\documentclass[11pt,twoside]{article}	
\usepackage{latex-macros/course}
\usepackage{textcomp} % to get \textcent

\newcommand{\cent}{\text{\textcent}}

\renewcommand{\reading}{
\href{http://theory.lcs.mit.edu/classes/6.042/spring02/lectures/ln1.pdf}{Week
1 Notes}, Rosen \S\S 1.1-6, 2.3, 3.1}
\renewcommand{\coursecopyrightnames}{Prof.~Albert~R.~Meyer}

%\hidesolutions
%\showsolutions

\begin{document}
\problemset{1}

%PROOF from Spring94
%%FALSE PROOF ARITHMETIC, revised ARM 9/3/01


\begin{problem}
Identify exactly where the bugs are in each of the following false
proofs.

\begin{problemparts}

\problempart
\begin{falsethm*} If $a$ and $b$ are two equal real numbers, then
$a=0$.
\end{falsethm*}

\begin{falseproof}
\begin{eqnarray}
a&=&b \\
a^2&=&ab \\
a^2-b^2&=&ab-b^2 \\
(a-b)(a+b)&=&(a-b)b \label{cancel}\\
a+b&=&b \label{bug}\\
a&=&0.
\end{eqnarray}
\end{falseproof}

\solution{The bug is at line~(\ref{bug}): one cannot cancel $(a-b)$ from
both sides of equation line~(\ref{cancel}) because $a-b = 0$. }

\problempart
You are richer than you think:\footnote{
Stueben, Michael and Diane Sandford. \emph{Twenty Years Before the
Blackboard}, Math. Assoc America, \copyright 1998, p.27.}

\begin{falsethm*}
\[1 \cent = \$0.01 = (\$0.1)^2 = (10\cent)^2 = 100\cent = \$1.\]
\end{falsethm*}

\solution{$\$0.01 = \$(0.1)^2 \neq (\$0.1)^2$ because the units $\$^2$ and
$\$$ don't match (just as in physics the difference between $sec^2$ and
$sec$ indicates the difference between acceleration and velocity).
Similarly, $(10\cent)^2 \neq 100\cent$.
}
\end{problemparts}
\end{problem}

\begin{problem}
Rosen, Ex 1.3.12  %English to quantifiers
\end{problem}


\begin{problem}
Rosen, Ex 1.4.14  %powerset
\end{problem}

\begin{problem}
Rosen, Ex 1.6.12  %create 1-1, onto examples
\end{problem}

\begin{problem}
Rosen, Ex 2.3.32  %gcd,lcm
\end{problem}

\begin{problem}
Rosen, Ex 2.3.34  % minus (mod m)
\end{problem}

\begin{problem}
Rosen, Ex 3.1.10  %rules of inference
\end{problem}

\iffalse

\begin{problem}
A self-proclaimed ``great logician'' has invented a new quantifier, on par
with $\exists$ (``there exists'') and $\forall$ (``for all'').  The new
quantifier is symbolized by $U$ and read ``there exists a unique''.
The proposition $Ux\, P(x)$ is true iff there is \emph{exactly one}
$x$ in the domain of discourse for which $P(x)$ is true.  The logician has
noted, ``There used to be two quantifiers, but now there are three!  I
have extended the whole field of mathematics by 50\%!''

\begin{problemparts}

\problempart Rewrite the proposition $U x\, P(x)$ using only the
$\exists$ quantifier.

\solution{
\[
\exists x (P(x) \wedge \neg (\exists y (x \neq y \wedge P(y)))
\]
}

\problempart Rewrite the proposition $U x\, P(x)$ using only the
$\forall$ quantifier.

\solution{
\[
\neg \forall x (\neg P(x) \vee \neg \forall y (x = y \vee \neg P(y))
\]
}
\end{problemparts}
\end{problem} 
\fi


\end{document}                            
