        \problemdata       % Takes 5 *mandatory* arguments
        {false-induction}             % latex-friendly label for the prob.
        {induction}  % The topic(s) of the problem content.
        {Rosen \S 3.2, Exercise 48}               % Source (if known)
        {S02 ps2-2} % Usage (list ones you are aware of).
        {Ashish, S02}        % Last revision info (author, date).



\begin{problem}
Rosen \S 3.2, Exercise 48.

Find the flaw in the following "proof" that $a^n = 1$ for all nonnegative integers $n$, whenever $a$ is a nonzero real number.

\begin{proof}
(by induction)

\textbf{Base Case:} $a^0 =1$ is true by definition of $a^0$.

\textbf{Inductive Step:} Assume that $a^k = 1$ for all nonnegative integers $k$ with $k \leq n$. Then note that
\[
a^{n+1} = \frac{a^n \cdot a^n}{a^{n-1}} = \frac{1 \cdot 1}{1} = 1.
\]
\end{proof}
\solution{
The flaw comes in the inductive step, where we implicitly assume
$n\geq 1$ in order to talk about $a^{n-1}$ in the denominator
(otherwise the exponent is not a nonnegative integer, so we cannot
apply the inductive hypothesis).  We checked the base case only for
$n=0$, so we are not justified in assuming that $n\geq 1$ when we try
to prove the statement for $n+1$ in the inductive step.  Indeed, the
proposition first breaks precisely at $n=1$.

}
\end{problem}
