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\def\HW{6}
\def\DUE{10/9/2018}
\def\NAME{YOUR NAME HERE}
\title[Homework \HW]{Homework \HW \\
Math 308\\
\small Due: \DUE}
\author{}
%\date{}                                           % Activate to display a given date or no date

\begin{document}
%\maketitle %Comment out this line, and uncomment the following for your assignment:
\begin{flushright}
\NAME\\
Math 308\\
Homework \HW\\
\DUE
\end{flushright}


\begin{enumerate}[1.]
\item Using what we learned from chapters 10 \& 11, rewrite you proofs of the following problems. 
\begin{enumerate}[(a)]
\item Let $a$ and $b$ be real numbers. Show that
$$\frac{a + b}{2} \geq \sqrt{ab} \qquad \text{ for all } \quad 0 < a \leq b.$$

%\begin{proof}
%
%\end{proof}


\item Show that 
$$a^2 + b^2 + c^2 \geq ab + bc + ca$$
for all positive integers $a$, $b$, and $c$. 

%\begin{proof}
%
%\end{proof}

\end{enumerate}
\item Submit a revision of your Proof Without Words from Homework 2.

\item Use a truth table to check that for statements $A$, $B$, and $C$, we have 
$$\big((A \Rightarrow B) \land (B \Rightarrow C) \land (C \Rightarrow A)\big)$$
$$\Leftrightarrow$$
$$\big((A \Leftrightarrow B) \land (B\Leftrightarrow C) \land (A \Leftrightarrow C) \big)$$
is a tautology.
(You might want to transpose your truth table, so that the columns are the T/F options, and the rows are the statements.)


%\begin{ans}
%{\def\arraystretch{1.2}
%\[
%\begin{array}{|r||c|c|c|c|c|c|c|c|}\hline
%A 	&	T &	T &	T &	T &	F &	F &	F &	F \\\hline
%B 	&	T &	T &	F &	F &	T &	T &	F &	F \\\hline
%C 	&	T &	F &	T &	F &	T &	F &	T &	F \\\hline\hline
%A \Rightarrow B
%	&	 &	 &	 &	 &	 &	 &	 & 	 \\\hline
%B \Rightarrow C
%	&	 &	 &	 &	 &	 &	 &	 & 	 \\\hline
%C \Rightarrow A
%	&	 &	 &	 &	 &	 &	 &	 & 	 \\\hline
%(A \Rightarrow B) \land (B \Rightarrow C) \land (C \Rightarrow A)
%	&	 &	 &	 &	 &	 &	 &	 & 	 \\\hline\hline
%A \Leftrightarrow B
%	&	 &	 &	 &	 &	 &	 &	 & 	 \\\hline
%B \Leftrightarrow C
%	&	 &	 &	 &	 &	 &	 &	 & 	 \\\hline
%A \Leftrightarrow C
%	&	 &	 &	 &	 &	 &	 &	 & 	 \\\hline
%(A \Leftrightarrow B) \land (B\Leftrightarrow C) \land (A \Leftrightarrow C)
%	&	 &	 &	 &	 &	 &	 &	 & 	 \\\hline\hline
%\begin{matrix}
%	\big((A \Rightarrow B) \land (B \Rightarrow C) \land (C \Rightarrow A)\big)\\
%	\Leftrightarrow\\
%	\big((A \Leftrightarrow B) \land (B\Leftrightarrow C) \land (A \Leftrightarrow C) \big)
%	\end{matrix}
%	&\bf	 &\bf	 &\bf	 &\bf	 &\bf	 &\bf	 &\bf	 &\bf 	 \\\hline
%\end{array}
%\]}
%\end{ans}

\item  Show that the sum of two consecutive odd numbers is a multiple of 4. 

%\begin{proof}
%
%\end{proof}


\item What is wrong with the following argument? 

\medskip
\begin{center}
\begin{minipage}{5in}
\sl We have
$$2=4 \Rightarrow 2\pi =4\pi \Rightarrow \sin(2\pi) = \sin(4\pi) \Rightarrow 0=0.$$
Therefore $2 = 4$.
\end{minipage}\end{center}
\medskip


\item Consider the following:
\begin{center} \textbf{Claim. } For all $n \in \NN$, the number $n^2 + 5n + 6$ is not prime. \end{center}

\noindent Find the mistake in the following `proofs'.

\begin{enumerate}[(a)]

\item ~

\begin{center}
\begin{minipage}{5in}
\sl For $n=2$, we have $n^2+5n+6=22+10+6=20$, which is not prime. Hence
the theorem is true.
\end{minipage}\end{center}
\medskip

%\begin{ans}
%
%\end{ans}

\item ~

\begin{center}
\begin{minipage}{5in}
\sl Suppose that $n$ is a natural number, so in particular $n > 0$. If $n^2 +5n+6$ is
not prime, then $n^2 + 5n + 6 = pq$ for some natural numbers $p$ and $q$ such that $0 < p < n^2 + 5n + 6$ and $0 < q < n^2 + 5n + 6$. Since $n^2 + 5n + 6 = pq$ and $p$ and $q$ do not equal $1$ or $n^2 +5n+6$, then $n^2 +5n+6$ is not prime, which was to be shown.
\end{minipage}\end{center}
\medskip


%\begin{ans}
%
%\end{ans}

\end{enumerate}


\item Find the error in the following proof that the sum of two rational numbers is rational. 

\medskip
\begin{center}
\begin{minipage}{5in}
\sl Suppose that $m$ and $n$ are rational numbers, so that we can write $m = p/q$ and $n = r/s$ where $p, q, r$, and  $s$ are integers (and $q$ and $s$ are non-zero). Then
$$m+n=\frac{p}{q} + \frac{r}{s}.$$
Since the sum of two fractions is a fraction, then $m + n$ must be a fraction. Hence,
$m + n$ is rational.
\end{minipage}\end{center}
\medskip


%\begin{ans}
%
%\end{ans}

\item Here�s another error that people make. What is it? ??

\medskip
\begin{center}
\begin{minipage}{5in}
\sl The following proves the statement 
	$\exists x \in \RR, \forall y \in \RR \left( (y+1)x^2 = \frac{1+y}{y^4}\right)$:
	
	\medskip
	
	Suppose that $x = 1/y^2$. Then 
	$$(y+1)x^2 = (y+1)\left( \frac{1}{y^2} \right)^2 = \frac{y+1}{y^4} = \frac{1+y}{y^4}.$$
\end{minipage}\end{center}
\medskip


%\begin{ans}
%
%\end{ans}

\item Write a basic plan/outline for a proof for each of the \emph{Proof Lab: Direct Proof} problems. 

\medskip

\noindent For example, suppose you had the following problem: \\
\begin{center}
\begin{minipage}{5in}
	\sl Let $x$ and $y$ be real numbers. If $y > x > 0$, then $$\frac{x+1}{y+1} > \frac{x}{y}.$$
\end{minipage}
\end{center}
Then your basic outline would be:
\begin{center}
\begin{minipage}{5in}
\begin{sketch}
	\begin{enumerate}[$\quad\circ$]
		\item {\bf Start with:} ``Let $x$ and $y$ be real numbers with  $y > x > 0$.''
		\item {\bf Goal:} Conclude $\frac{x+1}{y+1} > \frac{x}{y}.$
		\item {\bf To try:} Start with $\frac{x+1}{y+1} > \frac{x}{y}$ and work backwards? Don't forget to reorder computations in the end!
	\end{enumerate}
\end{sketch}
\end{minipage}
\end{center}


%\begin{enumerate}[I.]
%\item Show that for positive real numbers $x$ and $y$, we have $x > y$ if and only if $x^2 > y^2$. 
%
%	\begin{sketch}
%	
%	\end{sketch}
%
%
%\item Let $x$ be an integer. Prove that $x^n$ is odd if and only if $x$ is odd. 
%
%	\begin{sketch}
%	
%	\end{sketch}
%
%
%\item Let $f: X \to Y$ be a function between two sets $X$ and $Y$. Suppose $A$ and $B$ are subsets of $X$. Show that $f(A\cup B) = f(A) \cup f(B)$.
%
%	\begin{sketch}
%	
%	\end{sketch}
%
%
%\item Let $X$ be a set, and let $A$ and $B$ be subsets of $X$. Prove $B - A = (A \cup B) - A$. 
%
%	\begin{sketch}
%	
%	\end{sketch}
%
%
%\item Suppose $x$ and $y$ are rational numbers such that $x < y$. Prove that there exists $z \in \QQ$ such that $x < z < y$. That is, between every two rationals there exists another rational.
%
%	\begin{sketch}
%	
%	\end{sketch}
%
%
%\item Prove that the following statements are equivalent for $A$ and $B$ subsets of $X$. 
%\begin{enumerate}[(i)]
%\item  $A\subseteq B$,
%\item  $A \cap B^c = \emptyset$,
%\item $A^c \cup B=X$.
%\end{enumerate}
%
%	\begin{sketch}
%	
%	\end{sketch}
%
%
%\item Prove that the following statements are equivalent.
%\begin{enumerate}[(i)]
%\item There exists unique $x$ such that $P (x)$ is true.
%\item There exists $x$ such that (a) $P(x)$ holds and (b) for all  $y$, we have $P(y) \Rightarrow y = x$.
%\item There exists $x$ so that for all $y$, $P(y)$ is true if and only if $y=x$.
%\item There exists $x$ such that $P(x)$ holds, and for all $y$ and $z$, we have $(P(y) \land P(z)) \Rightarrow y = z$.
%\end{enumerate}
%
%
%
%	\begin{sketch}
%	
%	\end{sketch}
%
%
%\item Let $x_1$, $x_2$, $x_3$, \dots $x_n$ be an arrangement of the numbers $1, 2, 3, \dots n$.  Prove that if $n$ is odd, then $(x_1 - 1) (x_2 - 2)(x_3 - 3)\cdots(x_n - n)$ is even. 
%
%
%
%
%	\begin{sketch}
%	
%	\end{sketch}
%
%
%\end{enumerate}






\end{enumerate}



\end{document}