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

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

%\backwardscompatibility        % !! use for backwards compatibility
%\showsolutions                 % !! use to force solutions & skip query
%\hidesolutions                 % !! use to force no solutions & skip query
\renewcommand{\coursecopyrightnames}{Prof.~Albert~R.~Meyer}

\begin{document} 
\inclassproblems{1, Wed.}                    % !! start the paper

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\begin{problem}
Identify exactly where the bugs are in each of the following bogus
proofs.\footnote{From Stueben, Michael and Diane Sandford. \emph{Twenty
Years Before the Blackboard}, Math. Assoc America, \copyright 1998.}

\bparts

\problempart $1/8 > 1/4.$
\begin{bogusproof}
\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,
\end{align*}
and the claim now follows by the rules for multiplying fractions.
\end{bogusproof}

\solution{$\log x < 0$, for $0<x<1$, so since both sides of the inequality
``$3 > 2$'' are being multiplied by the negative quantity
$\log_{10}(1/2)$, the ``$>$'' in the second line should have been
``$<$.''}

\ppart $1 \cent = \$0.01 = (\$0.1)^2 = (10\cent)^2 = 100\cent = \$1.$

\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, $(10\cent)^2 \neq 100\cent$.
}

\eparts
\end{problem}


\instatements{\newpage}

\begin{problem}

\begin{proposition*}[Arithmetic-Geometric Mean Inequality]
For all nonnegative real numbers $a$ and $b$
\[
\frac{a + b}{2} \geq \sqrt{a b}.
\]
\end{proposition*}

What is wrong with the following proof of this proposition?

\begin{bogusproof}
\begin{align*}
\frac{a + b}{2} & \stackrel{?}{\geq} \sqrt{a b} \\
a + b  & \stackrel{?}{\geq} 2 \sqrt{a b} \\
a^2 + 2 a b + b^2  & \stackrel{?}{\geq} 4 a b \\
a^2 - 2 a b + b^2  & \stackrel{?}{\geq} 0 \\
(a - b)^2  & \geq 0
\end{align*}

The last statement is true because $a - b$ is a real number, and the
square of a real number is never negative.  This proves the claim.
\end{bogusproof}
\end{problem}

\solution{In this argument, we started with what we wanted to prove and
then reasoned until we reached a statement that is surely true.  The
little question marks presumably are supposed to indicate that we're not
quite certain that the inequalities are valid until we get down to the
last step.  At that step, the inequality checks out, \emph{but that
doesn't prove the claim}.  All we have proved is that \textbf{if} $(a + b)/2
\geq \sqrt{a b}$, \textbf{then} $(a - b)^2 \geq 0$, which is not very
interesting, since we already knew that the square of any nonnegative
number is nonnegative.

To be fair, this bogus proof is pretty good: if it was written in reverse
order -- or if ``is implied by'' was simply inserted after each line -- it
would actually prove the Arithmetic-Geometric Mean Inequality:

\begin{proof}

\begin{align*}
\frac{a + b}{2} & \geq \sqrt{a b} & \text{is implied by}\\
a + b  & \geq 2 \sqrt{a b},  & \text{which is implied by}\\
a^2 + 2 a b + b^2  & \geq 4 a b, & \text{which is implied by}\\
a^2 - 2 a b + b^2  & \geq 0, & \text{which is implied by}\\
(a - b)^2  & \geq 0.
\end{align*}

The last statement is true because $a - b$ is a real number, and the
square of a real number is never negative.  This proves the claim.
\end{proof}

But the problem with the bogus proof as written is that it reasons
backward, beginning with the proposition in question and reasoning to a
true conclusion.  This kind of backward reasoning can easily ``prove''
false statements.  Here's an example:

\begin{falseclm*}
\[
0 = 1.
\]
\end{falseclm*}

\begin{bogusproof}
\begin{align*}
0 & \stackrel{?}{=} 1 \\
1 & \stackrel{?}{=} 0 \\
0+1 & \stackrel{?}{=} 1+0 \\
1 & = 1
\end{align*}
and the last equality is trivially true.
\end{bogusproof}

We can also come up with very easy ``proofs'' of true theorems, for
example, here's an easy ``proof'' of the Arithmetic-Geometric Mean
Inequality:

\begin{bogusproof}
\begin{align*}
\frac{a + b}{2} & \stackrel{?}{\geq} \sqrt{a b} \\
0 \cdot \frac{a + b}{2} & \stackrel{?}{\geq} 0 \cdot \sqrt{a b}\\
0 & \geq 0 \qedhere
\end{align*}
\end{bogusproof}

So watch out for backward proofs!
}


\end{document}
