upload diverse filer

Diskret Mat/Primes and the Euclidean algorithm/Primes and the Eucledian algorithm.typ

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+#import "@local/dtu-template:0.5.1":*
+
+
+#show: dtu-note.with(
+  course: "01017",
+  course-name: "Discrete Mathematics",
+  title: "Primes and the Eucledian algorithm",
+  date: datetime(day: 27, month: 11, year: 2025),
+  author: "Rasmus Rosendahl-Kaa (S255955)",
+  semester: "2025 Fall",
+)
+
+= Primes
+#definition()[
+$
+a,b,c in ZZ, a in ZZ_+
+$
+If $a = b dot c$ then $a$ is _composite_ (sammensat), if not, then $a$ is a prime,
+
+]

Diskret Mat/Primes and the Euclidean algorithm/test.latex

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+\documentclass{article}
+\usepackage{amsmath}
+\usepackage{amssymb}
+\usepackage{amsthm}
+
+\newtheorem{theorem}{Theorem}
+\newtheorem{lemma}[theorem]{Lemma}
+\newtheorem{corollary}[theorem]{Corollary}
+\theoremstyle{definition}
+\newtheorem{definition}[theorem]{Definition}
+
+\title{Primes and the Euclidean Algorithm}
+\date{02-10-2025}
+
+\begin{document}
+
+\maketitle
+
+\section{Introduction}
+
+Let $a, b \in \mathbb{Z}$. We say that $a$ \textbf{divides} $b$, written $a \mid b$, if there exists $c \in \mathbb{Z}$ such that $b = ac$.
+
+\begin{definition}
+A positive integer $p > 1$ is called \textbf{prime} if its only positive divisors are $1$ and $p$ itself.
+\end{definition}
+
+\begin{definition}
+For integers $a$ and $b$, not both zero, the \textbf{greatest common divisor} $\gcd(a,b)$ is the largest positive integer that divides both $a$ and $b$.
+\end{definition}
+
+\section{The Euclidean Algorithm}
+
+The Euclidean algorithm is an efficient method for computing the greatest common divisor of two integers.
+
+\begin{theorem}[Division Algorithm]
+Let $a, b \in \mathbb{Z}$ with $b > 0$. Then there exist unique integers $q$ and $r$ such that
+\[
+a = bq + r \quad \text{with} \quad 0 \leq r < b.
+\]
+Here $q$ is called the \textbf{quotient} and $r$ is called the \textbf{remainder}.
+\end{theorem}
+
+\begin{theorem}
+If $a = bq + r$, then $\gcd(a,b) = \gcd(b,r)$.
+\end{theorem}
+
+\begin{proof}
+Let $d = \gcd(a,b)$. Then $d \mid a$ and $d \mid b$. Since $r = a - bq$, we have $d \mid r$. Thus $d$ is a common divisor of $b$ and $r$, so $d \leq \gcd(b,r)$.
+
+Conversely, let $d' = \gcd(b,r)$. Then $d' \mid b$ and $d' \mid r$. Since $a = bq + r$, we have $d' \mid a$. Thus $d'$ is a common divisor of $a$ and $b$, so $d' \leq \gcd(a,b) = d$.
+
+Therefore $d = \gcd(b,r)$.
+\end{proof}
+
+\subsection{The Algorithm}
+
+To compute $\gcd(a,b)$ where $a \geq b > 0$:
+\begin{enumerate}
+\item Apply the division algorithm repeatedly:
+\begin{align*}
+a &= bq_1 + r_1, \quad 0 \leq r_1 < b \\
+b &= r_1 q_2 + r_2, \quad 0 \leq r_2 < r_1 \\
+r_1 &= r_2 q_3 + r_3, \quad 0 \leq r_3 < r_2 \\
+&\vdots \\
+r_{n-2} &= r_{n-1} q_n + r_n, \quad 0 \leq r_n < r_{n-1} \\
+r_{n-1} &= r_n q_{n+1} + 0
+\end{align*}
+
+\item The last non-zero remainder $r_n$ is $\gcd(a,b)$.
+\end{enumerate}
+
+\begin{theorem}[Bézout's Identity]
+Let $a, b \in \mathbb{Z}$, not both zero, and let $d = \gcd(a,b)$. Then there exist integers $x$ and $y$ such that
+\[
+ax + by = d.
+\]
+\end{theorem}
+
+\begin{proof}
+Consider the set $S = \{ax + by : x, y \in \mathbb{Z} \text{ and } ax + by > 0\}$. This set is non-empty (contains $|a|$ or $|b|$) and bounded below by $1$, so by the well-ordering principle it has a smallest element, say $d' = ax_0 + by_0$.
+
+We claim that $d' = \gcd(a,b)$. First we show that $d' \mid a$. By the division algorithm, write $a = d'q + r$ with $0 \leq r < d'$. Then
+\[
+r = a - d'q = a - (ax_0 + by_0)q = a(1 - x_0q) + b(-y_0q).
+\]
+If $r > 0$, then $r \in S$ and $r < d'$, contradicting the minimality of $d'$. Thus $r = 0$ and $d' \mid a$. Similarly, $d' \mid b$.
+
+So $d'$ is a common divisor of $a$ and $b$, hence $d' \leq d = \gcd(a,b)$.
+
+Conversely, since $d \mid a$ and $d \mid b$, we have $d \mid (ax_0 + by_0) = d'$. Thus $d \leq d'$.
+
+Therefore $d = d'$, completing the proof.
+\end{proof}
+
+\end{document}

Diskret Mat/Primes and the Euclidean algorithm/test.typ

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+= Introduction
+<introduction>
+Let $a , b in bb(Z)$. We say that $a$ #strong[divides] $b$, written
+$a divides b$, if there exists $c in bb(Z)$ such that $b = a c$.
+
+#block[
+#strong[Definition 1];. A positive integer $p > 1$ is called
+#strong[prime] if its only positive divisors are $1$ and $p$ itself.
+
+Hello
+]
+#block[
+#strong[Definition 2];. For integers $a$ and $b$, not both zero, the
+#strong[greatest common divisor] $gcd (a , b)$ is the largest positive
+integer that divides both $a$ and $b$.
+
+]
+= The Euclidean Algorithm
+<the-euclidean-algorithm>
+The Euclidean algorithm is an efficient method for computing the
+greatest common divisor of two integers.
+
+#block[
+#strong[Theorem 3] (Division Algorithm). #emph[Let $a , b in bb(Z)$ with
+$b > 0$. Then there exist unique integers $q$ and $r$ such that
+$ a = b q + r quad upright("with") quad 0 lt.eq r < b . $ Here $q$ is
+called the #strong[quotient] and $r$ is called the #strong[remainder];.]
+
+]
+#block[
+#strong[Theorem 4];. #emph[If $a = b q + r$, then
+$gcd (a , b) = gcd (b , r)$.]
+
+]
+#block[
+#emph[Proof.] Let $d = gcd (a , b)$. Then $d divides a$ and
+$d divides b$. Since $r = a - b q$, we have $d divides r$. Thus $d$ is a
+common divisor of $b$ and $r$, so $d lt.eq gcd (b , r)$.
+
+Conversely, let $d' = gcd (b , r)$. Then $d' divides b$ and
+$d' divides r$. Since $a = b q + r$, we have $d' divides a$. Thus $d'$
+is a common divisor of $a$ and $b$, so $d' lt.eq gcd (a , b) = d$.
+
+Therefore $d = gcd (b , r)$.~◻
+
+]
+== The Algorithm
+<the-algorithm>
+To compute $gcd (a , b)$ where $a gt.eq b > 0$:
+
++ Apply the division algorithm repeatedly:
+  $ a & = b q_1 + r_1 , quad 0 lt.eq r_1 < b\
+  b & = r_1 q_2 + r_2 , quad 0 lt.eq r_2 < r_1\
+  r_1 & = r_2 q_3 + r_3 , quad 0 lt.eq r_3 < r_2\
+   & dots.v\
+  r_(n - 2) & = r_(n - 1) q_n + r_n , quad 0 lt.eq r_n < r_(n - 1)\
+  r_(n - 1) & = r_n q_(n + 1) + 0 $
+
++ The last non-zero remainder $r_n$ is $gcd (a , b)$.
+
+#block[
+#strong[Theorem 5] (Bézout’s Identity). #emph[Let $a , b in bb(Z)$, not
+both zero, and let $d = gcd (a , b)$. Then there exist integers $x$ and
+$y$ such that $ a x + b y = d . $]
+
+]
+#block[
+#emph[Proof.] Consider the set
+$S = { a x + b y : x , y in bb(Z) upright(" and ") a x + b y > 0 }$.
+This set is non-empty (contains $lr(|a|)$ or $lr(|b|)$) and bounded
+below by $1$, so by the well-ordering principle it has a smallest
+element, say $d' = a x_0 + b y_0$.
+
+We claim that $d' = gcd (a , b)$. First we show that $d' divides a$. By
+the division algorithm, write $a = d' q + r$ with $0 lt.eq r < d'$. Then
+$ r = a - d' q = a - (a x_0 + b y_0) q = a (1 - x_0 q) + b (- y_0 q) . $
+If $r > 0$, then $r in S$ and $r < d'$, contradicting the minimality of
+$d'$. Thus $r = 0$ and $d' divides a$. Similarly, $d' divides b$.
+
+So $d'$ is a common divisor of $a$ and $b$, hence
+$d' lt.eq d = gcd (a , b)$.
+
+Conversely, since $d divides a$ and $d divides b$, we have
+$d divides (a x_0 + b y_0) = d'$. Thus $d lt.eq d'$.
+
+Therefore $d = d'$, completing the proof.~◻
+
+]