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\documentclass[12pt,twoside]{article}
\usepackage{psfig}

\input{macros-course}

\newenvironment{claim}{\ \vspace{0.2cm} \\ {\bf Claim: \ }}{}

\begin{document}
\handout{proofs_gen}{Proofs:  Valid or Invalid?}{July 28, 2003}

\noindent Determine whether the following proofs are valid.  If a
proof is wrong, pinpoint the precise error.  If it is correct, justify
each step.

\begin{problems}

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\begin{problem}

% Proof by intimidation.

\begin{claim}
The number $\lg 3 = \log_2 3$ is irrational.
\end{claim}

\begin{proof}
A number is irrational if it is not the ratio of two integers.
Therefore, since $\lg 3$ cannot be written in the form $a/b$, where
$a$ and $b$ are integers, it is irrational.
\end{proof}

\end{problem}

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\begin{problem}

% Proof by contraposition.

\begin{claim}
If $n^2 \leq 9$ then, $n \leq 3$.
\end{claim}

\begin{proof}
If $n > 3$, then $n^2 > 9$.
\end{proof}

\end{problem}


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\begin{problem}

\begin{claim}
If $a$ and $b$ are two equal real numbers, then $2 = 1$.
\end{claim}

\begin{proof}
\begin{eqnarray*}
a & = & b \\
a^2 & = & a b \\
a^2 - b^2 & = & a b - b^2 \\
(a - b) (a + b) & = & (a - b) b \\
a + b & = & b \\
2b & = & b \\
2 & = & 1
\end{eqnarray*}
\end{proof}

\end{problem}

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\begin{problem}

% A cute, non-constructive proof.

\begin{claim}
There exist irrational numbers $a$ and $b$ such that $a^b$ is a
rational number.
\end{claim}

\begin{proof}
Note that $\sqrt{2}$ is irrational.  Now, consider the value
$\sqrt{2}^{\sqrt{2}}$.  If this quantity is rational, then the claim
holds for $a = \sqrt{2}$ and $b = \sqrt{2}$.  On the other hand, if
the quantity $\sqrt{2}^{\sqrt{2}}$ is not rational, then the claim
holds for $a = \sqrt{2}^{\sqrt{2}}$ and $b = \sqrt{2}$.
\end{proof}

\end{problem}

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\end{problems}

\end{document}
