\documentclass[11pt,pages]{exam} \usepackage{amsmath,amsfonts,times,amssymb,
  graphicx,multicol,setspace}

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  \newcommand{\een}{\end{enumerate}}

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  \newcommand{\ea}{\end{longanswer}}


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  \newcommand{\ep}{\end{problem}}

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\title{Exam 2} \target{Math 157}

\begin{document}


\begin{exam}{Thursday, October 17, 2013}


\bp Let $x \geq -1$.  Prove that for all $n\in \N$ we have $(1+x)^n\geq 1+xn$.
\ba
We use induction.  When $n=1$, the statement $(1+x)^n\geq 1+xn$ is the
statement $$(1+x)^1\geq 1+x,$$ which is true since both sides are equal.

Now, assume the statement is true for some natural number $k$.  That
is, assume $$(1+x)^k\geq 1+xk.$$
Consider $(1+x)^{k+1}$:
%We have to prove that $(1+x)^{k+1}\geq 1+x(k+1)$.  So, consider 
\begin{align*}
(1+x)^{k+1}&=(1+x)^k(1+x)\\
&\geq (1+kx)(1+x) & \text{ because $(1+x)^k\geq
1+xk$ and $(1+x)\geq 0$} \\
&=1+(k+1)x+kx^2\\
&\geq
1+(k+1)x &\text{ because $kx^2\geq 0.$}
\end{align*}
This proves that if the statement is true for $k$, then the statement
is true for
$k+1$, completing our proof by induction.
\ea
\ep


\bp Let $X$ and $Y$ be sets and $f:X\to Y$ be a function.  
Consider the following argument:

\begin{quote}\emph{
Suppose that $f:X\to Y$ is not injective.  Then there exist elements
$x,z\in X$ with $f(x)=f(z)$ and $x\neq z$.  
Let $A=\{x\}$.  Note that
$z\in X\setminus A$ so $$f(z)\in f(X\setminus A).$$  Since $x\in A$,
$f(x)\in f(A)$, so $$f(x)\notin Y\setminus f(A).$$
Since $f(x)=f(z)$, we've found an element in $f(X\setminus A)$ that is
not in $Y\setminus f(A)$.  Therefore, $f(X\setminus A) \nsubseteq
Y\setminus f(A)$.
}
\end{quote}

Which of the following propositions does the argument above prove?
\ben
\item If $f$ is injective, then for all $A\subseteq X$ we have $f(X\setminus A)\subseteq
  Y\setminus f(A)$.
\item If $f$ is injective, then there exists a set 
$A\subseteq X$ with $f(X\setminus A)\subseteq
  Y\setminus f(A)$.
\item If for all subsets $A\subseteq X$ we have $f(X\setminus A)\subseteq
  Y\setminus f(A)$, then $f$ is injective.
\item  If $f$ is injective, then for all $A\subseteq X$ we have $Y\setminus
  f(A)\subseteq f(X\setminus A).$
\item If for all sets $A\subseteq X$ we have $Y\setminus
  f(A)\subseteq f(X\setminus A)$, then $f$ is injective.
\item If $f$ is not injective, then for all $A\subseteq X$ we have $f(X\setminus A)\nsubseteq
  Y\setminus f(A)$.
\een
\ep  



\bp Give an example of a function $f:X\to Y$ and a set $A\subseteq X$
for which $$f^{-1}(f(A))\neq A.$$
\ba Let $X=\{1,2,3,4\}$ and $Y=\{a,b,c,d\}$ and define 
\begin{align*}
f:X&\to Y\\
1&\mapsto a\\
2&\mapsto b\\
3&\mapsto b\\
4&\mapsto c
\end{align*}
Let $A=\{1,2\}$.  Then $$f^{-1}(f(A))=f^{-1}(\{a,b\})=\{1,2,3\}\neq A.$$
\ea
\ep

% \bp The set $S=\{\heartsuit, \diamondsuit, \clubsuit\}$
% and the two binary operations $+$ and $\times$ defined below satisfy
% all of the field axioms.

% {%\Large
% \begin{center}
% \begin{tabular}{|l||c|c|c|c|}
% \hline 
% $+$ & $\heartsuit$ & $\diamondsuit$ & $\clubsuit$  \\
% \hline \hline
% $\heartsuit$ & $\heartsuit$ & $\diamondsuit$ & $\clubsuit$ \\
% \hline
% $\diamondsuit$ &$\diamondsuit$  & $\clubsuit$ & $\heartsuit$ \\
% \hline
% $\clubsuit$ & $\clubsuit$ & $\heartsuit$ & $\diamondsuit$ \\
% \hline
% \end{tabular}
% $\qquad$
% \begin{tabular}{|l||c|c|c|c|}
% \hline 
% $\times$ & $\heartsuit$ & $\diamondsuit$ & $\clubsuit$  \\
% \hline \hline
% $\heartsuit$ & $\heartsuit$ & $\heartsuit$ & $\heartsuit$ \\
% \hline
% $\diamondsuit$ & $\heartsuit$  & $\diamondsuit$ & $\clubsuit$ \\
% \hline
% $\clubsuit$ & $\heartsuit$ & $\clubsuit$ & $\diamondsuit$\\
% \hline
% \end{tabular}
% \end{center}
% } 
% \ben
% \addtolength{\itemsep}{2in}

% \item Which element of $S$ is an identity for the operation $\times$?  
% \item Which element of $S$ is $\frac{1}{\clubsuit}$?
% \een

% \ep


% \bp Let $A,B$ and $C$ be sets.  
% \ben
% \item One of the following
% conditions is sufficient for $A\setminus B = A
% \setminus C.$  Which one? 
% \begin{itemize}
% \item $A \subset (B\cup C)$
% \item $(B\cup C) \subset A$
% \item $A\cap B\cap C =\emptyset$
% \item $(B\cap C) \subset A$
% \item $A\cap C=A \cap B$
% \end{itemize}
% \item Prove that the condition you identified implies that $A\setminus
%   B = A
% \setminus C$.
% \vspace{4in}
% \item Give an example to show that the identified condition is not
%   \emph{necessary} for $A\setminus B = A
% \setminus C$.
% \een
% \ep

\bp True or False.  Right answer $+1$, wrong answer $-1$, no answer
$0$.
\ben
\addtolength{\itemsep}{2in}
\item The field axioms of $\R$ imply that $1+1\neq 0$.
\item For all functions $f:X\to Y$ and for all sets $C\subseteq
Y$, we have $f^{-1}(Y\setminus C)=X\setminus f^{-1}(C).$
\item For all functions $f:X\to Y$ and for all sets $C\subseteq
Y$, we have $f(f^{-1}(C))=C.$
\item For all injective functions  $f:X\to Y$ and for all sets $A\subseteq X$
we have $f^{-1}(f(A))=A.$
\een
\ep

% \bp Prove that for all functions $f:X\to Y$ and for all sets $C\subseteq
% Y$, we have $$f^{-1}(Y\setminus C)=X\setminus f^{-1}(C).$$
% \ba Let $C\subseteq Y$.  To show that $f^{-1}(Y\setminus C)\subseteq
% X\setminus f^{-1}(C)$, let $x\in f^{-1}(Y\setminus C).$  This means
% that $f(x)\in Y\setminus C$.  So, $f(x)\notin C$.  Since $f(x)\notin
% C$, we know $x\notin f^{-1}(C)$, implying that $x\in X\setminus
% f^{-1}(C)$.

% To show that $X\setminus f^{-1}(C)\subseteq f^{-1}(Y\setminus C)$, let
% $x\in X\setminus f^{-1}(C)$.  So $x\notin f^{-1}(C)$.  This implies
% that $f(x)\notin C$.  Therefore $f(x)\in Y\setminus C$.  This says
% that $x\in f^{-1}(Y\setminus C)$ as needed.
  

% \ea
% \ep


% \ben
% \item One of the following
% conditions is sufficient for $(A\setminus B)\setminus C = A
% \setminus (B\setminus C).$  Which one? 
% \begin{itemize}
% \item $A \subset (B\cup C)$
% \item $(B\cup C) \subset A$
% \item $A\cap B\cap C =\emptyset$
% \item $C \subset (B\setminus A)$
% \item $A\cap B=C \cap B$
% \end{itemize}
% \item Prove that the condition you identified implies that  $(A\setminus B)\setminus C = A
% \setminus (B\setminus C).$
% \vspace{5in}

% \item Give an example to show that the identified condition is not
%   \emph{necessary} for $A\setminus
%   B = A
% \setminus C$.

% \een
% \ep

% \bp True or False.  Give brief, but conclusive evidence, to support
% your answer.
% \ben
% \addtolength{\itemsep}{2in}
% \item For all sets $S$ and for all $A\subseteq S$ there exists a
%   unique set
%   $B\subseteq S$ with $A\cup B =S$.

% \item For all sets $A\subseteq \R$, either $A$ or $\R\setminus A$ is bounded
%   above.

% \item For all $x,y\in \R$, if $x^2<y^2$ then either $x<y$ or $-x<y$.

% \item For all $x\in \R$ and for all $n\in \N\left
%     (|2x-6|<\dfrac{1}{n}\Rightarrow x=3\right).$

% \een
% \ep

% \bp 
% \ben
% \item Use mathematical induction to prove that 
% $\ds \sum_{k=1}^n k^3 =\frac{(n)^2(n+1)^2}{4}$ for all $n\in \N$.
% \een
% \ep
% \addtocounter{problemnum}{-1}
% \bp
% \ben
% \addtocounter{enumi}{1}
% \item Use this result to compute the area of the region pictured below 
% (the
% vertical distance between the point $b$ units from $0$ is
% $\frac{1}{2}b^3$).
% \bigskip

% \begin{center}
% \includegraphics[scale=0.7]{exam1_pic.pdf}
% \end{center}
% \een
% \ep

\pagebreak



\end{exam}
\end{document}

