\documentclass[11pt]{article}
\usepackage{precalc}
\begin{document}
\assigntitle{26}{Number Theory, Part I: Divisibility and Primality}
This week, we will spend some time studying the basics of \term{number
theory}, which is essentially the study of the \term{natural
numbers} ($0$, $1$, $2$, $3$, \dots). How much could there be to say
about natural numbers, you ask? Plenty!
This is an important area of study for many reasons. At a purely
mathematical level, it gives rise to a plethora of interesting
questions which are easy to ask but difficult to answer. The study of
such questions has led to all kinds of insights in other areas of
mathematics. At a practical level, the insights of number theory
underlie all of modern cryptography. Every time you (say) make a
secure purchase from a website, your computer is doing number theory!
(Next week, we'll explore the relationship between number theory and
cryptography directly.)
\section{Natural numbers}
\label{sec:naturals}
\topic{naturals, everywhere!}
The \term{natural numbers}, denoted $\N$, are the set of positive
integers and zero. That is, \[ \N = \{0, 1, 2, 3, \dots \}. \] For
the next week or two, the natural numbers are all we will talk about!
No negative numbers, no fractions, no complex numbers. I'll try to
say things like ``$x$ is a natural number,'' but I'll probably forget
and say things like ``$x$ is a number,'' or just ``$x$,'' and you
should just assume that I meant $x$ is a \emph{natural} number, unless
I specifically say otherwise. Got it? Let's see if you were paying
attention:
\begin{problem}
Solve for $x$: \[ 3x = 2. \]
\end{problem}
\section{Divisibility}
\label{sec:divisibility}
You surely already know what is meant by \term{divisibility}: $b$ is
divisible by $a$ if dividing $b$ by $a$ leaves no remainder. But
there is a slightly different way of formulating this which is often
more useful: $b$ is divisible by $a$ if there is some natural number
$k$ such that $ak = b$. That is, $b$ is divisible by $a$ if $b$ is a
multiple of $a$.
\begin{defn}{divisibility}
If $a$ and $b$ are natural numbers, we say that $b$ is \term{divisible by} $a$
if there exists a natural number $k$ such that \[ ak = b. \] We
also say that \emph{$a$ divides $b$} and write $a \mid b$.
\end{defn}
You can typeset $a \mid b$ in \LaTeX\ as \verb|a \mid b|.
\begin{problem}
Which of the following are true statements? For each one that is
true, give the corresponding value of $k$.
\begin{subproblems}
\item $2 \mid 6$
\item $13 \mid 91$
\item $14 \mid 7$
\item $5 \mid 5$
\item $6 \mid 19$
\item $0 \mid 7$
\item $7 \mid 0$
\item $0 \mid 0$
\item For every natural number $n$, $1 \mid n$.
\end{subproblems}
\end{problem}
If $a$ does not divide $b$, we sometimes write $a \nmid b$. (The
``does not divide'' symbol can be typeset with \verb|\nmid|.)
\section{Prime numbers}
\label{sec:primes}
A \term{prime number} is one which is only divisible by itself and
one.
\begin{defn}{prime number}
A natural number $p \geq 2$ is \term{prime} if $k \mid p$ implies
that $k = 1$ or $k = p$.
\end{defn}
\topic{$1$ is not prime!}
An interesting point about this definition is that only numbers $p
\geq 2$ can be prime. We specifically exclude $0$ and $1$, even
though otherwise $1$ would be prime by this definition (the only
numbers which divide $1$ are $1$ and\dots $1$). There's a good reason
for this, which we'll see later.
\begin{problem}
Which of the following numbers are prime?
\begin{subproblems}
\item 2
\item 5
\item 15
\item 91
\item 379
\item 391
\item 3549874082
\item 90473512077
\end{subproblems}
\end{problem}
A natural number $n \geq 2$ which is not prime is
\term{composite}. Note that $0$ and $1$ are not composite! Every
number greater than $1$ is either prime or composite, but $0$ and $1$
are neither.
\begin{problem}
The following statement is true for every $n$: \emph{If $n$ is
composite, then it is divisible by some number $k$ where $2 \leq k
\leq \sqrt{n}$.} In other words, every composite number is
divisible by something less than (or equal to) its square root.
Why?
\end{problem}
\begin{problem}
\topic{twin primes}
If $p$ and $p + 2$ are both prime, they are called \emph{twin
primes}. For example, $5$ and $7$ are twin primes.
Interestingly, no one knows whether there are infinitely many pairs
of twin primes, or if there is some largest pair of twin primes
after which there aren't any more! Many mathematicians believe that
there are infinitely many; this is known as the \emph{Twin Prime
Conjecture}.
List five pairs of twin primes.
\end{problem}
\subsection{Prime factorization and the Fundamental Theorem of Arithmetic}
\label{sec:fta}
\topic{prime factorization}
As you probably know, every number greater than $1$ can be
\term{factored} into a product of prime numbers. For example, $2520 =
2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 5 \cdot 7$ (which we often
write as $2^3 \cdot 3^2 \cdot 5 \cdot 7$ to save space). Not only
that, but this factorization into prime numbers is always
\emph{unique}, if you ignore the order of the prime numbers (which
makes sense; $3 \cdot 7$ and $7 \cdot 3$ are obviously the \emph{same}
prime factorization of $21$, since multiplication is commutative).
This is called the \term{Fundamental Theorem of Arithmetic}.
\begin{defn}{Fundamental Theorem of Arithmetic}
Every natural number $n \geq 2$ can be factored \emph{uniquely} (up
to reordering) as the product of one or more prime numbers.
\end{defn}
Of course, you've been factoring numbers since something like third
grade, so you're probably quite familiar with this. But have you ever
stopped to think about how surprising this is? It's obvious, by
definition, that any number can be factored into primes (if $n$ isn't
prime, then by definition it must be equal to the product of two
numbers $ab$; then we repeat the argument for $a$ and $b$, and so on,
until we're left only with primes). But why should this factorization
be \emph{unique}? How do we know there aren't four primes $p$, $q$,
$r$, and $s$ for which $n = pq = rs$---so the number $n$ can be
factored in two different ways, as $n = pq$ or as $n = rs$? It turns out
there aren't, but this fact certainly isn't obvious. Euclid was the
first mathematician to think about this problem, but Gauss was the
first to rigorously prove it.
\begin{problem}
If $1$ were a prime number, the Fundamental Theorem of Arithmetic
would no longer be true. Explain why.
\end{problem}
\begin{problem}
Give the prime factorization of each number. (Of course, if a number
$p$ is prime, its prime factorization is just $p$---for example, the
prime factorization of $7$ is $7$).
\begin{subproblems}
\item $55$
\item $1404$
\item $1001$
\item $65536$
\item $577$
\item $6859$
\end{subproblems}
\end{problem}
\section{GCD and the Euclidean Algorithm}
\label{sec:gcd}
\topic{gcd by factoring}
The \term{greatest common divisor} of two numbers $a$ and $b$, written
$\gcd(a,b)$, is the largest natural number which is a divisor of both
$a$ and $b$. You were probably taught how to find the gcd of two
numbers in elementary school: if you list out all their prime factors,
you can just circle as many factors as possible that occur in both.
For example, to find the gcd of $84$ and $630$, we can first factor
them:
\begin{align*}
84 &= 2 \cdot 2 \cdot 3 \cdot 7 \\
630 &= 2 \cdot 3 \cdot 3 \cdot 5 \cdot 7
\end{align*}
Then we note that they share a $2$, a $3$, and a $7$, so their
greatest common divisor is $\gcd(84, 630) = 2 \cdot 3 \cdot 7 = 42$.
We can check: sure enough, $42 \mid 84$ ($k = 2$), and $42 \mid 630$
($k = 15$).
\topic{factoring is hard}
However, this grade-school method of finding the greatest common
divisor via factoring is not very efficient, because \emph{factoring
is hard}! For example, suppose you wanted to find the gcd of
$7013113$ and $2815433$.\footnote{Don't ask me why, just suppose you
wanted to, OK?} To use the above method, you would first have to
factor them. I don't know about you, but that doesn't sound like fun.
In fact, it turns out that their prime factorizations are
\begin{align*}
7013113 &= 383 \cdot 18311 && \text{and} \\
2815433 &= 383 \cdot 7351.
\end{align*}
Yikes. Now that we know the factorizations, we can see that the gcd
of these two numbers is $383$, but factoring those numbers by hand
would have taken forever.\footnote{In fact, I cheated: I didn't pick
these two numbers randomly and then factor them; I chose some big
primes first and multiplied them to get the numbers!}
\topic{the Euclidean Algorithm}
Well, it turns out that \emph{there's a better way}! This better way
is called the \emph{Euclidean Algorithm}, since it was invented by
Euclid, an influential Greek mathematician who lived around 300 BC.
In a sense, you can think of the Euclidean Algorithm as the
\emph{oldest known computer program}---an \term{algorithm} is just a
set of precise steps for solving a problem, which is exactly what a
computer program is. But back then, the computers were people.
So, how does it work? Let's say we have two numbers $a$ and $b$, and
we want to find their gcd. Suppose $a > b$. Now divide $a$ by $b$,
resulting in some quotient $q$ and remainder $r$. If $r = 0$, then
$a$ is divisible by $b$, and their gcd is just $b$ ($b$ clearly
divides itself, so if it divides $a$ too, it is a common divisor).
Otherwise---here is the clever part---$\gcd(a,b) = \gcd(b,r)$. That
is, if we throw away $a$ and replace it by the remainder when dividing
it by $b$, the gcd is still the same! So we can continue repeating
this process until we get a remainder of zero. Since we are always
replacing $a$ by something smaller, the process must eventually stop.
Let's try an example: suppose we want to find $\gcd(22,14)$. We
divide $22$ by $14$, which gives a remainder of $8$. The remainder
isn't zero, so we have to replace $22$ with $8$ and keep going; now we
are trying to find $\gcd(14,8)$. Well, $14$ divided by $8$ leaves a
remainder of $6$, so now we want to find $\gcd(8,6)$. $8$ divided by
$6$ leaves a remainder of $2$; finally, $6$ divided by $2$ is zero, so
the gcd is $2$. More succinctly, we calculated that \[ \gcd(22,14) =
\gcd(14,8) = \gcd(8,6) = \gcd(6,2) = 2. \] Pretty easy, huh? And no
factoring in sight!
\begin{problem}
Show how to compute each $\gcd$ using the Euclidean Algorithm. Don't just give
the final answer; be sure to show your work.
\begin{subproblems}
\item $\gcd(99,5)$
\item $\gcd(840, 720)$
\item $\gcd(42, 35)$
\item $\gcd(43, 35)$
\item $\gcd(7013113, 2815433)$
\end{subproblems}
\end{problem}
\end{document}