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\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}