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\begin{document}
\author{Jim Martens}
\title{Hausaufgaben zum 22./23. November}
\maketitle
\section{} %1
\subsection{} %a
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\\
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\subsection{} %b
Es müssen aufgrund der geforderten Reflexivität folgende Paare hinzugefügt werden:
(a,a), (b,b), (c,c), (d,d), (e,e), (f,f)

Wegen der geforderten Transitivität sind folgende Paare hinzuzufügen:
(a,c), (a,e), (a,f), (b,d), (b,f)
\subsection{} %c
$R^{+}$:\\
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\subsection{} %d
$R$:\\
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Um die Bedingungen einer Äquivalenzrelation zu erfüllen, müssen folgende Paare hinzugefügt werden:\\
reflexiv: (a,a), (b,b), (c,c), (d,d), (e,e)\\
symmetrisch und transitiv: (a,c), (a,d), (a,e), (b,a), (b,d), (c,a), (c,b), (d,a), (d,b), (d,c), (d,e), (e,a), (e,b), (e,c), (e,d)\\
Kurz geschrieben schreibt man $S = A \times A$.
\section{} %2
\subsubsection{} %(1)
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\\
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\subsubsection{} %(2)
Um die Bedingungen zu erfüllen müssen folgende Paare hinzugefügt werden:\\
reflexiv: (a,a), (b,b), (c,c), (d,d), (e,e), (f,f)\\
transitiv: (a,d)
\subsubsection{} %(3)
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\subsubsection{} %(4)
Um die Bedingungen zu erfüllen müssen folgende Paare hinzugefügt werden:\\
reflexiv: (a,a), (b,b), (c,c), (d,d), (e,e), (f,f)\\
symmetrisch: (b,a), (d,a), (d,b), (f,e)\\
transitiv: (a,d)\\
Verkürzt kann folgendes geschrieben werden: \\
\begin{equation*}
S = R \,\cup\, \{(a,a), (a,d), (b,a), (b,b), (c,c), (d,a), (d,b), (d,d), (e,e), (f,e), (f,f)\}
\end{equation*}
\section{} %3
\subsection{} %a
R = {(a,a), (a,b), (b,a), (b,b), (b,c), (c,b), (c,c), (d,d)}

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Es gibt eine Kante von a nach b und von b nach c, aber nicht von a nach c, also ist diese Relation nicht transitiv. Jedes Element der Grundmenge steht in Relation zu sich selbst, also ist die Relation reflexiv. Zu jeder Kante x nach y gibt es eine Rückkante y nach x, also ist R symmetrisch.
\subsection{} %b
R = {(a,a), (a,b), (b,b), (c,c), (d,d)}

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Man kommt von a nach b, aber nicht von b nach a, also ist die Relation nicht symmetrisch. Jedes Element der Grundmenge steht in Relation zu sich selbst, also ist die Relation reflexiv. Es gibt keine Kanten x nach y und y nach z, für die keine Kante x nach z existiert, also ist die Relation transitiv.
\subsection{} %c
R = {(a,a), (a,b), (b,a), (b,b)}

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Nicht jedes Element der Grundmenge steht in Relation mit sich selbst, also ist die Relation nicht reflexiv. Zu jeder Kante x nach y gibt es eine Rückkante y nach x, also ist die Relation symmetrisch. Es gibt keine Kanten x nach y und y nach z, für die keine Kante x nach z existiert, also ist die Relation transitiv.
\section{} %4
\subsection{} %a
R = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,2), (2,4), (2,6), (3,3), (3,6), (4,4), (5,5), (6,6)}

Graph:\\
\begin{tikzpicture}
	\node (1) {1};
	\node (2) [above left=0.4cm of 1] {2};
	\node (3) [above=0.4cm of 2] {3};
	\node (4) [above right=0.4cm of 3] {4};
	\node (5) [below right=0.4cm of 4] {5};
	\node (6) [above right=0.4cm of 1] {6};
	\path (1) edge[loop below] (1);
	\draw[->] (1) -- (2);
	\draw[->] (1) -- (3);
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	\draw[->] (1) -- (6);
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	\path (4) edge[loop above] (4);
	\path (5) edge[loop right] (5);
	\path (6) edge[loop right] (6);
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Hasse-Diagramm:\\
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	\node (1) {1};
	\node (2) [above left=0.4cm of 1] {2};
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	\draw (2) -- (4);
	\draw (2) -- (6);
	\draw (3) -- (6);
\end{tikzpicture}

\subsection{} %b
Graph:\\
\begin{tikzpicture}
	\node (0) {$\emptyset$};
	\node (1) [above left=2.5 of 0] {$\{1\}$};
	\node (2) [above right=2.5 of 0] {$\{2\}$};
	\node (3) [above=4 of 0] {$\{1,2\}$};
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	\path (2) edge[loop right] (2);
	\draw[->] (2) -- (3);
	\path (3) edge[loop above] (3);
\end{tikzpicture}

Hasse-Diagramm:\\
\begin{tikzpicture}
	\node (0) {$\emptyset$};
	\node (1) [above left=2.5 of 0] {$\{1\}$};
	\node (2) [above right=2.5 of 0] {$\{2\}$};
	\node (3) [above=4 of 0] {$\{1,2\}$};
	\draw (0) -- (1);
	\draw (0) -- (2);
	\draw (1) -- (3);
	\draw (2) -- (3);
\end{tikzpicture}
\subsection{} %c
$A = P(M) = \{\emptyset, \{1\}, \{2\}, \{3\}, \{1,2\}, \{1,3\}, \{2,3\}, \{1,2,3\}\}$
\begin{tikzpicture}
	\node (0) {$\emptyset$};
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	\node (4) [above=1.5 of 1] {$\{1,2\}$};
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\end{tikzpicture}
\end{document}