Problem_Set_02_Solutions

# Problem_Set_02_Solutions -...

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\documentclass[12pt]{amsart} \ \setlength{\parsep}{3pc} \setlength{\itemsep}{0.2in} \ \usepackage{fullpage} \usepackage{psfig} \ \title[Problem Set 2]{ArsDigita University\\Month 2: Discrete Mathematics\\Professor Shai Simonson\\Problem Set 2 SOLUTIONS --- Set, Functions, Big-O, Rates of Growth} B \begin{document} \ \maketitle \ \begin{enumerate} \item {\bf Prove by formal logic:} \begin{enumerate} \item {\bf The complement of the union of two sets equals the intersection of the complements.} c Let $x \in \overline{(A \cup B)}$. Then $x$ is not in $A$, and $x$ is not in $B$. Therefore $x \in \overline{A} \cap \overline{B}$. Since $x$ was an arbitrary element of $\overline{(A \cup B)}$, we conclude that $\overline{(A \cup B)} \subseteq \overline{A} \cap \overline{B}$. \ Now, suppose that $x \in \overline{A} \cap \overline{B}$. We can conclude that $x \in \overline{A}$, and that $x \in \overline{B}$. This impiles that $x \notin A$, and $x \notin B$, which implies $x \notin (A \cup B)$, which implies that $x \in \overline{(A \cup B)}$. Again, since $x$ was arbitrary, we conclude that $\overline{(A \cup B)} \subseteq \overline{A} \cap \overline{B}$. Two sets which are subsets of each other are equal (see below), so we can conclude that $\overline{(A \cup B)} = \overline{A} \cap \overline{B}$ $\item {\bf The complement of the interesection of two sets equals the union of the complements.} c Let$x \in \overline{(A \cap B)}$. Then$x$cannot be in both$A$and$B$--- at least one of the statements $x \in \overline{A}$'', $x \in \overline{B}$'' is true. Therefore$x \in \overline{A} \cup \overline{B}$, and$\overline{(A \cap B)} \subseteq \overline{A} \cup \overline{B}$.$ Now let $x \in \overline{A} \cup \overline{B}$. Again, $x$ cannot be in both $A$ and $B$, so $x \notin A \cap B$, so $x \in \overline(A \cap B)$, and $\overline{A} \cup \overline{B} \subseteq \overline{(A \cap B)}$. The two sets are equal. \ \item {\bf $(B-A) \cup (C-A) = (B \cup C) - A$.} \ Let $x \in (B-A) \cup (C-A)$. Then $x \in (B-A) \lor x \in (C-A)$. Assume without loss of generality that $x \in (B-A)$. This implies that $x \in B$ and $x \notin A$. Which implies that $x \in (B \cup C)$, and, combining this with the fact that $x \notin A$, we conclude that $x \in (B \cup C) - A$, and $(B-A) \cup (C-A) \subseteq (B \cup C) - A$. If $x \in (B \cup C) - A$, then $x \notin A$, and $x \in B \lor x \in C$. Therefore $x \in (B-A) \lor x \in (C-A)$, $x \in (B-A) \cup (C-A)$, and $(B \cup C) - A \subseteq (B-A) \cup (C-A)$,

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completing the proof. c \medskip \ \item {\bf If two sets are subsets of each other than they are equal.} \ If two sets $A$ and $B$ are not equal, then there exists some element that is in one set that is not in the other. Without loss of generality, denote such an element $x$ and assume that $x \in A$.
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Problem_Set_02_Solutions -...

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