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\documentclass[12pt]{article}
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\usepackage{ntnu}
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\usepackage{ntnu-math}
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\author{Øystein Tveit}
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\title{MA0301 Exercise 2}
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\begin{document}
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\ntnuTitle{}
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\break{}
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\begin{excs}
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\exc{}
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\exc{}
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\exc{}
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\begin{subexcs}
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\subexc{}
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\begin{ssubexcs}
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\ssubexc{}
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\begin{align*}
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{{2,3,5} \cup {6,4}} &\cap {4,6,8} \\
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{{2,4,6}} &\cap {4,6,8} \\
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\emptyset
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\end{align*}
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\ssubexc{}
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\begin{align*}
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P({7,8,9}) &- P({7,9}) \\
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{{7,8,9}, {7,8}, {8,9}, {7,9}, {7}, {8}, {9}, \emptyset} &- {{7,9}, {7}, {9}, \emptyset} \\
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{{7,8,9}, {7,8}, {8,9}, {8}}
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\end{align*}
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\ssubexc{}
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\begin{align*}
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P(\emptyset) \\
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{\emptyset}
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\end{align*}
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\ssubexc{}
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\begin{align*}
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{1, 3, 5} \times {0} \\
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{ \langle 1,0 \rangle, \langle 3,0 \rangle, \langle 5, 0 \rangle }
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\end{align*}
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\ssubexc{}
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\begin{align*}
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{2,4,6} \times \emptyset \\
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\emptyset
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\end{align*}
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\ssubexc{}
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\begin{align*}
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P({0}) &\times P({1}) \\
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{\emptyset, {0}} &\times {\emptyset, {1}} \\
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{\langle\emptyset,\emptyset\rangle, \langle\emptyset,{1}\rangle, \langle{0},\emptyset\rangle, \langle{0},{1}\rangle}
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\end{align*}
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\ssubexc{}
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\begin{align*}
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P(P({2})) \\
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P({\emptyset,{2}}) \\
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{ {{\emptyset}, {2}}, {{\emptyset}}, {{2}}, \emptyset }
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\end{align*}
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\end{ssubexcs}
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\subexc{}
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Because the elements in a power set can be represented as a binary tree where every leaf node is a set that has the cardinality of $1$, and that ${{x} : x \in A}$ would make up all the leaf nodes, we can reason that
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\[ |P(A) - {{x} : x \in A}| = \frac{n}{2} \]
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\end{subexcs}
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\exc{}
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\begin{subexcs}
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\subexc{}
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$\emptyset = {\emptyset}$ is {\color{red}False} because $|\emptyset| \neq |{\emptyset}|$
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\subexc{}
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$\emptyset = {0}$ is {\color{red}False} because $|\emptyset| \neq |{0}|$
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\subexc{}
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$|\emptyset| = 0$ is {\color{ForestGreen}True} because $\emptyset$ has $0$ elements
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\subexc{}
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$P(\emptyset)$ is {\color{red}False} because $P(\emptyset) = {{\emptyset}}$ has $1$ element
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\subexc{}
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$\emptyset = {}$ is {\color{ForestGreen}True} because the empty set is a subset of every possible set
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\subexc{}
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$\emptyset = {x \in \mathbb{N} : x \leq 0 and x > 0}$ is {\color{red}False} because $x \leq 0 \wedge x > 0 \equiv \F$, which means there are no such elements, and thus the set is empty
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\end{subexcs}
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\exc{}
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\begin{subexcs}
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\subexc{}
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\begin{align*}
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A \cap (\A \cup B) \\
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{x : x \in A \wedge x \in (A \cup B)} \\
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{x : x \in A \wedge (x \in A \or x \in B)} \\
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{x : x \in A} \\
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A
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\end{align*}
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\subexc{}
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\begin{align*}
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A-(B \cap C) \\
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{x : x \in A \wedge x \notin (B \cap C)} \\
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{x : x \in A \wedge (x \notin B \wedge x \notin C)} \\
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{x : (x \in A \wedge x \notin B) \vee (x \in A \wedge x \notin C)} \\
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{x : x \in (A - B) \vee x \in (A - C)} \\
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{x : x \in (A - B) \cup (A - C)} \\
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(A-B) \cup (A-C)
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\end{align*}
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\end{subexcs}
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\exc{}
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\begin{subexcs}
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\subexc{}
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\end{subexcs}
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\exc{}
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\begin{align*}
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X &= {{1,2,3}, {2,3}, {ef}} \cup {{e}} \\
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&= {{1,2,3}, {2,3}, {ef}, {e}} \\
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\\
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P(x) &= {
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{{1,2,3}, {2,3}, {ef}, {e}},
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{{1,2,3}, {2,3}, {ef}},
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{{1,2,3}, {2,3}, {e}},
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{{1,2,3}, {ef}, {e}},
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{{2,3}, {ef}, {e}},
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{{1,2,3}, {2,3}}
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{{1,2,3}, {e}}
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{{1,2,3}, {ef}}
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{{2,3}, {ef}}
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{{2,3}, {e}}
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{{ef}, {e}}
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{{e}}
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{{ef}},
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{{2,3}},
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{{1,2,3}}
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} \\
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\\
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P(X \cap Y) &= P({{1,2,3}, {2,3}, {ef}, {e}} \cap {{1,2,3,e,f}}) \\
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&= P(\emptyset) \\
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&= {\emptyset}
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\end{align*}
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\exc{}
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\begin{subexcs}
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\subexc{}
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Here, the exercise says ``[\ldots] four sets $A_1$, $A_2$, $A_3$''. I'm not sure if I'm supposed to do three or four, but I'll assume three.
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\begin{align*}
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A_1 \cap A_2 \cap A_3 \\
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A_1 \cap A_2 \cap \overline{A_3} \\
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A_1 \cap \overline{A_2} \cap A_3 \\
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A_1 \cap \overline{A_2} \cap \overline{A_3} \\
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\overline{A_1} \cap A_2 \cap A_3 \\
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\overline{A_1} \cap A_2 \cap \overline{A_3} \\
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\overline{A_1} \cap \overline{A_2} \cap A_3 \\
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\overline{A_1} \cap \overline{A_2} \cap \overline{A_3} \\
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\end{align*}
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\subexc{}
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For each set, the amount of fundamental products is multiplied by $2$. Therefore, the amount of fundamental sets of $n$ sets is $2^n$
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\end{subexcs}
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\exc{}
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\begin{align*}
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A\overline{(B\overline{C})}\overline{((A\overline{B})\overline{C})} \\
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A\overline{(B\overline{C})}\overline{((A\overline{B})\overline{C})} \\
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\end{align*}
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\end{excs}
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\end{document}
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