Assignment One

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\title{Assignment One}

\author{Ken Iannello}

\date{\today}

\begin{document}
\maketitle

\begin{abstract}
Modern Algebra HW
\end{abstract}

\section{Introduction}

In this assignment we will go over 6 proofs that we have recently derived from the Axioms of integer arithmetic off the assumption those Axioms are all we know. These Axioms can be found in the Notebook written by Dr. Oscar Chavez titled\textit{Modern Algebra}

\section{Problems}
\label{sec:examples}

\subsection{Prove that \textit{a\textit{*0 =0}}}

\begin{itemize}
\item a*0=0   ==
\end{itemize}
\begin{itemize}
\item a*0+0=0   ==
\end{itemize}
\begin{itemize}
\item a*0+(a-a)=0   ==
\end{itemize}
\begin{itemize}
\item (a*0+a)-a=0   ==
\end{itemize}
\begin{itemize}
\item a(0+1)-a =0   ==
\end{itemize}
\begin{itemize}
\item a-a=0   ==
\end{itemize}
\begin{itemize}
\item 0=0  QED
\end{itemize}
\todo[inline, color=green!40]{We have proven a*0=0 .}

\subsection{Prove that -(ab)=a(-b)}

\todo[inline, color=green!40]{We have proven a*0=0 .}
\begin{itemize}
\item a*0=0   ==
\end{itemize}
\begin{itemize}
\item a(b+(-b))=0    ==
\end{itemize}
\begin{itemize}
\item (ab)+a(-b)=0   ==
\end{itemize}
\begin{itemize}
\item ab+a(-b)-ab=0-ab   ==
\end{itemize}
\begin{itemize}
\item a(-b)=-(ab) QED
\end{itemize}
\todo[inline, color=green!40]{We have proven  a(-b)=-(ab) .}
\subsection{Prove that (-a)(-b)=ab}

\todo[inline, color=green!40]{We know a+(-a)=0 and b+(-b)=0 .}
\begin{itemize}
\item (a+(-a))(b+(-b))=0   ==
\end{itemize}
\begin{itemize}
\item ab+a(-b)+(-a)b+(-a)(-b)=0   ==
\end{itemize}
 \todo[inline, color=green!40]{We have proven  a(-b)=-(ab) .}
\begin{itemize}
\item ab+(-(ab))+(-(ab))+(-a)(-b)=0   ==
\end{itemize}
\begin{itemize}
\item (-(ab))+(-a)(-b)=0   ==
\end{itemize}
\begin{itemize}
\item(-(ab))+(ab)+(-a)(-b)=0+(ab   ==
\end{itemize}
\begin{itemize}
\item (-a)(-b)=ab  QED
\end{itemize}
\todo[inline, color=green!40]{We have proven  (-a)(-b)=ab .}

\subsection{ Prove For all integers \textit{a} and \textit{b}, if \textit{ab}=0 and \textit{a} does not = 0, then \textit{b}=0}
\todo[inline, color=green!40]{Case 1: if b is greater than 0}
\begin{itemize}
\item   If a does not equal zero and bis greater than 0 then ab does NOT equal 0
\end{itemize}
\todo[inline, color=green!40]{Case 2: if b is less than 0}
\begin{itemize}
\item   If a does not equal zero and b is less than 0 than ab does NOT equal 0
\end{itemize}
\todo[inline, color=green!40]{Case 3: if b = 0}
\begin{itemize}
\item   If a does not equal zero and b=0 than ab=0
\end{itemize}
\todo[inline, color=green!40]{Proposition proved}
\subsection{Prove that, for all integers \textit{a},\textit{b}, and \textit{c}, if \textit{ac}=\textit{bc} and \textit{c} does not = 0, then \textit{a}=\textit{b} }

\begin{itemize}
\item ac = bc   ==
\end{itemize}
\begin{itemize}
\item ac-(bc)=bc-(bc)   ==
\end{itemize}
\begin{itemize}
\item ac-bc=0    ==
\end{itemize}
\begin{itemize}
\item c(a-b)=0    ==
\end{itemize}
\todo[inline, color=green!40]{if c can not equal zero we have proven (a-b) must equal 0 .}
\begin{itemize}
\item a-b=0    ==
\end{itemize}
\begin{itemize}
\item a=b  QED
\end{itemize}
\todo[inline, color=green!40]{We have proven a=b.}
\subsection{Let \textit{a}.\textit{b}, and \textit{c} be integers such that \textit{a}+\textit{b}=\textit{a}+\textit{c}. Prove \textit{b}=\textit{c}}
\begin{itemize}
\item a+b=a+c   ==
\end{itemize}
\todo[inline, color=green!40]{a has such additive inverse such that a+(-a) =0. We will add it to both sides of the equation}
\begin{itemize}
\item a+b+(-a)=a+c+(-a)   ==
\end{itemize}
\begin{itemize}
\item b=c  QED
\end{itemize}
\todo[inline, color=green!40]{We have proven b=c.}
\end{document}