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\title{Recitation 5}
\author{Created by Taylor Spangler, Adapted by Beau Christ}
\date{}

\begin{document}
\maketitle

\begin{itemize}
\item (3 min max) Go over quiz from last week
\item (3 min max) Go over homework from last week.
\item Questions about lecture / homework so far?

\item To start, we'll look at some proofs, here we have 1.7.19:\\
\begin{itemize}
\item Let $P(n)$ be the proposition: \textit{If $a$ and $b$ are positive real numbers, then\\ $(a+b)^{n}\geq a^{n}+b^{n}$}.

\item Prove that $P(1)$ is true.

\begin{enumerate}
\item First let $a$ and $b$ be positive real numbers
\item $P(1)$ = $(a+b)^{1} \geq a^{1} + b^{1}$.
\item So $P(1) = a+b \geq a+b$.
\item Well this is clearly true, since $a+b = a+b$. 
\end{enumerate}

\item What kind of proof did we use? \textbf{direct proof}.
\end{itemize}

\item Now a more difficult proof 1.7.29:
\begin{itemize}
\item Prove or disprove: Given $m,n \in \mathbb{Z}$, and $mn =1$, either $m = 1$ and $n=1$, or $m = -1$ and $n = -1$.
\begin{enumerate}
\item Case 1: Suppose $\mid m\mid >1$. Regardless of what $n$ is, $mn = 1$, so $n = \frac{1}{\mid m\mid}$. But we said that both $m$ and $n$ are integers. If $\mid m\mid >1$, then $n$ here cannot be an integer. So, contradiction.
\item Case 2: Similar to case 1, but suppose $\mid  n\mid > 1$. Completely symmetric, works out the same.
\item Case 3: Suppose $m = 1$ and $n = -1$. Clearly then $mn = -1$. So, contradiction.
\item Case 4: Suppose $m = -1$ and $n = 1$. Again clearly $mn = -1$. So, contradiction.
\item Case 5: Suppose $m = 1$ and $n = 1$. Then $mn$ = 1.
\item Case 6: Suppose $m = 1$ and $n = 1$. Then $mn$ = 1.
\end{enumerate}
\item What techniques did we use? \textbf{cases and contradiction}
\end{itemize}

\item 1.8:Example 10: Show that there is a positive integer that can be written as the sum of cubes of positive integers, in two different ways:
\begin{itemize}
\item Here we're going to use a \textit{Constructive Existence} proof.
\item 1729 = $10^3 + 9^3 = 12^3 + 1^3$.
\item So, 1729 can be written of the sum of cubes of positive integers in two different ways.
\item Therefore there is a positive integer that can be written as the sum of cubes of positive integers, in two different ways.
\item These types of proof are basically finding an example.
\item Unfortunately, other than using your intuition, the only way to go about it is brute force.
\end{itemize}

\item 1.8:Example 20: Can we tile a standard chess board with opposite corners removed using dominos (e.g. upper left, lower right removed)?
\begin{itemize}
\item Suppose we can.
\item First, note that a standard chessboard can be covered by dominos, each having one white and one black square.
\item We know that the standard chess board has 64 square. Removing 2 gives 64-2 = 62 squares.
\item Tiling with dominos uses 62/2 = 31 dominos.
\item These dominos each have one white and one black square (like chessboards have white and black).
\item So using the tiles we have 31 white and 31 black squares.
\item However, when we remove two opposite corner squares, either 32 of the remaining squares are white and 30 are black, or 32 are black and 30 are white.
\item But we said we had 31 of each color...contradiction
\item Therefore we cannot tile a standard chessboard with two opposite corners removed, with dominos.
\end{itemize}

\pagebreak

\item Now moving on to sets, 2.1.11: Determine whether true or false:
\begin{itemize}
\item[a] $x \in \{x\}$
\begin{itemize}
\item true
\end{itemize}

\item[b] $\{x\} \subseteq \{x \}$
\begin{itemize}
\item true
\end{itemize}


\item[c] $\{x\} \in \{x\}$
\begin{itemize}
\item false
\end{itemize}

\item[d]  $\{x\} \in \{\{x\}\}$
\begin{itemize}
\item true
\end{itemize}

\item[e] $\emptyset \subseteq \{x\}$
\begin{itemize}
\item true
\end{itemize}

\item[f] $\emptyset \in \{x\}$
\begin{itemize}
\item false
\end{itemize}
\end{itemize}

\item 2.1.27(a) Let $A = \{a,b,c,d\}$, and $B = \{y,z\}$.
\begin{itemize}
\item[a] $A\times B$?
\begin{enumerate}
\item $|A\times B|$? As in, how big will it be? $4\times 2 = 8$
\item $\{(a,y),(b,y),(c,y),(d,y),(a,z),(b,z),(c,z),(d,z)\}$
\end{enumerate}

\item How about $\mathbb{Z}\times (\mathbb{Z}^{+}\setminus 0)$
\begin{enumerate}
\item What is it's \textit{cardinality}? $\infty$
\item What are some of it's elements? (1,2),(3,4),(-1,5).
\item Is $(2,0)\in \mathbb{Z}\times (\mathbb{Z}^{+}\setminus 0)$? No
\item Does this set remind you of anything? Could be used to represent the rational numbers, $\mathbb{Q}$.
\end{enumerate}

\end{itemize}

\item Computer the power set of the following set $S$=  $\{a,b,\{c\},\emptyset \}$
\begin{itemize}
\item First, how many elements are in $\mathcal{P}(S)$? $2^{4}$ = 16.
\item $\{ \emptyset , \{ \emptyset \}, \{\{c\}\}, \{a\},\{b\}, \{a,b\}, \{a,\{c\}\}, \{b,\{c\}\}, \{a,\emptyset\} , \{b, \emptyset \} , \{\{c\} , \emptyset \}, \\ \{a,b,\{c\}\},\{b,\{c\},\emptyset \}, \{a,\{c\}, \emptyset \}, \{a,b,\{c\},\emptyset \}  $
\end{itemize}

\item (Last 15 minutes) Give quiz

\end{itemize}

\end{document}