Merged changes

Advanced/Euler's Number/parts/1 limits.tex

@@ -59,7 +59,7 @@ and of a bounded sequence that does not have a limit.
 \definition{Limits (formal)}
 Let $a_n$ be a sequence.
 $L$ is the \textit{limit} of this sequence if for any $\varepsilon < 0$, \par
-we can find an $N$ so that $|a_n - L| < \varepsilon \forall n \geq N$.
+we can find an $N$ so that $|a_n - L| < \varepsilon ~~ \forall n \geq N$.
 
 \vfill
 

Advanced/Intro to Proofs/main.tex

@@ -52,6 +52,7 @@
 
 
 
+
 	\problem{}
 	Let $r \in \mathbb{R}$. We say $r$ is \textit{rational} if there exist $p, q \in \mathbb{Z}, q \neq 0$ so that $r = \frac{a}{b}$
 	
@@ -63,11 +64,13 @@
 		a proof.
 	\end{itemize}
 
-
 	\vfill
 	\pagebreak
 
 
+
+
+
 	\problem{}
 	Let $X = \{x \in \mathbb{Z} ~\bigl|~ n \geq 2 \}$. For $k \geq 2$, degine $X_k = \{kx ~\bigl|~ x \in X \}$. \par
 	What is $X - (X_2 \cup X_3 \cup X_4 \cup ...)$? Prove your claim.
@@ -77,6 +80,8 @@
 
 
 
+
+
 	\problem{}
 	For a set $X$, define its \textit{diagonal} as $\text{D}(X) = \{ (x, x) \in X \times X ~\bigl|~ x \in X \}$.
 
@@ -100,11 +105,14 @@
 		Friendship is always mutual.
 	\end{itemize}
 
-
 	\vfill
 	\pagebreak
 
 
+
+
+
+
 	\problem{}
 	Let $f$ be a function from a set $X$ to a set $Y$. We say $f$ is \textit{injective} if $f(x) = f(y) \implies x = y$. \par
 	We say $f$ is \textit{surjective} if for all $y \in Y$ there exists an $x \in X$ so that $f(x) = y$. \par
@@ -119,8 +127,138 @@
 	\vfill
 	\pagebreak
 
+
+
+
+
+
 	\problem{}
+	Let $X = \{1, 2, ..., n\}$ for some $n \geq 2$. Let $k \in \mathbb{Z}$ so that $1 \leq k \leq n - 1$. \par
+	Let $E = \{Y \subset X ~\bigl|~ |Y| = k\}$, $E_1 = \{Y \in E ~\bigl|~ 1 \in Y\}$, and $E_2 = \{Y \in E ~\bigl|~ 1 \notin Y\}$
+
+	\vspace{2mm}
+	\begin{itemize}[itemsep=4mm]
+		\item Show that $\{E_1, E_2\}$ is a partition of $E$. \par
+		In other words, show that $\varnothing \neq E_1$, $\varnothing \neq E_2$, $E_1 \cup E_2 = E$, and $E_1 \cap E_2 = \varnothing$. \par
+		\hint{What does this mean in English?}
+
+		\item Compute $|E_1|$, $|E_2|$, and $|E|$. \par
+		Recall that a set of size $n$ has $\binom{n}{k}$ subsets of size $k$.
+	
+		\item Conclude that for any $n$ and $k$ satisfying the conditions above,
+		$$
+			\binom{n-1}{k} + \binom{n-1}{k-1} = \binom{n}{k}
+		$$
+
+		\item For $t \in \mathbb{N}$, show that $\binom{2t}{t}$ is even.
+	
+	\end{itemize}
+
+
+	\vfill
+	\pagebreak
+
+
+
+
+
+
+	\problem{}
+	Let $x, y \in \mathbb{N}$ be natural numbers.
+	Consider the set $S = \{ax + by ~\bigl|~ a, b \in \mathbb{Z}, ax + by = 0\}$. \par
+	The well-ordering principle states that every nonempty subset of the natural numbers has a least element.
+	You many also need the division algorithm.
+
+	\vspace{4mm}
+	\begin{itemize}[itemsep=4mm]
+		\item Show that $S$ has a least element. Call it $d$.
+		\item Let $z = \text{gcd}(x, y)$. Show that $z$ divides $d$.
+		\item Show that $d$ divides $x$ and $d$ divides $y$.
+		\item Prove or disprove $\text{gcd}(x, y) \in S$.
+	\end{itemize}
+
+	\vfill
+	\pagebreak
+
+
+
+
+
+	\problem{}
+	
+	\begin{itemize}[itemsep=4mm]
+		\item Let $f: X \to Y$ be an injective function. Show that for any two functions $g: Z \to X$ and $h: Z \to X$,
+		if $f \circ g = f \circ h$ from $Z$ to $Y$ then $g = h$ from $Z$ to $X$. \par
+		By definition, functions are equal if they agree on every input in their domain. \par
+		\hint{This is a one-line proof.}
+
+
+		\item Let $f: X \to Y$ be a surjective function.
+		Show that for any two functions $g: Y \to W$ and $h: Y \to W$, if
+		$g \circ f = h \circ f \implies g = h$.
+
+		\item[\star] Let $f: X \to Y$ be a function where for any set $Z$ and functions $g: Z \to X$ and $h: Z \to X$,
+		$f \circ g = f \circ h \implies g = h$. Show that $f$ is injective.
+
+		\item[\star] Let $f: X \to Y$ be a function where for any set $W$ and functions $g: Y \to W$ and $h: Y \to W$,
+		$g \circ f = h \circ f \implies g = h$. Show f is surjective.
+	\end{itemize}
+
+	\vfill
+	\pagebreak
+
+
+
+
+
+
+	\problem{}
+	In this problem we prove the binomial theorem:
+	for $a, b \in \mathbb{R}$ and $n \in \mathbb{Z}^+$\hspace{-0.5ex}, we have
+	$$
+		(a + b)^n = \sum_{k=0}^n \binom{n}{k}a^kb^{N-k}
+	$$
+	In the proof below, we let $a$ and $b$ be arbitrary numbers.
+
+	\vspace{4mm}
+	\begin{itemize}
+		\item Check that this formula works for $n = 0$. Also, check a few small $n$
+		to get a sense of what's going on.
+
+		\item Let $N \in \mathbb{N}$. Suppose we know that for a specific value of $N$,
+		$$
+			(a + b)^N = \sum_{k=0}^N \binom{N}{k}a^kb^{N-k}
+		$$
+		Now, show that this formula also works for $N = N + 1$.
+
+		\item Conclude that this formula works for all $a, b \in \mathbb{R}$ and $n \in \mathbb{Z}^+$\hspace{-0.5ex}.
+	\end{itemize}
+
+	\vfill
+	\pagebreak
+
 	
 
 
+
+	\problem{}
+	A \textit{relation} on a set $X$ is an $R \subset X \times X$. \par
+	\begin{itemize}
+		\item We say $R$ is \textit{reflexive} if $(x,x) \in R$ for all $x \in X$.
+		\item We say $R$ is \textit{symmetric} if $(x, y) \in R \implies (y, x) \in R$.
+		\item We say $R$ is \textit{transitive} if $(x, y) \in R$ and $(y, z) \in R$ imply $(x, z) \in R$.
+		\item We say $R$ is an \textit{equivalence relation} if it is reflexive, symmetric, and transitive.
+	\end{itemize}
+	Say we have a set $X$ and an equivalence relation $R$. \par
+	The \textit{equivalence class} of an element $x \in X$ is the set $\{y \in X ~\bigl|~ (x, y) \in R\}$.
+
+
+	\vspace{4mm}
+
+	Let $R$ be an equivalence relation on a set $X$. \par
+	Show that the set of equivalence classes is a partition of $X$.
+
+	\vfill
+	\pagebreak
+
 \end{document}

Misc/Warm-Ups/nontransitive dice.tex

@@ -0,0 +1,128 @@
+\documentclass[
+	solutions,
+	hidewarning,
+	singlenumbering,
+	nopagenumber
+]{../../resources/ormc_handout}
+
+\usepackage{tikz}
+\usetikzlibrary{arrows.meta}
+\usetikzlibrary{shapes.geometric}
+
+% We put nodes in a separate layer, so we can
+% slightly overlap with paths for a perfect fit
+\pgfdeclarelayer{nodes}
+\pgfdeclarelayer{path}
+\pgfsetlayers{main,nodes}
+
+% Layer settings
+\tikzset{
+	% Layer hack, lets us write
+	% later = * in scopes.
+	layer/.style = {
+		execute at begin scope={\pgfonlayer{#1}},
+		execute at end scope={\endpgfonlayer}
+	},
+	%
+	% Arrowhead tweak
+	>={Latex[ width=2mm, length=2mm ]},
+	%
+	% Nodes
+	main/.style = {
+		draw,
+		circle,
+		fill = white,
+		line width = 0.35mm
+	}
+}
+
+\title{Warm Up: Odd dice}
+\subtitle{Prepared by Mark on \today}
+
+
+\begin{document}
+
+	\maketitle
+
+	\problem{}
+
+	We say a set of dice $\{A, B, C\}$ is \textit{nontransitive}
+	if, on average, $A$ beats $B$, $B$ beats $C$, and $C$ beats $A$.
+	In other words, we get a counterintuitive \say{rock - paper - scissors} effect.
+
+	\vspace{2mm}
+
+	Create a set of nontransitive six-sided dice. \par
+	\hint{All sides should be numbered with positive integers less than 10.}
+
+	\begin{solution}
+		One possible set can be numbered as follows:
+		\begin{itemize}
+			\item Die $A$: $2, 2, 4, 4, 9, 9$
+			\item Die $B$: $1, 1, 6, 6, 8, 8$
+			\item Die $C$: $3, 3, 5, 5, 7, 7$
+		\end{itemize}
+
+		\vspace{4mm}
+
+		Another solution is below:
+		\begin{itemize}
+			\item Die $A$: $3, 3, 3, 3, 3, 6$
+			\item Die $B$: $2, 2, 2, 5, 5, 5$
+			\item Die $C$: $1, 4, 4, 4, 4, 4$
+		\end{itemize}
+	\end{solution}
+
+	\vfill
+
+	\problem{}
+	Now, consider the set of six-sided dice below:
+	\begin{itemize}
+		\item Die $A$: $4, 4, 4, 4, 4, 9$
+		\item Die $B$: $3, 3, 3, 3, 8, 8$
+		\item Die $C$: $2, 2, 2, 7, 7, 7$
+		\item Die $D$: $1, 1, 6, 6, 6, 6$
+		\item Die $E$: $0, 5, 5, 5, 5, 5$
+	\end{itemize}
+	On average, which die beats each of the others? Draw a graph. \par
+
+	\begin{solution}
+		\begin{center}
+		\begin{tikzpicture}[scale = 0.5]
+			\begin{scope}[layer = nodes]
+				\node[main] (a) at (-2, 0.2) {$a$};
+				\node[main] (b) at (0, 2) {$b$};
+				\node[main] (c) at (2, 0.2) {$c$};
+				\node[main] (d) at (1, -2) {$d$};
+				\node[main] (e) at (-1, -2) {$e$};
+			\end{scope}
+		
+			\draw[->]
+				(a) edge (b)
+				(b) edge (c)
+				(c) edge (d)
+				(d) edge (e)
+				(e) edge (a)
+
+				(a) edge (c)
+				(b) edge (d)
+				(c) edge (e)
+				(d) edge (a)
+				(e) edge (b)
+			;
+		\end{tikzpicture}
+		\end{center}
+	\end{solution}
+
+	\vfill
+
+	Now, say we roll each die twice. What happens to the graph above?
+
+	\begin{solution}
+		The direction of each edge is reversed!
+	\end{solution}
+
+	\vfill
+	\pagebreak
+
+\end{document}