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\begin{document}
\section*{Math 221 Winter 2023, Lecture 5: 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}{8}

\subsection{Strong induction (cont'd)}

\subsubsection{Example: Paying with $3$-cent and $5$-cent coins}

Last time, we discussed the following variant(s) of mathematical induction:

\setcounter{theo}{2}

\begin{theorem}
[Principle of Strong Induction, standard form]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}

\begin{theorem}
[Principle of Strong Induction, baseless form]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}

\bigskip

\setcounter{theo}{6}

Here is another example of how strong induction can be used:

\begin{exercise}
Assume that you have $3$-cent coins and $5$-cent coins (each in infinite
supply). What denominations can you pay with these coins?
\end{exercise}

Let's make a table (\textquotedblleft yes\textquotedblright\ means that you
can pay it; \textquotedblleft no\textquotedblright\ means that you can't):%
\[%
\begin{tabular}
[c]{|c|c|}\hline
$0$ cents & yes\\\hline
$1$ cents & no\\\hline
$2$ cents & no\\\hline
$3$ cents & yes\\\hline
$4$ cents & no\\\hline
$5$ cents & yes\\\hline
$6$ cents & yes: $2\cdot3$\\\hline
$7$ cents & no\\\hline
$8$ cents & yes: $3+5$\\\hline
$9$ cents & yes: $3\cdot3$\\\hline
$10$ cents & yes: $2\cdot5$\\\hline
$11$ cents & yes: $2\cdot3+5$\\\hline
$12$ cents & yes: $4\cdot3$\\\hline
$13$ cents & yes: $3+2\cdot5$\\\hline
$\cdots$ & $\cdots$\\\hline
\end{tabular}
\ \ .
\]
Experimentally, we seem to observe that any denomination $\geq8$ cents can be
paid. Why?

We can notice that if a denomination $k$ (that is, $k$ cents) can be paid,
then so can $k+3$ (just add a $3$-cent coin). Thus, because we can pay $8$
cents, we can also pay $11,14,17,\ldots$ cents. Because we can pay $9$ cents,
we can also pay $12,15,18,\ldots$ cents. Because we can pay $10$ cents, we can
also pay $13,16,19,\ldots$ cents. Together, these three sequences account for
all the integers $\geq8$. Thus, any denomination of $\geq8$ cents can be paid.

Let us formalize this argument as an induction proof.

We define $\mathbb{N}$ to be the set of all nonnegative integers:%
\[
\mathbb{N}=\left\{  0,1,2,\ldots\right\}  .
\]


\begin{proposition}
For any integer $n\geq8$, we can pay $n$ cents with $3$-cent and $5$-cent
coins. In other words, any integer $n\geq8$ can be written as $n=3a+5b$ with
$a,b\in\mathbb{N}$.
\end{proposition}

\begin{proof}
We proceed by strong induction on $n$:

\textit{Base case:} For $n=8$, the claim is true, since $8=3\cdot1+5\cdot1$.

\textit{Induction step:} Fix an integer $n>8$. Assume that the proposition is
already proved for all the integers $8,9,\ldots,n-1$. We must prove that it
also holds for $n$. In other words, we must prove that we can pay $n$ cents
with $3$-cent and $5$-cent coins.

We are in one of the following three cases (since $n>8$):

\textit{Case 1:} We have $n=9$.

\textit{Case 2:} We have $n=10$.

\textit{Case 3:} We have $n\geq11$.

In Case 1, we are done, since $n=9=3\cdot3+0\cdot5$ (that is, $n$ can be paid
with three $3$-cent coins).

In Case 2, we are done, since $n=10=0\cdot3+2\cdot5$ (that is, $n$ can be paid
with two $5$-cent coins).

Now, consider Case 3. In this case, we have $n\geq11$. Hence, $n-3\geq8$. This
shows that $n-3$ is one of the numbers $8,9,\ldots,n-1$.

Thus, we can apply the induction hypothesis to $n-3$. We conclude that $n-3$
cents can be paid with $3$-cent and $5$-cent coins, i.e., we can write $n-3$
as $n-3=3c+5d$ with $c,d\in\mathbb{N}$. Using these $c,d\in\mathbb{N}$, we
therefore have%
\begin{align*}
n  &  =3+3c+5d\ \ \ \ \ \ \ \ \ \ \left(  \text{since }n-3=3c+5d\right) \\
&  =3\left(  c+1\right)  +5d,
\end{align*}
which shows that $n$ cents can also be paid with $3$-cent and $5$-cent coins.
This shows that the proposition is true for $n$, and thus the induction step
is complete.

The proposition is thus proved.
\end{proof}

Note that the above proof had one \textquotedblleft de-jure base
case\textquotedblright\ (the case $n=8$) and two \textquotedblleft de-facto
base cases\textquotedblright\ (the cases $n=9$ and $n=10$, which were formally
part of the induction step but had to be treated separately because $n-3$
would be smaller than $8$ in these cases). We could have just as well used the
baseless form of strong induction, in which case we would have to treat all
three of these cases as \textquotedblleft de-facto base
cases\textquotedblright. This would be a bit more uniform, although this is
entirely a matter of taste.

\bigskip

\section{Sums and products}

\subsection{Finite sums}

Previously, we have encountered sums such as%
\[
x^{n-1}+x^{n-2}y+x^{n-3}y^{2}+\cdots+x^{2}y^{n-3}+xy^{n-2}+y^{n-1}%
\]
(in Lecture 3). Such sums can be tricky to decipher: You need to guess the
pattern of the addends to understand what the \textquotedblleft$\cdots
$\textquotedblright\ means. There is a notation that makes such sums both
shorter and easier to understand. This is the \textbf{finite sum notation}
(also known as the \textbf{sigma notation}). In its simplest form, it is
defined as follows:

\begin{definition}
\label{def.finsum.k=u-v}Let $u$ and $v$ be two integers. Let $a_{u}%
,a_{u+1},\ldots,a_{v}$ be some numbers. Then,%
\[
\sum_{k=u}^{v}a_{k}%
\]
is defined to be the sum%
\[
a_{u}+a_{u+1}+\cdots+a_{v}.
\]
It is called the \textbf{sum of the numbers }$a_{k}$ \textbf{where }$k$
\textbf{ranges from }$u$ \textbf{to }$v$. When $v<u$, this sum is called
\textbf{empty} and defined to be $0$.
\end{definition}

For example:%
\begin{align*}
\sum_{k=5}^{10}k  &  =5+6+7+8+9+10=45;\\
\sum_{k=5}^{10}\dfrac{1}{k}  &  =\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}%
+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}=\dfrac{2131}{2520};\\
\sum_{k=5}^{10}k^{k}  &  =5^{5}+6^{6}+7^{7}+8^{8}+9^{9}+10^{10};\\
\sum_{k=5}^{5}k  &  =5;\\
\sum_{k=5}^{4}k  &  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{an empty sum}\right)
;\\
\sum_{k=5}^{3}k  &  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{an empty sum}\right)
;\\
\sum_{k=5}^{8}3  &  =3+3+3+3=12\ \ \ \ \ \ \ \ \ \ \left(  \text{a sum of four
equal terms}\right)  ;\\
\sum_{k=0}^{n-1}q^{k}  &  =q^{0}+q^{1}+\cdots+q^{n-1}%
\ \ \ \ \ \ \ \ \ \ \text{for any }n\in\mathbb{N}\text{ and any number }q;\\
\sum_{k=0}^{n-1}x^{k}y^{n-1-k}  &  =y^{n-1}+xy^{n-2}+x^{2}y^{n-3}%
+\cdots+x^{n-3}y^{2}+x^{n-2}y+x^{n-1}\\
&  =x^{n-1}+x^{n-2}y+x^{n-3}y^{2}+\cdots+x^{2}y^{n-3}+xy^{n-2}+y^{n-1}\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{for any }n\in\mathbb{N}\text{
and any numbers }x\text{ and }y.
\end{align*}
Thus, Theorem 1.6.2 from Lecture 3 is saying that%
\[
\left(  x-y\right)  \left(  \sum_{k=0}^{n-1}x^{k}y^{n-1-k}\right)
=x^{n}-y^{n}%
\]
for any numbers $x$ and $y$ and any $n\in\mathbb{N}$.

The variable $k$ is not set in stone; you can replace it by any other variable
(unless this other variable already stands for something else). For example,%
\[
\sum_{k=u}^{v}a_{k}=\sum_{i=u}^{v}a_{i}=\sum_{\circledS=u}^{v}a_{\circledS
}=\sum_{\spadesuit=u}^{v}a_{\spadesuit}.
\]
Just don't make it $\sum\limits_{u=u}^{v}a_{u}$.

Here are a couple more examples: For any $n\in\mathbb{N}$, we have%
\begin{align*}
\sum_{k=1}^{n}k  &  =1+2+\cdots+n=\dfrac{n\left(  n+1\right)  }{2}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Theorem 1.3.1 in Lecture 2}\right)  ;\\
\sum_{k=1}^{n}k^{2}  &  =1^{2}+2^{2}+\cdots+n^{2}\\
&  =\dfrac{n\left(  n+1\right)  \left(  2n+1\right)  }{6}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Theorem 1.3.2 in Lecture 2}\right)  ;\\
\sum_{k=1}^{n}1  &  =\underbrace{1+1+\cdots+1}_{n\text{ times}}=n\cdot1=n;\\
\sum_{k=1}^{n}\left(  2k-1\right)   &  =1+3+5+\cdots+\left(  2n-1\right) \\
&  =\left(  \text{the sum of the first }n\text{ odd positive integers}\right)
.
\end{align*}


We have not computed this last sum, so let us do this. I will use the
following \textquotedblleft laws of summation\textquotedblright:

\begin{itemize}
\item We have
\begin{equation}
\sum_{k=u}^{v}\left(  a_{k}-b_{k}\right)  =\sum_{k=u}^{v}a_{k}-\sum_{k=u}%
^{v}b_{k} \label{eq.sums.sum(a-b)}%
\end{equation}
for any integers $u,v$ and any numbers $a_{k},b_{k}$. Indeed, if you rewrite
this without finite sum notation, it takes the form%
\begin{align*}
&  \left(  a_{u}-b_{u}\right)  +\left(  a_{u+1}-b_{u+1}\right)  +\cdots
+\left(  a_{v}-b_{v}\right) \\
&  =\left(  a_{u}+a_{u+1}+\cdots+a_{v}\right)  -\left(  b_{u}+b_{u+1}%
+\cdots+b_{v}\right)  ,
\end{align*}
which is rather clear. (A formal proof can be given by induction on $v$.)

\item We have%
\begin{equation}
\sum_{k=u}^{v}\lambda a_{k}=\lambda\sum_{k=u}^{v}a_{k} \label{eq.sums.sum(la)}%
\end{equation}
for any integers $u,v$ and any numbers $\lambda,a_{k}$. Indeed, rewritten
without the use of finite sum notation, this is just saying that%
\[
\lambda a_{u}+\lambda a_{u+1}+\cdots+\lambda a_{v}=\lambda\left(
a_{u}+a_{u+1}+\cdots+a_{v}\right)  ,
\]
which is again clear (and can be proved by induction on $v$).
\end{itemize}

Rules like this are dime a dozen, and you should be able to come up with them
on the spot when you need them. (See \cite[\S 1.4.2]{detnotes} for these and
several others.)

Let us now compute our sum:%
\begin{align*}
\sum_{k=1}^{n}\left(  2k-1\right)   &  =\sum_{k=1}^{n}2k-\sum_{k=1}%
^{n}1\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{eq.sums.sum(a-b)})}\right) \\
&  =2\underbrace{\sum_{k=1}^{n}k}_{=\dfrac{n\left(  n+1\right)  }{2}%
}-\underbrace{\sum_{k=1}^{n}1}_{=n}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{eq.sums.sum(la)})}\right) \\
&  =2\cdot\dfrac{n\left(  n+1\right)  }{2}-n=n\left(  n+1\right)  -n=n^{2}.
\end{align*}


As another illustration of our method, we can rewrite Gauss's proof of the
equality
\begin{equation}
\sum_{k=1}^{n}k=1+2+\cdots+n=\dfrac{n\left(  n+1\right)  }{2}
\label{eq.sums.little-gauss}%
\end{equation}
(Theorem 1.3.2 in Lecture 2) using finite sum notation. We will need two new
rules this time:

\begin{itemize}
\item We have
\begin{equation}
\sum_{k=u}^{v}a_{k}+\sum_{k=u}^{v}b_{k}=\sum_{k=u}^{v}\left(  a_{k}%
+b_{k}\right)  \label{eq.sums.suma+sumb}%
\end{equation}
for any integers $u,v$ and any numbers $a_{k},b_{k}$. Indeed, if you rewrite
this without finite sum notation, it takes the form%
\begin{align*}
&  \left(  a_{u}+a_{u+1}+\cdots+a_{v}\right)  +\left(  b_{u}+b_{u+1}%
+\cdots+b_{v}\right) \\
&  =\left(  a_{u}+b_{u}\right)  +\left(  a_{u+1}+b_{u+1}\right)
+\cdots+\left(  a_{v}+b_{v}\right)  .
\end{align*}


\item We have%
\begin{equation}
\sum_{k=u}^{v}a_{k}=\sum_{k=u}^{v}a_{u+v-k} \label{eq.sums.revert}%
\end{equation}
for any integers $u,v$ and any numbers $a_{k}$. This is called
\textquotedblleft substituting $u+v-k$ for $k$ in the sum\textquotedblright%
\ or just \textquotedblleft turning the sum upside-down\textquotedblright, as
it amounts to reversing the order of the addends; restated without finite sum
notation, this is just saying that%
\[
a_{u}+a_{u+1}+\cdots+a_{v}=a_{v}+a_{v-1}+\cdots+a_{u},
\]
which is saying that a sum of a bunch of numbers does not change if we add its
addends together in reverse order.

\item For any integers $u\leq v$ and any number $\lambda$, we have
\begin{equation}
\sum_{k=u}^{v}\lambda=\left(  v-u+1\right)  \lambda. \label{eq.sums.lam}%
\end{equation}
(This is just saying that a sum of $v-u+1$ many equal addends $\lambda$ is
$\left(  v-u+1\right)  \lambda$. Note that the sum on the left hand side has
$v-u+1$ addends, because there are $v-u+1$ numbers in the set $\left\{
u,u+1,\ldots,v\right\}  $.)
\end{itemize}

Now, Gauss's proof of (\ref{eq.sums.little-gauss}) takes the following shape:%
\begin{align*}
2\sum_{k=1}^{n}k  &  =\sum_{k=1}^{n}k+\sum_{k=1}^{n}k\\
&  =\sum_{k=1}^{n}k+\sum_{k=1}^{n}\left(  n+1-k\right) \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }n+1-k\text{ for }k\text{ in the second}\\
\text{sum (i.e., rewrote it using (\ref{eq.sums.revert}))}%
\end{array}
\right) \\
&  =\sum_{k=1}^{n}\underbrace{\left(  k+\left(  n+1-k\right)  \right)
}_{=n+1}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{eq.sums.suma+sumb})}\right)
\\
&  =\sum_{k=1}^{n}\left(  n+1\right)  =n\cdot\left(  n+1\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{eq.sums.lam})}\right)  .
\end{align*}
Dividing both sides by $2$, we recover (\ref{eq.sums.little-gauss}) again.

\bigskip

We have found closed-form expressions (i.e., expressions without $\sum$ signs
or \textquotedblleft$\cdots$\textquotedblright s) for several sums. Not every
sum has a closed-form expression. For instance, there is no closed form for%
\[
\sum_{k=1}^{n}\dfrac{1}{k}\ \ \ \ \ \ \ \ \ \ \text{or for}%
\ \ \ \ \ \ \ \ \ \ \sum_{k=1}^{n}k^{k}.
\]


\bigskip

Some more terminology:

The notation $\sum\limits_{k=u}^{v}a_{k}$ is called \textbf{sigma notation} or
\textbf{finite sum notation}. The symbol $\sum$ itself is called the
\textbf{summation sign}. The numbers $u$ and $v$ are called the \textbf{lower
limit} and the \textbf{upper limit} of the summation\footnote{This use of the
word \textquotedblleft limit\textquotedblright\ is totally unrelated to the
way this word is used in analysis/calculus.}. The variable $k$ is called the
\textbf{summation index} or the \textbf{running index}, and is said to
\textbf{range} (or \textbf{run}) from $u$ to $v$. The numbers $a_{k}$ are
called the \textbf{addends} of the finite sum.

\bigskip

There are many similarities between finite sums $\sum_{k=u}^{v}a_{k}$ and
integrals $\int_{u}^{v}f\left(  x\right)  dx$, but the analogy should not be
taken too far (e.g., an integral $\int_{u}^{u}f\left(  x\right)  dx$ whose
upper and lower limit are equal will always be $0$, but an \textquotedblleft
analogous\textquotedblright\ finite sum $\sum_{k=u}^{u}a_{k}$ will be $a_{u}$).

\bigskip

We note two more rules for finite sums:

\begin{itemize}
\item The \textquotedblleft splitting-off rule\textquotedblright: For any
integers $u\leq v$ and any numbers $a_{u},a_{u+1},\ldots,a_{v}$, we have%
\[
\sum_{k=u}^{v}a_{k}=\sum_{k=u}^{v-1}a_{k}+a_{v}=a_{u}+\sum_{k=u+1}^{v}a_{k}.
\]
This is just saying that%
\[
a_{u}+a_{u+1}+\cdots+a_{v}=\left(  a_{u}+a_{u+1}+\cdots+a_{v-1}\right)
+a_{v}=a_{u}+\left(  a_{u+1}+a_{u+2}+\cdots+a_{v}\right)  .
\]
This rule allows us to split the first or the last addend out of a finite sum.
This is important for proofs by induction.

\item More generally, any finite sum $\sum_{k=u}^{v}a_{k}$ can be split at any
point: We have%
\[
\sum_{k=u}^{v}a_{k}=\sum_{k=u}^{w}a_{k}+\sum_{k=w+1}^{v}a_{k}%
\]
for any integers $u\leq w\leq v$ and any numbers $a_{k}$. This is just saying
that%
\[
a_{u}+a_{u+1}+\cdots+a_{v}=\left(  a_{u}+a_{u+1}+\cdots+a_{w}\right)  +\left(
a_{w+1}+a_{w+2}+\cdots+a_{v}\right)  .
\]
(Strictly speaking, this is true not just for $u\leq w\leq v$ but more
generally for $u-1\leq w\leq v$.)
\end{itemize}

\bigskip

Finite sum notation, in the form defined above, is helpful when the summation
index is running over an integer interval (i.e., a set of consecutive
integers). For more general situations, there is a more general version of
finite sum notation, e.g.:%
\[
\sum_{k\in\left\{  1,2,\ldots,n\right\}  \text{ is even}}k=2+4+6+\cdots+m,
\]
where $m$ is the largest even element of $\left\{  1,2,\ldots,n\right\}  $. We
won't use it much, but it is fairly self-explanatory; essentially, the writing
under the summation sign explains what $k$'s the sum is ranging over. See
\cite[\S 1.4.1]{detnotes} for a more precise explanation.

\subsection{Finite products}

Finite products are analogous to finite sums, just using multiplication
instead of addition:

\begin{definition}
\label{def.finprod.k=u-v}Let $u$ and $v$ be two integers. Let $a_{u}%
,a_{u+1},\ldots,a_{v}$ be some numbers. Then,%
\[
\prod_{k=u}^{v}a_{k}%
\]
is defined to be the product%
\[
a_{u}a_{u+1}\cdots a_{v}.
\]
It is called the \textbf{product of the numbers }$a_{k}$ \textbf{where }$k$
\textbf{ranges from }$u$ \textbf{to }$v$. When $v<u$, this product is called
\textbf{empty} and defined to be $1$.
\end{definition}

For example:%
\begin{align*}
\prod_{k=5}^{10}k  &  =5\cdot6\cdot7\cdot8\cdot9\cdot10=151\ 200;\\
\prod_{k=1}^{5}\dfrac{1}{k}  &  =\dfrac{1}{1}\cdot\dfrac{1}{2}\cdot\dfrac
{1}{3}\cdot\dfrac{1}{4}\cdot\dfrac{1}{5}=\dfrac{1}{120};\\
\prod_{k=1}^{n}a  &  =\underbrace{aa\cdots a}_{n\text{ times}}=a^{n}%
\ \ \ \ \ \ \ \ \ \ \text{for any fixed number }a;\\
\prod_{k=1}^{n}a^{k}  &  =a^{1}a^{2}\cdots a^{n}\\
&  =a^{1+2+\cdots+n}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by one of the laws of exponents:}\\
\text{namely, the law }a^{i_{1}}a^{i_{2}}\cdots a^{i_{n}}=a^{i_{1}%
+i_{2}+\cdots+i_{n}}%
\end{array}
\right) \\
&  =a^{n\left(  n+1\right)  /2}\ \ \ \ \ \ \ \ \ \ \text{for any fixed number
}a\text{ and any }n\in\mathbb{N}\text{.}%
\end{align*}


\bigskip

In a finite product $\prod\limits_{k=u}^{v}a_{k}$, the $k$ is called the
\textbf{product index} or the \textbf{running index}, and the symbol $\prod$
is called the \textbf{product sign}. The numbers $a_{k}$ are called the
\textbf{factors} of the product. Other terminology is analogous to the case of
a finite sum (e.g., lower limit, upper limit). Almost all rules for finite
sums have analogues for finite products. Let me only state the analogue of the
\textquotedblleft splitting-off rule\textquotedblright:

\begin{itemize}
\item The \textquotedblleft splitting-off rule\textquotedblright\ for
products: For any integers $u\leq v$ and any numbers $a_{u},a_{u+1}%
,\ldots,a_{v}$, we have%
\[
\prod_{k=u}^{v}a_{k}=\left(  \prod_{k=u}^{v-1}a_{k}\right)  a_{v}=a_{u}%
\prod_{k=u+1}^{v}a_{k}.
\]
This is just saying that%
\[
a_{u}a_{u+1}\cdots a_{v}=\left(  a_{u}a_{u+1}\cdots a_{v-1}\right)
a_{v}=a_{u}\left(  a_{u+1}a_{u+2}\cdots a_{v}\right)  .
\]
This rule allows us to split the first or the last factor out of a finite
product. This is important for proofs by induction.
\end{itemize}

\subsection{Factorials}

Now, we define a sequence of integers that appears all over mathematics.
Recall that $\mathbb{N}=\left\{  0,1,2,\ldots\right\}  $.

\begin{definition}
\label{def.fac}For any $n\in\mathbb{N}$, we define the positive integer $n!$
(called the \textbf{factorial} of $n$, and often pronounced \textquotedblleft%
$n$ \textbf{factorial}\textquotedblright) by%
\[
n!=\prod_{k=1}^{n}k=1\cdot2\cdot\cdots\cdot n.
\]

\end{definition}

For example,%
\begin{align*}
0!  &  =\left(  \text{empty product}\right)  =1;\\
1!  &  =1=1;\\
2!  &  =1\cdot2=2;\\
3!  &  =1\cdot2\cdot3=6;\\
4!  &  =1\cdot2\cdot3\cdot4=24;\\
5!  &  =1\cdot2\cdot3\cdot4\cdot5=120;\\
6!  &  =1\cdot2\cdot3\cdot4\cdot5\cdot6=720;\\
7!  &  =5\ 040;\\
8!  &  =40\ 320;\\
9!  &  =362\ 880;\\
10!  &  =3\ 628\,800.
\end{align*}
Note the following:

\begin{proposition}
[recursion of the factorials]For any positive integer $n$, we have%
\[
n!=\left(  n-1\right)  !\cdot n.
\]

\end{proposition}

\begin{proof}
Let $n$ be a positive integer. Then,%
\[
n!=1\cdot2\cdot\cdots\cdot n=\underbrace{\left(  1\cdot2\cdot\cdots
\cdot\left(  n-1\right)  \right)  }_{=\left(  n-1\right)  !}\cdot n=\left(
n-1\right)  !\cdot n.
\]

\end{proof}

\subsection{Binomial coefficients: Definition}

We shall now define one of the most important families of numbers in mathematics:

\begin{definition}
\label{def.binom}Let $n$ and $k$ be any numbers. Then, we define a number
$\dbinom{n}{k}$ as follows:

\begin{itemize}
\item If $k\in\mathbb{N}$, then we set%
\[
\dbinom{n}{k}:=\dfrac{n\left(  n-1\right)  \left(  n-2\right)  \cdots\left(
n-k+1\right)  }{k!}%
\]
(where the numerator is the product of $k$ consecutive integers, the largest
of which is $n$; you can also write it as $\prod\limits_{i=0}^{k-1}\left(
n-i\right)  $).

\item If $k\notin\mathbb{N}$, then we set%
\[
\dbinom{n}{k}:=0.
\]

\end{itemize}

The number $\dbinom{n}{k}$ is called \textquotedblleft$n$ \textbf{choose }%
$k$\textquotedblright, and is known as the \textbf{binomial coefficient} of
$n$ and $k$. Do not mistake the notation $\dbinom{n}{k}$ for a vector $\left(
%
\begin{array}
[c]{c}%
n\\
k
\end{array}
\right)  $.
\end{definition}

\begin{example}
For any number $n$, we have%
\begin{align*}
\dbinom{n}{3}  &  =\dfrac{n\left(  n-1\right)  \left(  n-2\right)  }%
{3!}=\dfrac{n\left(  n-1\right)  \left(  n-2\right)  }{6};\\
\dbinom{n}{2}  &  =\dfrac{n\left(  n-1\right)  }{2!}=\dfrac{n\left(
n-1\right)  }{2};\\
\dbinom{n}{1}  &  =\dfrac{n}{1!}=n;\\
\dbinom{n}{0}  &  =\dfrac{\left(  \text{empty product}\right)  }{0!}=\dfrac
{1}{1}=1;\\
\dbinom{n}{2.5}  &  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }2.5\notin%
\mathbb{N}\right)  ;\\
\dbinom{n}{-1}  &  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }-1\notin%
\mathbb{N}\right)  .
\end{align*}
For any $k\in\mathbb{N}$, we have%
\begin{align*}
\dbinom{0}{k}  &  =\dfrac{0\left(  0-1\right)  \left(  0-2\right)
\cdots\left(  0-k+1\right)  }{k!}=%
\begin{cases}
1, & \text{if }k=0;\\
0, & \text{if }k\neq0;
\end{cases}
\\
\dbinom{-1}{k}  &  =\dfrac{\left(  -1\right)  \left(  -2\right)  \left(
-3\right)  \cdots\left(  -k\right)  }{k!}=\left(  -1\right)  ^{k}%
\cdot\underbrace{\dfrac{1\cdot2\cdot\cdots\cdot k}{k!}}_{=1}=\left(
-1\right)  ^{k}.
\end{align*}

\end{example}

Let us tabulate the values of $\dbinom{n}{k}$ for nonnegative integers $n$ and
$k$:%
\[%
\begin{tabular}
[c]{|c||c|c|c|c|c|c|c|}\hline
& $k=0$ & $k=1$ & $k=2$ & $k=3$ & $k=4$ & $k=5$ & $k=6$\\\hline\hline
$n=0$ & $1$ & $0$ & $0$ & $0$ & $0$ & $0$ & $0$\\\hline
$n=1$ & $1$ & $1$ & $0$ & $0$ & $0$ & $0$ & $0$\\\hline
$n=2$ & $1$ & $2$ & $1$ & $0$ & $0$ & $0$ & $0$\\\hline
$n=3$ & $1$ & $3$ & $3$ & $1$ & $0$ & $0$ & $0$\\\hline
$n=4$ & $1$ & $4$ & $6$ & $4$ & $1$ & $0$ & $0$\\\hline
$n=5$ & $1$ & $5$ & $10$ & $10$ & $5$ & $1$ & $0$\\\hline
$n=6$ & $1$ & $6$ & $15$ & $20$ & $15$ & $6$ & $1$\\\hline
\end{tabular}
\ \ \ \ .
\]


What patterns can we spot in this table? (We are ignoring negative and
non-integer $n$'s for now.)

\begin{proposition}
\label{prop.binom.0}Let $n\in\mathbb{N}$ and $k>n$. Then, $\dbinom{n}{k}=0$.
\end{proposition}

\begin{proof}
If $k\notin\mathbb{N}$, then this is clear by definition. Otherwise, again by
definition, we have%
\[
\dbinom{n}{k}=\dfrac{n\left(  n-1\right)  \left(  n-2\right)  \cdots\left(
n-k+1\right)  }{k!}=\dfrac{0}{k!}%
\]
(since the product $n\left(  n-1\right)  \left(  n-2\right)  \cdots\left(
n-k+1\right)  $ has a factor of $n-n=0$, and thus is $0$). For example, for
$n=3$ and $k=6$, we have%
\[
\dbinom{3}{6}=\dfrac{3\cdot2\cdot1\cdot0\cdot\left(  -1\right)  \cdot\left(
-2\right)  }{6!}=\dfrac{0}{6!}=0.
\]

\end{proof}

\begin{remark}
Note that Proposition \ref{prop.binom.0} would \textbf{not} hold without the
$n\in\mathbb{N}$ assumption. For example,%
\[
\dbinom{1.5}{3}=\dfrac{1.5\cdot0.5\cdot\left(  -0.5\right)  }{3!}%
\neq0\ \ \ \ \ \ \ \ \ \ \text{even though }3>1.5.
\]
The product in the numerator is not $0$, since it \textquotedblleft
misses\textquotedblright\ the $0$ factor.
\end{remark}

Proposition \ref{prop.binom.0} explains why our above table of $\dbinom{n}{k}$
has so many zeroes in it. More precisely, it tells us that all entries above
the main diagonal of the table are zeroes (no matter how many more rows and
columns we add). Thus, we can redraw our table as a triangular table (and fill
in a few more rows while at that):%
\[
\hspace{-1pc}\hspace{-0.3cm}%
\begin{tabular}
[c]{ll|ccccccccccccccccc}%
\vphantom{$\overset{k=0}{\swarrow}$} &  & $\phantom{20}$ & $\phantom{20}$ &
$\phantom{20}$ & $\phantom{20}$ & $\phantom{20}$ & $\phantom{20}$ &  &  &  &
$\overset{k=0}{\swarrow}$ & $\phantom{20}$ & $\phantom{20}$ & $\phantom{20}$ &
$\phantom{20}$ & $\phantom{20}$ & $\phantom{20}$ & $\phantom{20}$\\
\vphantom{$\overset{k=0}{\swarrow}$}$n=0$ & $\rightarrow$ &  &  &  &  &  &  &
&  & $1$ &  & $\overset{k=1}{\swarrow}$ &  &  &  &  &  & \\
\vphantom{$\overset{k=0}{\swarrow}$}$n=1$ & $\rightarrow$ &  &  &  &  &  &  &
& $1$ &  & $1$ &  & $\overset{k=2}{\swarrow}$ &  &  &  &  & \\
\vphantom{$\overset{k=0}{\swarrow}$}$n=2$ & $\rightarrow$ &  &  &  &  &  &  &
$1$ &  & $2$ &  & $1$ &  & $\overset{k=3}{\swarrow}$ &  &  &  & \\
\vphantom{$\overset{k=0}{\swarrow}$}$n=3$ & $\rightarrow$ &  &  &  &  &  & $1$
&  & $3$ &  & $3$ &  & $1$ &  & $\overset{k=4}{\swarrow}$ &  &  & \\
\vphantom{$\overset{k=0}{\swarrow}$}$n=4$ & $\rightarrow$ &  &  &  &  & $1$ &
& $4$ &  & $6$ &  & $4$ &  & $1$ &  & $\overset{k=5}{\swarrow}$ &  & \\
\vphantom{$\overset{k=0}{\swarrow}$}$n=5$ & $\rightarrow$ &  &  &  & $1$ &  &
$5$ &  & $10$ &  & $10$ &  & $5$ &  & $1$ &  & $\overset{k=6}{\swarrow}$ & \\
\vphantom{$\overset{k=0}{\swarrow}$}$n=6$ & $\rightarrow$ &  &  & $1$ &  & $6$
&  & $15$ &  & $20$ &  & $15$ &  & $6$ &  & $1$ &  & $\overset{k=7}{\swarrow}%
$\\
\vphantom{$\overset{k=0}{\swarrow}$}$n=7$ & $\rightarrow$ &  & $1$ &  & $7$ &
& $21$ &  & $35$ &  & $35$ &  & $21$ &  & $7$ &  & $1$ & \\
\vphantom{$\overset{k=0}{\swarrow}$}$n=8$ & $\rightarrow$ & $1$ &  & $8$ &  &
$28$ &  & $56$ &  & $70$ &  & $56$ &  & $28$ &  & $8$ &  & $1$%
\end{tabular}
\ \ \ \ \ \ \ \ \ \
\]
This table is known as \textbf{Pascal's triangle}, and has a variety of
wonderful properties. Here are just a few:

\begin{proposition}
[Pascal's identity, aka the recurrence of the binomial coefficients]For any
numbers $n$ and $k$, we have%
\[
\dbinom{n}{k}=\dbinom{n-1}{k-1}+\dbinom{n-1}{k}.
\]

\end{proposition}

\begin{proposition}
[symmetry of binomial coefficients]For any $n\in\mathbb{N}$ and any $k$, we
have $\dbinom{n}{k}=\dbinom{n}{n-k}$.
\end{proposition}

\begin{proposition}
We have $\dbinom{n}{n}=1$ for each $n\in\mathbb{N}$.
\end{proposition}

\begin{proposition}
[integrality of binomial coefficients]For any $n\in\mathbb{Z}$ and any $k$, we
have $\dbinom{n}{k}\in\mathbb{Z}$.
\end{proposition}

Next time, we will prove these four propositions and more.

\begin{thebibliography}{99999999}                                                                                         %


\bibitem[Grinbe15]{detnotes}\href{https://arxiv.org/abs/2008.09862v3}{Darij
Grinberg, \textit{Notes on the combinatorial fundamentals of algebra}, 15
September 2022, arXiv:2008.09862v3.}
\end{thebibliography}


\end{document}