%%% Preamble starts here.
\documentclass{math215}

% Include any special packages you might use.  Uncomment the
% following to use Times as the default font insteand of
% TeX's default font of Computer Modern.
\usepackage{times,txfonts}

% The following commands set up the material that appears
% in the header.
\doclabel{Math 215: Homework 11}
\docauthor{Your name here}
\docdate{April 19, 2013}

% I've provided a file (math215extras.tex) with some commonly used extra 
% commands. If you've downloaded it, you can include it in your document
% by uncommenting the line below.  Feel free to make changes to that file.
\input{math215extras.tex}

%%%% Main document starts here.

\begin{document}

This looks long, but most of the proofs are very short!

\begin{proposition}{8.A} 
The number $0\in\Reals$ does not have a multiplicative inverse.
\end{proposition}
\begin{proof}
\end{proof}

\begin{proposition}{8.B} If $c,x\in\Reals$ and $cx=1$, then $x\neq  0$ and $c=x^{-1}$.
\end{proposition}
\begin{proof}
\end{proof}

\begin{proposition}{8.C} If $x,y\in\Reals$ and $x\neq 0$ and $y\neq 0$, then
$xy\neq0$ and $(xy)^{-1}=x^{-1}y^{-1}$.
\end{proposition}
\begin{proof}
\end{proof}

\begin{proposition}{8.D} If $x\in\Reals$ and $x\neq 0$, then $x^{-1}\neq 0$
        and $(x^{-1})^{-1}=x$.
\end{proposition}
\begin{proof}
\end{proof}

\begin{proposition}{8.E} If $x\in\Reals$ and $x>0$, then $x^{-1}> 0$.
\end{proposition}
\begin{proof}
\end{proof}

\begin{corollary}{8.F} If $x\in\Reals$ and $x\neq 0$, if $x^{-1}> 0$ then $x>0$.
\end{corollary}
\begin{proof}
\end{proof}

\begin{proposition}{8.40}\noindent
  \begin{itemize}
    \item[(ii)] Let $x,y\in\Reals$ such that $0<x<y$.  Then $0<1/y<1/x$.
\end{itemize}    
\end{proposition}
\begin{proof}
Be sure to take advantage of Propositions 2A-2H as well as the work you've just done to make for a very short proof.
\end{proof}

\begin{proposition}{8.43}  Let $x,y\in\Reals$ such that $x<y$.
  Then there exists $z\in\Reals$ such that $x<z<y$.
\end{proposition}
\begin{proof}
\end{proof}

\begin{proposition}{8.45} If $x_1$ and $x_2$ are least upper bounds for $A\subseteq\Reals$, then $x_1=x_2$.
\end{proposition}
\begin{proof}
\end{proof}

\begin{proposition}{8.45} $\sup((-\infty,0))=0$
\end{proposition}
\begin{proof}
\end{proof}

\begin{lemma}{9.A} Suppose $f:A\rightarrow B$ and $g:B\rightarrow C$.
\begin{enumerate} 
\item If $g\circ f$ is injective, then $f$ is injective.
\item If $g\circ f$ is surjective, then $g$ is surjective.
\item If $g\circ f$ is bijective, then $f$ is injective and $g$ is surjective.
\end{enumerate}           
\end{lemma}
\begin{proof}
\end{proof}

\begin{proposition}{9.7} (ii) If $f:A\rightarrow B$ is surjective, and $G:B\rightarrow C$ is
        surjective, then $g\circ f:A\rightarrow C$ is surjective.
\end{proposition}
\begin{proof}
\end{proof}

\begin{proposition}{9.11}  If $f:A\rightarrow B$ has an inverse
function, the inverse function is unique.
\end{proposition}
\begin{proof}
\end{proof}

\begin{proposition}{11.3} If $x,y,z\in\Reals$ with $y\neq 0$ and $z\neq 0$,
  then 
\[
\frac{xz}{yz}=\frac{x}{y}.
\]
\end{proposition}
\begin{proof}
Your proof goes here.
\end{proof}


\end{document}