### 数学代写|解析数论作业代写Analytic Number Theory代考|MATH3170

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• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|解析数论作业代写Analytic Number Theory代考|The Euclidean Algorithm

In this section, we prove a result that allows us to compute the greatest common divisor of two integers. First, we need a lemma.
Lemma 1.11. Let a, $b, q$, and $r$ be integers such that
$$a=b q+r,$$
then
$$(a, b)=(b, r) .$$
Proof. Let $d=(a, b)$ and $d^{\prime}=(b, r)$. Note that since $d \mid a$ and $d \mid b$, we find that $d \mid(a-b q)$ by Theorem $1.6$ (c). Hence, $d \mid r$ and $d$ is a common divisor of $b$ and $r$. By Definition $1.4$ (c), $d \mid d^{\prime}$ since $d^{\prime}=(b, r)$. Similarly, $d^{\prime} \mid b$ and $d^{\prime} \mid r$ implies that $d^{\prime} \mid(b q+r)$ by Theorem $1.6$ (c) and consequently, $d^{\prime} \mid a$. By Definition $1.4$ (c), $d^{\prime} \mid d$ since $d=(a, b)$. Therefore, by Theorem $1.6$ (i), $d=d^{\prime}$.

Theorem 1.12 (The Euclidean Algorithm). Given positive integers a and $b$, where $b \nmid a$. Let $r_0=a, r_1=b$, and apply the division algorithm repeatedly to obtain a set of remainders $r_2, r_3, \ldots, r_n, r_{n+1}$ defined successively by the relations
$$\begin{array}{ccc} r_0 & =r_1 q_1+r_2 & 0<r_2<r_1 \ r_1 & =r_2 q_2+r_3 & 0<r_3<r_2 \ \vdots & \ r_{n-2}=r_{n-1} q_{n-1}+r_n & 0<r_n<r_{n-1} \ r_{n-1}=r_n q_n+r_{n+1} & r_{n+1}=0 . \end{array}$$
Then $r_n$, the last nonzero remainder in this process is $(a, b)$, the greatest common divisor of $a$ and $b$.

Proof. There is a stage at which $r_{n+1}=0$ because the $r_i$ are decreasing and non-negative. Next, applying Lemma 1.11, we find that
$$(a, b)=\left(r_0, r_1\right)=\left(r_1, r_2\right)=\cdots=\left(r_n, r_{n+1}\right)=\left(r_n, 0\right)=r_n .$$
This completes the proof of Theorem 1.12.

## 数学代写|解析数论作业代写Analytic Number Theory代考|Fundamental Theorem of Arithmetic

Theorem $1.17$ (Fundamental Theorem of Arithmetic). Every positive integer $n>1$ can be expressed as a product of primes; this representation is unique apart from the order in which the factors occur.

Proof. We first show that $n$ can be expressed as a prime or a product of primes. We use induction on $n$. The statement is clearly true for $n=2$ since 2 is a prime. Suppose $m$ is a prime or a product of primes for $2 \leq m \leq n-1$. If $n$ is a prime then we are done. Suppose $n$ is composite then $n=a b$, where $11$ is a prime or a product of primes.

To prove uniqueness, we use induction on $n$ again. If $n=2$ then the representation of $n$ as a product of primes is clearly unique. Assume, then that it is true for all integers greater than 1 and less than $n$. We shall prove that it is also true for $n$. If $n$ is prime, then there is nothing to prove. Assume, then, that $n$ is composite and that $n$ has two factorizations, say,
$$n=p_1 p_2 \cdots p_s=q_1 q_2 \cdots q_t .$$
Since $p_1$ divides the product $q_1 q_2 \cdots q_t$, it must divide at least one factor by Corollary 1.16. Relabel $q_1, q_2, \ldots, q_t$ so that $p_1 \mid q_1$. Then $p_1=q_1$ since both $p_1$ and $q_1$ are primes. In (1.4), we may cancel $p_1$ on both sides to obtain
$$n / p_1=p_2 \cdots p_s=q_2 \cdots q_t .$$
Now the induction hypothesis implies that the two factorizations of $n / p_1$ must be the same, apart from the order of the factors. Therefore, $s=t$ and the factorizations in (1.4) are also identical, apart from order. This completes the proof.

# 解析数论代写

## 数学代写|解析数论作业代写Analytic Number Theory代考|The Euclidean Algorithm

$$a=b q+r$$

$$(a, b)=(b, r) .$$

(C)。因此， $d \mid r$ 和 $d$ 是公约数 $b$ 和 $r$. 根据定义 $1.4$ (C) ， $d \mid d^{\prime}$ 自从 $d^{\prime}=(b, r)$. 相似地， $d^{\prime} \mid b$ 和 $d^{\prime} \mid r$ 暗示 $d^{\prime} \mid(b q+r)$ 通过定理1.6(c) 因此， $d^{\prime} \mid a$. 根据定义 $1.4$ (C)， $d^{\prime} \mid d$ 自从 $d=(a, b)$. 因 此，由定理 $1.6$ (一世)， $d=d^{\prime}$.

$$r_0=r_1 q_1+r_2 \quad 0<r_2<r_1 r_1=r_2 q_2+r_3 \quad 0<r_3<r_2 \vdots \quad r_{n-2}=r_{n-1} q_{n-1}+r_n$$

$$(a, b)=\left(r_0, r_1\right)=\left(r_1, r_2\right)=\cdots=\left(r_n, r_{n+1}\right)=\left(r_n, 0\right)=r_n .$$

## 数学代写|解析数论作业代写Analytic Number Theory代考|Fundamental Theorem of Arithmetic

$$n=p_1 p_2 \cdots p_s=q_1 q_2 \cdots q_t$$

$$n / p_1=p_2 \cdots p_s=q_2 \cdots q_t$$

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## MATLAB代写

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