\documentclass{beamer}
\title{Math 23a - Proof 1.1 }
\author{Paul Bamberg}
\date{September 9, 2016}
\begin{document}
\maketitle
\setbeamercovered{dynamic}
\begin{frame}
\begin{itemize}
\item Suppose that $a$ and $b$ are two elements of a field $F$. Using only the axioms for a field, prove that $\forall a \in F$, $0a = 0.$
\pause
\item $0+0 = 0$
\pause
\item $(0+0)a = 0a$
\pause
\item $0a + 0a = 0a$
\pause
\item $0a + 0a + (-0a) = 0a + (-0a)$
\pause
\item $0a + 0 = 0$
\pause
\item $0a = 0$
\end{itemize}
\end{frame}

\begin{frame}
\begin{itemize}
\item Suppose that $a$ and $b$ are two elements of a field $F$. Using only the axioms for a field, prove that if $ab =0$, then either $a$ or $b$ must be 0.
\item Either $a = 0$ or $a \neq 0$
\item ...
\end{itemize}

\end{frame}

\begin{frame}
\begin{itemize}
\item Suppose that $a$ and $b$ are two elements of a field $F$. Using only the axioms for a field, prove the additive inverse of $a$ is unique.
\item...
\end{itemize}

\end{frame}





\end{document}
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