![为什么 Tex 不显示多于一页?](https://linux22.com/image/348663/%E4%B8%BA%E4%BB%80%E4%B9%88%20Tex%20%E4%B8%8D%E6%98%BE%E7%A4%BA%E5%A4%9A%E4%BA%8E%E4%B8%80%E9%A1%B5%EF%BC%9F.png)
这是我第一次使用 tex,我的教授给了我们一份家庭作业问题集的模板。我用 LaTeXiT(他告诉我们使用的软件)打开了它,但不知何故它不会显示第一页之后的内容。我尝试将其导出为 pdf,但仍然只显示第一页。抱歉包含了这么多代码,因为我说我是新手,我不知道是什么原因导致的。我能做些什么来解决这个问题?
\documentclass[11pt]{amsart} %superior to the ``article'' documentclass, I think
%packages give you more functionality
\usepackage[
margin=1in,
%total={6.5in, 9in},
%left=1in,
footskip=25pt]{geometry} %controls the page layout
%\usepackage{fullpage} %to get full 8.5-by-11 inch margins, can use instead of the geometry package
\usepackage{amssymb} %gives you most of the symbols you could want
\usepackage{mathabx} %more symbols
\usepackage{graphicx} %helpful for putting in pictures
\usepackage{enumerate} %lets me control list environments precisely how I want them
%\usepackage{enumitem} %another list control package
\usepackage{color} %lets me use colored text
\usepackage{tikz} %nifty math drawing package
\usepackage{multicol} %for multi-column text
\usepackage{fancyhdr} %to put the nice headers and footers on each page
\usepackage{hyperref} %allows embedding of links in pdf files
%\thispagestyle{empty} %this command removes headers and footers from the first page
%these are commands I use a lot so want shorthand for, which \newcommand lets me do.
%mostly the ``blackboard bold'' letters, I like those a lot
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\N}{\mathbb{N}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\C}{\mathbb{C}}
\newcommand{\F}{\mathbb{F}}
\newcommand{\B}{\mathcal{B}}
\newcommand{\1}{\mathbbm{1}}
\newcommand{\mb}[1]{\vec{\mathbf{#1}}}
%These change the text color, but you need to put everything within curly braces {} or else all the text following will be that color.
\newcommand{\blue}{\color{blue}}
\newcommand{\red}{\color{red}}
%I'm a weirdo and like to control the numbering and list environments to within an inch of their lives.
%\resumeenum and \pauseenum let me start a list, pause the list and go out of the list environment (which has its own vagaries) for awhile, then come back and resume the numbering from where I left off
\newcommand{\resumeenum}
{
\begin{enumerate}[\bf 1.]
\setlength{\parsep}{0.07in}
\setlength{\parskip}{0.06in}
\setlength{\parindent}{0mm}
\setcounter{enumi}{\thesaveenumi}
}
\newcommand{\pauseenum}
{
\setcounter{saveenumi}{\theenumi}
\end{enumerate}
}
\newcounter{saveenumi}
\setcounter{saveenumi}{0}
%I want to write certain math expressions in Roman font instead of the italicized math font. Especially functions, but things like sine and cosine have built in \sin and \cos commands. \DeclareMathOperator lets you make similar function expressions of your own.
\DeclareMathOperator{\grad}{grad}
\DeclareMathOperator{\GL}{GL}
\DeclareMathOperator{\Span}{Span}
\DeclareMathOperator{\Aut}{Aut}
\DeclareMathOperator{\Perm}{Perm}
\DeclareMathOperator{\Gal}{Gal}
\DeclareMathOperator{\im}{im}
\DeclareMathOperator{\ch}{ch}
\DeclareMathOperator{\ord}{ord}
\DeclareMathOperator{\curl}{curl}
\DeclareMathOperator{\mydiv}{div}
%\DeclareMathOperator{\gcd}{gcd}
%I like to control the spacing between lines and between paragraphs. I don't particularly like the defaults so I use these. They are evolving in that sometimes I decide that something is a little too close together or far apart, so I end up tweaking these values slightly.
\setlength{\parindent}{0mm}
\setlength{\parskip}{0.07in}
%Spacing for List Environments
%\setlength{\parsep}{0.07in}
%\setlength{\parskip}{0.07in}
%The following makes the nice headers and footers
\pagestyle{fancy}
\fancyhf{}
\lhead{Math 2502, Number Theory and Cryptography}
\rhead{Spring 2017, Prof King} %maybe replace this header with your name?
\lfoot{Problem Set \#1}
\cfoot{Due February 1st}
\rfoot{Page \thepage}
\renewcommand{\headrulewidth}{0.7pt}
\renewcommand{\footrulewidth}{0.7pt}
%Everything above is called the ``preamble''.
%Hey look, we're finally starting the document!
\begin{document}
%The first page doesn't have the headers and footers unless you put this in.
\thispagestyle{fancy}
\ %This creates a blank line, putting a little space at the top of the page
%Make sure to put your name on the problem set!
{\Large\textbf{Name: } }
Who I worked with: ---\\
\begin{center}
{\Large\textbf{\blue Problem Set \#1}}\\
\textsc{due Wednesday, February 1, 2017}
\end{center}
\vspace{0.3in} %force vertical space
%Leaving a blank line in the code will create a line break. Otherwise it doesn't know you want to move onto the next line. E.g. look how I write the next sentence on multiple lines but it all appears on one line in the pdf.
Please turn in solutions to the
following problems at the beginning of class
on the date above.\\
%The double backslash creates a larger break in the text.
%\textbf{\LaTeX\ Instructions:}
%You may submit any handwritten computations (or code if you had a computer do it for you) for \#4(a), 6, 7, and 8, but you should also type up a summary of your results (and any conjectures!) in your \TeX. Everything else should be \TeX'ed.\\
{\large\textbf{\blue Axiomatic Ring Proofs}}
%\large makes the text a bit bigger, \textbf is bold font, which should be contained in braces or else the whole document will be bold from this point on!
Prove the following using only the ring axioms (Axioms 0--6) or the previous problems. Be rigorous!
\resumeenum
\item %This command is for a single list item, so you don't have to keep track of the numbering yourself
(Handout 1, \#1)
(Additive Cancellation Law)
For all $a,b,c \in \Z$,\ \ $a+b = a+c \implies b=c$.\\
\textbf{Proof: $a+b = a+c \implies b=c$. Let $a=0$ so $0+b = 0+c$. By Axiom 4, $b=c$.}
\qed
\item
(Handout 1, \#5)
Every $a \in \Z$ has a \emph{unique} additive inverse. In other words, the axiom guarantees there is at least one additive inverse for any chosen integer. Show that there is also at most one additive inverse! Hint: suppose $a$ had two additive inverses. Call them\ldots\ I don't know\ldots\ $b$ and $c$. Show $b=c$.\\
\textbf{Proof: Let $b$ and $c$ be additive inverses for $a$. So $a+b = 0$ and $a+c = 0$. So $a+b = a+c$. Using the above proof, we can reduce to $b = c$. Thus, the additive inverse for $a$ is unique.}
\qed
\item
(Handout 1, \#6)
For all $a \in \Z$, $-(-a) = a$. Hint: what are the possible inverses of $-a$?\\
\textbf{Proof: By Axiom 6, we say $-a+-(-a) = 0$. Since the additive inverse is unique, we can say $-(-a) = a$.}
\pauseenum
{\large\textbf{\blue Finite Rings}}
Let's explore arithmetic and equations in the finite rings $\Z/m\Z$.
\resumeenum
\item
(Handout 2, \#13)
Which of the following can you find in $\Z/11\Z$? In $\Z/35\Z$? In $\Z/9\Z$?
\
\[
7+8,\ 4-9,\ 3\cdot 5,\ 5^2,\ 5^3,\ 1/5,\ 3/8,\ \sqrt{3},\ \sqrt{-2},\ \sqrt{-6}.\\
\]
%the commands \[ and \] start and end displayed math equations. You can also use \begin{displaymath} and \end{displaymath}. \qquad inserts some horizontal space.
\textbf{Answer:\\
In $\Z/11\Z$: \[7+8,\ 4-9,\ 3\cdot 5,\ 5^2,\ 5^3,\ 1/5,\ 3/8,\ \sqrt{3},\ \sqrt{-2}, \ \sqrt{-6}\]\\
In $\Z/35\Z$: \[7+8,\ 4-9,\ 3\cdot 5,\ 5^2,\ 5^3,\ 3/8\]\\
In $\Z/9\Z$: \[7+8,\ 4-9,\ 3\cdot 5,\ 5^2,\ 5^3,\ 3/8\]
}\\
\item
(Handout 2, \#15)
Make a list of the perfect squares in each of $\Z/5\Z$, $\Z/7\Z$, $\Z/11\Z$, $\Z/13\Z$, $\Z/17\Z$, and $\Z/19\Z$. I.e., square every element in $\Z/5\Z$ and write down the resulting list. How many \emph{distinct} squares are there in each case? Any conjectures?\\
%\textbf{Answer:}
\item
(Handout 2, \#16)
Solve $7x = 2$ in $\Z/17\Z$.\\
%\textbf{Answer:}
\item
(Handout 2, \#17)
Find all $x$ in $\Z/12\Z$ which solve the equation $8x = 4$ in $\Z/12\Z$. Find all $x$ in $\Z/15\Z$ which solve the equation $9x = 10$ in $\Z/15\Z$. Find all $x$ in $\Z/55\Z$ which solve the equation $15x = 35$ in $\Z/55\Z$. \\
%\textbf{Answer:}
\item
(Multiplicative Inverses)
The integers are boring; only $\pm 1$ have multiplicative inverses. But in $\Z/5\Z$, say, $3$ has a multiplicative inverse, because $3 \cdot 2 = 1$ in $\Z/5\Z$. So we could say $\frac{1}{3} = 2$ in $\Z/5\Z$!
An element of a ring which has a multiplicative inverse is called a \textbf{\blue unit}. For example, $\pm 1$ are the only units of $\Z$, whereas every nonzero real number is a unit in $\R$.
For each of $m = 9, 10, 11, 12, 13, 14, 15$, find all the units in $\Z/m\Z$. Do you see a pattern?\\
%\textbf{Answer:}
\pauseenum
{\large\textbf{\blue Applying modular thinking to get results about the integers}}
\resumeenum
\item
(Divisibility Rules)
Given $n \in \N$, we all know the divisibility rule: $n$ is a multiple of $3$ if and only if the sum of $n$'s digits is a multiple of $3$. But why is this true?
First, how do we formulate ``digits'' mathematically? Well, 154 means $1 \cdot 10^2 + 5 \cdot 10^1 + 4$. So we write $n = d_kd_{k-1} \cdots d_1d_0$ in digits iff
\[
n = d_k \cdot 10^k + d_{k-1} \cdot 10^{k-1} + \cdots + d_1 \cdot 10^1 + d_0 = \sum_{i=0}^k d_i \cdot 10^i,\quad 0 \leq d_i \leq 9, d_k \neq 0.
\]
%You can see how summation notation works in this example.
%Also, notice that if you want more than one symbol in an exponent, you need to enclose the exponent in curly braces.
\begin{enumerate}[\bf a.]
\setlength{\parindent}{0mm}
\setlength{\parskip}{0.09in}
\item
Prove the rule for divisibility by $3$ by reducing mod 3.
%\textbf{Proof:}
\item
Why does the same rule work for divisibility by $9$?
%\textbf{Answer:}
\item
Formulate and prove a rule for divisibility by 11.\\
%\textbf{Proof:}
\end{enumerate}
\pauseenum