%2multibyte Version: 5.50.0.2953 CodePage: 1252 \documentclass{article} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \usepackage{amsfonts, amssymb} \usepackage{amsmath} \usepackage{latexsym} \setcounter{MaxMatrixCols}{10} %TCIDATA{OutputFilter=LATEX.DLL} %TCIDATA{Version=5.50.0.2953} %TCIDATA{Codepage=1252} %TCIDATA{} %TCIDATA{BibliographyScheme=Manual} %TCIDATA{LastRevised=Thursday, February 15, 2018 15:34:46} %TCIDATA{} \newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{axiom}[theorem]{Axiom} \addtolength{\textwidth}{3.5cm} \addtolength{\oddsidemargin}{-2cm} \addtolength{\topmargin}{-3.5cm} \addtolength{\textheight}{5.25cm} \newcommand{\mbf}{\mathbf{#1}} \newcommand{\mbb}{\mathbb{#1}} \DeclareMathOperator{\D}{D} \DeclareMathOperator{\Mat}{Mat} \DeclareMathOperator{\im}{im} \pagestyle{empty} \input{tcilatex} \begin{document} \hfill \textbf{February 14, 2015} \vspace{1cm} \noindent\textbf{Name:}\hfill \vspace{0.5 cm} \centerline{ \textbf{\large{Test 1 \hspace{1 cm} Math 5330}}} \vspace{0.5 cm} \noindent You have \textbf{90} minutes to complete the test. You cannot use any books or notes. \vspace{1pt} \begin{enumerate} \item State the Well-Ordering Principle for $\mathbb{Z}^{+}$. Answer: For the test, study first and second form of induction. \item Prove by induction that $1+3+5+\cdots +(2n+1)=(n+1)^{2}$. \textbf{% Answer:}\ This is obvious for $n=1:1+3=4=2^{2}.$Assume the claim for $n$ \ Then:\ $1+3+5+\cdots +(2n+1)+(2(n+1)+1)=(n+1)^{2}+2n+3=n^{2}+2n+1+2n+3=n^{2}+4n+4=(n+2)^{2}.$ \item Prove by Mathematical Induction that if a set $S$ has $n$ elements then $S$ has $2^{n}-$many subsets. Answer:\ If $S$ has no element, then there is only one subset, namely the empty set. Assume the claim for any set $S$ with $n-$elements. Add any element $c$ to the set $S.$Then the subsets of $S\cup \{c\}$ are the subsets $A$ of $S$ plus all sets $A\cup \{c\}.$ Thus $S\cup \{c\}$ has twice as many subsets as $S$ has. $S$ has by induction hypothesis $2^{n}$ many subsets thus $S\cup \{c\}$ has $2\times 2^{n}=2^{n+1}-$many subsets. \item Is division a commutative operation on the set $\mathbb{R}^{+}$ of positive numbers? Is it associative? Explain your answers. Answer: of course, division is not commutative, $1/2\neq 2/1.$ It is not associative, $% ((1/2)/3)\neq 1/(2/3)$ \item Define that $(G,\ast )$ is a group. You can also state that $(G,\ast ,{}^{-1},e)$ is a group. Answer: RTead this in the book. \item Define the additive group $(\mathbb{Z}_{n},\oplus )$ of integers modulo $n$. You have to state exactly what its elements are, how $\oplus$ is defined, what the identity is, and how the inverse of an element is defined. \textbf{Answer:} Read this in the book. \item \begin{enumerate} \item Find the additive inverse of $40$ modulo $48$. \textbf{Answer:} $-40% \func{mod}48=8\func{mod}48$ \item Solve $15+x=7$ modulo $48$. Answer:\ $x=7-15=-8\func{mod}48=-8+48=40% \func{mod}48$ \end{enumerate} \item Let $(G,\ast )$ be a group such that $x^{2}=e$ for all $x\in G$. Show that $G$ is abelian. Answer: $x^{2}=e$ is the same as $x=x^{-1}$ for every $% x.$Thus $xy=(xy)^{-1}=y^{-1}x^{-1}=yx$ \item Let $G$ be a finite group, and consider the multiplication table for multiplication of $G$. Prove that every element of $G$ occurs precisely once in each row and once in each column. \textbf{Answer:} This is just that given some $x$ and $z$ you can find some $y$ such that $xy=z.$ This is clear, $y=x^{-1}z.$ This takes care for rows. For columns you argue that given any $y$ and $z$ you can finds some $x$ such that $xy=z.$ Here $% x=y^{-1}z.$ \item Find a multiplication table on the set $A=\{a,b,c,d\}$ where every element of $A$ occurs precisely once in each row and once in each column but where $A$ is \textbf{not }a group. You have to explain why your algebra $% A=(\{a,b,c,d\},\ast )$ is not a group. \textbf{Answer:} Optional puzzle. Not needed for the test. \end{enumerate} \end{document}