\item If $\phi = R\psi$, where $\psi$ is the self-conjugate strain
+$\sqrt{\phi^{\prime}\phi}$, then $R$ is a rotation. So $\phi =
+\chi{R}$, where $\chi = R\psi{R}^{-1} = \sqrt{\phi\phi^\prime}$.
+
+\item Show that $\phi^{\prime} \cdot V\phi\beta\phi\gamma =
+V\beta\gamma \cdot \text{mod }\phi$. [56 \textit{j}.]
+
+\item Show that the strain $\phi\rho = \rho - a\alpha S\beta\rho$,
+where $\alpha$, $\beta$, are perpendicular unit vectors, consists
+of a shearing of all planes perpendicular to $\beta$, the amount
+and direction of sliding of each plane being $a\alpha$ per unit
+distance of the plane from $O$.
+
+\item Determine $\psi$ and $R$ of Ex.\ 22 for the strain of Ex.\ 24,
+and find the latent directions and roots of $\psi$.
+\end{enumerate}
+
+%%%%%%%%%%%%%%%%%%%%%%%%% GUTENBERG LICENSE %%%%%%%%%%%%%%%%%%%%%%%%%%
+\iffalse %%%%% Start of original license %%%%
+
+\newpage
+\small
+\chapter{PROJECT GUTENBERG "SMALL PRINT"}
+\pagenumbering{gobble}
+\begin{verbatim}
+
+\end{verbatim}
+\normalsize
+\fi
+%%%%% End of original license %%%%
+
+\PGLicense
+\begin{PGtext}
+End of Project Gutenberg's A Primer of Quaternions, by Arthur S. Hathaway
+
+*** END OF THIS PROJECT GUTENBERG EBOOK A PRIMER OF QUATERNIONS ***
+
+***** This file should be named 9934-pdf.pdf or 9934-pdf.zip *****
+This and all associated files of various formats will be found in:
+        http://www.gutenberg.org/9/9/3/9934/
+
+Produced by Cornell University, Joshua Hutchinson, John
+Hagerson, and the Online Distributed Proofreading Team
+
+
+Updated editions will replace the previous one--the old editions
+will be renamed.
+
+Creating the works from public domain print editions means that no
+one owns a United States copyright in these works, so the Foundation
+(and you!) can copy and distribute it in the United States without
+permission and without paying copyright royalties.  Special rules,
+set forth in the General Terms of Use part of this license, apply to
+copying and distributing Project Gutenberg-tm electronic works to
+protect the PROJECT GUTENBERG-tm concept and trademark.  Project
+Gutenberg is a registered trademark, and may not be used if you
+charge for the eBooks, unless you receive specific permission.  If you
+do not charge anything for copies of this eBook, complying with the
+rules is very easy.  You may use this eBook for nearly any purpose
+such as creation of derivative works, reports, performances and
+research.  They may be modified and printed and given away--you may do
+practically ANYTHING with public domain eBooks.  Redistribution is
+subject to the trademark license, especially commercial
+redistribution.
+
+
+
+*** START: FULL LICENSE ***
+
+THE FULL PROJECT GUTENBERG LICENSE
+PLEASE READ THIS BEFORE YOU DISTRIBUTE OR USE THIS WORK
+
+To protect the Project Gutenberg-tm mission of promoting the free
+distribution of electronic works, by using or distributing this work
+(or any other work associated in any way with the phrase "Project
+Gutenberg"), you agree to comply with all the terms of the Full Project
+Gutenberg-tm License available with this file or online at
+  www.gutenberg.org/license.
+
+
+Section 1.  General Terms of Use and Redistributing Project Gutenberg-tm
+electronic works
+
+1.A.  By reading or using any part of this Project Gutenberg-tm
+electronic work, you indicate that you have read, understand, agree to
+and accept all the terms of this license and intellectual property
+(trademark/copyright) agreement.  If you do not agree to abide by all
+the terms of this agreement, you must cease using and return or destroy
+all copies of Project Gutenberg-tm electronic works in your possession.
+If you paid a fee for obtaining a copy of or access to a Project
+Gutenberg-tm electronic work and you do not agree to be bound by the
+terms of this agreement, you may obtain a refund from the person or
+entity to whom you paid the fee as set forth in paragraph 1.E.8.
+
+1.B.  "Project Gutenberg" is a registered trademark.  It may only be
+used on or associated in any way with an electronic work by people who
+agree to be bound by the terms of this agreement.  There are a few
+things that you can do with most Project Gutenberg-tm electronic works
+even without complying with the full terms of this agreement.  See
+paragraph 1.C below.  There are a lot of things you can do with Project
+Gutenberg-tm electronic works if you follow the terms of this agreement
+and help preserve free future access to Project Gutenberg-tm electronic
+works.  See paragraph 1.E below.
+
+1.C.  The Project Gutenberg Literary Archive Foundation ("the Foundation"
+or PGLAF), owns a compilation copyright in the collection of Project
+Gutenberg-tm electronic works.  Nearly all the individual works in the
+collection are in the public domain in the United States.  If an
+individual work is in the public domain in the United States and you are
+located in the United States, we do not claim a right to prevent you from
+copying, distributing, performing, displaying or creating derivative
+works based on the work as long as all references to Project Gutenberg
+are removed.  Of course, we hope that you will support the Project
+Gutenberg-tm mission of promoting free access to electronic works by
+freely sharing Project Gutenberg-tm works in compliance with the terms of
+this agreement for keeping the Project Gutenberg-tm name associated with
+the work.  You can easily comply with the terms of this agreement by
+keeping this work in the same format with its attached full Project
+Gutenberg-tm License when you share it without charge with others.
+
+1.D.  The copyright laws of the place where you are located also govern
+what you can do with this work.  Copyright laws in most countries are in
+a constant state of change.  If you are outside the United States, check
+the laws of your country in addition to the terms of this agreement
+before downloading, copying, displaying, performing, distributing or
+creating derivative works based on this work or any other Project
+Gutenberg-tm work.  The Foundation makes no representations concerning
+the copyright status of any work in any country outside the United
+States.
+
+1.E.  Unless you have removed all references to Project Gutenberg:
+
+1.E.1.  The following sentence, with active links to, or other immediate
+access to, the full Project Gutenberg-tm License must appear prominently
+whenever any copy of a Project Gutenberg-tm work (any work on which the
+phrase "Project Gutenberg" appears, or with which the phrase "Project
+Gutenberg" is associated) is accessed, displayed, performed, viewed,
+copied or distributed:
+
+This eBook is for the use of anyone anywhere at no cost and with
+almost no restrictions whatsoever.  You may copy it, give it away or
+re-use it under the terms of the Project Gutenberg License included
+with this eBook or online at www.gutenberg.org
+
+1.E.2.  If an individual Project Gutenberg-tm electronic work is derived
+from the public domain (does not contain a notice indicating that it is
+posted with permission of the copyright holder), the work can be copied
+and distributed to anyone in the United States without paying any fees
+or charges.  If you are redistributing or providing access to a work
+with the phrase "Project Gutenberg" associated with or appearing on the
+work, you must comply either with the requirements of paragraphs 1.E.1
+through 1.E.7 or obtain permission for the use of the work and the
+Project Gutenberg-tm trademark as set forth in paragraphs 1.E.8 or
+1.E.9.
+
+1.E.3.  If an individual Project Gutenberg-tm electronic work is posted
+with the permission of the copyright holder, your use and distribution
+must comply with both paragraphs 1.E.1 through 1.E.7 and any additional
+terms imposed by the copyright holder.  Additional terms will be linked
+to the Project Gutenberg-tm License for all works posted with the
+permission of the copyright holder found at the beginning of this work.
+
+1.E.4.  Do not unlink or detach or remove the full Project Gutenberg-tm
+License terms from this work, or any files containing a part of this
+work or any other work associated with Project Gutenberg-tm.
+
+1.E.5.  Do not copy, display, perform, distribute or redistribute this
+electronic work, or any part of this electronic work, without
+prominently displaying the sentence set forth in paragraph 1.E.1 with
+active links or immediate access to the full terms of the Project
+Gutenberg-tm License.
+
+1.E.6.  You may convert to and distribute this work in any binary,
+compressed, marked up, nonproprietary or proprietary form, including any
+word processing or hypertext form.  However, if you provide access to or
+distribute copies of a Project Gutenberg-tm work in a format other than
+"Plain Vanilla ASCII" or other format used in the official version
+posted on the official Project Gutenberg-tm web site (www.gutenberg.org),
+you must, at no additional cost, fee or expense to the user, provide a
+copy, a means of exporting a copy, or a means of obtaining a copy upon
+request, of the work in its original "Plain Vanilla ASCII" or other
+form.  Any alternate format must include the full Project Gutenberg-tm
+License as specified in paragraph 1.E.1.
+
+1.E.7.  Do not charge a fee for access to, viewing, displaying,
+performing, copying or distributing any Project Gutenberg-tm works
+unless you comply with paragraph 1.E.8 or 1.E.9.
+
+1.E.8.  You may charge a reasonable fee for copies of or providing
+access to or distributing Project Gutenberg-tm electronic works provided
+that
+
+- You pay a royalty fee of 20% of the gross profits you derive from
+     the use of Project Gutenberg-tm works calculated using the method
+     you already use to calculate your applicable taxes.  The fee is
+     owed to the owner of the Project Gutenberg-tm trademark, but he
+     has agreed to donate royalties under this paragraph to the
+     Project Gutenberg Literary Archive Foundation.  Royalty payments
+     must be paid within 60 days following each date on which you
+     prepare (or are legally required to prepare) your periodic tax
+     returns.  Royalty payments should be clearly marked as such and
+     sent to the Project Gutenberg Literary Archive Foundation at the
+     address specified in Section 4, "Information about donations to
+     the Project Gutenberg Literary Archive Foundation."
+
+- You provide a full refund of any money paid by a user who notifies
+     you in writing (or by e-mail) within 30 days of receipt that s/he
+     does not agree to the terms of the full Project Gutenberg-tm
+     License.  You must require such a user to return or
+     destroy all copies of the works possessed in a physical medium
+     and discontinue all use of and all access to other copies of
+     Project Gutenberg-tm works.
+
+- You provide, in accordance with paragraph 1.F.3, a full refund of any
+     money paid for a work or a replacement copy, if a defect in the
+     electronic work is discovered and reported to you within 90 days
+     of receipt of the work.
+
+- You comply with all other terms of this agreement for free
+     distribution of Project Gutenberg-tm works.
+
+1.E.9.  If you wish to charge a fee or distribute a Project Gutenberg-tm
+electronic work or group of works on different terms than are set
+forth in this agreement, you must obtain permission in writing from
+both the Project Gutenberg Literary Archive Foundation and Michael
+Hart, the owner of the Project Gutenberg-tm trademark.  Contact the
+Foundation as set forth in Section 3 below.
+
+1.F.
+
+1.F.1.  Project Gutenberg volunteers and employees expend considerable
+effort to identify, do copyright research on, transcribe and proofread
+public domain works in creating the Project Gutenberg-tm
+collection.  Despite these efforts, Project Gutenberg-tm electronic
+works, and the medium on which they may be stored, may contain
+"Defects," such as, but not limited to, incomplete, inaccurate or
+corrupt data, transcription errors, a copyright or other intellectual
+property infringement, a defective or damaged disk or other medium, a
+computer virus, or computer codes that damage or cannot be read by
+your equipment.
+
+1.F.2.  LIMITED WARRANTY, DISCLAIMER OF DAMAGES - Except for the "Right
+of Replacement or Refund" described in paragraph 1.F.3, the Project
+Gutenberg Literary Archive Foundation, the owner of the Project
+Gutenberg-tm trademark, and any other party distributing a Project
+Gutenberg-tm electronic work under this agreement, disclaim all
+liability to you for damages, costs and expenses, including legal
+fees.  YOU AGREE THAT YOU HAVE NO REMEDIES FOR NEGLIGENCE, STRICT
+LIABILITY, BREACH OF WARRANTY OR BREACH OF CONTRACT EXCEPT THOSE
+PROVIDED IN PARAGRAPH 1.F.3.  YOU AGREE THAT THE FOUNDATION, THE
+TRADEMARK OWNER, AND ANY DISTRIBUTOR UNDER THIS AGREEMENT WILL NOT BE
+LIABLE TO YOU FOR ACTUAL, DIRECT, INDIRECT, CONSEQUENTIAL, PUNITIVE OR
+INCIDENTAL DAMAGES EVEN IF YOU GIVE NOTICE OF THE POSSIBILITY OF SUCH
+DAMAGE.
+
+1.F.3.  LIMITED RIGHT OF REPLACEMENT OR REFUND - If you discover a
+defect in this electronic work within 90 days of receiving it, you can
+receive a refund of the money (if any) you paid for it by sending a
+written explanation to the person you received the work from.  If you
+received the work on a physical medium, you must return the medium with
+your written explanation.  The person or entity that provided you with
+the defective work may elect to provide a replacement copy in lieu of a
+refund.  If you received the work electronically, the person or entity
+providing it to you may choose to give you a second opportunity to
+receive the work electronically in lieu of a refund.  If the second copy
+is also defective, you may demand a refund in writing without further
+opportunities to fix the problem.
+
+1.F.4.  Except for the limited right of replacement or refund set forth
+in paragraph 1.F.3, this work is provided to you 'AS-IS', WITH NO OTHER
+WARRANTIES OF ANY KIND, EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO
+WARRANTIES OF MERCHANTABILITY OR FITNESS FOR ANY PURPOSE.
+
+1.F.5.  Some states do not allow disclaimers of certain implied
+warranties or the exclusion or limitation of certain types of damages.
+If any disclaimer or limitation set forth in this agreement violates the
+law of the state applicable to this agreement, the agreement shall be
+interpreted to make the maximum disclaimer or limitation permitted by
+the applicable state law.  The invalidity or unenforceability of any
+provision of this agreement shall not void the remaining provisions.
+
+1.F.6.  INDEMNITY - You agree to indemnify and hold the Foundation, the
+trademark owner, any agent or employee of the Foundation, anyone
+providing copies of Project Gutenberg-tm electronic works in accordance
+with this agreement, and any volunteers associated with the production,
+promotion and distribution of Project Gutenberg-tm electronic works,
+harmless from all liability, costs and expenses, including legal fees,
+that arise directly or indirectly from any of the following which you do
+or cause to occur: (a) distribution of this or any Project Gutenberg-tm
+work, (b) alteration, modification, or additions or deletions to any
+Project Gutenberg-tm work, and (c) any Defect you cause.
+
+
+Section  2.  Information about the Mission of Project Gutenberg-tm
+
+Project Gutenberg-tm is synonymous with the free distribution of
+electronic works in formats readable by the widest variety of computers
+including obsolete, old, middle-aged and new computers.  It exists
+because of the efforts of hundreds of volunteers and donations from
+people in all walks of life.
+
+Volunteers and financial support to provide volunteers with the
+assistance they need are critical to reaching Project Gutenberg-tm's
+goals and ensuring that the Project Gutenberg-tm collection will
+remain freely available for generations to come.  In 2001, the Project
+Gutenberg Literary Archive Foundation was created to provide a secure
+and permanent future for Project Gutenberg-tm and future generations.
+To learn more about the Project Gutenberg Literary Archive Foundation
+and how your efforts and donations can help, see Sections 3 and 4
+and the Foundation information page at www.gutenberg.org
+
+
+Section 3.  Information about the Project Gutenberg Literary Archive
+Foundation
+
+The Project Gutenberg Literary Archive Foundation is a non profit
+501(c)(3) educational corporation organized under the laws of the
+state of Mississippi and granted tax exempt status by the Internal
+Revenue Service.  The Foundation's EIN or federal tax identification
+number is 64-6221541.  Contributions to the Project Gutenberg
+Literary Archive Foundation are tax deductible to the full extent
+permitted by U.S. federal laws and your state's laws.
+
+The Foundation's principal office is located at 4557 Melan Dr. S.
+Fairbanks, AK, 99712., but its volunteers and employees are scattered
+throughout numerous locations.  Its business office is located at 809
+North 1500 West, Salt Lake City, UT 84116, (801) 596-1887.  Email
+contact links and up to date contact information can be found at the
+Foundation's web site and official page at www.gutenberg.org/contact
+
+For additional contact information:
+     Dr. Gregory B. Newby
+     Chief Executive and Director
+     gbnewby@pglaf.org
+
+Section 4.  Information about Donations to the Project Gutenberg
+Literary Archive Foundation
+
+Project Gutenberg-tm depends upon and cannot survive without wide
+spread public support and donations to carry out its mission of
+increasing the number of public domain and licensed works that can be
+freely distributed in machine readable form accessible by the widest
+array of equipment including outdated equipment.  Many small donations
+($1 to $5,000) are particularly important to maintaining tax exempt
+status with the IRS.
+
+The Foundation is committed to complying with the laws regulating
+charities and charitable donations in all 50 states of the United
+States.  Compliance requirements are not uniform and it takes a
+considerable effort, much paperwork and many fees to meet and keep up
+with these requirements.  We do not solicit donations in locations
+where we have not received written confirmation of compliance.  To
+SEND DONATIONS or determine the status of compliance for any
+particular state visit www.gutenberg.org/donate
+
+While we cannot and do not solicit contributions from states where we
+have not met the solicitation requirements, we know of no prohibition
+against accepting unsolicited donations from donors in such states who
+approach us with offers to donate.
+
+International donations are gratefully accepted, but we cannot make
+any statements concerning tax treatment of donations received from
+outside the United States.  U.S. laws alone swamp our small staff.
+
+Please check the Project Gutenberg Web pages for current donation
+methods and addresses.  Donations are accepted in a number of other
+ways including checks, online payments and credit card donations.
+To donate, please visit:  www.gutenberg.org/donate
+
+
+Section 5.  General Information About Project Gutenberg-tm electronic
+works.
+
+Professor Michael S. Hart was the originator of the Project Gutenberg-tm
+concept of a library of electronic works that could be freely shared
+with anyone.  For forty years, he produced and distributed Project
+Gutenberg-tm eBooks with only a loose network of volunteer support.
+
+Project Gutenberg-tm eBooks are often created from several printed
+editions, all of which are confirmed as Public Domain in the U.S.
+unless a copyright notice is included.  Thus, we do not necessarily
+keep eBooks in compliance with any particular paper edition.
+
+Most people start at our Web site which has the main PG search facility:
+
+     www.gutenberg.org
+
+This Web site includes information about Project Gutenberg-tm,
+including how to make donations to the Project Gutenberg Literary
+Archive Foundation, how to help produce our new eBooks, and how to
+subscribe to our email newsletter to hear about new eBooks.
+\end{PGtext}
+
+% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %
+%                                                                         %
+% End of Project Gutenberg's A Primer of Quaternions, by Arthur S. Hathaway
+%                                                                         %
+% *** END OF THIS PROJECT GUTENBERG EBOOK A PRIMER OF QUATERNIONS ***     %
+%                                                                         %
+% ***** This file should be named 9934-t.tex or 9934-t.zip *****          %
+% This and all associated files of various formats will be found in:      %
+%         http://www.gutenberg.org/9/9/3/9934/                            %
+%                                                                         %
+% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %
+
+\end{document}
+
+This is pdfTeX, Version 3.1415926-2.5-1.40.14 (TeX Live 2013/Debian) (format=pdflatex 2014.9.6)  25 APR 2015 04:54
+entering extended mode
+ %&-line parsing enabled.
+**9934-t.tex
+(./9934-t.tex
+LaTeX2e <2011/06/27>
+Babel <3.9h> and hyphenation patterns for 78 languages loaded.
+(/usr/share/texlive/texmf-dist/tex/latex/base/book.cls
+Document Class: book 2007/10/19 v1.4h Standard LaTeX document class
+(/usr/share/texlive/texmf-dist/tex/latex/base/bk12.clo
+File: bk12.clo 2007/10/19 v1.4h Standard LaTeX file (size option)
+)
+\c@part=\count79
+\c@chapter=\count80
+\c@section=\count81
+\c@subsection=\count82
+\c@subsubsection=\count83
+\c@paragraph=\count84
+\c@subparagraph=\count85
+\c@figure=\count86
+\c@table=\count87
+\abovecaptionskip=\skip41
+\belowcaptionskip=\skip42
+\bibindent=\dimen102
+) (/usr/share/texlive/texmf-dist/tex/latex/amsmath/amsmath.sty
+Package: amsmath 2013/01/14 v2.14 AMS math features
+\@mathmargin=\skip43
+For additional information on amsmath, use the `?' option.
+(/usr/share/texlive/texmf-dist/tex/latex/amsmath/amstext.sty
+Package: amstext 2000/06/29 v2.01
+(/usr/share/texlive/texmf-dist/tex/latex/amsmath/amsgen.sty
+File: amsgen.sty 1999/11/30 v2.0
+\@emptytoks=\toks14
+\ex@=\dimen103
+)) (/usr/share/texlive/texmf-dist/tex/latex/amsmath/amsbsy.sty
+Package: amsbsy 1999/11/29 v1.2d
+\pmbraise@=\dimen104
+) (/usr/share/texlive/texmf-dist/tex/latex/amsmath/amsopn.sty
+Package: amsopn 1999/12/14 v2.01 operator names
+)
+\inf@bad=\count88
+LaTeX Info: Redefining \frac on input line 210.
+\uproot@=\count89
+\leftroot@=\count90
+LaTeX Info: Redefining \overline on input line 306.
+\classnum@=\count91
+\DOTSCASE@=\count92
+LaTeX Info: Redefining \ldots on input line 378.
+LaTeX Info: Redefining \dots on input line 381.
+LaTeX Info: Redefining \cdots on input line 466.
+\Mathstrutbox@=\box26
+\strutbox@=\box27
+\big@size=\dimen105
+LaTeX Font Info:    Redeclaring font encoding OML on input line 566.
+LaTeX Font Info:    Redeclaring font encoding OMS on input line 567.
+\macc@depth=\count93
+\c@MaxMatrixCols=\count94
+\dotsspace@=\muskip10
+\c@parentequation=\count95
+\dspbrk@lvl=\count96
+\tag@help=\toks15
+\row@=\count97
+\column@=\count98
+\maxfields@=\count99
+\andhelp@=\toks16
+\eqnshift@=\dimen106
+\alignsep@=\dimen107
+\tagshift@=\dimen108
+\tagwidth@=\dimen109
+\totwidth@=\dimen110
+\lineht@=\dimen111
+\@envbody=\toks17
+\multlinegap=\skip44
+\multlinetaggap=\skip45
+\mathdisplay@stack=\toks18
+LaTeX Info: Redefining \[ on input line 2665.
+LaTeX Info: Redefining \] on input line 2666.
+) (/usr/share/texlive/texmf-dist/tex/latex/amsfonts/amssymb.sty
+Package: amssymb 2013/01/14 v3.01 AMS font symbols
+(/usr/share/texlive/texmf-dist/tex/latex/amsfonts/amsfonts.sty
+Package: amsfonts 2013/01/14 v3.01 Basic AMSFonts support
+\symAMSa=\mathgroup4
+\symAMSb=\mathgroup5
+LaTeX Font Info:    Overwriting math alphabet `\mathfrak' in version `bold'
+(Font)                  U/euf/m/n --> U/euf/b/n on input line 106.
+)) (/usr/share/texlive/texmf-dist/tex/latex/amscls/amsthm.sty
+Package: amsthm 2004/08/06 v2.20
+\thm@style=\toks19
+\thm@bodyfont=\toks20
+\thm@headfont=\toks21
+\thm@notefont=\toks22
+\thm@headpunct=\toks23
+\thm@preskip=\skip46
+\thm@postskip=\skip47
+\thm@headsep=\skip48
+\dth@everypar=\toks24
+)
+
+LaTeX Warning: You have requested, on input line 76, version
+               `2009/07/02' of package amsthm,
+               but only version
+               `2004/08/06 v2.20'
+               is available.
+
+(/usr/share/texlive/texmf-dist/tex/latex/base/alltt.sty
+Package: alltt 1997/06/16 v2.0g defines alltt environment
+) (/usr/share/texlive/texmf-dist/tex/latex/graphics/graphicx.sty
+Package: graphicx 1999/02/16 v1.0f Enhanced LaTeX Graphics (DPC,SPQR)
+(/usr/share/texlive/texmf-dist/tex/latex/graphics/keyval.sty
+Package: keyval 1999/03/16 v1.13 key=value parser (DPC)
+\KV@toks@=\toks25
+) (/usr/share/texlive/texmf-dist/tex/latex/graphics/graphics.sty
+Package: graphics 2009/02/05 v1.0o Standard LaTeX Graphics (DPC,SPQR)
+(/usr/share/texlive/texmf-dist/tex/latex/graphics/trig.sty
+Package: trig 1999/03/16 v1.09 sin cos tan (DPC)
+) (/usr/share/texlive/texmf-dist/tex/latex/latexconfig/graphics.cfg
+File: graphics.cfg 2010/04/23 v1.9 graphics configuration of TeX Live
+)
+Package graphics Info: Driver file: pdftex.def on input line 91.
+(/usr/share/texlive/texmf-dist/tex/latex/pdftex-def/pdftex.def
+File: pdftex.def 2011/05/27 v0.06d Graphics/color for pdfTeX
+(/usr/share/texlive/texmf-dist/tex/generic/oberdiek/infwarerr.sty
+Package: infwarerr 2010/04/08 v1.3 Providing info/warning/error messages (HO)
+) (/usr/share/texlive/texmf-dist/tex/generic/oberdiek/ltxcmds.sty
+Package: ltxcmds 2011/11/09 v1.22 LaTeX kernel commands for general use (HO)
+)
+\Gread@gobject=\count100
+))
+\Gin@req@height=\dimen112
+\Gin@req@width=\dimen113
+) (./9934-t.aux)
+\openout1 = `9934-t.aux'.
+
+LaTeX Font Info:    Checking defaults for OML/cmm/m/it on input line 127.
+LaTeX Font Info:    ... okay on input line 127.
+LaTeX Font Info:    Checking defaults for T1/cmr/m/n on input line 127.
+LaTeX Font Info:    ... okay on input line 127.
+LaTeX Font Info:    Checking defaults for OT1/cmr/m/n on input line 127.
+LaTeX Font Info:    ... okay on input line 127.
+LaTeX Font Info:    Checking defaults for OMS/cmsy/m/n on input line 127.
+LaTeX Font Info:    ... okay on input line 127.
+LaTeX Font Info:    Checking defaults for OMX/cmex/m/n on input line 127.
+LaTeX Font Info:    ... okay on input line 127.
+LaTeX Font Info:    Checking defaults for U/cmr/m/n on input line 127.
+LaTeX Font Info:    ... okay on input line 127.
+(/usr/share/texlive/texmf-dist/tex/context/base/supp-pdf.mkii
+[Loading MPS to PDF converter (version 2006.09.02).]
+\scratchcounter=\count101
+\scratchdimen=\dimen114
+\scratchbox=\box28
+\nofMPsegments=\count102
+\nofMParguments=\count103
+\everyMPshowfont=\toks26
+\MPscratchCnt=\count104
+\MPscratchDim=\dimen115
+\MPnumerator=\count105
+\makeMPintoPDFobject=\count106
+\everyMPtoPDFconversion=\toks27
+) (/usr/share/texlive/texmf-dist/tex/generic/oberdiek/pdftexcmds.sty
+Package: pdftexcmds 2011/11/29 v0.20 Utility functions of pdfTeX for LuaTeX (HO
+)
+(/usr/share/texlive/texmf-dist/tex/generic/oberdiek/ifluatex.sty
+Package: ifluatex 2010/03/01 v1.3 Provides the ifluatex switch (HO)
+Package ifluatex Info: LuaTeX not detected.
+) (/usr/share/texlive/texmf-dist/tex/generic/oberdiek/ifpdf.sty
+Package: ifpdf 2011/01/30 v2.3 Provides the ifpdf switch (HO)
+Package ifpdf Info: pdfTeX in PDF mode is detected.
+)
+Package pdftexcmds Info: LuaTeX not detected.
+Package pdftexcmds Info: \pdf@primitive is available.
+Package pdftexcmds Info: \pdf@ifprimitive is available.
+Package pdftexcmds Info: \pdfdraftmode found.
+)
+LaTeX Font Info:    Try loading font information for U+msa on input line 157.
+(/usr/share/texlive/texmf-dist/tex/latex/amsfonts/umsa.fd
+File: umsa.fd 2013/01/14 v3.01 AMS symbols A
+)
+LaTeX Font Info:    Try loading font information for U+msb on input line 157.
+(/usr/share/texlive/texmf-dist/tex/latex/amsfonts/umsb.fd
+File: umsb.fd 2013/01/14 v3.01 AMS symbols B
+) [1
+
+
+{/var/lib/texmf/fonts/map/pdftex/updmap/pdftex.map}] [2
+
+] [3
+
+] [1] [2] (./9934-t.toc [3
+
+])
+\tf@toc=\write3
+\openout3 = `9934-t.toc'.
+
+[4]
+Chapter 1.
+<images/Image1.png, id=37, 672.26157pt x 234.1247pt>
+File: images/Image1.png Graphic file (type png)
+<use images/Image1.png>
+Package pdftex.def Info: images/Image1.png used on input line 306.
+(pdftex.def)             Requested size: 227.62204pt x 79.27267pt.
+[1
+
+
+ <./images/Image1.png>] <images/Image2.png, id=42, 277.035pt x 284.56313pt>
+File: images/Image2.png Graphic file (type png)
+<use images/Image2.png>
+Package pdftex.def Info: images/Image2.png used on input line 318.
+(pdftex.def)             Requested size: 227.62204pt x 233.80786pt.
+<images/Image3.png, id=43, 858.20625pt x 429.605pt>
+File: images/Image3.png Graphic file (type png)
+<use images/Image3.png>
+Package pdftex.def Info: images/Image3.png used on input line 331.
+(pdftex.def)             Requested size: 227.62204pt x 113.9431pt.
+[2 <./images/Image2.png> <./images/Image3.png>] <images/Image4.png, id=50, 359.
+84438pt x 289.83281pt>
+File: images/Image4.png Graphic file (type png)
+<use images/Image4.png>
+Package pdftex.def Info: images/Image4.png used on input line 347.
+(pdftex.def)             Requested size: 227.62204pt x 183.34303pt.
+[3 <./images/Image4.png>]
+LaTeX Font Info:    Try loading font information for OMS+cmr on input line 387.
+
+(/usr/share/texlive/texmf-dist/tex/latex/base/omscmr.fd
+File: omscmr.fd 1999/05/25 v2.5h Standard LaTeX font definitions
+)
+LaTeX Font Info:    Font shape `OMS/cmr/m/n' in size <12> not available
+(Font)              Font shape `OMS/cmsy/m/n' tried instead on input line 387.
+<images/Image5.png, id=55, 730.22812pt x 210.7875pt>
+File: images/Image5.png Graphic file (type png)
+<use images/Image5.png>
+Package pdftex.def Info: images/Image5.png used on input line 405.
+(pdftex.def)             Requested size: 227.62204pt x 65.70367pt.
+[4 <./images/Image5.png>] <images/Image6.png, id=61, 658.71094pt x 395.22656pt>
+File: images/Image6.png Graphic file (type png)
+<use images/Image6.png>
+Package pdftex.def Info: images/Image6.png used on input line 426.
+(pdftex.def)             Requested size: 227.62204pt x 136.5704pt.
+[5 <./images/Image6.png>] <images/Image7.png, id=66, 682.55pt x 509.905pt>
+File: images/Image7.png Graphic file (type png)
+<use images/Image7.png>
+Package pdftex.def Info: images/Image7.png used on input line 484.
+(pdftex.def)             Requested size: 227.62204pt x 170.04312pt.
+[6 <./images/Image7.png>] <images/Image8.png, id=72, 610.28pt x 414.54875pt>
+File: images/Image8.png Graphic file (type png)
+<use images/Image8.png>
+Package pdftex.def Info: images/Image8.png used on input line 510.
+(pdftex.def)             Requested size: 227.62204pt x 154.62042pt.
+[7 <./images/Image8.png>] [8] <images/Image9.png, id=81, 668.4975pt x 777.65532
+pt>
+File: images/Image9.png Graphic file (type png)
+<use images/Image9.png>
+Package pdftex.def Info: images/Image9.png used on input line 587.
+(pdftex.def)             Requested size: 227.62204pt x 264.7909pt.
+[9 <./images/Image9.png>] [10] <images/Image10.png, id=91, 491.58656pt x 548.04
+75pt>
+File: images/Image10.png Graphic file (type png)
+<use images/Image10.png>
+Package pdftex.def Info: images/Image10.png used on input line 669.
+(pdftex.def)             Requested size: 227.62204pt x 253.76904pt.
+[11 <./images/Image10.png>] <images/Image11.png, id=96, 529.98pt x 461.725pt>
+File: images/Image11.png Graphic file (type png)
+<use images/Image11.png>
+Package pdftex.def Info: images/Image11.png used on input line 708.
+(pdftex.def)             Requested size: 227.62204pt x 198.30537pt.
+[12 <./images/Image11.png>] <images/Image12.png, id=102, 500.62032pt x 502.8787
+5pt>
+File: images/Image12.png Graphic file (type png)
+<use images/Image12.png>
+Package pdftex.def Info: images/Image12.png used on input line 737.
+(pdftex.def)             Requested size: 227.62204pt x 228.64905pt.
+[13 <./images/Image12.png>] [14]
+Underfull \hbox (badness 2591) in paragraph at lines 817--821
+[]\OT1/cmr/m/n/10.95 Show that $[]$ is bi-sected by $[]$, and tri-sected by
+ []
+
+[15] [16] [17]
+Chapter 2.
+<images/Image13.png, id=120, 496.85625pt x 357.58594pt>
+File: images/Image13.png Graphic file (type png)
+<use images/Image13.png>
+Package pdftex.def Info: images/Image13.png used on input line 978.
+(pdftex.def)             Requested size: 227.62204pt x 163.82039pt.
+[18
+
+ <./images/Image13.png>] <images/Image14.png, id=126, 700.6175pt x 542.025pt>
+File: images/Image14.png Graphic file (type png)
+<use images/Image14.png>
+Package pdftex.def Info: images/Image14.png used on input line 1012.
+(pdftex.def)             Requested size: 227.62204pt x 176.09813pt.
+[19 <./images/Image14.png>] <images/Image15.png, id=131, 645.41125pt x 595.2237
+5pt>
+File: images/Image15.png Graphic file (type png)
+<use images/Image15.png>
+Package pdftex.def Info: images/Image15.png used on input line 1047.
+(pdftex.def)             Requested size: 227.62204pt x 209.92085pt.
+[20 <./images/Image15.png>] <images/Image16.png, id=136, 582.175pt x 534.99875p
+t>
+File: images/Image16.png Graphic file (type png)
+<use images/Image16.png>
+Package pdftex.def Info: images/Image16.png used on input line 1082.
+(pdftex.def)             Requested size: 227.62204pt x 209.17113pt.
+[21 <./images/Image16.png>] <images/Image17.png, id=141, 605.26125pt x 519.9425
+pt>
+File: images/Image17.png Graphic file (type png)
+<use images/Image17.png>
+Package pdftex.def Info: images/Image17.png used on input line 1119.
+(pdftex.def)             Requested size: 227.62204pt x 195.53331pt.
+[22 <./images/Image17.png>] <images/Image18.png, id=146, 642.4pt x 572.1375pt>
+File: images/Image18.png Graphic file (type png)
+<use images/Image18.png>
+Package pdftex.def Info: images/Image18.png used on input line 1162.
+(pdftex.def)             Requested size: 227.62204pt x 202.72174pt.
+[23] <images/Image19.png, id=150, 487.8225pt x 462.97969pt>
+File: images/Image19.png Graphic file (type png)
+<use images/Image19.png>
+Package pdftex.def Info: images/Image19.png used on input line 1182.
+(pdftex.def)             Requested size: 227.62204pt x 216.03215pt.
+[24 <./images/Image18.png>] [25 <./images/Image19.png>] [26] [27]
+Chapter 3.
+[28
+
+] <images/Image20.png, id=170, 592.2125pt x 260.975pt>
+File: images/Image20.png Graphic file (type png)
+<use images/Image20.png>
+Package pdftex.def Info: images/Image20.png used on input line 1380.
+(pdftex.def)             Requested size: 227.62204pt x 100.30644pt.
+[29 <./images/Image20.png>] [30] [31] [32] <images/Image21.png, id=187, 480.294
+37pt x 307.1475pt>
+File: images/Image21.png Graphic file (type png)
+<use images/Image21.png>
+Package pdftex.def Info: images/Image21.png used on input line 1605.
+(pdftex.def)             Requested size: 227.62204pt x 145.56381pt.
+[33 <./images/Image21.png>]
+Overfull \hbox (3.6522pt too wide) in paragraph at lines 1635--1637
+[]$[]\OML/cmm/m/it/12 q[] \OT1/cmr/m/n/12 = []\OML/cmm/m/it/12 q \OT1/cmr/m/n/1
+2 + []\OML/cmm/m/it/12 q \OT1/cmr/m/n/12 + []\OML/cmm/m/it/12 q [] \OT1/cmr/m/n
+/12 = \OML/cmm/m/it/12 n[]q \OMS/cmsy/m/n/12 ^^F \OT1/cmr/m/n/12 ([])$. 
+ []
+
+[34] <images/Image22.png, id=197, 442.65375pt x 327.47343pt>
+File: images/Image22.png Graphic file (type png)
+<use images/Image22.png>
+Package pdftex.def Info: images/Image22.png used on input line 1697.
+(pdftex.def)             Requested size: 227.62204pt x 168.39836pt.
+[35 <./images/Image22.png>] <images/Image23.png, id=202, 536.75531pt x 308.6531
+2pt>
+File: images/Image23.png Graphic file (type png)
+<use images/Image23.png>
+Package pdftex.def Info: images/Image23.png used on input line 1750.
+(pdftex.def)             Requested size: 227.62204pt x 130.8909pt.
+[36 <./images/Image23.png>]
+Overfull \hbox (10.74892pt too wide) in paragraph at lines 1810--1812
+[]\OT1/cmr/m/n/10.95 Show that $[]$; $[]$; $[]$; $[]$. 
+ []
+
+[37] <images/Image24.png, id=211, 457.71pt x 521.69907pt>
+File: images/Image24.png Graphic file (type png)
+<use images/Image24.png>
+Package pdftex.def Info: images/Image24.png used on input line 1863.
+(pdftex.def)             Requested size: 227.62204pt x 259.43988pt.
+[38]
+Underfull \hbox (badness 1661) in paragraph at lines 1866--1873
+\OT1/cmr/m/n/12 let $\OML/cmm/m/it/12 p[]$\OT1/cmr/m/n/12 , $\OML/cmm/m/it/12 q
+[]$\OT1/cmr/m/n/12 , and $[]$, then $(\OML/cmm/m/it/12 p \OT1/cmr/m/n/12 +
+ []
+
+<images/Image25.png, id=215, 451.6875pt x 191.9672pt>
+File: images/Image25.png Graphic file (type png)
+<use images/Image25.png>
+Package pdftex.def Info: images/Image25.png used on input line 1885.
+(pdftex.def)             Requested size: 227.62204pt x 96.742pt.
+[39 <./images/Image24.png>] <images/Image26.png, id=220, 472.76625pt x 502.8787
+5pt>
+File: images/Image26.png Graphic file (type png)
+<use images/Image26.png>
+Package pdftex.def Info: images/Image26.png used on input line 1905.
+(pdftex.def)             Requested size: 227.62204pt x 242.12337pt.
+[40 <./images/Image25.png> <./images/Image26.png>] <images/Image27.png, id=226,
+ 529.98pt x 672.5125pt>
+File: images/Image27.png Graphic file (type png)
+<use images/Image27.png>
+Package pdftex.def Info: images/Image27.png used on input line 1934.
+(pdftex.def)             Requested size: 227.62204pt x 288.83609pt.
+[41 <./images/Image27.png>] <images/Image28.png, id=231, 547.04375pt x 461.725p
+t>
+File: images/Image28.png Graphic file (type png)
+<use images/Image28.png>
+Package pdftex.def Info: images/Image28.png used on input line 1974.
+(pdftex.def)             Requested size: 227.62204pt x 192.11957pt.
+[42] [43 <./images/Image28.png>] [44] [45] [46] <images/Image29.png, id=251, 40
+8.77719pt x 421.575pt>
+File: images/Image29.png Graphic file (type png)
+<use images/Image29.png>
+Package pdftex.def Info: images/Image29.png used on input line 2234.
+(pdftex.def)             Requested size: 227.62204pt x 234.74881pt.
+[47 <./images/Image29.png>] <images/Image30.png, id=256, 501.875pt x 371.3875pt
+>
+File: images/Image30.png Graphic file (type png)
+<use images/Image30.png>
+Package pdftex.def Info: images/Image30.png used on input line 2257.
+(pdftex.def)             Requested size: 227.62204pt x 168.4432pt.
+[48 <./images/Image30.png>] <images/Image31.png, id=262, 484.81125pt x 396.4812
+5pt>
+File: images/Image31.png Graphic file (type png)
+<use images/Image31.png>
+Package pdftex.def Info: images/Image31.png used on input line 2306.
+(pdftex.def)             Requested size: 227.62204pt x 186.1526pt.
+[49 <./images/Image31.png>] [50] [51] [52]
+Chapter 4.
+<images/Image32.png, id=276, 484.05844pt x 344.78813pt>
+File: images/Image32.png Graphic file (type png)
+<use images/Image32.png>
+Package pdftex.def Info: images/Image32.png used on input line 2587.
+(pdftex.def)             Requested size: 227.62204pt x 162.13461pt.
+[53
+
+] [54 <./images/Image32.png>] <images/Image33.png, id=285, 750.805pt x 389.455p
+t>
+File: images/Image33.png Graphic file (type png)
+<use images/Image33.png>
+Package pdftex.def Info: images/Image33.png used on input line 2638.
+(pdftex.def)             Requested size: 227.62204pt x 118.0675pt.
+[55 <./images/Image33.png>] [56] [57] [58] <images/Image34.png, id=299, 451.687
+5pt x 442.65375pt>
+File: images/Image34.png Graphic file (type png)
+<use images/Image34.png>
+Package pdftex.def Info: images/Image34.png used on input line 2862.
+(pdftex.def)             Requested size: 227.62204pt x 223.0757pt.
+[59 <./images/Image34.png>] [60] [61] [62] [63] [64] [65]
+Overfull \hbox (2.34082pt too wide) in paragraph at lines 3244--3248
+[]\OT1/cmr/m/n/12 Conversely, this func-tion sat-is-fies (\OT1/cmr/m/it/12 a\OT
+1/cmr/m/n/12 ), for we have $\OML/cmm/m/it/12 S^^[^^^[]^^Z \OT1/cmr/m/n/12 = \O
+ML/cmm/m/it/12 S^^Z^^^\OT1/cmr/m/n/12 (\OMS/cmsy/m/n/12 ^^@\OT1/cmr/bx/n/12 i\O
+ML/cmm/m/it/12 S\OT1/cmr/bx/n/12 i\OML/cmm/m/it/12 ^^[ \OMS/cmsy/m/n/12 ^^@
+ []
+
+[66] [67] [68] [69] [70] [71] [1
+
+] [2] [3] [4] [5] [6] [7] (./9934-t.aux)
+
+ *File List*
+    book.cls    2007/10/19 v1.4h Standard LaTeX document class
+    bk12.clo    2007/10/19 v1.4h Standard LaTeX file (size option)
+ amsmath.sty    2013/01/14 v2.14 AMS math features
+ amstext.sty    2000/06/29 v2.01
+  amsgen.sty    1999/11/30 v2.0
+  amsbsy.sty    1999/11/29 v1.2d
+  amsopn.sty    1999/12/14 v2.01 operator names
+ amssymb.sty    2013/01/14 v3.01 AMS font symbols
+amsfonts.sty    2013/01/14 v3.01 Basic AMSFonts support
+  amsthm.sty    2004/08/06 v2.20
+   alltt.sty    1997/06/16 v2.0g defines alltt environment
+graphicx.sty    1999/02/16 v1.0f Enhanced LaTeX Graphics (DPC,SPQR)
+  keyval.sty    1999/03/16 v1.13 key=value parser (DPC)
+graphics.sty    2009/02/05 v1.0o Standard LaTeX Graphics (DPC,SPQR)
+    trig.sty    1999/03/16 v1.09 sin cos tan (DPC)
+graphics.cfg    2010/04/23 v1.9 graphics configuration of TeX Live
+  pdftex.def    2011/05/27 v0.06d Graphics/color for pdfTeX
+infwarerr.sty    2010/04/08 v1.3 Providing info/warning/error messages (HO)
+ ltxcmds.sty    2011/11/09 v1.22 LaTeX kernel commands for general use (HO)
+supp-pdf.mkii
+pdftexcmds.sty    2011/11/29 v0.20 Utility functions of pdfTeX for LuaTeX (HO)
+ifluatex.sty    2010/03/01 v1.3 Provides the ifluatex switch (HO)
+   ifpdf.sty    2011/01/30 v2.3 Provides the ifpdf switch (HO)
+    umsa.fd    2013/01/14 v3.01 AMS symbols A
+    umsb.fd    2013/01/14 v3.01 AMS symbols B
+images/Image1.png
+images/Image2.png
+images/Image3.png
+images/Image4.png
+  omscmr.fd    1999/05/25 v2.5h Standard LaTeX font definitions
+images/Image5.png
+images/Image6.png
+images/Image7.png
+images/Image8.png
+images/Image9.png
+images/Image10.png
+images/Image11.png
+images/Image12.png
+images/Image13.png
+images/Image14.png
+images/Image15.png
+images/Image16.png
+images/Image17.png
+images/Image18.png
+images/Image19.png
+images/Image20.png
+images/Image21.png
+images/Image22.png
+images/Image23.png
+images/Image24.png
+images/Image25.png
+images/Image26.png
+images/Image27.png
+images/Image28.png
+images/Image29.png
+images/Image30.png
+images/Image31.png
+images/Image32.png
+images/Image33.png
+images/Image34.png
+ ***********
+
+ ) 
+Here is how much of TeX's memory you used:
+ 2474 strings out of 493304
+ 32167 string characters out of 6139872
+ 93173 words of memory out of 5000000
+ 5868 multiletter control sequences out of 15000+600000
+ 21250 words of font info for 79 fonts, out of 8000000 for 9000
+ 957 hyphenation exceptions out of 8191
+ 26i,11n,26p,264b,378s stack positions out of 5000i,500n,10000p,200000b,80000s
+</usr/share/texlive/texmf-dist/fonts/type1/public/amsfonts/cm/cmbx10.pfb></us
+r/share/texlive/texmf-dist/fonts/type1/public/amsfonts/cm/cmbx12.pfb></usr/shar
+e/texlive/texmf-dist/fonts/type1/public/amsfonts/cm/cmbx8.pfb></usr/share/texli
+ve/texmf-dist/fonts/type1/public/amsfonts/cm/cmcsc10.pfb></usr/share/texlive/te
+xmf-dist/fonts/type1/public/amsfonts/cm/cmex10.pfb></usr/share/texlive/texmf-di
+st/fonts/type1/public/amsfonts/cm/cmmi10.pfb></usr/share/texlive/texmf-dist/fon
+ts/type1/public/amsfonts/cm/cmmi12.pfb></usr/share/texlive/texmf-dist/fonts/typ
+e1/public/amsfonts/cm/cmmi6.pfb></usr/share/texlive/texmf-dist/fonts/type1/publ
+ic/amsfonts/cm/cmmi8.pfb></usr/share/texlive/texmf-dist/fonts/type1/public/amsf
+onts/cm/cmr10.pfb></usr/share/texlive/texmf-dist/fonts/type1/public/amsfonts/cm
+/cmr12.pfb></usr/share/texlive/texmf-dist/fonts/type1/public/amsfonts/cm/cmr17.
+pfb></usr/share/texlive/texmf-dist/fonts/type1/public/amsfonts/cm/cmr6.pfb></us
+r/share/texlive/texmf-dist/fonts/type1/public/amsfonts/cm/cmr7.pfb></usr/share/
+texlive/texmf-dist/fonts/type1/public/amsfonts/cm/cmr8.pfb></usr/share/texlive/
+texmf-dist/fonts/type1/public/amsfonts/cm/cmsy10.pfb></usr/share/texlive/texmf-
+dist/fonts/type1/public/amsfonts/cm/cmsy6.pfb></usr/share/texlive/texmf-dist/fo
+nts/type1/public/amsfonts/cm/cmsy7.pfb></usr/share/texlive/texmf-dist/fonts/typ
+e1/public/amsfonts/cm/cmsy8.pfb></usr/share/texlive/texmf-dist/fonts/type1/publ
+ic/amsfonts/cm/cmti10.pfb></usr/share/texlive/texmf-dist/fonts/type1/public/ams
+fonts/cm/cmti12.pfb></usr/share/texlive/texmf-dist/fonts/type1/public/amsfonts/
+cm/cmtt10.pfb></usr/share/texlive/texmf-dist/fonts/type1/public/amsfonts/cm/cmt
+t9.pfb></usr/share/texlive/texmf-dist/fonts/type1/public/amsfonts/symbols/msam1
+0.pfb>
+Output written on 9934-t.pdf (85 pages, 621107 bytes).
+PDF statistics:
+ 444 PDF objects out of 1000 (max. 8388607)
+ 262 compressed objects within 3 object streams
+ 0 named destinations out of 1000 (max. 500000)
+ 171 words of extra memory for PDF output out of 10000 (max. 10000000)
+
diff --git a/9934-t/images/Image1.png b/9934-t/images/Image1.png
new file mode 100644
index 0000000..90ba3e8
--- /dev/null
+++ b/9934-t/images/Image1.png
diff --git a/9934-t/images/Image10.png b/9934-t/images/Image10.png
new file mode 100644
index 0000000..72d4bd1
--- /dev/null
+++ b/9934-t/images/Image10.png
diff --git a/9934-t/images/Image11.png b/9934-t/images/Image11.png
new file mode 100644
index 0000000..68ef978
--- /dev/null
+++ b/9934-t/images/Image11.png
diff --git a/9934-t/images/Image12.png b/9934-t/images/Image12.png
new file mode 100644
index 0000000..ad99ead
--- /dev/null
+++ b/9934-t/images/Image12.png
diff --git a/9934-t/images/Image13.png b/9934-t/images/Image13.png
new file mode 100644
index 0000000..6c02e98
--- /dev/null
+++ b/9934-t/images/Image13.png
diff --git a/9934-t/images/Image14.png b/9934-t/images/Image14.png
new file mode 100644
index 0000000..5cf7eaa
--- /dev/null
+++ b/9934-t/images/Image14.png
diff --git a/9934-t/images/Image15.png b/9934-t/images/Image15.png
new file mode 100644
index 0000000..0f5ec49
--- /dev/null
+++ b/9934-t/images/Image15.png
diff --git a/9934-t/images/Image16.png b/9934-t/images/Image16.png
new file mode 100644
index 0000000..1a68700
--- /dev/null
+++ b/9934-t/images/Image16.png
diff --git a/9934-t/images/Image17.png b/9934-t/images/Image17.png
new file mode 100644
index 0000000..c4cf264
--- /dev/null
+++ b/9934-t/images/Image17.png
diff --git a/9934-t/images/Image18.png b/9934-t/images/Image18.png
new file mode 100644
index 0000000..d64f08d
--- /dev/null
+++ b/9934-t/images/Image18.png
diff --git a/9934-t/images/Image19.png b/9934-t/images/Image19.png
new file mode 100644
index 0000000..7911261
--- /dev/null
+++ b/9934-t/images/Image19.png
diff --git a/9934-t/images/Image2.png b/9934-t/images/Image2.png
new file mode 100644
index 0000000..9268334
--- /dev/null
+++ b/9934-t/images/Image2.png
diff --git a/9934-t/images/Image20.png b/9934-t/images/Image20.png
new file mode 100644
index 0000000..445bca5
--- /dev/null
+++ b/9934-t/images/Image20.png
diff --git a/9934-t/images/Image21.png b/9934-t/images/Image21.png
new file mode 100644
index 0000000..e976c8c
--- /dev/null
+++ b/9934-t/images/Image21.png
diff --git a/9934-t/images/Image22.png b/9934-t/images/Image22.png
new file mode 100644
index 0000000..a507b30
--- /dev/null
+++ b/9934-t/images/Image22.png
diff --git a/9934-t/images/Image23.png b/9934-t/images/Image23.png
new file mode 100644
index 0000000..cb3ba78
--- /dev/null
+++ b/9934-t/images/Image23.png
diff --git a/9934-t/images/Image24.png b/9934-t/images/Image24.png
new file mode 100644
index 0000000..3a118c5
--- /dev/null
+++ b/9934-t/images/Image24.png
diff --git a/9934-t/images/Image25.png b/9934-t/images/Image25.png
new file mode 100644
index 0000000..d53aea3
--- /dev/null
+++ b/9934-t/images/Image25.png
diff --git a/9934-t/images/Image26.png b/9934-t/images/Image26.png
new file mode 100644
index 0000000..2b00f89
--- /dev/null
+++ b/9934-t/images/Image26.png
diff --git a/9934-t/images/Image27.png b/9934-t/images/Image27.png
new file mode 100644
index 0000000..2157f74
--- /dev/null
+++ b/9934-t/images/Image27.png
diff --git a/9934-t/images/Image28.png b/9934-t/images/Image28.png
new file mode 100644
index 0000000..37cb93e
--- /dev/null
+++ b/9934-t/images/Image28.png
diff --git a/9934-t/images/Image29.png b/9934-t/images/Image29.png
new file mode 100644
index 0000000..43b2f04
--- /dev/null
+++ b/9934-t/images/Image29.png
diff --git a/9934-t/images/Image3.png b/9934-t/images/Image3.png
new file mode 100644
index 0000000..9e2d8c0
--- /dev/null
+++ b/9934-t/images/Image3.png
diff --git a/9934-t/images/Image30.png b/9934-t/images/Image30.png
new file mode 100644
index 0000000..e6731fd
--- /dev/null
+++ b/9934-t/images/Image30.png
diff --git a/9934-t/images/Image31.png b/9934-t/images/Image31.png
new file mode 100644
index 0000000..58af316
--- /dev/null
+++ b/9934-t/images/Image31.png
diff --git a/9934-t/images/Image32.png b/9934-t/images/Image32.png
new file mode 100644
index 0000000..80e42e9
--- /dev/null
+++ b/9934-t/images/Image32.png
diff --git a/9934-t/images/Image33.png b/9934-t/images/Image33.png
new file mode 100644
index 0000000..7c6422d
--- /dev/null
+++ b/9934-t/images/Image33.png
diff --git a/9934-t/images/Image34.png b/9934-t/images/Image34.png
new file mode 100644
index 0000000..a054ae2
--- /dev/null
+++ b/9934-t/images/Image34.png
diff --git a/9934-t/images/Image4.png b/9934-t/images/Image4.png
new file mode 100644
index 0000000..92327e9
--- /dev/null
+++ b/9934-t/images/Image4.png
diff --git a/9934-t/images/Image5.png b/9934-t/images/Image5.png
new file mode 100644
index 0000000..2853b37
--- /dev/null
+++ b/9934-t/images/Image5.png
diff --git a/9934-t/images/Image6.png b/9934-t/images/Image6.png
new file mode 100644
index 0000000..ab7eabe
--- /dev/null
+++ b/9934-t/images/Image6.png
diff --git a/9934-t/images/Image7.png b/9934-t/images/Image7.png
new file mode 100644
index 0000000..7dd051a
--- /dev/null
+++ b/9934-t/images/Image7.png
diff --git a/9934-t/images/Image8.png b/9934-t/images/Image8.png
new file mode 100644
index 0000000..3c6e308
--- /dev/null
+++ b/9934-t/images/Image8.png
diff --git a/9934-t/images/Image9.png b/9934-t/images/Image9.png
new file mode 100644
index 0000000..6d93dfa
--- /dev/null
+++ b/9934-t/images/Image9.png
diff --git a/LICENSE.txt b/LICENSE.txt
new file mode 100644
index 0000000..6312041
--- /dev/null
+++ b/LICENSE.txt
@@ -0,0 +1,11 @@
+This eBook, including all associated images, markup, improvements,
+metadata, and any other content or labor, has been confirmed to be
+in the PUBLIC DOMAIN IN THE UNITED STATES.
+
+Procedures for determining public domain status are described in
+the "Copyright How-To" at https://www.gutenberg.org.
+
+No investigation has been made concerning possible copyrights in
+jurisdictions other than the United States. Anyone seeking to utilize
+this eBook outside of the United States should confirm copyright
+status under the laws that apply to them.
diff --git a/README.md b/README.md
new file mode 100644
index 0000000..e3ea5e8
--- /dev/null
+++ b/README.md
@@ -0,0 +1,2 @@
+Project Gutenberg (https://www.gutenberg.org) public repository for
+eBook #9934 (https://www.gutenberg.org/ebooks/9934)
diff --git a/old/9934-t.zip b/old/9934-t.zip
new file mode 100644
index 0000000..a1c2882
--- /dev/null
+++ b/old/9934-t.zip
diff --git a/old/pqtrn10p.zip b/old/pqtrn10p.zip
new file mode 100644
index 0000000..837ea8c
--- /dev/null
+++ b/old/pqtrn10p.zip