handouts/Advanced/Intro to Proofs/main.tex

% use [nosolutions] flag to hide solutions.
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\documentclass[
	solutions,
	singlenumbering
]{../../resources/ormc_handout}
\usepackage{../../resources/macros}


\uptitlel{Advanced 2}
\uptitler{Fall 2023}
\title{Intro to Proofs}
\subtitle{Prepared by \githref{Mark} on \today{}}


\begin{document}

	\maketitle





	\problem{}
	We say an integer $x$ is \textit{even} if $x = 2k$ for some $k \in \mathbb{Z}$.
	We say $x$ is \textit{odd} if $x = 2k + 1$ for some $k \in \mathbb{Z}$. \par
	Assume that every integer is even or odd, and never both.

	\vspace{2mm}
	\begin{itemize}[itemsep=4mm]
		\item
		Show that the product of two odd integers is odd.
		
		\item
		Let $a, b \in \mathbb{Z}, a \neq 0$.
		We say $a$ \textit{divides} $b$ and write $a~|~b$ if there is a $k \in \mathbb{Z}$ so that $ak = b$.
	
		Show that $a~|~b \implies a~|~2b$

		\item
		Show that $5n^2 + 3n + 7$ is odd for any $n \in \mathbb{Z}$.

		\item
		Let $a, b, c$ be integers so that $a^2 + b^2 = c^2$. \par
		Show that one of $a, b$ is even.

		\item
		Show that every odd integer is the difference of two squares.

		\item
		Prove the assumption in the statement of this problem.
	\end{itemize}


	\vfill
	\pagebreak







	\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 $r = \frac{a}{b}$
	
	\vspace{2mm}
	\begin{itemize}[itemsep=4mm]
		\item Show that $\sqrt{2}$ is irrational.
		\item Show that the product of two rational numbers must be rational, while the product of
		irrational numbers may be rational or irrational. If you claim a number is irrational, provide
		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.

	\vfill
	\pagebreak









	\problem{}
	Show that there are infinitely may primes. \par
	You may use the fact that every integer has a prime factorization.


	\vfill
	\pagebreak





	\problem{}
	For a set $X$, define its \textit{diagonal} as $\text{D}(X) = \{ (x, x) \in X \times X ~\bigl|~ x \in X \}$.

	\vspace{2mm}

	An \textit{undirected graph} $G$ is an ordered pair $(V, E)$, where $V$ is a set, and $E \subset V \times V$
	satisfies $(a, b) \in E \iff (b, a) \in E$ and $E \cap \text{D}(X) = \varnothing$.

	The elements of $V$ are called \textit{vertices}; the elements of $E$ are called \textit{edges}.

	\vspace{2mm}
	\begin{itemize}[itemsep=4mm]

		\item Make sense of the conditions on $E$.
		
		\item The \textit{degree} of a vertex $a$ is the number of edges connected to that vertex. \par
		We'll denote this as $d(a)$. Write a formal definition of this function using set-builder notation and the definitions above.
		Recall that $|X|$ denotes the size of a set $X$.

		\item There are 9 people at a party. Show that they cannot each have 3 friends. \par
		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
	Let $A, B, C$ be sets, and let $f: A \to B$, $g: B \to C$ be functions. Let $h = g \circ f$.

	\vspace{2mm}
	\begin{itemize}
		\item Show that if $h$ is injective, $f$ must be injective and $g$ may not be injective.
		\item Show that if $h$ is surjective, $g$ must be surjective and $f$ may not be surjective.
	\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 $X$.

	\vfill
	\pagebreak

\end{document}