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\handout{lec2-sup}{Supplement to Lecture 2}{September 29, 1998}

\section{An Induction Proof Format}

An induction argument has a regular, but somewhat intricate structure.
You can use the following format to ensure that you miss no steps;
this format is not a requirement, just something to retreat to if you
are a bit confused.

\bigskip

First, define the induction hypothesis $P(n)$ carefully.  Then proceed
as shown below in the style where an assertion is made first, and then
the justification follows in a sequence of indented steps.

\bigskip

$\forall n \geq 0 \ P(n)$.

\begin{enumerate}
\item (base case) $P(0)$.

	(This is is usually easy, but not always.)

\item (inductive step) $\forall n \geq 0 \ (P(n) \rightarrow P(n+1))$
\begin{enumerate}
  \item Fix $n \geq 0$.
  \item $P(n) \rightarrow P(n+1)$.
  \begin{enumerate}
    \item Assume $P(n)$.
    \item $P(n+1)$.

	(This is usually the hard part of an induction proof.)

    \item Q. E. D. (by implication rule)
  \end{enumerate}
  \item Q. E D. (by definition of $\forall$)
  \end{enumerate}
\item Q. E. D. (by induction)
\end{enumerate}

\bigskip

\newpage

{\bf Example.}  We cast a standard induction proof into the above
format as a model.

\begin{theorem}
For all $n \geq 0$, $1 + 2 + \ldots + n = \frac{n(n+1)}{2}$.
\end{theorem}

\begin{proof}
Let $P(n)$ be the proposition that $1 + 2 + \ldots + n =
\frac{n(n+1)}{2}$.

\bigskip

$\forall n \geq 0 \ P(n)$.
\begin{enumerate}
\item (base case) $P(0)$ holds because $1 = \frac{1(1+1)}{2}$.
\item (inductive step) $\forall n \geq 0 \ (P(n) \rightarrow P(n+1))$
\begin{enumerate}
  \item Fix $n \geq 0$.
  \item $P(n) \rightarrow P(n+1)$.
  \begin{enumerate}
    \item Assume $1 + 2 + \ldots + n = \frac{n(n+1)}{2}$.
    \item $1 + 2 + \ldots + n + (n+1) = \frac{(n+1)(n+2)}{2}$
    \begin{enumerate}
      \item $1 + 2 + \ldots + n = \frac{n(n+1)}{2}$ (by assumption)
      \item $1 + 2 + \ldots + n + (n+1) = \frac{n(n+1)}{2} + (n+1)$ (by adding $n+1$ to each side)
      \item $1 + 2 + \ldots + n + (n+1) = \frac{(n+1)(n+2)}{2}$ (by simplifying)
    \end{enumerate}
    \item Q. E. D. (by implication rule)
  \end{enumerate}
  \item Q. E D. (by definition of $\forall$)
  \end{enumerate}
\item Q. E. D. (by induction)
\end{enumerate}

\end{proof}

\section{Double Induction}

Suppose that the function $f(n, k)$ is define for all non-negative
integers $n$ and $k$ as follows:

\begin{itemize}
\item For all $k \geq 0$, $f(0, k) = 1$.
\item For all $n \geq 0$, $f(n, 0) = 1$.
\item For all $n, k \geq 0$, $f(n + 1, k + 1) = f(n, k + 1) + f(n + 1, k)$.
\end{itemize}

This description is sufficient to calculate $f(n, k)$ for specific
values of $n$ and $k$ as shown in the table below.

\[
\begin{array}{c|ccccc}
	& k = 0	& 1	& 2	& 3	& 4	\\ \hline
n = 0	& 1	& 1	& 1	& 1	& 1	\\
1	& 1	& 2	& 3	& 4	& 5	\\
2	& 1	& 3	& 6	& 10	& 15	\\
3	& 1	& 4	& 10	& 20	& 35
\end{array}
\]

The following theorem gives the value of $f(n, k)$ directly.

\begin{theorem}
For all $n, k \geq 0$,

\begin{eqnarray*}
f(n, k) & = & \frac{(n+k)!}{n!\ k!}.
\end{eqnarray*}
\end{theorem}

For example, the theorem states the $f(3, 2) = \frac{5!}{3! \ 2!} = 10$,
which matches the value computed in the table above.

\medskip

There is a difficulty with trying to prove this theorem by induction;
namely, there are two variables involved, $n$ and $k$.  One solution
is to use a {\em double induction} argument.  The overall structure of
the argument is an induction on $n$, but to complete the induction
step, we use another whole induction argument on $k$.

\bigskip

\begin{proof}
The proof is by double induction.  Let $P(n)$ be the proposition that
for all $k \geq 0$, $f(n, k) = \frac{(n+k)!}{n!\ k!}$.  In the base
case, $P(0)$ is true because, by part one of the definition of $f$,
$f(0, k) = 1 = \frac{(0+k)!}{0!\ k!}$.  In the inductive step, for $n
\geq 0$ assume that for all $k \geq 0$, $f(n, k) = \frac{(n+k)!}{n!\
k!}$ in order to prove that that for all $k \geq 0$, $f(n+1, k) =
\frac{(n+1+k)!}{(n+1)!\ k!}$.

We complete the inductive step by using another whole induction
argument, this time on $k$.  Let $Q(k)$ be the proposition that
$f(n+1, k) =
\frac{(n+1+k)!}{(n+1)!\ k!}$.  In the base case, $Q(0)$ is true
because, by part two of the definition of $f$, $f(n + 1, 0) = 1 =
\frac{(n+1+0)!}{(n+1)!\ 0!}$.

In the inductive step, for $k \geq 0$ assume that $f(n+1, k) =
\frac{(n+1+k)!}{(n+1)!\ k!}$ in order to prove that $f(n+1, k+1) =
\frac{(n+1+k+1)!}{(n+1)!\ (k+1)!}$.  We reason as follows:

\begin{eqnarray*}
f(n + 1, k + 1)
	& = &	f(n, k + 1) + f(n + 1, k) \\
	& = &	\frac{(n + k + 1)!}{n!\ (k+1)!} +
		\frac{(n + 1 + k)!}{(n+1)!\ k!} \\
	& = &	\frac{(n+1)(n + k + 1)!}{(n+1)!\ (k+1)!} +
		\frac{(k+1)(n + 1 + k)!}{(n+1)!\ (k+1)!} \\
	& = &	\frac{(n+1+k+1)!}{(n+1)!\ (k+1)!}
\end{eqnarray*}

The first step uses the third part of the definition of $f$.  The
second step uses the induction hypotheses from both the induction on
$n$ and the induction on $k$.  The remaining steps are
simplifications.
\end{proof}

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