MA0301/exercise3/main2.tex

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\documentclass[12pt]{article}
\usepackage{ntnu}
\usepackage{ntnu-math}
\author{Øystein Tveit}
\title{MA0301 Exercise 2}
\begin{document}
\ntnuTitle{}
\break{}
\begin{excs}
\exc{}
\exc{}
\exc{}
\begin{subexcs}
\subexc{}
\begin{ssubexcs}
\ssubexc{}
\begin{align*}
{{2,3,5} \cup {6,4}} &\cap {4,6,8} \\
{{2,4,6}} &\cap {4,6,8} \\
\emptyset
\end{align*}
\ssubexc{}
\begin{align*}
P({7,8,9}) &- P({7,9}) \\
{{7,8,9}, {7,8}, {8,9}, {7,9}, {7}, {8}, {9}, \emptyset} &- {{7,9}, {7}, {9}, \emptyset} \\
{{7,8,9}, {7,8}, {8,9}, {8}}
\end{align*}
\ssubexc{}
\begin{align*}
P(\emptyset) \\
{\emptyset}
\end{align*}
\ssubexc{}
\begin{align*}
{1, 3, 5} \times {0} \\
{ \langle 1,0 \rangle, \langle 3,0 \rangle, \langle 5, 0 \rangle }
\end{align*}
\ssubexc{}
\begin{align*}
{2,4,6} \times \emptyset \\
\emptyset
\end{align*}
\ssubexc{}
\begin{align*}
P({0}) &\times P({1}) \\
{\emptyset, {0}} &\times {\emptyset, {1}} \\
{\langle\emptyset,\emptyset\rangle, \langle\emptyset,{1}\rangle, \langle{0},\emptyset\rangle, \langle{0},{1}\rangle}
\end{align*}
\ssubexc{}
\begin{align*}
P(P({2})) \\
P({\emptyset,{2}}) \\
{ {{\emptyset}, {2}}, {{\emptyset}}, {{2}}, \emptyset }
\end{align*}
\end{ssubexcs}
\subexc{}
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
\[ |P(A) - {{x} : x \in A}| = \frac{n}{2} \]
\end{subexcs}
\exc{}
\begin{subexcs}
\subexc{}
$\emptyset = {\emptyset}$ is {\color{red}False} because $|\emptyset| \neq |{\emptyset}|$
\subexc{}
$\emptyset = {0}$ is {\color{red}False} because $|\emptyset| \neq |{0}|$
\subexc{}
$|\emptyset| = 0$ is {\color{ForestGreen}True} because $\emptyset$ has $0$ elements
\subexc{}
$P(\emptyset)$ is {\color{red}False} because $P(\emptyset) = {{\emptyset}}$ has $1$ element
\subexc{}
$\emptyset = {}$ is {\color{ForestGreen}True} because the empty set is a subset of every possible set
\subexc{}
$\emptyset = {x \in \mathbb{N} : x \leq 0 and x > 0}$ 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
\end{subexcs}
\exc{}
\begin{subexcs}
\subexc{}
\begin{align*}
A \cap (\A \cup B) \\
{x : x \in A \wedge x \in (A \cup B)} \\
{x : x \in A \wedge (x \in A \or x \in B)} \\
{x : x \in A} \\
A
\end{align*}
\subexc{}
\begin{align*}
A-(B \cap C) \\
{x : x \in A \wedge x \notin (B \cap C)} \\
{x : x \in A \wedge (x \notin B \wedge x \notin C)} \\
{x : (x \in A \wedge x \notin B) \vee (x \in A \wedge x \notin C)} \\
{x : x \in (A - B) \vee x \in (A - C)} \\
{x : x \in (A - B) \cup (A - C)} \\
(A-B) \cup (A-C)
\end{align*}
\end{subexcs}
\exc{}
\begin{subexcs}
\subexc{}
\end{subexcs}
\exc{}
\begin{align*}
X &= {{1,2,3}, {2,3}, {ef}} \cup {{e}} \\
&= {{1,2,3}, {2,3}, {ef}, {e}} \\
\\
P(x) &= {
{{1,2,3}, {2,3}, {ef}, {e}},
{{1,2,3}, {2,3}, {ef}},
{{1,2,3}, {2,3}, {e}},
{{1,2,3}, {ef}, {e}},
{{2,3}, {ef}, {e}},
{{1,2,3}, {2,3}}
{{1,2,3}, {e}}
{{1,2,3}, {ef}}
{{2,3}, {ef}}
{{2,3}, {e}}
{{ef}, {e}}
{{e}}
{{ef}},
{{2,3}},
{{1,2,3}}
} \\
\\
P(X \cap Y) &= P({{1,2,3}, {2,3}, {ef}, {e}} \cap {{1,2,3,e,f}}) \\
&= P(\emptyset) \\
&= {\emptyset}
\end{align*}
\exc{}
\begin{subexcs}
\subexc{}
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.
\begin{align*}
A_1 \cap A_2 \cap A_3 \\
A_1 \cap A_2 \cap \overline{A_3} \\
A_1 \cap \overline{A_2} \cap A_3 \\
A_1 \cap \overline{A_2} \cap \overline{A_3} \\
\overline{A_1} \cap A_2 \cap A_3 \\
\overline{A_1} \cap A_2 \cap \overline{A_3} \\
\overline{A_1} \cap \overline{A_2} \cap A_3 \\
\overline{A_1} \cap \overline{A_2} \cap \overline{A_3} \\
\end{align*}
\subexc{}
For each set, the amount of fundamental products is multiplied by $2$. Therefore, the amount of fundamental sets of $n$ sets is $2^n$
\end{subexcs}
\exc{}
\begin{align*}
A\overline{(B\overline{C})}\overline{((A\overline{B})\overline{C})} \\
A\overline{(B\overline{C})}\overline{((A\overline{B})\overline{C})} \\
\end{align*}
\end{excs}
\end{document}