Abstract Algebra Homework #1
Autor
Chas Lewis
Letzte Aktualisierung
vor 9 Jahren
Lizenz
LaTeX Project Public License 1.3c
Abstrakt
First abstract algebra home work.
First abstract algebra home work.
\documentclass[12pt,twoside]{amsart}
\special{papersize=8.5in,11in}
\usepackage{amssymb}
\usepackage[margin=1in]{geometry}
\usepackage{graphicx}
\usepackage[pdfpagemode=UseOutlines,bookmarksopen=false,letterpaper,bookmarksnumbered=true,pdfborder={0 0 0},colorlinks=false]{hyperref}
% -- Symbols
\newcommand{\CC}{\mathbb{C}}
\newcommand{\NN}{\mathbb{N}}
\newcommand{\QQ}{\mathbb{Q}}
\newcommand{\RR}{\mathbb{R}}
\newcommand{\ZZ}{\mathbb{Z}}
\DeclareMathOperator{\lcm}{lcm}
\newcommand{\st}{\ensuremath{\colon\,}} % such that
\newcommand{\dcup}{\ensuremath{\,\mathaccent\cdot\cup\,}} % disjoint union
\newcommand{\ddd}{\ensuremath{\dcup \cdots \dcup}} % many disjoint unions
\def\urltilda{\kern -.15em\lower .7ex\hbox{\~{}}\kern .04em}
% -- Alterations
\renewcommand{\labelenumi}{(\roman{enumi})}
\numberwithin{figure}{section}
\numberwithin{equation}{section}
% -- Theorem-like Environments
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{claim}[theorem]{Claim}
\newtheorem{assumption}[theorem]{Assumption}
\newtheorem{question}[theorem]{Question}
\theoremstyle{definition}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{notation}[theorem]{Notation}
\newtheorem{example}[theorem]{Example}
\newtheorem*{acknowledgement}{Acknowledgement}
%------------
% -- Document
\begin{document}
\setcounter{section}{1} % Homework Number
\begin{center}
{\Large MAT 3530 Homework \#1} \\
due 28.\ August 2015 \\
Chas Lewis
\end{center}~\\
\begin{question}
The \emph{power set} of a finite set $A$ is the set $\mathcal{P}(A)$ of all subsets of $A$.
Determine a formula for $\#\mathcal{P}(A)$ in terms of $\#A$. Prove your formula is correct.
\end{question}
\begin{proof}
If the power set of $\mathcal{P}$(A) contains the totality of all elements in A ie a$\in$A and A$\in$${P}$(A), then there also exists a relation between the cardinality of ${P}$ and the cardinality of A. We can show this by example A=\{1,2\} therefore ${P}$(A)=[\{0\},\{1\}] $\cup$ ${P}$(2)=[\{2\}]. So the cardinality of ${P}$(1) equals the cardinality of A, therefore we can show the cardinality of a added to the cardinality of ${P}$(1) equals the cardinality of ${P}$(A).
\end{proof}
\begin{question}
Let $f: X \to Y$ be a function, and let $A, B \subset X$.
\begin{enumerate}
\item Prove $f(A \cup B) = f(A) \cup f(B)$.\\
We first need to show f(A$\cup$B) $\subset$ f(A) $\cup$f(B) \
Let's say y$\in$f(A$\cup$B), and there is x$\in$f(A) \
or x$\in$B so f(x)$\in$f(B) \
in either case y=f(x)$\in$f(A)$\cup$f(B) \
secondly we need to show f(A)$\cup$f(B) $\subset$ f(A$\cup$B) \
y$\in$f(A)$\cup$f(B), so y$\in$f(A) or y$\in$f(B) \
if y$\in$f(A), then there is a$\in$A so f(a)=y such that y=f(A)$\in$f(A$\cup$B) since a$\in$f(A$\cup$B)\
also if y=f(B), then y$\in$f(A$\cup$B) and the equality is complete.\
\item Prove $f(A \cap B) \subset f(A) \cap f(B)$. \
Say y$\in$f(A$\cap$B) and x$\in$A$\cap$B \
so I can then say that x$\in$A, y$\in$f(A) and x$\in$b and y$\in$f(B). \
So y$\in$$\subset$f(A)$\cap$f(B).
\item Find a counterexample to $f(A \cap B) = f(A) \cap f(B)$.\
\end{enumerate}
\end{question}
\begin{proof}
Let's say A = \{1,2\} and b = \{1,3\}. Now we multiply A, which yields A*A=\{1,2,2,4\} and B*A=\{1,3,3,9\}. Now, f(A$\cap$B)=\{1,2\} but f(A)$\cap$f(B)=\{1,0\}.
\end{proof}
\begin{question}
Define the relation $\sim$ on $\QQ$ by $r \sim s$ if and only if $r - s \in \ZZ$. Prove $\sim$ is an equivalence relation.
\end{question}
\begin{proof}
We know that $\mathbb{Z}$$\subset$$\mathbb{Q}$ and if c$\in$r-s then c$\in$$\mathbb{Q}$ and c$\in$$\mathbb{Z}$ so r$\sim$s, r$\sim$c, and s$\sim$c so it is transitive. also, r-r$\in\ZZ,\QQ$ and s-s$\in\ZZ,\QQ$ so it is reflexive. Finally since r-s$\in\ZZ$ then s-r$\in\ZZ$ because r$\sim$s so it is symmetric.
\end{proof}
\begin{question}
Let $f: X \to Y$ be a function. Prove that the relation $u \sim v$ if and only if $f(u) = f(v)$ is an equivalence relation.
\end{question}
\begin{proof}
Say u$\in$x and v$\in$y if f(u)=f(v) and v$\in$f(v) and u$\in$f(u) then f(v)$\sim$f(u) and this show reflexive as well as symmetric. If c$\in$y then c$\in$f(v),f(u) thus c$\sim$u,c$\sim$v, and of course u$\sim$v which is transitive.
\end{proof}
%---------------------------
% -- Happy, Happy, Joy, Joy!
\end{document}