\problemdata       % Takes 5 *mandatory* arguments
  {sqrt-2    }             % latex-friendly label for the prob.
  {least number principle} % The topic(s) of the problem content.
  {}                       % Source (if known)
  {S99 tut; S02 cp2F-3}    % Usage (list ones you are aware of).
  {Nick Matsakis, S02}     % Last revision info (author, date).
%%Example from Spring99, tutorial 2, Least Number Principle
\begin{problem}

Here's a different proof that the square root of 2 is irrational, this one
based on the Least Number Principle.

\begin{proof}
The proof is by contradiction.

Assume that the square root of 2 is rational.  Then $\sqrt{2}$ is equal to
$m / n$ for some positive integers $m$ and $n$.  So $n$ has the property
that
\begin{equation}\label{ns2}
n \text{ and } n\sqrt{2} \text{ are positive integers}.
\end{equation}
Now by the Least Number Principle, there must be a \emph{smallest}
positive integer, $n$, satisfying~(\ref{ns2}).  For this $n$, let $n_0
\eqdef n (\sqrt{2} - 1)$.  But then $n_0$ is smaller than $n$, and $n_0$
and $n_0\sqrt{2}$ are positive integers, a contradiction.
\end{proof}

\begin{problemparts}

\problempart Are you convinced by this proof?  Briefly explain.

In particular, why is $n_0$ an integer?

\solution{Because it equals $n\sqrt{2} - n$
and $n$ and  $n\sqrt{2}$ are integers by hypothesis.}

Why is $n_0$ positive?

\solution{Because $(\sqrt{2} -1) \approx 0.414\dots$ is positive.}

Why is $n_0$ smaller than $n$?

\solution{Because $(\sqrt{2}-1) < 0.415 < 1$.}

Why is $n_0\sqrt{2}$ a positive integer?

\solution{It's an integer because $n_0\sqrt{2} = (n\sqrt{2} - n)\sqrt{2}
= 2n - n\sqrt{2}$, which is again an integer because both terms are
integers.  It's positive because $2n - n\sqrt{2} = (2 - \sqrt{2})n > 0.58
n > 0$.}

\end{problemparts}

We can generalize this proof to prove that $\sqrt{k}$ is irrational for
integers $k$ other than 2.  Simply revise the next to last sentence to
read, ``\dots and let $n_0 \eqdef (\sqrt{k} - \floor{\sqrt{k}})$''.  (For
any real number, $r$, the expression $\floor{r}$ denotes $r$ rounded
\emph{down} to the nearest integer.)

\begin{problemparts}
\problempart Are you convinced by this argument?  Explain.

\solution{You shouldn't be convinced because the conclusion is wrong,
For example, $\sqrt{k}$ is certainly rational if $k=4$.

The proof works for $\sqrt{k}$ as long as $k$ is not a perfect square.
But if $k$ is a perfect square, then the proof breaks down in the next to
last sentence: the number $n_0 (\sqrt{k} - \lfloor \sqrt{k} \rfloor)$ is
not positive in this case, but zero.}
\end{problemparts}

\end{problem}
