\documentclass[12pt]{amsart}

\usepackage{amsmath, amssymb, amsthm}
\usepackage[pdflang={en-US}]{hyperref}

\usepackage[margin=.75in]{geometry} % Adjust margins to your preference

%%% The following sets the numbering scheme.
\renewcommand{\thesubsection}{\Alph{subsection}}

%% This sets up some definition and theorem environments.
\newtheorem*{thm}{Theorem}
\newtheorem*{lem}{Lemma}
\newtheorem*{stm}{Statement}
\theoremstyle{definition}
\newtheorem{dfn}{Definition}

%%% Add any useful abbreviations here
\newcommand{\lcm}{\text{lcm}}

\title{Math 301 Portfolio}
\author{YOUR NAME}
\date{\today}

\begin{document} 

\maketitle

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\begin{section}{Expository entries}
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\begin{subsection}{Response to \textbf{TITLE} by AUTHOR}

Paste your first portfolio entry here.

\end{subsection}

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\begin{subsection}{Response to \textbf{TITLE} by AUTHOR}

Paste your second portfolio entry here.

\end{subsection}

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\begin{subsection}{Response to WHATEVER YOU DID FOR THE THIRD EXPOSITORY PORTFOLIO}

Paste your third portfolio entry here.

\end{subsection}
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\end{section}

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\begin{section}{Proofs}

\begin{subsection}{Direct and contrapositive proofs}

Paste your first proof here.
I recommend putting your proof in the following format.

If you wish to include a relevant definition, use a definition environment:

\begin{dfn} \label{limit} % This label allows me to easily refer to the definition
 $\lim_{x \to a} f(x) = L$ if for every number $\epsilon > 0$ there is a  number $\delta > 0$ such that \[ 0 < \lvert x - a \rvert < \delta \implies \lvert f(x) - L \rvert < \epsilon. \]
\end{dfn}

State the result you proved as a theorem, then give the proof.
If you'd like to include a lemma, then use  a lemma environment.

\begin{lem}
 $\lim_{x \to 0} f(x) \neq 0$ if there is a number $\epsilon > 0$ such that for every number $\delta > 0$ there is a number $x$ such that \[ 0 < \lvert x \rvert < \delta \text{ and } \lvert f(x) \rvert \geq \epsilon. \]
\end{lem}

\begin{proof}
This is the negation of the definition of the limit (definition \ref{limit}) with $a = 0$ and $L=0$.
All quantifiers have been swapped, even the implicit quantifier for the conditional statement at the end of the definition.
\end{proof}

\begin{thm}
$\displaystyle \lim_{x \to 0}  \sin\left(1 / x \right) \neq 0$.
\end{thm}

\begin{proof}
Let $\delta$ be a real number greater than $0$.
Let $n$ be an odd natural number such that $\frac{1}{n} < \delta$.
Because $0 < \frac{2}{\pi} < 1$, it follows that $0 < \frac{2}{ n \pi} < \delta$.
Moreover, because $n$ is odd \[
 \Big{\lvert} \sin \left( \frac{1}{\left( 2/ ( n \pi) \right) } \right) \Big{\rvert} =    \Big{\lvert} \sin \left( \frac{n \pi}{2} \right) \Big{\rvert} = 1 .
 \]
 Thus we have shown that there is a number $x = \frac{2}{ n \pi}$, such that \[ 
 0 < \lvert x \rvert < \delta \text{ and }  \lvert \sin\left(1 / x \right) \rvert \geq 1 .
 \]
 Since $\delta$ was arbitrary, we can conclude that this holds for every positive real number $\delta$.
 Therefore, by the lemma, $\displaystyle \lim_{x \to 0}  \sin\left(1 / x \right) \neq 0$
\end{proof}

\end{subsection}

\vspace{.1in}
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\begin{subsection}{Proofs by contradiction and non-conditional statements}

Paste your second proof here.

\end{subsection}

\vspace{.1in}
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\begin{subsection}{Proofs and disproofs}

Paste your third proof (or disproof) here.
It may not be be appropriate to call the statement in question a theorem, since you may be providing a disproof.
I suggest the following. 

\begin{stm}
Suppose $A$, $B$, and $C$ are sets. If $A \times C \subseteq  B \times C$, then $A \subseteq B$.
\end{stm}

The statement is \textbf{false}.

\begin{proof}
To prove that the statement is false, we produce sets $A$, $B$, and $C$, such that $A \times C \subseteq  B \times C$ and $A \nsubseteq B$.
Et cetera.
\end{proof}

\end{subsection}

\vspace{.1in}
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\begin{subsection}{Induction I}

Paste your fourth proof here.

\end{subsection}

\vspace{.1in}
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\begin{subsection}{Induction II}

Paste your fifth proof here.

\end{subsection}

\vspace{.1in}
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\begin{subsection}{Uncategorized proofs}

Paste your sixth proof here.

\end{subsection}

\end{section}

\end{document}