From a07393114556e859e3ed42b09211833890c9e766 Mon Sep 17 00:00:00 2001 From: mark Date: Fri, 6 Oct 2023 08:40:37 -0700 Subject: [PATCH] Added proof problems --- Advanced/Intro to Proofs/main.tex | 114 ++++++++++++++++++++++++++++++ 1 file changed, 114 insertions(+) diff --git a/Advanced/Intro to Proofs/main.tex b/Advanced/Intro to Proofs/main.tex index ba22a3e..943f9a9 100755 --- a/Advanced/Intro to Proofs/main.tex +++ b/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} \ No newline at end of file