Added proof problems

Advanced/Intro to Proofs/main.tex

@@ -101,4 +101,118 @@
 	\end{itemize}
 
 
+	\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 $X$.
+
 \end{document}