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\documentclass[10pt]{article}
\usepackage{fullpage}
\usepackage{amsmath,amsfonts,amsthm}
% these are compressed lists to help fit into a 1 page limit
\newenvironment{enumerate*}%
{\vspace{-2ex} \begin{enumerate} %
\setlength{\itemsep}{-1ex} \setlength{\parsep}{0pt}}%
{\end{enumerate}}
\newenvironment{itemize*}%
{\vspace{-2ex} \begin{itemize} %
\setlength{\itemsep}{-1ex} \setlength{\parsep}{0pt}}%
{\end{itemize}}
\newenvironment{description*}%
{\vspace{-2ex} \begin{description} %
\setlength{\itemsep}{-1ex} \setlength{\parsep}{0pt}}%
{\end{description}}
\DeclareMathOperator*{\E}{\mathbb{E}}
\let\Pr\relax
\DeclareMathOperator*{\Pr}{\mathbb{P}}
\newcommand{\inprod}[1]{\left\langle #1 \right\rangle}
\newcommand{\eqdef}{\mathbin{\stackrel{\rm def}{=}}}
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}{Lemma}
\author{Jelani Nelson}
\title{CS 229r Problem Set 1 -- Solutions}
\begin{document}
\maketitle
Here is a theorem with proof.
\begin{theorem}
The square root of $2$ is irrational.
\end{theorem}
\begin{proof}
For the sake of contradiction suppose $\sqrt{2}$ is rational. Write $\sqrt{2} = a/b$ with $a,b$ positive integers with gcd $1$. Then $2 = a^2/b^2$, so $a = 2k$ is even. Then $2 = 4k^2/b^2$ so that $b = 2k^2$, implying $b$ is even. This contradicts that $a,b$ have gcd $1$.
\end{proof}
Some random facts in a list:
\begin{itemize*}
\item Compared with the ``itemize'' environment in \LaTeX, itemize$^*$ has smaller separation between bullet points.
\item The $n$th Catalan number is $C_n \eqdef \frac{1}{n+1} \binom{2n}{n}$.
\item If $\pi(x)$ is the number of primes less than or equal to $x$, then
$$\lim_{x\rightarrow\infty} \frac{\pi(x)}{x/\ln(x)} = 1 . $$
\item $$ \sum_{\substack{1\le i\le 2n\\i\ \mathrm{even}}} i = \frac{2n(n+1)}2 .$$
\end{itemize*}
Have fun on your problem sets.
\end{document}
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