\documentclass{article}

\usepackage{amsmath, amssymb, amsthm}
\usepackage[margin=0.5in]{geometry}

\newcommand{\scs}[1]{\begin{scshape}#1\end{scshape}}
%\newcommand{\inv}{^{-1}}
\newcommand{\ZZ}{\mathbb{Z}}
\newcommand{\NN}{\mathbb{N}}
\newcommand{\RR}{\mathbb{R}}

\newtheorem{lemma}{Lemma}
\theoremstyle{definition}
\newtheorem{definition}{Definition}
\newtheorem{problem}{}

\newcommand{\lem}[1]{\begin{lemma}#1\end{lemma}}
\newcommand{\prob}[1]{\begin{problem}#1\end{problem} \vfill}
\newcommand{\dfn}[1]{\begin{definition}#1\end{definition}}

\newcommand{\abs}[1]{\left| #1 \right|}
\newcommand{\m}[1][n]{\left(\text{mod}\ #1\right)}

\begin{document} \pagestyle{empty} \begin{flushleft}

\scs{Name(s):} 

\scs{Math 301 \hfill Contradictions and Biconditionals \hfill February 25, 2015}

\lem{
Let $n,p \in \NN$ with $p$ prime.
Then for any integer $a$ it follows that $p|a$ if and only if $p|a^n$.
}
Recall that a biconditional statement like this is really two implications:
\begin{enumerate}
\item If $p|a$, then $p|a^n$;
\item If $p|a^n$, then $p|a$.
\end{enumerate}
Proving the lemma requires two proofs, one for each of these implications.
Each implication can be proved using any of our proof techniques: direct, contrapositive, or contradiction.
For now we won't prove the lemma (the first implication is easy to prove directly but the second is more complicated) but you may use it anyway.

\prob{
Prove that $\sqrt[3]{3}$ is irrational.
}


\newpage

\prob{
Let $a \in \ZZ$.
Prove that $a$ is even if and only if $a^3$ is even.
}


\prob{
Let $a \in \ZZ$.
Prove that $6|a$ if and only if $2|a$ and $3|a$.
}

\end{flushleft} 
\end{document}