Lecture 1 - 2025-01-06

chapters/chapter1.tex

@@ -1 +1,218 @@
 \chapter{Introduction}
+
+A number is a way of identifying what is \textit{common} between different collections of objects. What is common is that 3 of anything can be put in correspondence with each other.
+
+\section{Notations}
+
+\begin{definition}
+    A set \( S \) is has a \term{size} \( |S| \).
+\end{definition}
+
+\begin{remark}
+    If \( S = \{ a, b, c \} \) and \( T = \{ 1, 2, 3 \} \), then \( |S| = |T| = 3 \).
+\end{remark}
+
+\subsubsection{Some Operations on Sets}
+
+% Here we ignore the fact that some objects may be in both sets. \textbf{Only the size of the sets are considered.}
+
+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}.
+
+\begin{itemize}
+    \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 \( T \). \[
+              | S \cup T | = |S| + |T|
+          \]
+
+    \item For \( T \subseteq S \), the \term{difference}\index{Set!Difference} of \( S \) and \( T \), \( S \setminus T \), is removing \( T \) from \( S \), \[
+              | S \setminus T | = |S| - |T|
+          \]
+
+    \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 \( T \). \[
+              | S \times T | = |S| \cdot |T|
+          \]
+\end{itemize}
+
+\section{National Numbers}
+
+\begin{definition}\index{National Number}
+    \term{National numbers} are a way of identifying the size of a set. The set of all natural numbers is denoted by \( \N \).
+\end{definition}
+
+\begin{remark}[Notation]
+    To disambiguate between whether 0 is included in the set of natural numbers, we use
+    \begin{itemize}
+        \item \( \N_{\geq 0} \) to include 0,
+        \item \( \N_{> 0} \) to exclude 0.
+    \end{itemize}
+\end{remark}
+
+\begin{remark}[Observation]
+    Every \( |S| \) appears somewhere in this list. Size gets \textbf{smaller} as you remove things.
+\end{remark}
+
+\begin{axiom}[Peano Axioms]\index{Peano Axioms}
+    (for \( \N_{\geq 0} \))
+
+    \begin{enumerate}%[label=(\Roman*)]
+        \item There exists an element \( 0 \).
+        \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.
+        \item \textbf{Induction Schema} For any property \( P \)  on \( N \), if
+              \begin{itemize}
+                  \item \( P(0) \) is true, and
+                  \item \( P(n) \implies P(S(n)) \)
+              \end{itemize}
+              Then \( P(n) \) is true for all \( n \in \N_{\geq 0} \).
+    \end{enumerate}
+\end{axiom}
+
+\begin{example}[System Where III Fails]
+    Take \( \Z \in \N_{\geq 0} \cup -\N_{\geq 0} \), the set of all integers. \( S(n) = n + 1 \).
+\end{example}
+
+\section{Multiplication and Division}
+
+\begin{definition}[Divisibility]\index{Divisibility}
+    \( n \) divides \( m \) if there exists \( u \) such that \( m = nu \). We write \( a \mid b \).
+\end{definition}
+
+\begin{definition}[Prime Number (Definition I - Irreducible Number)]\index{Prime Number}\index{Irreducible Number}
+    A number \( p \) is \term{prime} if \( p \) is not 1 and the only divisors of \( p \) are 1 and \( p \).
+\end{definition}
+
+\begin{definition}[Prime Number (Definition II - Prime Number)]\index{Prime Number}
+    A number \( p \) is prime if \( \forall a, b \) such that \( p \mid ab \), \( p \mid a \) or \( p \mid b \).
+\end{definition}
+
+\begin{theorem}
+    Prime $\implies$ reducible
+\end{theorem}
+
+\begin{proof}
+    WTS \( n \) is reducible \( \implies n \) is not prime.
+
+    Assume \( n \) is reducible, \( n = ab \) with \( a, b \neq 1 \).
+
+    Then, \( n \mid ab \), but \( n \nmid a \) and \( n \nmid b \) as \( n > a, b \).
+\end{proof}
+
+\begin{remark}[Irreducible $\not\implies$ Prime]
+    Take \( \Z \left[ \sqrt{-5} \right] = \{ a + b\sqrt{-5}, {~~~} a, b \in \Z \} \)
+
+    Let \( a = 1 + \sqrt{-5} \), \( b = 1 - \sqrt{-5} \).
+
+    \begin{align*}
+        a + b & = 3 + 2 \sqrt{5}                 \\
+        a b   & = (1 + \sqrt{-5})(1 - \sqrt{-5}) \\
+              & = 1 - \sqrt{-5} + \sqrt{-5} + 5  \\
+              & = -4
+    \end{align*}
+
+    Now consider the number $6$. \[
+        6 = 2 \cdot 3 = (1 + \sqrt{-5})(1 - \sqrt{-5})
+    \]
+
+    Turns out \( 2, 3, 1 \pm \sqrt{-5} \) are all irreducible but not prime.
+
+    For example, $2 \mid (1 + \sqrt{-5})(1 - \sqrt{-5})$ but $2 \nmid 1 \pm \sqrt{-5}$.
+\end{remark}
+
+\begin{definition}[Greatest Common Divisor]\index{Greatest Common Divisor}
+    The \term{greatest common divisor} of \( a \) and \( b \) is the largest number that divides both \( a \) and \( b \).
+
+    \( d = gcd(a, b) \) if \( d \mid a \), \( d \mid b \), and for any \( c \) such that \( c \mid a, b \), we have \( c \mid d \).
+\end{definition}
+
+\begin{remark}
+    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 \( c \nmid d \).
+\end{remark}
+
+\begin{theorem}\label{thm:gcd-exists}
+    For any \( a, b \in \N_{> 0} \), \( gcd(a, b) \) exists.
+\end{theorem}
+
+\begin{remark}
+    \begin{minipage}[t]{0.45\linewidth}
+        Consider \( a = 30 \) and \( b = 12 \).
+
+        \begin{center}
+            \begin{tikzpicture}[
+                    every node/.style = {draw, circle, minimum size = 0.75cm},
+                ]
+                \node (1) at (1, 0) {\( 1 \)};
+                \node (2) at (0, 1) {\( 2 \)};
+                \node (3) at (2, 1) {\( 3 \)};
+                \node (6) at (1, 2) {\( 6 \)};
+
+                \draw (1) -- (2);
+                \draw (1) -- (3);
+                \draw (2) -- (6);
+                \draw (3) -- (6);
+            \end{tikzpicture}
+        \end{center}
+    \end{minipage}
+    \begin{minipage}[t]{0.45\linewidth}
+        Consider \( a = 60 \) and \( b = 15 \).
+
+        \begin{center}
+            \begin{tikzpicture}[
+                    every node/.style = {draw, circle, minimum size = 0.75cm},
+                ]
+                \node (1) at (1, 0) {\( 1 \)};
+                \node (3) at (0, 1) {\( 3 \)};
+                \node (5) at (2, 1) {\( 5 \)};
+                \node (15) at (1, 2) {\( 15 \)};
+
+                \draw (1) -- (3);
+                \draw (1) -- (5);
+                \draw (3) -- (15);
+                \draw (5) -- (15);
+            \end{tikzpicture}
+        \end{center}
+    \end{minipage}
+\end{remark}
+
+\begin{theorem}[Bezout's Identity]\index{Bezout's Identity}
+    For any \( a, b \in \N_{> 0} \), there exist \( x, y \in \Z \) such that \( ax + by = gcd(a, b) \).
+\end{theorem}
+
+\begin{example}
+    Consider $a = 7$ and $b = 9$, $gcd(7, 9) = 1$.
+
+    We have $7 \cdot 5 + 9 \cdot (-4) = 1$.
+\end{example}
+
+Define \[ S = \{ ax + by \mid x, y \in \Z, ax + by > 0 \}. \]
+
+\begin{lemma}
+    \( gcd \) is the smallest (positive) element of \( S \).
+\end{lemma}
+
+\begin{proof}
+    WTS \( gcd(a, b) = d \iff d \mid a, b \) and \( d \) for any \( c \mid a, b \), we have \( c \mid d \).
+
+    Let $m$ be the smallest element of $S$. $m = ax + by$ for some $x, y \in \Z$.
+
+    \begin{enumerate}
+        \item WTS \( m \mid a, b \)
+        
+        Let \( a = km + r \) where \( 0 \leq r < m \).
+
+        Then, \( r = a - km = a(1 - kx) + b(ky) \) is in \( S \cup \{ 0 \} \), but \( m \) is the smallest element in \( S \) so \( r < m \). 
+        
+        Therefore, \( r = 0 \), \( a = km \), and $m \mid a$.
+
+        The same argument can be applied to \( b \).
+
+        \item WTS for any \( c \mid a, b \), we have \( c \mid m \)
+        
+        Let \( c \mid a, b \). 
+        
+        Then, \( a = cu \) and \( b = cv \) for some \( u, v \in \Z \).
+
+        Then, \( m = ax + by = cux + cvy = c(ux + vy) \).
+        
+        Therefore, \( c \mid m \).
+    \end{enumerate}
+
+    Thus, the smallest element of \( S \) is \( gcd(a, b) \).
+\end{proof}

config/config.tex

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+\newcommand{\Z}{\mathbb{Z}}
 
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+
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+
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