generated from lance1416/Template-LectureNotes
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18 KiB
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520 lines
18 KiB
TeX
\chapter{Unique Factorization}
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A number is a way of identifying what is \textit{common} between different collections of objects. For example, 3 sheep, 3 coins, and 3 apples are all sets of 3 objects. The number 3 is a way of identifying the size of these sets. What is common is that \textit{3 of anything} can be put in correspondence with each other.
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\section{Introduction}
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\subsection{Notations}
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\begin{definition}
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A set \( S \) is has a \term{size} \( |S| \).
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\end{definition}
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\begin{remark}
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If \( S = \{ a, b, c \} \) and \( T = \{ 1, 2, 3 \} \), then \( |S| = |T| = 3 \).
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\end{remark}
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\subsubsection{Some Operations on Sets}
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% Here we ignore the fact that some objects may be in both sets. \textbf{Only the size of the sets are considered.}
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Note that we are only considering the \textit{size} of the sets, not the objects themselves. We are only considering sets that are \textit{finite}.
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\begin{itemize}
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\item The \term{union}\index{Set!Union} of disjoint sets \( S \) and \( T \), \( S \cup T \), is the arbitrary collection of objects in either \( S \) or \( T \). \[
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| S \cup T | = |S| + |T|
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\]
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\item For \( T \subseteq S \), the \term{difference}\index{Set!Difference} of \( S \) and \( T \), \( S \setminus T \), is removing \( T \) from \( S \), \[
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| S \setminus T | = |S| - |T|
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\]
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\item The \term{product} of two sets \( S \) and \( T \)\index{Set!Product}\footnote{\( S \times T = \left\{ (s, t), s \in S, t \in T \right\} \)}, \( S \times T \), is the set of all pairs of objects where the first object is in \( S \) and the second object is in \( T \). \[
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| S \times T | = |S| \cdot |T|
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\]
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\end{itemize}
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\subsection{National Numbers}
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\begin{definition}\index{National Number}
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\term{National numbers} are a way of identifying the size of a set. The set of all natural numbers is denoted by \( \N \).
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\end{definition}
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\begin{remark}[Notation]
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To disambiguate between whether 0 is included in the set of natural numbers, we use
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\begin{itemize}
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\item \( \N_{\geq 0} \) to include 0,
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\item \( \N_{> 0} \) to exclude 0.
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\end{itemize}
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\end{remark}
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\begin{remark}[Observation]
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Every \( |S| \) appears somewhere in this list. Size gets \textbf{smaller} as you remove things.
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\end{remark}
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\begin{axiom}[Peano Axioms (for \( \N_{\geq 0} \))]\index{Peano Axioms}
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\begin{enumerate}[label=(\Roman*)]
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\item There exists an element \( 0 \).
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\item For each \( n \in \N_{\geq 0} \), there exists \( S(n) \in \N_{\geq 0} \) such that \( S(n) = S(m) \implies n = m \). \( S \) is an operator called the \term{successor} operator.
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\item \textbf{Induction Schema} For any property \( P \) on \( N \), if
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\begin{itemize}
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\item \( P(0) \) is true, and
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\item \( P(n) \implies P(S(n)) \)
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\end{itemize}
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Then \( P(n) \) is true for all \( n \in \N_{\geq 0} \).
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\end{enumerate}
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\end{axiom}
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\begin{example}[System Where III Fails]
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Take \( \Z \in \N_{\geq 0} \cup -\N_{\geq 0} \), the set of all integers. \( S(n) = n + 1 \).
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\end{example}
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\section{Prime and Irreducible Numbers}
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\subsection{Multiplication and Division}
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\begin{definition}[Divisibility]\index{Divisibility}
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\( n \) divides \( m \) if there exists \( u \) such that \( m = nu \). We write \( a \mid b \).
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\end{definition}
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\begin{definition}[Prime Number (Definition I - Irreducible Number)]\index{Prime Number}\index{Irreducible Number}
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A number \( p \neq 1 \) is \term{irreducible} if \( p \) is not 1 and the only divisors of \( p \) are 1 and \( p \). In other words, \[
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p = uv \implies u = 1 \text{ or } v = 1.
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\]
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\end{definition}
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\begin{definition}[Prime Number (Definition II - Prime Number)]\index{Prime Number}
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A number \( p \) is \term{prime} if \( \forall a, b \) such that \( p \mid ab \), \( p \mid a \) or \( p \mid b \).
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\end{definition}
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\begin{theorem}
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If \( p \) is prime, then \( p \) is irreducible.
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\end{theorem}
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\begin{proof}
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WTS \( n \) is reducible \( \implies n \) is not prime.
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Assume \( n \) is reducible, \( n = ab \) with \( a, b \neq 1 \).
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Then, \( n \mid ab \), but \( n \nmid a \) and \( n \nmid b \) as \( n > a, b \).
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\end{proof}
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\begin{note}[System Where Irreducible $\not\implies$ Prime]
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In this system, the Bezout's Identity does not hold.
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{~~~}
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Take \( \Z \left[ \sqrt{-5} \right] = \{ a + b\sqrt{-5}, {~~~} a, b \in \Z \} \)
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Let \( a = 1 + \sqrt{-5} \), \( b = 1 - \sqrt{-5} \).
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\begin{align*}
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a + b & = 3 + 2 \sqrt{5} \\
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a b & = (1 + \sqrt{-5})(1 - \sqrt{-5}) \\
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& = 1 - \sqrt{-5} + \sqrt{-5} + 5 \\
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& = 6
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\end{align*}
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Now consider the number $6$. \[
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6 = 2 \cdot 3 = (1 + \sqrt{-5})(1 - \sqrt{-5})
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\]
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Turns out \( 2, 3, 1 \pm \sqrt{-5} \) are all irreducible but not prime.
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For example, $2 \mid (1 + \sqrt{-5})(1 - \sqrt{-5})$ but $2 \nmid 1 \pm \sqrt{-5}$.
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\end{note}
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\subsection{GCD and Bezout's Identity}
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\begin{definition}[Greatest Common Divisor]\index{Greatest Common Divisor}
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The \term{greatest common divisor} of \( a \) and \( b \) is the largest number that divides both \( a \) and \( b \).
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\( d = gcd(a, b) \) if \( d \mid a \), \( d \mid b \), and for any \( c \) such that \( c \mid a, b \), we have \( c \mid d \).
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\end{definition}
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\begin{remark}
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Note that if we change the definition to \( c \leq d \), it is clear that $gcd$ exists. However, with the given definition, we need to prove that $gcd$ exists, as we may land in the case where \( c < d \) but \( c \nmid d \).
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\end{remark}
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\begin{theorem}\label{thm:gcd-exists}
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For any \( a, b \in \N_{> 0} \), \( gcd(a, b) \) exists.
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\end{theorem}
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\begin{remark}
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\begin{minipage}[t]{0.45\linewidth}
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Consider \( a = 30 \) and \( b = 12 \).
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\begin{center}
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\begin{tikzpicture}[
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every node/.style = {draw, circle, minimum size = 0.75cm},
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]
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\node (1) at (1, 0) {\( 1 \)};
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\node (2) at (0, 1) {\( 2 \)};
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\node (3) at (2, 1) {\( 3 \)};
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\node (6) at (1, 2) {\( 6 \)};
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\draw (1) -- (2);
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\draw (1) -- (3);
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\draw (2) -- (6);
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\draw (3) -- (6);
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\end{tikzpicture}
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\end{center}
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\end{minipage}
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\begin{minipage}[t]{0.45\linewidth}
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Consider \( a = 60 \) and \( b = 15 \).
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\begin{center}
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\begin{tikzpicture}[
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every node/.style = {draw, circle, minimum size = 0.75cm},
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]
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\node (1) at (1, 0) {\( 1 \)};
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\node (3) at (0, 1) {\( 3 \)};
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\node (5) at (2, 1) {\( 5 \)};
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\node (15) at (1, 2) {\( 15 \)};
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\draw (1) -- (3);
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\draw (1) -- (5);
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\draw (3) -- (15);
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\draw (5) -- (15);
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\end{tikzpicture}
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\end{center}
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\end{minipage}
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\end{remark}
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\begin{lemma}[Bezout's Identity]\index{Bezout's Identity}
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For any \( a, b \in \N_{> 0} \), there exist \( x, y \in \Z \) such that \( ax + by = gcd(a, b) \).
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\end{lemma}
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\begin{example}
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Consider $a = 7$ and $b = 9$, $gcd(7, 9) = 1$.
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We have $7 \cdot 5 + 9 \cdot (-4) = 1$.
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\end{example}
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Define \[ S = \{ ax + by \mid x, y \in \Z, ax + by > 0 \}. \]
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\begin{lemma}
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\( gcd \) is the smallest (positive) element of \( S \).
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\end{lemma}
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\begin{proof}
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WTS \( gcd(a, b) = d \iff d \mid a, b \) and \( d \) for any \( c \mid a, b \), we have \( c \mid d \).
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Let $m$ be the smallest element of $S$. $m = ax + by$ for some $x, y \in \Z$.
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\begin{enumerate}
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\item WTS \( m \mid a, b \)
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Let \( a = km + r \) where \( 0 \leq r < m \).
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Then, \( r = a - km = a(1 - kx) + b(ky) \) is in \( S \cup \{ 0 \} \), but \( m \) is the smallest element in \( S \) so \( r < m \).
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Therefore, \( r = 0 \), \( a = km \), and $m \mid a$.
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The same argument can be applied to \( b \).
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\item WTS for any \( c \mid a, b \), we have \( c \mid m \)
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Let \( c \mid a, b \).
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Then, \( a = cu \) and \( b = cv \) for some \( u, v \in \Z \).
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Then, \( m = ax + by = cux + cvy = c(ux + vy) \).
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Therefore, \( c \mid m \).
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\end{enumerate}
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Thus, the smallest element of \( S \) is \( gcd(a, b) \).
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\end{proof}
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\begin{theorem}
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If \( p \) is irreducible, then \( p \) is prime.
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\end{theorem}
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\begin{proof}
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Let \( p \) be irreducible.
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Suppose \( p | ab \). That is, \( pu = ab \) for some \( u \).
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WTS \( p | a \) or \( p | b \).
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Assume \( p \nmid a \). Then, \( gcd(p, a) = 1 \) since \( p \) irreducible.
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By Bezout's Identity, \( 1 = px + ay \) for some \( x, y \in \Z \).
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Consider \( d = gcd(p, b) \).
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Then, \( b = b(px + ay) = pbx + aby = pbx + puy = p(bx + uy) \).
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In other words, \( p | b \).
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\end{proof}
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\section{Unique Factorization}
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\begin{definition}[Factorization]\index{Factorization}
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A \term{factorization} of \( a \) is a representation of \( a \) as a product of irreducible numbers.
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\end{definition}
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\begin{definition}[Prime (More General)]\index{Prime Number}
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A number \( p \) is \term{prime} if \( p \neq 1 \) and \( p \mid a_1 a_2 \cdots a_k \) implies \( p \mid a_i \) for some \( i \).
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\end{definition}
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\begin{theorem}
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\( \N \) has unique factorization.
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That is, if \( a = p_1 p_2 \cdots p_k = p_k = q_1 q_2 \cdots q_l \) with \( p_i, q_j \) irreducible., then \( k = l \) and \( p_i \) and \( q_j \) are the same up to reordering.
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\end{theorem}
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% \begin{proof}
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% WTS \( k = l \) and \( p_i = q_j \) up to reordering.
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% We will prove this by induction on \( a \).
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% \begin{enumerate}
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% \item Base Case: \( a = 1 \)
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% \( 1 = 1 \) is the only factorization.
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% \item Inductive Step: Assume true for all \( b < a \).
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% Let \( a = p_1 p_2 \cdot p_k = q_1 q_2 \cdot q_l \).
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% WTS \( k = l \) and \( p_i = q_j \) up to reordering.
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% WLOG, assume \( p_1 = q_1 \).
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% Then, \( p_2 \cdot p_3 \cdot p_k = q_2 \cdot q_3 \cdot q_l \).
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% By induction, \( k - 1 = l - 1 \) and \( p_2 \cdot p_3 \cdot p_k = q_2 \cdot q_3 \cdot q_l \).
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% Therefore, \( k = l \) and \( p_i = q_j \) up to reordering.
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% \end{enumerate}
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% \end{proof}
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\begin{proof}
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Since \( p_1 \) irreducible, \( p_1 \) is prime.
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Therefore, \( p_1 | q_1 q_2 \cdots q_l \) implies \( p_1 | q_i \) for some \( i \).
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Say \( p_1 | q_1 \).
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Then, \( q_1 = p_1 u \) for some \( u \).
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Say \( p_1 = q_1 \).
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Then, \( p_2 \cdots p_k = q_2 \cdots q_l \).
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By induction, \( k = l \) and \( p_i = q_j \) up to reordering.
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Therefore, \( \N \) has unique factorization.
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\end{proof}
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\begin{remark}
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Up to \( N \), how many prime numbers are there? \[
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\frac{N}{\log N} + \mathcal{O}\left( N^{1/2} \right)
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\]
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where the \( \mathcal{O} \) term is the Riemann Hypothesis.
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\end{remark}
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\section{Distribution of Primes}
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Are there infinitely many primes? Yes! Euclid proved this 2000 years ago.
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\subsection{Euclid's Algorithm}
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How to show that there are infinitely many primes? Given a list of primes \[
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S = \{ p_1, p_2, \dots, p_k \},
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\]we can construct a new prime \( p \notin S \).
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\begin{example}
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Consider \( S = \{ 2, 3, 5 \} \).
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We suggest \( N = 2 \times 3 \times 5 + 1 = 31 \).
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We know that \( 31 \) is not divisible by \( 2, 3, 5 \), but it has a prime factorization, \( 31 = 1 \times 31 \).
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Thus, there must be some prime \( p = 31 \) that is not in \( S \).
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\end{example}
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\begin{theorem}
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There are infinitely many primes.
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\end{theorem}
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\begin{proof}[Proof (Euclid's Algorithm)]
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Let \( S = \{ p_1, p_2, \dots, p_k \} \) be a list of primes.
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Consider \( N = \left( p_1 \cdot p_2 \cdots p_k \right) + 1 \).
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\( N \) has a remainder of \( 1 \) when divided by any of the primes in \( S \), so it must have a prime factor distinct from the primes in \( S \).
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\end{proof}
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\begin{remark}
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The proof is constructive, but it does not give us a way to find the next prime.
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\end{remark}
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\subsection{Prime Number Theorem}
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How many primes are there up to \( N \)? Denote this as \( \Pi(N) \). As \( N \to \infty \), what is the behavior of \( \Pi(N) \)?
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\subsubsection{Method of Inclusion-Exclusion}
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\begin{remark}
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An obvious statement would be \( \Pi(N) \leq N \).
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\end{remark}
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\begin{itemize}
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\item \textbf{Divisible by 2:} \( \Pi(x) \leq \frac{1}{2} (x-2) + 1 \) because half of the numbers are even.
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The \( x - 2 \) comes from excluding 1 and 2, while \( +1 \) is for 2.
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\item \textbf{Divisible by 3:} \( \Pi(x) \leq \frac{2}{3} (x-3) + 1 \) is a better bound.
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\end{itemize}
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We observe that the divisible by \( 2 \) case is a closer bound \( \sim \frac{x}{2} \). However, we know that numbers divisible by both \( 2 \) and \( 3 \) are also not primes. Using inclusion-exclusion, we can get a better bound \[
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\Pi(x) \leq \text{divisible by 2} + \text{divisible by 3} - \text{divisible by 6}.
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\] This idea is known as the \term{Sieve of Eratosthenes}.
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\begin{remark}[Sieve of Eratosthenes]\index{Sieve of Eratosthenes}
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The Sieve of Eratosthenes is a way to find all primes up to a certain number.
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\begin{enumerate}
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\item Write down all numbers from 2 to \( N \).
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\item Circle 2 and cross out all multiples of 2.
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\item Circle the next number that is not crossed out and cross out all multiples of that number.
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\item Repeat until all numbers are either circled or crossed out.
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\end{enumerate}
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\end{remark}
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Thus, we have the bound \[
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\Pi(x) \leq x \left( 1 - \frac{1}{2} \right) \left( 1 - \frac{1}{3} \right) \left( 1 - \frac{1}{5} \right) \cdots
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\]
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\begin{remark}
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Note that \[
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x \left( 1 - \frac{1}{2} \right) \left( 1 - \frac{1}{3} \right) = x \left( 1 - \frac{1}{2} - \frac{1}{3} + \frac{1}{6} \right),
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\] which is precisely what we get from inclusion-exclusion.
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\end{remark}
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\subsubsection{Method of Prime Factorization}
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Recall that any number has a unique prime factorization \[
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n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}
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\] where \[
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p_i \leq n \qquad \text{and} \qquad e_i \leq \log_{p_i} n \leq \log_2 n.
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\]
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Suppose \( \left( p_1, \dots, p_k \right) \) are all the primes up to \( x \), \( k = \Pi(x) \). There will be \[
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\left( \log_2 x + 1 \right)^k
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\] possible numbers up to \( x \).
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\begin{remark}
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Each \( p_i \) could have power \( 0, 1, \dots, \log_2 x \). Thus, We have \( k \) choices of primes, and each prime has \( \log_2 x + 1 \) choices of powers. This gives us \( \left( \log_2 x + 1 \right)^k \) combinations.
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\end{remark}
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Thus, \begin{align*}
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\left( \log_2 x + 1 \right)^k & \geq x \\
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k \log_2 (\log_2 x + 1) & \geq \log_2 x \\
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k & \geq \frac{\log_2 x}{\log_2 (\log_2 x + 1)}.
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\end{align*}
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Since \( k = \Pi(x) \), we have \[
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\Pi(x) \geq \frac{\log_2 x}{\log_2 (\log_2 x + 1)} \approx \frac{\log_2 x}{\log_2 \log_2 x}.
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\]
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\begin{theorem}[Prime Number Theorem]
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\[
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\lim_{x \to \infty} \Pi(x) - \frac{x}{\ln x} = 0.
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\]
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\end{theorem}
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\subsection{Riemann Zeta Function}
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\begin{definition}[Riemann Zeta Function]\index{Riemann Zeta Function}
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The \term{Riemann Zeta Function} is defined as \[
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\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}.
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\]
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\end{definition}
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\begin{remark}
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For what values of \( s \) does the series converge?
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If \( s > 1 \), then the series converges.
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Recall the integral test \[
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\sum_{n=1}^{\infty} \frac{1}{n^s} \sim \int_1^{\infty} \frac{1}{x^s} \, dx
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\] in terms of convergence and divergence.
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\begin{align*}
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\int_1^\infty \frac{1}{x^s} \, dx
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& = \left. \frac{x^{1-s}}{1-s} \right|_1^\infty \\
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& = \frac{1}{s - 1}
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\end{align*} which is finite if \( 1 - s < 0 \iff s > 1 \).
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\end{remark}
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Why is the Riemann Zeta Function important? It is related to the distribution of primes.
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Consider \[
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\prod_{p \text{ prime}} \left( \sum_{n=1}^{\infty} \frac{1}{p^{n}} \right).
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\] Each term in the expansion looks like \[
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\frac{1}{2^{n_2}} 3^{n_3} 5^{n_5} \cdots
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\] By Unique Factorization, each number of from \( \frac{1}{n} \) appears exactly once in the expansion, so the product is \[
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\prod_{p \text{ prime}} \left( \sum_{n=1}^{\infty} \frac{1}{p^{n}} \right) = \sum_{n=1}^{\infty} \frac{1}{n}.
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\] Similarly, \[
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\prod_{p \text{ prime}} \left( \sum_{n=1}^{\infty} \frac{1}{p^{ns}} \right) = \sum_{n=1}^{\infty} \frac{1}{n^s}.
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\]
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\begin{proof}[Proof due to Euler]
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WTS there are infinitely many primes.
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\[
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\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{1 - \left( \frac{1}{n} \right)^s} = \prod_{p \text{ prime}} \left( \frac{1}{1 - p^{-s}} \right).
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\] If there were finitely many primes, then \[
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\frac{1}{1 - p^{-1}} \neq \infty
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\] so finite product of such terms would be finite. However, we know that \( \zeta(1) \) diverges, so there must be infinitely many primes.
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\end{proof}
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\begin{remark}
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\[
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\zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}
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\] which is irrational.
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If there were finitely many primes, then \[
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\zeta(2) = \prod_{i=1}^k \left( 1 - \frac{1}{{p_i}^2} \right)^{-1}
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\] would be rational, which is a contradiction.
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\end{remark}
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Alternatively, consider the harmonic series \[
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H(x) = \sum_{n=1}^{x} \frac{1}{n} \sim \ln x \sim \int_2^x \frac{1}{t} \, dt = \frac{\ln x}{\ln 2}.
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\]
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\begin{remark}
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\[
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H(x) \sim \prod_{p \leq x} \left( \frac{1}{1 - p^{-1}} \right).
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\]
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We can use this to bound the number of primes.
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\end{remark}
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Analyzing the zeta function on the complex plain is very important in bounding the number of primes. We will see this at the end of the course.
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\begin{example}
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Are there infinitely many primes of the form \( 4k + 3, k \in \N \)?
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Examples include \( 3, 7, 11, 19, 23, \dots \).
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Yes.
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\begin{proof}
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Suppose there are finitely many primes of the form \( 4k + 3 \), \( p_1, p_2, \dots, p_k \).
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Consider \( N = 4 \left( \prod_{i=1}^k p_i \right) + 3 \), \( N \) coprime to all \( p_i \).
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\( N \) must have a prime factor of the form \( 4k + 3 \).
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Suppose not. Then, all prime factors of \( N \) are of the form \( 4k + 1 \).
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Consider \( 4m + 1 \) and \( 4n + 1 \).
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Then, \( (4m+1) \cdot (4n+1) = 4(4mn + m + n) + 1 \) is also of the form \( 4k + 1 \). This is a contradiction.
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\end{proof}
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\end{example} |