%EXERCISE TEN
          
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\begin{center}
	\textbf{Exercise Ten: Tables}
\end{center}  

Below is the operation table (or Cayley table) for 
$(\mathbb{Z}_4,\cdot_4)$, that is, the integers \emph{modulo} 4 
under multiplication. Ask yourself whether or not this forms a group.

\begin{center}
\begin{tabular}{c||cccc}
$\cdot_4$ & 0 & 1 & 2 & 3 \\ \hline 
0 & 0 & 0 & 0 & 0 \\
1 & 0 & 1 & 2 & 3 \\
2 & 0 & 2 & 0 & 2 \\
3 & 0 & 3 & 2 & 1 \\
\end{tabular}
\end{center}

% It does not form a group. The elements 0 and 2 do not have inverses.

\vspace{20pt}
Ah, truth tables ... good times. 

This one demonstrates why ``proof by contrapositive'' works by 
showing that the statements 
$P\Rightarrow Q$ and $(\mathord{\sim}Q)\Rightarrow\mathord({\sim}P)$ have the
same truth values.
\begin{table}[h]
\begin{center}
\begin{tabular}{cccccc}\hline
$P$ & $Q$ & $P\Rightarrow Q$ & $\mathord{\sim}P$ & $\mathord{\sim}Q$ 
& $(\mathord{\sim}Q)\Rightarrow\mathord({\sim}P)$ \\ \hline
T & T & T & F & F & T \\
T & F & F & F & T & F \\
F & T & T & T & F & T \\
F & F & T & T & T & T \\ \hline
\end{tabular}
\caption{Proof by contrapositive}
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\end{table}

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