我似乎收到了与其他几个人相同的错误消息，但我不知道如何将解决方案应用到我自己的工作中。我知道错误发生在这里的某个地方：
\begin{solution}
By Trichonomy, we know that x $>$ 0, x $=$ 0, or x $<$ 0.
$x\not=0$ is given so we are left with two cases:
1--If x$>$0 then $x^2$ $=$ $x\ x$
By order axiom six--if 0$\leq $x and 0$\leq $y then 0$\leq $ $x\ y$--$x^2$ $>$ 0.
2--If x$<$0 then $x^2$ $=$ $$$-1\ x$\ -1$\ x$
$=$ $$-1^2$\ $x^2$$
$=$ $1\ $x^2$$
$=$ $x^2$ $>$ 0, completing the proof.
\end{solution}
我遇到了Display math should end with $$错误，然后还有一个错误告诉我$在最后一行（以“完成证明”结尾的那行）的开头插入另一个，这对我来说也没有意义。您能否提供一些见解，说明我做错了什么？
\documentclass[12pt]{exam}
\usepackage[dvips]{graphicx}
\usepackage{amsfonts,amssymb,amsmath,latexsym}
\pagestyle{empty}
\newcommand{\n}{\noindent}
\newcommand{\pp}{\parskip=0.25cm}
\newtheorem{tem}{{\sc Theorem}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\N}{\mathbb{N}}
\begin{document}
\title{{\bf MA410.F13--Problem Set 1}}
\author{{\bf Lisa Snay }}
\date{}
\maketitle
\begin{questions}\pp
\question[10] Prove that if $a\in\R$ and $a\not=0$, then $a^2>0$
\end{questions}
\begin{solution}
By Trichonomy, we know that $x > 0$, $x = 0$, or $x < 0$.
$x\not=0$ is given so we are left with two cases:
1--If $x>0$ then $x^2 = x\ x$
By order axiom six--if $0\leq x$ and $0\leq y$ then $0\leq x\ y$ so $x^2 > 0$.
2--If $x<0$ then $x^2 = -1\ x\ -1\ x
= -1^2\ x^2
= 1\ x^2
= x^2 > 0$, completing the proof.
\end{solution}
\end{document}
答案1
默认情况下（它在课堂评论中说）解决方案不会被打印，您需要该answers选项。错误是因为您不能在行数学中有段落中断（空行），我修复了一些（不是全部：-）其他问题，例如过时的字体命令\bf和图形包选项。
\documentclass[12pt,answers]{exam}
\usepackage{graphicx}
\usepackage{amsfonts,amssymb,amsmath,latexsym}
\pagestyle{empty}
\newcommand{\n}{\noindent}
\newcommand{\pp}{\parskip=0.25cm}
\newtheorem{tem}{\textsc{Theorem}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\N}{\mathbb{N}}
\begin{document}
\title{\textbf{MA410.F13--Problem Set 1}}
\author{\textbf{Lisa Snay }}
\date{}
\maketitle
\begin{questions}\pp
\question[10] Prove that if $a\in\R$ and $a\not=0$, then $a^2>0$
\end{questions}
\begin{solution}
By Trichonomy, we know that $x > 0$, $x = 0$, or $x < 0$.
$x\not=0$ is given so we are left with two cases:
1--If $x>0$ then $x^2 = x\ x$
By order axiom six--if $0\leq x$ and $0\leq y$ then $0\leq x\ y$ so $x^2 > 0$.
2--If $x<0$ then
\begin{align}x^2 &= -1 x -1 x\\
&= (-1)^2 x^2\\
&= 1 x^2\\
& = x^2 > 0
\end{align}, completing the proof.
\end{solution}
\end{document}