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\begin{document}
\section*{Math 221 Winter 2023, Lecture 4: Induction}

\textbf{website:}
\texttt{\href{https://www.cip.ifi.lmu.de/~grinberg/t/23wd}{\texttt{https://www.cip.ifi.lmu.de/\symbol{126}%
grinberg/t/23wd}}}

\section{Induction and recursion (cont'd)}

\setcounter{subsection}{7}

\subsection{More on the Fibonacci numbers (cont'd)}

Recall the Fibonacci sequence, which we defined in Lecture 2:

\begin{definition}
The \textbf{Fibonacci sequence} is the sequence $\left(  f_{0},f_{1}%
,f_{2},\ldots\right)  $ of nonnegative integers defined recursively by setting%
\begin{align*}
f_{0}  &  =0,\ \ \ \ \ \ \ \ \ \ f_{1}=1,\ \ \ \ \ \ \ \ \ \ \text{and}\\
f_{n}  &  =f_{n-1}+f_{n-2}\ \ \ \ \ \ \ \ \ \ \text{for each }n\geq2.
\end{align*}

\end{definition}

The entries of the Fibonacci sequence are called the \textbf{Fibonacci
numbers}. Here are the first few:%
\[%
\begin{tabular}
[c]{|c||c|c|c|c|c|c|c|c|c|c|c|c|c|c|}\hline
$n$ & $0$ & $1$ & $2$ & $3$ & $4$ & $5$ & $6$ & $7$ & $8$ & $9$ & $10$ & $11$
& $12$ & $13$\\\hline
$f_{n}$ & $0$ & $1$ & $1$ & $2$ & $3$ & $5$ & $8$ & $13$ & $21$ & $34$ & $55$
& $89$ & $144$ & $233$\\\hline
\end{tabular}
\ \ \ \ .
\]


\bigskip

\setcounter{theo}{5}

Is there an explicit formula for $f_{n}$, that is, a formula that does not
rely on the previous entries of the Fibonacci sequence?

Yes, there is one; it is known as \textbf{Binet's formula}:

\begin{theorem}
[Binet's formula]\label{thm.fib.binet}Let%
\[
\varphi=\dfrac{1+\sqrt{5}}{2}\approx1.618\ldots\ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ \psi=\dfrac{1-\sqrt{5}}{2}\approx-0.618\ldots.
\]
Then,%
\[
f_{n}=\dfrac{\varphi^{n}-\psi^{n}}{\sqrt{5}}\ \ \ \ \ \ \ \ \ \ \text{for
every integer }n\geq0.
\]

\end{theorem}

Some remarks:

\begin{itemize}
\item The number $\varphi$ is called the
\textbf{\href{https://en.wikipedia.org/wiki/Golden_ratio}{\textbf{golden
ratio}}}, and is famous for many properties, including the fact that
$\varphi^{2}=\varphi+1$ (which you can easily check by expanding both
sides\footnote{Namely: From $\varphi=\dfrac{1+\sqrt{5}}{2}$, we obtain
\[
\varphi^{2}=\left(  \dfrac{1+\sqrt{5}}{2}\right)  ^{2}=\dfrac{1+2\sqrt{5}%
+5}{4}=\dfrac{6+2\sqrt{5}}{4}=\dfrac{3+\sqrt{5}}{2}=1+\dfrac{1+\sqrt{5}}%
{2}=1+\varphi.
\]
}). The number $\psi$ is its so-called conjugate and also satisfies $\psi
^{2}=\psi+1$.

\item The numbers $f_{n}$ are integers, but Binet's formula expresses them in
terms of two irrational numbers $\varphi$ and $\psi$. This should be rather unexpected.

\item As $n$ grows large, $\psi^{n}$ approaches $0$ (since $-1<\psi<1$),
whereas $\varphi^{n}$ grows exponentially (since $\varphi>1$). So $f_{n}$ also
grows exponentially (according to Binet's formula), with growth rate
$\varphi\approx1.618\ldots$.
\end{itemize}

Questions:

\begin{enumerate}
\item How do we prove Binet's formula?

\item How could we find Binet's formula if we didn't already know it?
\end{enumerate}

We will answer Question 1 today. Question 2 is significantly trickier and will
not be answered in this course\footnote{Answers at different levels of
generality can be found in:
\par
\begin{itemize}
\item Subsection 4.9.2 of
\href{http://www.cip.ifi.lmu.de/~grinberg/t/20f/mps.pdf}{my notes \textit{Math
235: Mathematical Problem Solving}} (which solves any linear recurrence of the
form $x_{n}=ax_{n-1}+bx_{n-2}$ for constant numbers $a$ and $b$ in an explicit
and elementary way);
\par
\item
\href{http://matematicas.uam.es/~mavi.melian/CURSO_15_16/web_Discreta/recurrence.pdf}{Mar\'{\i}a
Victoria Meli\'{a}n, \textit{Linear recurrence relations with constant
coefficients}} and
\href{https://nikolaivivanov.files.wordpress.com/2014/02/ivanov2008arecurrence.pdf}{Nikolai
V. Ivanov, \textit{Linear Recurrences}} (which solve the more general version
$x_{n}=a_{1}x_{n-1}+a_{2}x_{n-2}+\cdots+a_{k}x_{n-k}$ in terms of the
eigenvalues of a matrix).
\end{itemize}
\par
Textbooks on combinatorics or advanced linear algebra also tend to discuss
such sequences (called \textbf{linearly recurrent sequences}).}.

\bigskip

Let us try to prove Binet's formula by induction on $n$:

\begin{proof}
[Attempted proof of Binet's formula.]We induct on $n$:

\textit{Base case:} For $n=0$, we have $f_{n}=f_{0}=0$ and
\[
\dfrac{\varphi^{n}-\psi^{n}}{\sqrt{5}}=\dfrac{\varphi^{0}-\psi^{0}}{\sqrt{5}%
}=\dfrac{1-1}{\sqrt{5}}=0.
\]
Thus, Binet's formula holds for $n=0$.

\textit{Induction step:} Let $n\geq0$ be an integer.

Assume (as induction hypothesis) that Binet's formula holds for $n$; we must
prove that it holds for $n+1$.

So we must prove that%
\[
f_{n+1}=\dfrac{\varphi^{n+1}-\psi^{n+1}}{\sqrt{5}}.
\]
The recursive definition of the Fibonacci sequence yields%
\[
f_{n+1}=f_{n}+f_{n-1}=\dfrac{\varphi^{n}-\psi^{n}}{\sqrt{5}}+f_{n-1}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the induction hypothesis}\right)  .
\]
So far so good, but how can we simplify $f_{n-1}$ ? Our induction hypothesis
only tells us that $f_{n}=\dfrac{\varphi^{n}-\psi^{n}}{\sqrt{5}}$, but it says
nothing about $f_{n-1}$.
\end{proof}

So this induction proof does not work.\footnote{There is also one more little
(fixable) gap in the above attempted proof. Do you see it?}

Let us see how to fix this by introducing a more advanced version of induction.

\subsection{Strong induction}

\subsubsection{Reminder on regular induction}

Recall the (original) principle of mathematical induction:

\begin{theorem}
[Principle of Mathematical Induction]Let $b$ be an integer.

Let $P\left(  n\right)  $ be a mathematical statement defined for each integer
$n\geq b$.

Assume the following:

\begin{enumerate}
\item \textquotedblleft\textbf{Base case}\textquotedblright: The statement
$P\left(  b\right)  $ holds.

\item \textquotedblleft\textbf{Induction step}\textquotedblright: For each
integer $n\geq b$, the implication $P\left(  n\right)  \Longrightarrow
P\left(  n+1\right)  $ holds.
\end{enumerate}

Then, the statement $P\left(  n\right)  $ holds for every integer $n\geq b$.
\end{theorem}

We can restate this principle slightly by renaming the $n$ in the induction
step as $n-1$ (so that the implication $P\left(  n\right)  \Longrightarrow
P\left(  n+1\right)  $ turns into $P\left(  n-1\right)  \Longrightarrow
P\left(  n\right)  $). Thus, it takes the following form:

\begin{theorem}
[Principle of Mathematical Induction, restated]Let $b$ be an integer.

Let $P\left(  n\right)  $ be a mathematical statement defined for each integer
$n\geq b$.

Assume the following:

\begin{enumerate}
\item \textquotedblleft\textbf{Base case}\textquotedblright: The statement
$P\left(  b\right)  $ holds.

\item \textquotedblleft\textbf{Induction step}\textquotedblright: For each
integer $n>b$, the implication $P\left(  n-1\right)  \Longrightarrow P\left(
n\right)  $ holds.
\end{enumerate}

Then, the statement $P\left(  n\right)  $ holds for every integer $n\geq b$.
\end{theorem}

The idea behind the principle (in either form) is that the base case gives us
$P\left(  b\right)  $ whereas the induction step gives us the implications%
\begin{align*}
P\left(  b\right)   &  \Longrightarrow P\left(  b+1\right)  ,\\
P\left(  b+1\right)   &  \Longrightarrow P\left(  b+2\right)  ,\\
P\left(  b+2\right)   &  \Longrightarrow P\left(  b+3\right)  ,\\
&  \ldots.
\end{align*}
In the domino metaphor, the base case tips over the first domino, and the
induction step ensures that each domino falls from the impact of the previous
domino's falling.

\subsubsection{Strong induction}

Now, assume that the $b+2$-domino (i.e., $P\left(  b+2\right)  $) falls not
from the impact of the previous domino $P\left(  b+1\right)  $, but rather
from the combined force of the dominos $P\left(  b\right)  $ and $P\left(
b+1\right)  $. This would still suffice, because the latter two dominos have
already fallen. In other words, instead of the implication $P\left(
b+1\right)  \Longrightarrow P\left(  b+2\right)  $, we could just as well
prove the implication%
\[
\left(  P\left(  b\right)  \text{ AND }P\left(  b+1\right)  \right)
\Longrightarrow P\left(  b+2\right)  ,
\]
which is somewhat weaker (since it assumes more to get to the same conclusion)
but nevertheless gives the same result. Likewise, we could just as well
replace the implication $P\left(  b+2\right)  \Longrightarrow P\left(
b+3\right)  $ by the weaker implication%
\[
\left(  P\left(  b\right)  \text{ AND }P\left(  b+1\right)  \text{ AND
}P\left(  b+2\right)  \right)  \Longrightarrow P\left(  b+3\right)  .
\]
More generally, for each $n>b$, instead of proving the implication $P\left(
n-1\right)  \Longrightarrow P\left(  n\right)  $, it will suffice to prove the
weaker implication%
\[
\underbrace{\left(  P\left(  b\right)  \text{ AND }P\left(  b+1\right)  \text{
AND }P\left(  b+2\right)  \text{ AND }\cdots\text{ AND }P\left(  n-1\right)
\right)  }_{\text{i.e., the statement }P\left(  k\right)  \text{ holds for
each }k\in\left\{  b,b+1,\ldots,n-1\right\}  }\Longrightarrow P\left(
n\right)
\]
(so that the domino $P\left(  n\right)  $ is tipped over by the combined force
of all the preceding dominos, not just the one domino directly to its left).

This induction principle is called \textbf{strong induction}. Explicitly, it
says the following:

\begin{theorem}
[Principle of Strong Induction]Let $b$ be an integer.

Let $P\left(  n\right)  $ be a mathematical statement defined for each integer
$n\geq b$.

Assume the following:

\begin{enumerate}
\item \textquotedblleft\textbf{Base case}\textquotedblright: The statement
$P\left(  b\right)  $ holds.

\item \textquotedblleft\textbf{Induction step}\textquotedblright: For each
integer $n>b$, the implication
\[
\left(  P\left(  b\right)  \text{ AND }P\left(  b+1\right)  \text{ AND
}P\left(  b+2\right)  \text{ AND }\cdots\text{ AND }P\left(  n-1\right)
\right)  \Longrightarrow P\left(  n\right)
\]
holds.
\end{enumerate}

Then, the statement $P\left(  n\right)  $ holds for every integer $n\geq b$.
\end{theorem}

Proofs using this principle are called \textbf{proofs by strong induction} (or
\textbf{strong induction proofs}). They differ from proofs by (regular)
induction as follows: In the induction step of a strong induction proof, you
can use not just the preceding statement $P\left(  n-1\right)  $, but also all
the statements before it ($P\left(  n-2\right)  $ and $P\left(  n-3\right)  $
and so on, all the way down to $P\left(  b\right)  $). In other words, the
induction hypothesis is now stronger (thus the name \textquotedblleft strong
induction\textquotedblright). Roughly speaking, strong induction is
\textquotedblleft induction with a long memory\textquotedblright\ (as opposed
to regular induction, whose memory only is $1$ step long).

(We will later see a slightly nicer form of strong induction, in which the
base case is incorporated in the induction step.)

\bigskip

Before we see an example of a strong induction proof, let me explain why it
works. Let's say you have proved a statement $P\left(  n\right)  $ for all
$n\geq0$ by strong induction. Thus,

\begin{itemize}
\item you have proved $P\left(  0\right)  $ (this is the base case);

\item you have proved the implication $P\left(  0\right)  \Longrightarrow
P\left(  1\right)  $ (this is the induction step for $n=1$), so you conclude
that $P\left(  1\right)  $ holds (since $P\left(  0\right)  $ holds);

\item you have proved the implication $\left(  P\left(  0\right)  \text{ AND
}P\left(  1\right)  \right)  \Longrightarrow P\left(  2\right)  $ (this is the
induction step for $n=2$), so you conclude that $P\left(  2\right)  $ holds
(since $P\left(  0\right)  $ and $P\left(  1\right)  $ hold);

\item you have proved the implication $\left(  P\left(  0\right)  \text{ AND
}P\left(  1\right)  \text{ AND }P\left(  2\right)  \right)  \Longrightarrow
P\left(  3\right)  $ (this is the induction step for $n=3$), so you can
conclude that $P\left(  3\right)  $ holds (since $P\left(  0\right)  $ and
$P\left(  1\right)  $ and $P\left(  2\right)  $ hold);

\item and so on.
\end{itemize}

\subsubsection{Example: Proof of Binet's formula}

Let us now prove Binet's formula by strong induction:

\begin{proof}
[Proof of Theorem \ref{thm.fib.binet} (i.e., of Binet's formula).]We strongly
induct on $n$ (i.e., we use strong induction on $n$):

\textit{Base case:} As above, we check that the formula holds for $n=0$.

\textit{Induction step:} Let $n>0$ be an integer.

We assume that Binet's formula holds for $0$, for $1$, for $2$, and so on, all
the way up to $n-1$. (In other words, we assume that $f_{k}=\dfrac{\varphi
^{k}-\psi^{k}}{\sqrt{5}}$ for each $k\in\left\{  0,1,\ldots,n-1\right\}  $.)

We have to prove that Binet's formula also holds for $n$. In other words, we
have to prove that $f_{n}=\dfrac{\varphi^{n}-\psi^{n}}{\sqrt{5}}$.

We assumed that Binet's formula holds for $n-1$. That is, we have
$f_{n-1}=\dfrac{\varphi^{n-1}-\psi^{n-1}}{\sqrt{5}}$.

We assumed that Binet's formula holds for $n-2$. That is, we have
$f_{n-2}=\dfrac{\varphi^{n-2}-\psi^{n-2}}{\sqrt{5}}$.

As we have seen above, we have $\varphi^{2}=\varphi+1$ and $\psi^{2}=\psi+1$.

But the recursive definition of the Fibonacci sequence yields%
\begin{align*}
f_{n}  &  =f_{n-1}+f_{n-2}=\dfrac{\varphi^{n-1}-\psi^{n-1}}{\sqrt{5}}%
+\dfrac{\varphi^{n-2}-\psi^{n-2}}{\sqrt{5}}\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \text{since }f_{n-1}%
=\dfrac{\varphi^{n-1}-\psi^{n-1}}{\sqrt{5}}\text{ and }f_{n-2}=\dfrac
{\varphi^{n-2}-\psi^{n-2}}{\sqrt{5}}\right) \\
&  =\dfrac{1}{\sqrt{5}}\left(  \varphi^{n-1}-\psi^{n-1}+\varphi^{n-2}%
-\psi^{n-2}\right) \\
&  =\dfrac{1}{\sqrt{5}}\left(  \underbrace{\varphi^{n-1}+\varphi^{n-2}%
}_{=\varphi^{n-2}\left(  \varphi+1\right)  }-\underbrace{\left(  \psi
^{n-1}+\psi^{n-2}\right)  }_{=\psi^{n-2}\left(  \psi+1\right)  }\right) \\
&  =\dfrac{1}{\sqrt{5}}\left(  \varphi^{n-2}\underbrace{\left(  \varphi
+1\right)  }_{=\varphi^{2}}-\,\psi^{n-2}\underbrace{\left(  \psi+1\right)
}_{=\psi^{2}}\right) \\
&  =\dfrac{1}{\sqrt{5}}\left(  \underbrace{\varphi^{n-2}\varphi^{2}}%
_{=\varphi^{n}}-\underbrace{\psi^{n-2}\psi^{2}}_{=\psi^{n}}\right)  =\dfrac
{1}{\sqrt{5}}\left(  \varphi^{n}-\psi^{n}\right)  =\dfrac{\varphi^{n}-\psi
^{n}}{\sqrt{5}}.
\end{align*}
So we have proved Binet's formula for $n$. Right?

$\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots$

Wait a moment! We have assumed (as the induction hypothesis) that Binet's
formula holds for each of the numbers $0,1,\ldots,n-1$. But then we have used
it for $n-2$ and for $n-1$. This tacitly relied on the fact that $n-2$ and
$n-1$ are among the numbers $0,1,\ldots,n-1$. However, this fact is only true
if $n\geq2$. If $n=1$, then $n-2$ is not among the numbers $0,1,\ldots,n-1$
(because it is negative).

So our induction step worked for $n=2,3,4,\ldots$ but not for $n=1$. What can
we do?

We can fix this by just proving the claim for $n=1$ by hand. So we must prove
that $f_{1}=\dfrac{\varphi^{1}-\psi^{1}}{\sqrt{5}}$. This can be checked by a
direct computation:%
\[
\dfrac{\varphi^{1}-\psi^{1}}{\sqrt{5}}=\dfrac{\varphi-\psi}{\sqrt{5}}%
=\dfrac{\dfrac{1+\sqrt{5}}{2}-\dfrac{1-\sqrt{5}}{2}}{\sqrt{5}}=\dfrac{\sqrt
{5}}{\sqrt{5}}=1=f_{1}.
\]


Now our induction step is really complete, and Binet's formula is proved.
\end{proof}

\bigskip

Let us summarize: We have used strong induction in our above proof of Theorem
\ref{thm.fib.binet}, because the \textquotedblleft extra
memory\textquotedblright\ in a strong induction step allowed us to express not
just $f_{n-1}$ but also $f_{n-2}$ via the induction hypothesis.

Note that we have had to handle the two cases $n=0$ and $n=1$ by hand in our
above proof, because we had to reach \textquotedblleft$2$ steps
back\textquotedblright\ in memory in the induction step (i.e., we had to apply
the induction hypothesis both to $n-1$ and to $n-2$).\ \ \ \ \footnote{Had we
reached further back, we would have needed extra cases (e.g., if we had
applied the induction hypothesis to $n-5$, then we would have to handle all
the cases $n=0,1,2,3,4$ by hand).} The case $n=0$ was our base case, whereas
the case $n=1$ was part of the induction step, but nevertheless had to be
singled out for special treatment (since $n-2$ is negative for $n=1$).
Nevertheless, it makes sense to think of the $n=1$ case as a \textquotedblleft
second base case\textquotedblright, even if it is de-jure part of the
induction step.

\subsubsection{Baseless strong induction}

You can actually reformulate the principle of strong induction in a form that
does not have a de-jure base case at all:

\begin{theorem}
[Principle of Strong Induction, restated]Let $b$ be an integer.

Let $P\left(  n\right)  $ be a mathematical statement defined for each integer
$n\geq b$.

Assume the following:

\begin{itemize}
\item \textquotedblleft\textbf{Induction step\textquotedblright:}\ For each
integer $n\geq b$, the implication
\[
\left(  P\left(  b\right)  \text{ AND }P\left(  b+1\right)  \text{ AND
}P\left(  b+2\right)  \text{ AND }\cdots\text{ AND }P\left(  n-1\right)
\right)  \Longrightarrow P\left(  n\right)
\]
holds.
\end{itemize}

Then, the statement $P\left(  n\right)  $ holds for every integer $n\geq b$.
\end{theorem}

How does this restated principle work without a base case? Easy: We have just
repackaged the base case into the induction step. Indeed, note that the
induction step now says \textquotedblleft$n\geq b$\textquotedblright, not
\textquotedblleft$n>b$\textquotedblright. In particular, this means that the
implication%
\[
\left(  P\left(  b\right)  \text{ AND }P\left(  b+1\right)  \text{ AND
}P\left(  b+2\right)  \text{ AND }\cdots\text{ AND }P\left(  n-1\right)
\right)  \Longrightarrow P\left(  n\right)
\]
has to hold for $n=b$. However, for $n=b$, the antecedent (= if-part) of this
implication is a tautology (i.e., is an empty statement that is automatically
true by dint of its emptiness\footnote{Don't believe it? Observe that this
antecedent%
\[
\left(  P\left(  b\right)  \text{ AND }P\left(  b+1\right)  \text{ AND
}P\left(  b+2\right)  \text{ AND }\cdots\text{ AND }P\left(  n-1\right)
\right)
\]
is a conjunction of $n-b$ statements (since there are $n-b$ numbers between
$b$ and $n-1$ inclusive). If $n=b$, this means that it is a conjunction of
$b-b=0$ statements, i.e., of no statements whatsoever. So it is an empty
statement, automatically true.}), and thus proving this implication is
tantamount to just unconditionally proving $P\left(  b\right)  $, which was
what we previously viewed as our base case. So we have not magically removed
the need for a base case; we just have merged it into the induction step.
Nevertheless, this makes for a slightly cleaner version of strong induction.

\begin{noncompile}
So a strong induction needs no base case.

In practice, in the induction step, some values of $n$ will require extra
treatment, because there are not enough induction hypotheses in these cases;
thus, we might end up doing de-facto base cases (often more than one) inside
our induction step.
\end{noncompile}

\subsubsection{Example: Prime factorizations exist}

Another example of a strong induction proof comes from elementary number
theory. We recall a basic definition (more on this later, when we cover number theory):

\begin{definition}
A \textbf{prime} (or \textbf{prime number}) means an integer $p>1$ whose only
positive divisors are $1$ and $p$.
\end{definition}

So the primes (in increasing order) are%
\[
2,\ \ 3,\ \ 5,\ \ 7,\ \ 11,\ \ 13,\ \ 17,\ \ 19,\ \ 23,\ \ 29,\ \ \ldots.
\]
There are infinitely many primes, as we will show later.

\begin{theorem}
\label{thm.ent.primefac.exists}Every positive integer is a product of finitely
many primes.
\end{theorem}

Here and in the following, I understand an empty product (i.e., a product of
no numbers whatsoever) to be $1$. Thus, Theorem \ref{thm.ent.primefac.exists}
does hold for $1$, since $1$ is a product of no primes.

Here are more interesting examples:

\begin{itemize}
\item $2023=\allowbreak7\cdot17\cdot17$ is a product of three primes.

\item $2024=\allowbreak2\cdot2\cdot2\cdot11\cdot23$ is a product of five primes.

\item $2=2$ is a product of one prime (namely, $2$ itself).
\end{itemize}

How do we prove Theorem \ref{thm.ent.primefac.exists} in general?

\begin{proof}
[Proof of Theorem \ref{thm.ent.primefac.exists}.]We must prove the statement%
\[
P\left(  n\right)  =\left(  \text{\textquotedblleft}n\text{ is a product of
finitely many primes\textquotedblright}\right)
\]
for each integer $n\geq1$.

We shall prove this by strong induction on $n$. (We use the original variant
of strong induction, with a base case.)

\textit{Base case:} $P\left(  1\right)  $ is true, since $1$ is a product of
finitely many primes (specifically, of $0$ primes, as we saw).

\textit{Induction step:} Let $n>1$. We must prove the implication%
\[
\left(  P\left(  1\right)  \text{ AND }P\left(  2\right)  \text{ AND }%
\cdots\text{ AND }P\left(  n-1\right)  \right)  \Longrightarrow P\left(
n\right)  .
\]


So we assume that $P\left(  1\right)  $ AND $P\left(  2\right)  $
AND\ $\cdots$ AND $P\left(  n-1\right)  $ holds. We must prove that $P\left(
n\right)  $ holds.

In other words, we must prove that $n$ is a product of finitely many primes.

We are in one of the following two cases:

\textit{Case 1:} The only positive divisors of $n$ are $1$ and $n$.

\textit{Case 2:} There is a positive divisor $d$ of $n$ that is neither $1$
nor $n$.

(Other cases are not possible, since $1$ and $n$ always are positive divisors
of $n$.)

Consider Case 1 first. In this case, $n$ itself is a prime (by the definition
of a prime), and thus is a product of finitely many primes (namely, of just
$1$ prime: itself). Thus, $P\left(  n\right)  $ holds in Case 1.

Now, consider Case 2. In this case, there is a positive divisor $d$ of $n$
that is neither $1$ nor $n$. Consider such a $d$ (you might have to choose
one, but any choice is fine). Since $d$ is a positive divisor of $n$, we have
$1\leq d\leq n$ (strictly speaking, this needs to be proved, but we take this
for granted here). Therefore, $1<d<n$ (since $d$ is neither $1$ nor $n$).
Hence, $d$ is one of the numbers $1,2,\ldots,n-1$ (actually $2,3,\ldots,n-1$,
but we don't care).

Furthermore, $\dfrac{n}{d}$ is an integer (since $d$ is a divisor of $n$) and
positive (since $n$ and $d$ are positive). Multiplying the inequality $1<d$ by
$\dfrac{n}{d}$, we obtain $1\cdot\dfrac{n}{d}<d\cdot\dfrac{n}{d}$ (since we
can always divide an inequality by a positive number\footnote{This is a basic
fact that we are taking for granted.}). In other words, $\dfrac{n}{d}<n$.
Since $\dfrac{n}{d}$ is a positive integer, we thus conclude that $\dfrac
{n}{d}$ is one of the numbers $1,2,\ldots,n-1$.

Now, our induction hypothesis says that $P\left(  1\right)  $ AND $P\left(
2\right)  $ AND\ $\cdots$ AND $P\left(  n-1\right)  $ holds. In particular,
$P\left(  d\right)  $ holds (since $d$ is one of the numbers $1,2,\ldots
,n-1$). In other words, $d$ is a product of primes. That is, we can write $d$
as%
\[
d=p_{1}p_{2}\cdots p_{k}\ \ \ \ \ \ \ \ \ \ \text{for some primes }p_{1}%
,p_{2},\ldots,p_{k}.
\]
Consider these primes $p_{1},p_{2},\ldots,p_{k}$.

Again, our induction hypothesis says that $P\left(  1\right)  $ AND $P\left(
2\right)  $ AND\ $\cdots$ AND $P\left(  n-1\right)  $ holds. In particular,
$P\left(  \dfrac{n}{d}\right)  $ holds (since $\dfrac{n}{d}$ is one of the
numbers $1,2,\ldots,n-1$). In other words, $\dfrac{n}{d}$ is a product of
primes. That is, we can write $\dfrac{n}{d}$ as%
\[
\dfrac{n}{d}=q_{1}q_{2}\cdots q_{\ell}\ \ \ \ \ \ \ \ \ \ \text{for some
primes }q_{1},q_{2},\ldots,q_{\ell}.
\]
Consider these primes $q_{1},q_{2},\ldots,q_{\ell}$.

Now,%
\[
n=d\cdot\dfrac{n}{d}=p_{1}p_{2}\cdots p_{k}\cdot q_{1}q_{2}\cdots q_{\ell}%
\]
(since $d=p_{1}p_{2}\cdots p_{k}$ and $\dfrac{n}{d}=q_{1}q_{2}\cdots q_{\ell}%
$). This shows that $n$ is a product of primes (since $p_{1},p_{2}%
,\ldots,p_{k}$ as well as $q_{1},q_{2},\ldots,q_{\ell}$ are primes). In other
words, $P\left(  n\right)  $ holds. Thus, we have proved $P\left(  n\right)  $
in Case 2.

Now, we have proved $P\left(  n\right)  $ both in Case 1 and Case 2.
Therefore, $P\left(  n\right)  $ always holds. Thus, the induction step is
complete, and Theorem \ref{thm.ent.primefac.exists} is proven.
\end{proof}

The above proof is just reflecting the elementary recursive algorithm for
factoring an integer $n$ into a product of primes: We search for a positive
divisor $d$ of $n$ that is neither $1$ nor $n$. If such a $d$ does not exist,
then $n$ itself is a prime. If it does, then we are reduced to the simpler
problems of factoring $d$ and $\dfrac{n}{d}$, and just have to multiply the
resulting factorizations at the end.

\bigskip

\begin{noncompile}
Here is another property of the Fibonacci sequence provable by either strong
or regular induction:

For any $n\geq0$, we have%
\begin{align*}
&  f_{3n}\text{ is even;}\\
&  f_{3n+1}\text{ is odd;}\\
&  f_{3n+2}\text{ is odd.}%
\end{align*}

\end{noncompile}

\bigskip

\begin{noncompile}
The name \textquotedblleft strong induction\textquotedblright\ refers to the
fact that it \textbf{feels} stronger than regular induction: After all, in the
induction step of a strong induction, we can use not just the single previous
claim $P\left(  n-1\right)  $, but all previously proved claims $P\left(
0\right)  ,\ P\left(  1\right)  ,\ \ldots,\ P\left(  n-1\right)  $.
Nevertheless, \textbf{logically} the principle of strong induction is
\textbf{equivalent} to the principle of regular induction. In other words,
anything you can prove by regular induction you can also prove by strong
induction, and vice versa. The first part of this is obvious, but the second
might appear surprising.

For example, let me explain how to prove Binet's formula by regular induction.
This is despite our first attempt at doing so being unsuccessful!
Nevertheless, this can be done. The trick is to define $P\left(  n\right)  $
to be \textbf{not} the statement \textquotedblleft$f_{n}=\dfrac{\varphi
^{n}-\psi^{n}}{\sqrt{5}}$\textquotedblright\ (for a specific $n$), but rather
the statement \textquotedblleft$f_{k}=\dfrac{\varphi^{k}-\psi^{k}}{\sqrt{5}}$
for all $k\in\left\{  0,1,\ldots,n\right\}  $\textquotedblright. In other
words, we \textbf{incorporate} all the statements $P\left(  0\right)
,P\left(  1\right)  ,\ldots,P\left(  n-1\right)  $ into our new statement
$P\left(  n\right)  $. Then, the induction hypothesis will include Binet's
formula not just for $n-1$ but also for $n-2$ (and for all the other numbers
before it). So we get all the advantages of strong induction from regular
induction by just pushing all the necessary memory into the claim that we are proving.

For another example, instead of proving the claim%
\[
\text{\textquotedblleft}n\text{ is a product of primes\textquotedblright}%
\]
by strong induction, we can prove the claim%
\[
\text{\textquotedblleft each integer between }1\text{ and }n\text{ is a
product of primes\textquotedblright}%
\]
by regular induction.

This all is an aside; it's perfectly fine to use strong induction.
\end{noncompile}


\end{document}