我正在为 baby rudin 写解决方案手册，经过一番研究，我偶然发现了考试包，它可以帮助我正确地编写问题和答案，但我遇到了一些问题，因为我对 latex 了解不多，所以我逐一介绍。附件：我的代码和结果的屏幕截图，以便更好地理解。
(1) 如您所见，我使用 documentclass 下的 exam 并编写了所有问题，但与 amsart 相比，行距太大。我可以更改行距，以便两个问题的间距更小吗？
(2) 观察问题 1 和“Ans”之间的距离。每当我写答案时，它都会从新行开始。我怎样才能将其带回到“Ans”的同一行？
(3) 页码看起来像“第 2 页”。 有什么方法可以将其更改为仅“2”。
(4) 我想给这个解决方案手册起个标题。在 amsart 中我按如下方式操作“ \title[]{Algebra} \author[]{Ripan Das} \begin{document} \maketitle \newpage”，但这个东西在这里不起作用。请解决这些问题。我还附上了代码。
\documentclass[answers, 11pt]{exam}
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\begin{document}
\newpage
\tableofcontents
\newpage
\section{The real and complex number system}
Unless the contrary explicitly stated, all numbers that are mentioned in these exercises are understood to be real.
\begin{questions}
\question If $r$ is rational $(r\neq 0)$ and $x$ is irrational, prove that $r+r$ and $rx$ is irrational.
\begin{solution}
jfgjdfs;gj;jbbnggtggguyy
\end{solution}
\question Prove that there is no rational number whose square is 12.
\begin{solution}
\end{solution}
\question Prove Proposition 1.15\footnote{Proposition 1.15 The axioms for multiplication imply the following statements.\\
(a) If $n\neq 0$ and $xy=xz$ then $y=z.$\\
(b) If $x\neq 0$ and $xy=x$ then $y=1.$\\
(c) If $x\neq 0$ and $xy=1$ then $y=1/x.$\\
(d) If $x\neq 0$ then $1/(1/x)=x.$}
\question Let $E$ be a non empty subset of an ordered set; suppose $\alpha$ is a lower bound of $E$ and $\beta$ is a upper bound of $E$. Prove that $\alpha\leq \beta.$
\question Let $A$ be a non empty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x\in A.$ Prove that $$\inf A=-\sup(-A).$$
\question Fix $b>1.$
\begin{parts}
\part If $m,n,p,q$ are integers, $n>0,q>0$ and $r=m/n=p/q$, prove that $$(b^m)^{1/q}=(b^p)^{1/q}.$$
Hence it make sense to define $b^r=(b^m)^{1/n}$.
\part Prove that $b^{r+s}=b^rb^s$ if $r$ and $s$ is rational.
\part If $x$ is real, define $B(x)$ to be the set of all numbers $b^t$, where $t$ is rational and $t\leq x$. Prove that $$b^r=\sup B(r)$$ when $r$ is rational. Hence it make sense to define $$b^x=\sup B(x)$$ for every real $x.$
\part Prove that $b^{x+y}=b^xb^y$ for all real $x$ and $y.$
\end{parts}
\question Fix $b>1,y>0$ and prove that there is a unique real $x$ such that $b^x=y$, by completing the following outline. (This $x$ is called the \textit{logarithm} of $y$ to the \textit{base} $b.$)
\begin{parts}
\part For any positive integer $n, b^n-1\geq n(b-1).$
\part Hence $b-1\geq n(b^{1/n}-1).$
\part If $t>1$ and $n>(b-1)/(t-1),$ then $b^{1/n}<t.$
\part If $w$ is such that $b^w<y$ then $b^{w+(1/n)}<y$ for sufficiently large $n;$ to see this, apply part (c) with $t=y\cdot b^{-w}.$
\part If $b^w>y$ then $b^{w-(1/n)}>y$ for sufficiently large $n$.
\part Let $A$ be the set of all $w$ such that $b^w<y$, and show that $x=\sup A$ satisfies $b^x=y.$
\part Prove that this $x$ is unique.
\end{parts}
\question Prove that no order can be defined in the complex field that turns it into an ordered field. Hint: $-1$ is a square.
\question Suppose, $z=a+bi,w=c+di$. Define $z<w$ if $a<c$, and also if $a=c$ but $b<d$. Prove that this turns the set of all complex numbers into an ordered set. (This type of order relation is called a \textit{dictionary order,} or \textit{lexicographic order,} for obvious reason.) Does this ordered set have least upper bound property?
\question Suppose $z=a+bi,w=u+iv$, and \begin{align*}
a=\left(\dfrac{|w|+u}{2}\right)^{1/2},\qquad b=\left(\dfrac{|w|-u}{2}\right)^{1/2}.
\end{align*}
Prove that $z^2=w$ if $v\geq 0$ and $(\bar{z}^2=w$ if $v\leq 0$. Conclude that every complex number (with one exception!) has two complex square roots.
\question If $z$ is a complex number, prove that there is an $r$ and a complex number $w$ with $|w|=1$ such that $z=rw.$ Are $w$ and $r$ is always uniquely determined by $z$?
\question If $z_1,\cdots,z_n$ are complex, prove that $$|z_1+z_2+\cdots+z_n|\leq |z_1|+|z_2|+\cdots+|z_n|.$$
\question If $x,y$ are complex, prove that $$||x|-|y||\leq |x-y|.$$
\question If $z$ is a complex number such that $|z|=1$, that is such that $z\bar{z}=1$, compute $$|1+z|^2+|1-z|^2.$$
\question Under what conditions does the equality hold in the Schwarz inequality?
\question Suppose, $k\geq 3,\textbf{x,y}\in {\RR}^k, |\textbf{x-y}|=d>0,$ and $r>0$. Prove:
\begin{parts}
\part If $2r>d$, there are infinitely many $z\in {\RR}^k$ such that $$|\textbf{z-x}|=|\textbf{z-y}|=r.$$
\part If $2r=d$, there is exactly one such $\textbf{z}.$
\part If $2r<d,$ there is no such $\textbf{z}.$
\end{parts}
How must these statements be modified if $k$ is 2 or 1.
\question Prove that $$|\textbf{x+y}|^2+|\textbf{x-y}|^2=2|\textbf{x}|^2+2|\textbf{y}|^2$$ if $\textbf{x}\in {\RR}^k$ and $\textbf{y}\in {\RR}^k$. Interpret this geometrically, as a statement about parallelograms.
\question If $k\geq 2$ and $\textbf{x}\in {\RR}^k,$ prove that there exists $\textbf{y}\in {\RR}^k$ such that $\textbf{y}\neq \textbf{0}$ but $\textbf{x $\cdot$ y}=\textbf{0}.$ Is this also true for $k=1?$
\question Suppose $\textbf{a}\in {\RR}^k$, $\textbf{b}\in {\RR}^k$. Find $\textbf{c}\in {\RR}^k$ and $r\geq 0$ such that $$|\textbf{x}-\textbf{a}|=2|\textbf{x}-\textbf{b}|$$ if and only if $|\textbf{x}-\textbf{c}|=r.$\\
(\textit{Solution: $3\textbf{c}=4\textbf{b}-\textbf{a},3r=2|\textbf{b}-\textbf{a}|$})
\question With reference to the Appendix, suppose that property $(\mathrm{III})$ were omitted from the definition of a cut. Keep the same definition of order and addition. Show that the resulting ordered set has the least-upper-bound property, that satisfies axiom (A1) to (A4)(with slightly different zero element!) but that (A5) fails.
\end{questions}
\end{document}