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\fancyhead[L]{Math 221 - 01 (Alvarado)}
\fancyhead[C]{Homework 1}
\fancyhead[R]{Allison Tanguay}
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\begin{document}
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	\begin{center}
		{\Large Math 221: Transition to Theoretical Mathematics - 01 (Professor Alvarado)}
	\end{center}
\vskip-.05in
	{\large Homework 1}\hfill{Allison Tanguay}
	
	\noindent\rule{6.8in}{1pt}
	\vskip.2in
	
	\vskip.25in
	\textbf{Print Name} (first, last)\textbf{:} Allison Tanguay 
	
	\vskip.25in
	
	\textbf{I worked with/used the following sources:} Ryan Alvarado, textbook, \dots
	
	\begin{enumerate}
		\item Let $U:=\{0,1,2,3,4,5,6,7,8,9,10\}$. List out the elements in each set (write its elements within braces).

		\begin{enumerate}
			\item $S:=\{x\in U:\,x^2<46\}$\\
			\emph{Answer:} $S=\{ 0, 1, 2, 3, 4, 5, 6 \}$
			
			\item $T:=\{x\in U:\,x^2-3x=-2\}$ \\
			\emph{Answer:}
				\[x^2-3x=-2\]
					\[x^2-3x+2=0\]
					\[(x-2)(x-1)=0\]
					\[ x=2 \; \text{      or      } \;  x=1\]
				\[T=\{1,2\}\]
			
		\end{enumerate}
	
		
		\item Let $A:=\{1,2,3,4,5\}$. Write each of the following sets in the form $\{x\in A:\,p(x)\}$,
		where $p(x)$ is a condition on $x$.	(See problem 1 to get an idea of what $p(x)$ could be.)
		%
		\begin{enumerate}
			\item $B:=\{2,3,4\}$
			
					\emph{Answer:} $B=\{x\in A:\, (x-2)(x-3)(x-4)=0\}$
			
			\item $C:=\{1,2,4,5\}$
			
				\emph{Answer:} $C=\{x\in A:\, x\ne3 \}$
		\end{enumerate}
		
	 	\item Let $A:=\big\{a,\{a\},\{a,\{a\}\}\big\}$. How many elements are in the set $A$?
	 	For each of the following parts, answer true or false.	Explain briefly your answer.
	 	
	 		\emph{Answer:} There are three elements in the set $A$. 
	 	
	 		 	 \begin{enumerate}
	 		 	 	\item $a\in A$
	 		 	 	
	 		 	 	\emph{Answer:} True since the element $a$ is in the set $A$. 	
	 		 	 	
	 		 	 	\item $a\subseteq A$
	 		 	 	
	 		 	 		\emph{Answer:} False because $a$ is not a set and therefore cannot be a subset of $A$. 
	 		 	 	
	 		 	 	\item $\{a\}\subseteq A$
	 		 	 	
	 		 	 		\emph{Answer:} True since the element $a$ is in the set $A$, so $\{a\}\subset A$. 
	 		 	 	
	 		 	 	\item $\{a\}\in A$
	 		 	 	
	 		 	 		\emph{Answer:} True since the set $\{a\}$ is an element of the set $A$.
	 		 	 	
	 		 	 \end{enumerate}
		
\item Give an example of three sets, $A$, $B$, and $C$, such that $A\in B$, $B\in C$ and
$A\not\in C$.

	\emph{Answer:} Let $$A=\{a\}$$ 
	$$B=\{\{a\}\}=\{A\}$$
		$$C=\{\{\{a\}\}\}=\{B\}$$

\item Let $U:=\{-3,-2,-1,0,1,2,3\}$. Which of the following sets are equal?

\begin{enumerate}
	\item $A:=\{n\in U:\, |n|<2\}$
	\[ A= \{ -1, 0, 1\} \]
	
	\item $B:=\{n\in U:\, n^2\leq 1\}$
		\[ B= \{ -1, 0, 1\} \]
	
	\item $C:=\{n\in U:\, n^3 =n\}$
		\[ C= \{ -1, 0, 1\} \]
	
	\item $D:=\{-1,0,1\}$
	\item $E:=\{n\in U:\, n^2\leq n\}$
		\[ E= \{ 0, 1\} \]
		
		Therefore, the first four sets are equal, i.e. $A=B=C=D$, but $E$ is not equal to the others. 
	
\end{enumerate}
		
		\newpage

		
		\item Let $S = \{1,2,3,4,5\}$,\,\, $T = \{3,4,5,7,8,9\}$,\,\, $U = \{1,2,3,4,9\}$,\,\, $V = \{2,4,6,8\}$.
		Find each of the following sets.
		%
		\begin{enumerate}
			\item $S\cap U $\\
			\emph{Answer:} $S\cap U = \{ 1, 2, 3, 4\}$.
			
			\item $(S\cap T)\cup U$\\
			\emph{Answer:} $(S\cap T)\cup U = \{3,4,5\} \cup U = \{1,2,3,4,5,9\}$. 
			
			\item $(S\cup V)\setminus U$\\
				\emph{Answer:}  $(S\cup V)\setminus U = \{1,2,3,4,5,6,8\}\setminus U = \{5,6,8\}$. 
			
			\item $(S\setminus U)\cup (V\setminus U)$\\
			\emph{Answer:} $(S\setminus U)\cup (V\setminus U)=\{5\} \cup \{6,8\} = \{5,6,8\}$. 
			
			\item $(S\cup V)\setminus (T\cap U)$\\
			\emph{Answer:} $(S\cup V)\setminus (T\cap U)=\{1,2,3,4,5,6,8\} \setminus \{3,4,9\}=\{1,2,5,6,8\}$
			
			\item $(S\times V)\setminus(T\times U)$\\
		\emph{Answer:}
		\begin{multline*}
		(S\times V)=\Big\{(1,2),(1,4),(1,6),(1,8),(2,2),(2,4),(2,6),(2,8),
		(3,2),(3,4),(3,6),(3,8),
		\\
		 (4,2),(4,4),(4,6),(4,8),(5,2),(5,4),(5,6),(5,8)\Big\}
		\end{multline*}
	
			\begin{multline*}
		(T\times U)=\Big\{ (3,1),(3,2),(3,3),(3,4),(3,9),(4,1),(4,2),(4,3),(4,4),(4,9),\\
		(5,1),(5,2),(5,3),(5,4),(5,9),(7,1),(7,2),(7,3),(7,4),(7,9)\\
			(8,1),(8,2),(8,3),(8,4),(8,9),(9,1),(9,2),(9,3),(9,4),(9,9)\Big\}
		\end{multline*}
		
		\begin{multline*} 
		(S\times V)\setminus(T\times U) = \Big \{ (1,2),(1,4),(1,6),(1,8),(2,2),(2,4),(2,6),(2,8),\\
		(3,6),(3,8),(4,6),(4,8),(5,6),(5,8)\Big\}
		\end{multline*}
			
			\item $(V\setminus T)\times(U\setminus S)$\\
			\emph{Answer:}	$ V\setminus T=\{ 2,6 \}$ and $ U\setminus S=\{ 9 \}$. Then
	\begin{align*}
	    (V\setminus T)\times(U\setminus S) = \Big\{ (2,9),(6,9) \Big\}
	    \end{align*}
	    
		
		\end{enumerate}

		
		
		   
		\item Let $U:=\{-4,-3,-2,-1,0,1,2,3,4\}$. If $X:=\big\{n\in U:\,n^2<11\big\}$ and $A:=\big\{n\in X:\,\frac{n}{2}\in\mathbb{N}\big\}$,
		then find $X\setminus A$. \\
		\emph{Answer:} $X=\{-3,-2,-1,0,1,2,3\}$ and $A=\{-2,2\}$. Then $X \setminus A = \{ -3,-1,0,1,3\}$.  
	
	\newpage
		
		\item Which of the following sets are partitions of $A = \{1, 2, 3, 4, 5\}$? Explain briefly why or why not.
		%
		\begin{enumerate}
			\item $A_1 := \{\{1,3\},\{2,5\}\}$\\
			\emph{Answer:} $A_1$ is not a partition of $A$ because $4\in A$ and $4$ is not an element of either of the sets in $A_1$. (The union of the sets in $A_1$ is not equal to $A$.) 
			
			\item $A_2 := \{\{1,2\},\{3,4,5\}\}$\\
			\emph{Answer:} $A_2$ is a partition of $A$ because $\{1,2\} \cup \{3,4,5\} = A$ and $\{1,2\} \cap \{3,4,5\} = \emptyset$. 
			
			\item $A_3 := \{\{1,2\},\{2,3\},\{3,4\},\{4,5\}\}$\\
			\emph{Answer:} $A_3$ is not a partition of $A$ because the sets in $A_3$ are not mutually disjoint. For example, $\{1,2\}\cap\{2,3\}\ne \emptyset$. 
			
		\end{enumerate}

		\item Suppose $A := \{e,0\}$ and $B := \{0,1\}$. Write out the indicated sets by listing their elements.

\begin{enumerate}
	\item $A\times B$ \\
	\emph{Answer:} $A\times B = \Big\{ (e,0),(e,1),(0,0),(0,1)\Big\}$
	\item $B\times A$\\
		\emph{Answer:} $B\times A = \Big\{ (0,e),(0,0),(1,e),(1,0)\Big\}$
	\item $A\times A$\\
		\emph{Answer:} $A\times A = \Big\{ (e,e),(e,0),(0,e),(0,0)\Big\}$
	\item $A\times\emptyset$\\
		\emph{Answer:} $A\times \emptyset = \emptyset$
\end{enumerate}



	\end{enumerate}
	
	
	
	
	
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