\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}