\documentclass[11pt,twoside]{article}
\input{macros-course}

\newtheorem{falsethm}[theorem]{False Theorem}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{method}{Method}

\def\cent{{\not\!\mathrm{c}}}

\usepackage{amsmath}

\begin{document}
\handout{false-proofs}{False Proofs}{February 2, 1999}

\begin{enumerate}

\item 
\begin{eqnarray*}
1 & = & \sqrt{1} \\
& = & \sqrt{(-1)(-1)} \\
& = & \sqrt{-1} \sqrt{-1} \\
& = & \sqrt{-1}^2 \\
& = & -1.\\
\mbox{So, }1 & = & -1! \qquad \mbox{(From 6.042, Fall 97 Lecture 1 notes.)}
\end{eqnarray*}


\iffalse
\emph{Explanation}: The step $\sqrt{(-1)(-1)} = \sqrt{-1} \sqrt{-1}$ is invalid.
Generally, $\sqrt{u v} = \sqrt{u} \sqrt{v}$ does not hold over complex
numbers.
\fi

\item\label{log10}
\begin{align*}
    3 &> 2 \\
    3 \log_{10} (1/2) &> 2 \log_{10}(1/2) \\
    \log_{10} (1/2)^3 &> \log_{10} (1/2)^2 \\
    (1/2)^3 &> (1/2)^2 \\
    \mbox{Therefore, } 1/8 &> 1/4!
\end{align*}

\item\label{cents}
You are richer than you think:
\[1 \cent = \$0.01 = (\$0.1)^2 = (10\cent)^2 = \$1!\]


\iffalse

\item
$(x + y)(x^2 - y^2) = (x + y)^2 (x - y)$\\

\fi

\item Prof. Meyer announces that he plans to give a surprise 6.042 quiz
next week.  Can he give the exam Friday of next week?  Obviously not,
because if he hadn't given it before Friday, everyone would know that he
had only Friday left on which to give it, so it wouldn't be a surprise any
more.

OK, could he give it Thursday?  Well, if he hadn't given it before
Thursday, and we already know he can't give it Friday, he would have to
give it on the Thursday.  But now that we've figured that out, it wouldn't
be a surprise if he gave it on Thursday either.  Similarly, he can't give
it Wednesday, Tuesday, or Monday.  Namely, it's impossible for him to give
a surprise quiz next week, and all the students can relax knowing that
Prof. Meyer must have been bluffing.

And since no one expects the quiz, that's why, when Prof. Meyer gives it
on Tuesday next week, it really is a surprise!

\medskip
Mathematicians and philosophers have had a lot to say about what might be
wrong with the students' reasoning, see Chow, Timothy Y. \emph{The
surprise examination or unexpected hanging paradox}, American
Math. Monthly (January 1998), pp.41--51.

\item We know that every nonempty set of nonnegative integers has a least
member --- this is a special case of a mathematical result called the
\emph{Well-Ordering Principle}.

So there must be a least nonnegative integer, $m$, which cannot be defined
in fewer than fourteen words.  But ``The least nonnegative integer which
cannot be defined in fewer than fourteen words,'' is a thirteen word
definition of $m$.  Is there something wrong with the Well-Ordering
Principle?

\item Simpler variation of the above: ``This statement is false.''

\item More serious version of the above is \emph{Russell's paradox}:

Let $S$ be a variable ranging over all sets, and define
\[W ::= \set{S \mid S \not\in S}.\]
So by definition,
\[S \in W  \mbox{  iff  } S \not\in S,\]
for every set $S$.  In particular, we can let $S$ be $W$, and obtain
the contradictory result that
\[W \in W  \mbox{  iff  } W \not\in W!\]

\item\label{millions} Consider the following scheme to make millions.
Obtain a slab of gold $10'' \times 11'' \times 1''$.  Score it diagonally
as shown in the Figure attached.  Then cut out two little one-inch
triangles at the corners.  These triangles are the profit.  Finally, slide
the two slabs together again and note that the rectangle is now $11''
\times 10'' \times 1''$---the same size as it was before we removed the
small triangles.  By repeating the process, we can produce an unlimited
supply of gold!

(This illustrates why ``proof by diagram'' is mathematically suspicious.)

\end{enumerate}

Examples~\ref{log10},~\ref{cents}, ~\ref{millions} are from Stueben
Michael and Diane Sandford. \emph{Twenty Years Before the Blackboard},
Math. Assoc America, \copyright 1998, p.27.

\end{document}
