\documentclass{report}
-% We have to load this before mjotex so that mjotex knows to define
-% its glossary entries.
+% Setting hypertexnames=false forces hyperref to use a consistent
+% internal counter for proposition/equation references rather than
+% being clever, which doesn't work after we reset those counters.
+\usepackage[hypertexnames=false]{hyperref}
+\hypersetup{
+ colorlinks=true,
+ linkcolor=blue,
+ citecolor=blue
+}
+
+% We have to load this after hyperref, so that links work, but before
+% mjotex so that mjotex knows to define its glossary entries.
\usepackage[nonumberlist]{glossaries}
\makenoidxglossaries
\end{section}
\begin{section}{Arrow}
- The identity operator on $V$ is $\identity{V}$. The composition of
- $f$ and $g$ is $\compose{f}{g}$. The inverse of $f$ is
+ The constant function that always returns $a$ is $\const{a}$. The
+ identity operator on $V$ is $\identity{V}$. The composition of $f$
+ and $g$ is $\compose{f}{g}$. The inverse of $f$ is
$\inverse{f}$. If $f$ is a function and $A$ is a subset of its
domain, then the preimage under $f$ of $A$ is $\preimage{f}{A}$.
\end{section}
\begin{section}{Common}
The function $f$ applied to $x$ is $f\of{x}$. We can group terms
like $a + \qty{b - c}$ or $a + \qty{b - \sqty{c - d}}$. Here's a
- set $\set{1,2,3} = \setc{n \in \Nn[1]}{ n \le 3 }$. Here's a pair
- of things $\pair{1}{2}$ or a triple of them $\triple{1}{2}{3}$,
- and the factorial of the number $10$ is $\factorial{10}$.
+ set $\set{1,2,3} = \setc{n \in \Nn[1]}{ n \le 3 }$. The tuples go
+ up to seven, for now:
+ %
+ \begin{itemize}
+ \begin{item}
+ Pair: $\pair{1}{2}$,
+ \end{item}
+ \begin{item}
+ Triple: $\triple{1}{2}{3}$,
+ \end{item}
+ \begin{item}
+ Quadruple: $\quadruple{1}{2}{3}{4}$,
+ \end{item}
+ \begin{item}
+ Qintuple: $\quintuple{1}{2}{3}{4}{5}$,
+ \end{item}
+ \begin{item}
+ Sextuple: $\sextuple{1}{2}{3}{4}{5}{6}$,
+ \end{item}
+ \begin{item}
+ Septuple: $\septuple{1}{2}{3}{4}{5}{6}{7}$.
+ \end{item}
+ \end{itemize}
+ %
+ The factorial of the number $10$ is $\factorial{10}$.
The Cartesian product of two sets $A$ and $B$ is
$\cartprod{A}{B}$; if we take the product with $C$ as well, then
$\boundedops[W]{V}$. If $W = V$, then we write $\boundedops{V}$
instead.
+ If you want to solve a system of equations, try Cramer's
+ rule~\cite{ehrenborg}.
+
The direct sum of $V$ and $W$ is $\directsum{V}{W}$, of course,
but what if $W = V^{\perp}$? Then we wish to indicate that fact by
writing $\directsumperp{V}{W}$. That operator should survive a
\glsaddall
\printnoidxglossaries
+ \bibliographystyle{mjo}
+ \bibliography{local-references}
+
\printindex
\end{document}