## 数学代写|数论作业代写number theory代考|MATH4307

statistics-lab™ 为您的留学生涯保驾护航 在代写数论number theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写数论number theory代写方面经验极为丰富，各种代写数论number theory相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|数论作业代写number theory代考|Answers to Selected Supplementary Problems

1.13. (a) True since $644=7(92)$. (b) True since $644=-7(-92)$.
(c) False since the remainder is 4 , not $0 . \quad$ (d) True since $2916=$ $243(12)$
(e) True since $16 m-12=4(4 m-3)$. (f) True since $-7 m=$ $m(-7)$.
(g) False since the remainder is $3 m$, not 0 . (h) True since $-k^{3}+$ $2 k=k\left(-k^{2}+2\right)$.
1.15. (a) $487=14(34)+11$, i.e., $q=34$ and $r=11$.
(b) $-386=27(-15)+19$, i.e., $q=-15$ and $r=19$.
1.16. Since $486=15 a+6$, we get $a=480 / 15=32$.
1.17. $44=2^{2} \cdot 11,111=3 \cdot 37$, so $\operatorname{gcd}(44,111)=1$.
1.19. (a) $F_{9}=34, F_{10}=55, F_{11}=89, F_{12}=144$.
1.20. (a) $84 . \quad$ (b) $525 .$
1.22. (a) $44=\left(2^{2}\right)(11)$ and $111=(3)(37)$, so $\operatorname{gcd}(44,111)=1$.
(b) $111=44(2)+23,44=23(1)+21,23=21(1)+2,21=2(10)+1$.
1.23. By the Euclidean Algorithm, we have $71=23(3)+2$ and $23=2(11)+1$.

Hence $1=23-2(11)=23-(71-23(3))(11)=23(34)+71(-11)$, so $x=34$ and $y=-11$.

## 数学代写|数论作业代写number theory代考|Listing Primes: The Sieve of Eratosthenes

A second question we now ask is, How can we list all of the prime numbers up to some positive value $n \geq 2$ ? A method to do this is known as the Sieve of Eratosthenes, named in honor of Eratosthenes $(276 \mathrm{BC}-194 \mathrm{BC})$, who appears to be the first to make use of this process. The process, described below, is quite efficient as long as $n$ isn’t too large.

We begin by listing all of the numbers from 2 to $n$. Then since 2 is prime, we leave it in the list and delete all multiples of 2 (except 2 itself) up to and including $n$. That knocks out all the even numbers in our list larger than 2 . We then leave 3 and delete all larger multiples of 3 . The next value not already deleted is 5 , so we leave it and delete all multiples of 5 . We continue this process with 7 which is yet to be deleted, then 11, ctc. The numbers remaining in the list give all primes up to $n$. A question you might have is, When can we stop this process so that we have indeed listed all the primes up to $n$ ? You are asked in Problem $2.2$ to show that we need only process primes which are less than or equal to the square root of $n$.

Example 2.2. We illustrate the Sieve of Eratosthenes by finding all primes up to $n=50$. We begin by listing all of the positive integers from 2 through 50 . By what we just stated, we need only process $2,3,5$, and 7 since $11>\sqrt{50}$.
$\begin{array}{rrrrrrrrrr} & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \ 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 \ 21 & 22 & 23 & 24 & 25 & 26 & 27 & 28 & 29 & 30 \ 31 & 32 & 33 & 34 & 35 & 36 & 37 & 38 & 39 & 40 \ 41 & 42 & 43 & 44 & 45 & 46 & 47 & 48 & 49 & 50\end{array}$
We boldface the number 2 as the first item in this list (since we know it is prime), and then cross out each multiple of 2 that is greater than 2. It is important to note that no actual arithmetic must be done here! We simply start at 2, skip by the amount of 2 (which gets us to the number 4), cross out the 4 , then skip by another 2 to get to 6 , cross out the 6 , and so on. This stage of the process is quite straightforward. This now leaves us with the following table.

## 数学代写|数论作业代写number theory代考|Answers to Selected Supplementary Problems

1.13。 (a) 真因为 $644=7(92)$. (b) 真因为 $644=-7(-92)$.
(c) 假，因为余数是 4 ，不是 $0 . \quad$ (d) 真因为 $2916=243(12)$
(e) 真因为 $16 m-12=4(4 m-3)$. (f) 真因为 $-7 m=m(-7)$.
(g) 假，因为余数是 $3 m$ ，而不是 0 。(h) 真因为 $-k^{3}+2 k=k\left(-k^{2}+2\right)$.
1.15。(一个) $487=14(34)+11$ ，那是， $q=34$ 和 $r=11$.
(二) $-386=27(-15)+19$ ， 那是， $q=-15$ 和 $r=19$.
1.16。自从 $486=15 a+6$ ，我们得到 $a=480 / 15=32$.
1.17. $44=2^{2} \cdot 11,111=3 \cdot 37$ ，所以 $\operatorname{gcd}(44,111)=1$.
1.19。 (-个) $F_{9}=34, F_{10}=55, F_{11}=89, F_{12}=144$.
1.20。(一个) 84 . (b) 525 .
1.22。(一个) $44=\left(2^{2}\right)(11)$ 和 $111=(3)(37)$ ，所以 $\operatorname{gcd}(44,111)=1$.
(二) $111=44(2)+23,44=23(1)+21,23=21(1)+2,21=2(10)+1$.
1.23。根据欧几里得算法，我们有 $71=23(3)+2$ 和 $23=2(11)+1$.

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 数学代写|数论作业代写number theory代考|MATH1001

statistics-lab™ 为您的留学生涯保驾护航 在代写数论number theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写数论number theory代写方面经验极为丰富，各种代写数论number theory相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|数论作业代写number theory代考|Greatest Common Divisors

1.17. Find $\operatorname{gcd}(44,111)$ using factorization into powers of prime numbers.
1.18. For a positive integer $a$, what are the possibilities for the quantity $\operatorname{gcd}(a+3, a)$ ? Find specific examples to demonstrate each possibility. Now prove your conjecture. (Hint: Suppose $d$ divides both $a$ and $a+3$, then by Lemma $1.1$ Part (ii) $\cdots$.)
1.19. The sequence of numbers $1,1,2,3,5,8,13,21, \ldots$ is known as the sequence of Fibonacci number. After the first two values, a given number is obtained as the sum of the previous two numbers. We denote this sequence of positive integers by $F_{1}, F_{2}, F_{3}, \ldots$, in honor of Fibonacci who first wrote about these numbers in his book “Liber Abaci,” which was published in $1202 .$
(a) Above we have written $F_{1}$ through $F_{8}$. Write down $F_{y}$ through $F_{12}$.
(b) Prove that for any positive integer $k \geq 1, \operatorname{gcd}\left(F_{k}, F_{k+1}\right)=1$, i.e., prove that any two consecutive Fibonacci numbers are relatively prime. (Hint: Suppose $d$ divides both $F_{k+1}$ and $F_{k}$; then by Lemma 1.1 Part (ii) (or (iii)) it divides their difference, which is what by the definition of these numbers? Hence $d$ must divide $F_{k}$ and $F_{k-1}$. Continue this process, concluding finally that we must have $d=1$.
1.20. (a) What is $\operatorname{lcm}(21,28)$ ?
(b) What is $\operatorname{lcm}(21,25)$ ? (See Solved Problem 1.8.)
1.21. Prove that $\operatorname{lcm}(a, b)=\frac{a \cdot b}{\operatorname{gcd}(a, b)}$.

## 数学代写|数论作业代写number theory代考|Euclidean Algorithm

1.22. Find $\operatorname{gcd}(44,111)$ using the Euclidean Algorithm.
1.23. The numbers 23 and 71 are relatively prime since both are themselves prime. Find integers $(x, y)$ such that $23 x+71 y=1$.
1.24. (a) Find $\operatorname{gcd}(381,3837)$ using the Euclidean Algorithm.
(b) Find integers $(x, y)$ such that $\operatorname{gcd}(381,3837)=381 x+3837 y$.
1.25. How many steps must the Euclidean Algorithm take to find the gcd of two positive integers $a$ and $b$ ? We have seen through examples and problems that it can vary, but what is the “worst case?” It can be shown that if $a$ is the first divisor, then the total number of steps can be no more than 7 times the number of decimal digits of $a$.
(a) Suppose the first divisor $a$ is between a billion and 9 billion (and the dividend $b$ is larger). What is the maximum number of steps to discover $\operatorname{gcd}(a, b)$ by the Euclidean Algorithm?
(b) Returning to the Fibonacci numbers (Supplementary Problem 1.19), we saw that $F_{11}=89$ and $F_{12}=144$. According to the above information, what is the maximum number of steps needed to compute $\operatorname{gcd}\left(F_{11}, F_{12}\right)$ using the Euclidean Algorithm? Now do the actual computation and check the number of steps. (Note: This part illustrates that computing the gcd of two adjacent Fibonacci numbers using the Euclidean Algorithm goes about as slowly as possible.)

## 数学代写|数论作业代写number theory代考|Greatest Common Divisors

1.17。寻找 $\operatorname{acd}(44,111)$ 使用因式分解为素数的幂。
1.18。对于一个正整数 $a$, 数量的可能性是什么 $\operatorname{gcd}(a+3, a)$ ? 找到具体的例子来展示每一种可能性。现在证明你的 猜想。（提示：假设 $d$ 将两者分开 $a$ 和 $a+3$ ，然后由引理 $1.1$ 第 (ii) 部分 $\cdots$ )
1.19。数字序列 $1,1,2,3,5,8,13,21, \ldots$ 被称为斐波那契数列。在前两个值之后，得到一个给定的数字作为前两 个数字的总和。我们将这个正整数序列表示为 $F_{1}, F_{2}, F_{3}, \ldots$, 以纪念斐波那契，他在他的书“Liber Abaci”中首次 写到这些数字，该书发表于 1202 .
(a) 上面我们写了 $F_{1}$ 通过 $F_{8}$. 写下 $F_{y}$ 通过 $F_{12}$.
(b) 证明对于任何正整数 $k \geq 1, \operatorname{gcd}\left(F_{k}, F_{k+1}\right)=1$ ，即证明任何两个连续的斐波那契数都是互质的。（提示: 假 设 $d$ 将两者分开 $F_{k+1}$ 和 $F_{k}$; 然后通过引理 $1.1$ 第 (ii) 部分 (或 (iii)) 它划分它们的差异，这些数字的定义是什么? 因此 $d$ 必须分开 $F_{k}$ 和 $F_{k-1}$. 继续这个过程，最后得出结论，我们必须有 $d=1$.
1.20。(一) 什么是 $\operatorname{lcm}(21,28)$ ?
(b) 什么是 $\operatorname{lcm}(21,25)$ ? (见已解决的问题 1.8。)
1.21。证明 $\operatorname{lcm}(a, b)=\frac{a \cdot b}{\operatorname{gcd}(a, b)}$.

## 数学代写|数论作业代写number theory代考|Euclidean Algorithm

1.22。寻找gcd $(44,111)$ 使用欧几里得算法。
1.23。数字 23 和 71 是相对质数，因为它们本身都是质数。查找整数 $(x, y)$ 这样 $23 x+71 y=1$.
1.24。(a) 查找 $\operatorname{gcd}(381,3837)$ 使用欧几里得算法。
(b) 求整数 $(x, y)$ 这样 $\operatorname{gcd}(381,3837)=381 x+3837 y$.
1.25。欧几里得算法需要多少步才能找到两个正整数的 $g c d a$ 和 $b$ ? 我们已经通过例子和问题看到了它可能会有所不 同，但“最坏的情况”是什么? 可以证明，如果 $a$ 是第一个除数，那么总步数不能超过小数位数的7倍 $a$.
(a) 假设第一个除数 $a$ 介于 10 亿和 90 亿之间（以及股息 $b$ 更大）。发现的最大步数是多少 $\operatorname{gcd}(a, b)$ 欧几里得算法?
(b) 回到斐波那契数列（补充问题 1.19），我们看到 $F_{11}=89$ 和 $F_{12}=144$. 根据以上信息，计算需要的最大步数 是多少 $\operatorname{gcd}\left(F_{11}, F_{12}\right)$ 使用欧几里得算法? 现在进行实际计算并检查步数。（注意：这部分说明了使用欧几里得算 法计算两个相邻斐波那契数的 gcd 尽可能慢。)

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 数学代写|数论作业代写number theory代考|MATH2088

statistics-lab™ 为您的留学生涯保驾护航 在代写数论number theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写数论number theory代写方面经验极为丰富，各种代写数论number theory相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|数论作业代写number theory代考|The Euclidean Algorithm

Given two positive integers $a$ and $b$, how do we compute their greatest common divisor? If $a$ and $b$ are relatively small, one way (which we shall discuss in some detail in our next chapter) is to factor them both into their representation as a product of powers of prime numbers and observe what factors are in common. However, if at least one of the integers $a$ and $b$ is large, this method can be difficult and relatively inefficient. Euclid, in Book VII of his Elements, describes a very efficient procedure for computing greatest common divisors. Before stating his method, let us look at a couple of examples.

Example 1.4. (a) What is $\operatorname{gcd}(420,378)$ ? As suggested above, one way is to factor them both: $420=2^{2} \cdot 3 \cdot 5 \cdot 7$ and $378=2 \cdot 3^{3} \cdot 7$. These factorizations have in common 2,3 and 7 , so $\operatorname{gcd}(420,378)=$ $2 \cdot 3 \cdot 7=42$.
(b) What is $\operatorname{gcd}(858,1092)$ ? We could factor again, but it does not look easy. We are seeing that the larger the numbers become, the more difficult the factorization method becomes. Here is a better method.

Theorem 1.3. (Euclidean Algorithm) Let a and b be positive integers. If $a$ divides $b$, then the $\operatorname{gcd}(a, b)=a$. Otherwise, repeatedly using the Division Algorithm, there exists a strictly decreasing sequence of positive integers $r_{1}, \ldots, r_{n}$ so that
\begin{aligned} b &=a q_{1}+r_{1} \ a &=r_{1} q_{2}+r_{2} \ r_{1} &=r_{2} q_{3}+r_{3} \ & \vdots \ r_{n-2} &=r_{n-1} q_{n}+r_{n} \ r_{n-1} &=r_{n} q_{n+1}+0 . \end{aligned}

## 数学代写|数论作业代写number theory代考|Division Algorithm

1.4. (a) Using the Division Algorithm, find the quotient $q$ and the remainder $r$ when $b=711$ is divided by $a=23$.
(b) Do the same as in Part (a) for dividing $b=-135$ by $a=31$.
Solution:
(a) $711=23(30)+21$, i.e., $q=30$ and $r=21$.
(b) $-135=31(-5)+20$, i.e., $q=-5$ and $r=20$.
1.5. In the Division Algorithm, suppose that the dividend $b=259$, the quotient $q=12$, and the remainder $r=7$. What is the divisor $a$ ?
Solution:
Since $259=12 a+7$, we get $a=252 / 12=21$.
Greatest Common Divisors
1.6. Find $\operatorname{gcd}(35,180)$ using factorization into powers of prime numbers.
Solution:
$35=5 \cdot 7,180=2^{2} \cdot 3^{2} \cdot 5$, so $\operatorname{gcd}(35,180)=5$.
1.7. Find $\operatorname{gcd}(224,468)$ using factorization into powers of prime numbers.
Solution:
$$221=2^{5} \cdot 7,168=2^{2} \cdot 3^{2} \cdot 13 \text {, so } \operatorname{gcd}(221,168)=2^{2}=1 \text {. }$$

## 数学代写|数论作业代写number theory代考|The Euclidean Algorithm

(b) 什么是 $\operatorname{gcd}(858,1092)$ ? 我们可以再次考虑因素，但这看起来并不容易。我们看到数字越大，分解方法就越困 难。这里有一个更好的方法。

$$b=a q_{1}+r_{1} a \quad=r_{1} q_{2}+r_{2} r_{1}=r_{2} q_{3}+r_{3} \quad \vdots r_{n-2}=r_{n-1} q_{n}+r_{n} r_{n-1} \quad=r_{n} q_{n+1}+0 .$$

## 数学代写|数论作业代写number theory代考|Division Algorithm

1.4. (a) 使用除法算法，求商 $q$ 和其余的 $r$ 什么时候 $b=711$ 被除以 $a=23$.
(b) 做与 $(\mathrm{a})$ 部分相同的除法 $b=-135$ 经过 $a=31$.

(二) $-135=31(-5)+20$ ， 那是， $q=-5$ 和 $r=20$.
1.5。在除法算法中，假设被除数 $b=259$ ，商 $q=12$ ，和余数 $r=7$. 什么是除数 $a$ ?

1.6。寻找 $\operatorname{gcd}(35,180)$ 使用因式分解为素数的幂。

$35=5 \cdot 7,180=2^{2} \cdot 3^{2} \cdot 5$ ，所以 $\operatorname{gcd}(35,180)=5$.
1.7. 寻找 $\operatorname{gcd}(224,468)$ 使用因式分解为素数的幂。

$$221=2^{5} \cdot 7,168=2^{2} \cdot 3^{2} \cdot 13, \text { so } \operatorname{gcd}(221,168)=2^{2}=1$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 数学代写|数论作业代写number theory代考|MAST90136

statistics-lab™ 为您的留学生涯保驾护航 在代写数论number theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写数论number theory代写方面经验极为丰富，各种代写数论number theory相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|数论作业代写number theory代考|Divisibility

Definition: Suppose $b$ is an integer and $a$ is a non-zero integer. We say that $a$ divides $b$ if there is an integer $q$ so that $b=a q$. If there are such integers, we denote the fact that $a$ divides $b$ by using the notation $a \mid b$.

Be aware that the notation $a \mid b$ is a sentence with the verb being “divides.” Contrast this with the notation $\frac{a}{b}$, which is an element of the rational numbers $\mathbb{Q}$ (see Appendix $B$, not a sentence.

Example 1.1. Clearly $2 \mid 8$ since $8=2(4) ; 36 \mid 108$ since $108=$ $36(3) ; 3 \mid(-36)$ since $-36=3(-12)$; and for any integer $m, 3 \mid(15 m+$ 3) since $15 m+3=3(5 m+1)$. On the other hand, 3 does not divide 13 as there is no integer $q$ with $13=3 q$.

In the following lemma, we provide a few basic properties involving the divisibility of integers. (Note: A “lemma” is a “helping theorem,” i.e., an often easily proved result which is then used to establish bigger results.)
Lemma 1.1. Let $a, b, c, d$ be integers with $a>0$ and $d>0$.
(i) If $a \mid b$ and $a \mid c$, then $a \mid(b+c)$;
(ii) If $a \mid b$ and $a \mid c$, then $a \mid(b-c)$;
(iii) If $a \mid b$ and $a \mid c$, then $a \mid(m b+n c)$ for any integers $m$ and $n$;
(iv) If $d \mid a$ and $a \mid b$, then $d \mid b$.
Proof. To prove Part (i), we may assume that $b=a q$ and $c=a s$ where $q$ and $s$ are integers. Then $b+c=a q+a s=a(q+s)$ so that $a$ divides $b+c$ (since $q+s$ is an integer). The proof of Part (ii) is similar and hence omitted. Proofs of the remaining parts are left to the reader in Supplementary Problem 1.14.

## 数学代写|数论作业代写number theory代考|The Division Algorithm

What if a positive integer $a$ does not divide an integer $b$ ? Here is where the seemingly simple but very important Division Algorithm comes into play when doing computations in $\mathbb{Z}$.

Theorem 1.2. (Division Algorithm) Let $a$ and $b$ be integers with $a>0$. Then there are integers $q$ and $r$ with $0 \leq r<a$ so that $b=a q+r$.

This is simply a formal statement of the long division process. The integer $b$ is often called the dividend, $a$ the divisor, $q$ the quotient, and $r$ the remainder. The key is that the remainder $r$ must be non-negative and must be less than the divisor $a$. It is clear that $a \mid b$ if and only if $r=0$.

Example 1.2. Given $b=436$ and $a=17$, we can compute by long division that $436=17(25)+11$. Note that, as required, the remainder 11 is greater than or equal to 0 and is less than the divisor 17. Given $b=-67$ and $a=12$, we get $-67=12(-6)+5$, so in this case the quotient $-6$ is negative, but again the remainder 5 must be non-negative and below the divisor 12 .

## 数学代写|数论作业代写number theory代考|Divisibility

(一) 如果 $a \mid b$ 和 $a \mid c$ ，然后 $a \mid(b+c)$ ；
(ii) 如果 $a \mid b$ 和 $a \mid c$ ，然后 $a \mid(b-c)$;
(iii) 如果 $a \mid b$ 和 $a \mid c$ ，然后 $a \mid(m b+n c)$ 对于任何整数 $m$ 和 $n$;
(iv) 如果 $d \mid a$ 和 $a \mid b$ ，然后 $d \mid b$.

$b+c=a q+a s=a(q+s)$ 以便 $a$ 划分 $b+c$ (自从 $q+s$ 是一个整数)。(ii) 部分的证明类似，因此省略。其余 部分的证明留给读者在补充问题 $1.14$ 中。

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 数学代写|数论作业代写number theory代考|MATH 1001

statistics-lab™ 为您的留学生涯保驾护航 在代写数论number theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写数论number theory代写方面经验极为丰富，各种代写数论number theory相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|数论作业代写number theory代考|Coprime Numbers

Natural numbers $m$ and $n$ are called coprime, if $(m, n)=1$. An interesting real-life technical application of coprime numbers is depicted in Fig. 1.5. It shows two gear ratios. In the left part of the figure, the larger wheel has 20 teeth and the smaller one 10 teeth. If there is one tooth slightly damaged on the larger wheel (it is marked with a dot in Fig. 1.5), then it fits into exactly the same gap in the smaller wheel after each turn of the larger wheel. Just at this gap, the smaller wheel will be very quickly worn out. In the right part of Fig. $1.5$ we see two wheels with 25 and 12 teeth. Since $(25,12)=1$, there will be completely uniform wear. Let us still note that the ratio of the teeth is actually almost the same in both cases: 2 and $2.083 .$

Here is another practical use of coprime numbers. The fixed part of the caliper is equipped with a scale in which each centimeter is divided into $10 \mathrm{~mm}$. On the moving part, the so-called vernier is divided into 10 equally long pieces, which together have $9 \mathrm{~mm}$ (see Fig. 1.6). The fact that 9 and 10 are coprime allows us to determine the dimensions of small objects with precision to the nearest tenth of a millimeter. When measuring, we determine which line of the vernier merges with some line on the millimeter scale of the caliper. So many tenths of a millimeter is then added to the measured millimeters (i.e. to their largest integer value). If we were to choose instead of $9 \mathrm{~mm}$ another length that divides $10 \mathrm{~mm}$, then several lines could merge so we would not know which data applies.

## 数学代写|数论作业代写number theory代考|Euclidean Algorithm

To calculate the greatest common divisor $(m, n)$ of two large natural numbers $m \geq n$ the well-known Euclidean algorithm is often used. It can be briefly characterized as follows:

If $n$ divides $m$, then $(m, n)=n$, otherwise we have
$$(m, n)=(n, z),$$
where $z \geq 1$ is the remainder when dividing the number $m$ by the number $n$. Since $z<m$, larger problem is thus converted to a smaller one. The next steps of the algorithm then proceed similarly. The original problem is thus reduced to smaller and smaller parts until we get the remainder 0 .
For instance, if $m=54$ and $n=16$, then by the Euclidean algorithm we get
$$(54,16)=(16,6)=(6,4)=(4,2)=2 .$$
Now let us imagine that we have a squared paper with dimensions $54 \times 16$ (see Fig. 1.7). From this we will gradually cut off squares as large as possible and we will perform this as long as possible (see [188]), i.e., in the first step we cut off 3 squares $16 \times 16$, in the next step we cut 2 squares $6 \times 6$, etc. The length of the side of the square that we have left, is the result of the Euclidean algorithm, i.e. the largest common divisor of numbers 54 and 16 .

If the numbers $m$ and $n$ are coprime, the Euclidean algorithm ends with the least possible square $1 \times 1$. For example, two consecutive natural numbers are always coprime.

To calculate the least common multiple of $[m, n]$, it is also useful to apply the Euclidean algorithm first, because $(m, n) \leq[m, n]$, and then use the relation (1.2). We return to the Euclidean algorithm in Theorem 7.7.

## 数学代写|数论作业代写number theory代考|Euclidean Algorithm

$$(m, n)=(n, z)$$

$$(54,16)=(16,6)=(6,4)=(4,2)=2 .$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 数学代写|数论作业代写number theory代考|MATH2088

statistics-lab™ 为您的留学生涯保驾护航 在代写数论number theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写数论number theory代写方面经验极为丰富，各种代写数论number theory相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|数论作业代写number theory代考|Simple Criteria of Divisibility

We say that $d$ divides (without remainder) a natural number $n$, if there exists $k \in \mathbb{N}$ such that $n=d \cdot k$. In this case we shall write
$$d \mid n$$
for instance, $3 \mid 6$. The number $d$ is called a divisor of the number $n$ and the numbers 1 and $n$ are called trivial divisors of $n$. If $1<d<n$, then $d$ is called a nontrivial divisor, and if $d<n$, then $d$ is called a proper divisor of $n$. If $m$ does not divide $n$, we shall write $m \nmid n$, for instance, $5 \nmid 6$. Similar definitions can also be introduced for integer numbers when $m \neq 0$. An integer divisible by 2 is called even, otherwise odd.
Theorem 1.1 A natural number $n$ written in the decimal system is divisible by
(a) two, if its last digit is even,
(b) three, if the sum of all its digits is divisible by 3 ,
(c) four, if the number formed by the last two digits of $n$ is divisible by 4 ,
(d) five, if its last digit is 0 or 5 ,
(e) $\operatorname{six}$, if it is even and divisible by 3 ,
(f) seven, if twice the number of hundreds increased by the number formed by the last two digits is divisible by 7,
(g) eight, if the number formed by the last three digits of $n$ is divisible by 8 ,
(h) nine, if the sum of all its digits is divisible by 9 ,
(i) ten, if its last digit is 0 .

## 数学代写|数论作业代写number theory代考|The Least Common Multiple and the Greatest

Let $m$ and $n$ be arbitrary natural numbers. Denote by $M \subset \mathbb{N}$ a subset of all common multiples of $m$ and $n$. The set $M$ is clearly nonempty, because it contains e.g. the product $m n$. Since $\mathbb{N}$ is well ordered, $M$ must contain a smallest element, which we denote by $[m, n]$ and call the least common multiple of the numbers $m, n \in \mathbb{N}$. Thus, it is the smallest natural number divisible by both $m$ and $n$.

Similarly, the greatest common divisor of two integer numbers $m$ and $n$, which are not zero at the same time, is the largest integer that divides both $m$ and $n$. The greatest common divisor of numbers $m$ and $n$ will be denoted by $(m, n)$.

A basic property of the largest common divisor and least common multiple is obviously
$$(m, n)=(n, m), \quad[m, n]=[n, m] .$$
For $k, m, n \in \mathbb{N}$ the following distributive properties hold
$$[k,(m, n)]=([k, m],[k, n]) \text { and }(k,[m, n])=[(k, m),(k, n)] .$$

Theorem $1.2$ For any natural numbers $m$ and $n$ we have
$$m n=(m, n)[m, n] .$$
Proof Denote by $d \geq 1$ an arbitrary common divisor of $m$ and $n$. Then $\frac{m}{d}$ and $\frac{n}{d}$ are natural numbers, $\frac{m}{d} n$ is an integer multiple of the number $n$, and $\frac{n}{d} m$ is an integer multiple of the number $m$. Therefore, $\frac{m n}{d}$ is a common multiple of the numbers $m$ and $n$. Now if $d$ is the greatest common divisor of the numbers $m$ and $n$, then $\frac{m n}{d}$ has to be the least common multiple $m$ and $n$.

## 数学代写|数论作业代写number theory代考|Simple Criteria of Divisibility

$$d \mid n$$

(a) 2整除，如果它的最后一位是偶数，
(b) 三，如果所有数字的总和可以被 3 整除，
(c) 四，如果最后两位组成的数字的位数 $n$ 能被 4 整除，
(d) 5 ，如果最后一位是 0 或 5 ，
(e)six，如果它是偶数并且能被 3 整除，
(f) 七，如果百位数的两倍加上最后两位数字组成的数字可以被 7 整除，
(g) 八，如果最后三位数字组成的数字的 $n$ 可被 8 整除，
(h) 九，如果其所有数字的总和可被 9 整除，
(i) 十，如果其最后一位为 0 。

## 数学代写|数论作业代写number theory代考|The Least Common Multiple and the Greatest

$$(m, n)=(n, m), \quad[m, n]=[n, m] .$$

$$[k,(m, n)]=([k, m],[k, n]) \text { and }(k,[m, n])=[(k, m),(k, n)] .$$

$$m n=(m, n)[m, n]$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 数学代写|数论作业代写number theory代考|MAST90136

statistics-lab™ 为您的留学生涯保驾护航 在代写数论number theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写数论number theory代写方面经验极为丰富，各种代写数论number theory相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|数论作业代写number theory代考|Divisibility and Congruence

In one of the oldest Chinese books I-Ching (Book of Changes), which dates approximately from the 8 th century $\mathrm{BC}$, there is a picture (so-called hexagram) containing $8 \times 8$ boxes. Each box contains 6 broken or full horizontal lines (see Fig. 1.1). The broken line indicates the old Chinese principle yin and the full principle of yang, which are in opposition. Yin is associated with the Moon, humidity, darkness, Earth, woman, and passivity, yang on the other hand with the Sun, drought, light, heaven, man, and activity.

The prominent German mathematician Gottfried Wilhelm Leibniz (1646-1716) associated this hexagram with the discovery of a binary system. Considering zero instead of the broken line and one instead of the full line, the symbols in particular boxes from left to right (starting from the top line) can be interpreted as the numbers $0,1,2,3, \ldots$ written in the binary system. The first number in the upper left corner is therefore zero, even though this notation was not used for operations with numbers in the 8th century BC. The last number in the lower right corner corresponds to 63 , which is written as 111111 in the binary system. The use of zero nowadays seems completely natural, but its discovery and in particular, its symbolic representation signified great progress in mathematics over the entire world (cf. Fig. 1.2).

Although the ancient Chinese did not perform with the symbols yin-yang any arithmetic operations, we cannot deny they were the first to represent numbers by the binary system. The discovery of the binary system found practical application only in today’s computer age, i.e. almost three thousand years later. Computers display and process all information (including numbers) in the binary system. This is the easiest way in electronic circuits of a computer to process data. Thus, the functioning of e-mail, scanners, copiers, digital cameras, compact disks $\mathrm{CD}$ and DVD, cell phones, and the worldwide network of Internet is actually based on the ancient Chinese principles of yin $(=0)$ and yang $(=1)$.

## 数学代写|数论作业代写number theory代考|Natural Numbers

From ancient times people used the numbers $1,2,3, \ldots$ to express the number of some objects. The oldest use of zero was recorded in India. For a long time, zero was not even considered to be a number. Moreover, at present historians still do not have a year zero (but it is used by astronomers).

Sometimes we encounter the question of whether zero is or is not a natural number. Unfortunately, it is not possible to give a clear answer to this question of the YES/NO type, since whether or not we consider zero to be a natural number is a matter of definition. It is advisable to include zero in the set of natural numbers, for example, when determining the number of elements of finite sets, because the number of elements of the empty set is zero.

On the other hand, there are good reasons why it is sometimes advantageous not to include zero in the set of natural numbers. This is, for example, to avoid division by zero or when raising natural numbers to a natural power. In particular, the symbol $0^{0}$ cannot be unambiguously assigned to one value that would naturally correspond to standard arithmetical operations with real numbers. For example, for $n=1,2, \ldots$ we have $0^{n}=0$, while $n^{0}=1$. Archimedes’ axiom presented below could not be applied if 0 would be a natural number. It is also not possible to define reasonably the least common multiple of, for example, the numbers 0 and 3 , as we shall see in Sect. 1.4. Therefore, more often zero is not considered to be a natural number.
The set of natural numbers (positive integers) will be denoted by
$$\mathbb{N}={1,2,3, \ldots}$$
It took a long time for mathematicians to figure out how in fact, natural numbers should be introduced. Among several options, the following four axioms formulated around 1891 hy the Italian mathematician Giuseppe Peano (1858-1939) were defined. ‘They use a special function “successor”, truthfully characterize the set of natural numbers and are called Peano’s axioms after him:
(A1) There exists a unique natural number that is not a successor of any natural numbers. We will denote this number by the symbol 1 .
(A2) Each natural number has exactly one successor.

## 数学代写|数论作业代写number theory代考|Natural Numbers

ñ=1,2,3,…

(A1) 存在一个唯一的自然数，它不是任何自然数的后继。我们将用符号 1 来表示这个数字。
(A2) 每个自然数只有一个后继。

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 数学代写|数论作业代写number theory代考|PMATH650

statistics-lab™ 为您的留学生涯保驾护航 在代写数论number theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写数论number theory代写方面经验极为丰富，各种代写数论number theory相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|数论作业代写number theory代考|SOME ELEMENTARY NUMBER THEORY

This section contains some basic number-theoretic definitions and results which you ought to know. Proofs in this section are abbreviated or omitted, and you should be able to supply proofs for yourself. If necessary, this material can be found in any work on elementary number theory. The most popular of the classic texts are regularly revised, thereby offering a proven exposition together with additions which bring the content and presentation up to date. From a very crowded field we mention Hardy and Wright $[28],[29]$, Niven and Zuckerman $[45],[46]$ and Baker [10].

Lemma 1.10. The division algorithm. If $a$ and $b$ are integers with $b>0$, then there exist integers $q$ and $r$ such that $a=b q+r$ and $0 \leq r<b$.

Using the division algorithm recursively gives the Euclidean algorithm for computing the greatest common divisor of two integers, not both zero.

Lemma 1.11. The Bézout property. If $a$ and $b$ are integers, not both zero, and $g$ is the greatest common divisor of $a$ and $b$, then there exist integers $x$ and $y$ such that $a x+b y=g$.

Given specific $a$ and $b$, you should know how to use the Euclidean algorithm to find $g, x$ and $y$.

Lemma 1.12. If $a$ and $m$ have no common factor and $a \mid m n$, then $a \mid n$.
Definition 1.4. Let $m$ be a positive integer. We say that integers a and b are congruent modulo $m$, written $a \equiv b(\bmod m)$, if $m \mid a-b$.

To “reduce an integer $a$ modulo $m$ ” means to find an integer $b$ such that $a \equiv b(\bmod m)$ and $b$ lies in a “suitable” range, usually $0 \leq b<m$. That this can always be done is a consequence of the division algorithm. Although congruence notation is just another way of expressing a divisibility relation, and in that sense “nothing new”, it is very useful because congruence shares many of the basic properties of equality.

## 数学代写|数论作业代写number theory代考|IRRATIONALITY OF er

In the actual details of the final proof, Hermite’s method is (at least for the earlier results) not too difficult. However, the motivation behind the proof can be obscure. Therefore, instead of giving the proofs straight away, we shall start by trying to explain the aims and ideas behind a relatively simple case. We wish to generalise results of Chapter 1 by showing that if $r$ is rational then $e^{r}$ is irrational, with the obvious exception that $e^{0}=1$.

As usual we seek a proof by contradiction: take $r=a / b$ with $a \neq 0$, and suppose that $e^{r}=p / q$. Following the method of Theorem $1.9$, we try to obtain a contradiction by constructing an integer that lies between 0 and 1 . Hermite’s idea, which originated in his study of approximations to $e^{x}$, was to consider the definite integral
$$\int_{0}^{r} f(x) e^{x} d x$$
and to identify a function $f$ which will give us what we want. Integrating by parts yields
$$\int_{0}^{r} f(x) e^{x} d x=\left(f(r) e^{r}-f(0)\right)-\int_{0}^{r} f^{\prime}(x) e^{x} d x$$
and since the integral on the right-hand side has very much the same form as that on the left, we may apply the same procedure repeatedly to obtain
$$\int_{0}^{r} f(x) e^{x} d x=\left(f(r)-f^{\prime}(r)+f^{\prime \prime}(r)-\cdots\right) e^{r}-\left(f(0)-f^{\prime}(0)+f^{\prime \prime}(0)-\cdots\right)$$
Here the right-hand side purports to contain two infinite series and therefore must be treated with caution, but if we choose $f$ to be a polynomial, then the sums will actually involve a finite number of terms only, and we shall have no convergence problems. We write
$$F(x)=f(x)-f^{\prime}(x)+f^{\prime \prime}(x)-f^{\prime \prime \prime}(x)+\cdots,$$
so that
$$\int_{0}^{r} f(x) e^{x} d x=F(r) e^{r}-F(0)$$
and the next step is to make some sort of evaluation of the right-hand side. An idea that will help is to notice that $F(0)$ will be simple if $f$ has a large number of derivatives that vanish at $x=0$; that is, $f(x)$ should have many factors of $x$. Similarly, $f(x)$ should have many factors of $r-x$ in order to keep $F(r)$ simple. So we set
$$f(x)=c x^{n}(r-x)^{n}$$
where $c$, a constant, and $n$ are yet to be chosen. Now
\begin{aligned} f^{(k)}(0) &=k ! \times\left{\text { coefficient of } x^{k}\right} \ &=k ! c\left(\begin{array}{c} n \ k-n \end{array}\right) r^{2 n-k}(-1)^{k-n} \end{aligned}
if $n \leq k \leq 2 n$, and $f^{(k)}(0)=0$ otherwise. Recall that our aim is to make the integral (2.1), or something similar, an integer. The expression for $f^{(k)}(0)$ contains a factor $r^{2 n-k}$, and this could have a denominator as big as $b^{n}$. Therefore, we choose $c=b^{n} ;$ then $f^{(k)}(0)$ is always an integer, and so is $F(0)$. Either by invoking the symmetry of $f$ or by direct calculation, we note that
\begin{aligned} f(r-x)=f(x) & \Rightarrow(-1)^{k} f^{(k)}(r-x)=f^{(k)}(x) \ & \Rightarrow f^{(k)}(r)=(-1)^{k} f^{(k)}(0) \end{aligned}
Therefore, $F(r)$ is an integer too, and so is
$$q \int_{0}^{r} f(x) e^{x} d x=p F(r)-q F(0)$$

## 数学代写|数论作业代写number theory代考|SOME ELEMENTARY NUMBER THEORY

$0 \leq b<m$. 总能做到这一点是除法算法的结果。尽管全等表示法只是表示可分关系的另一种方式，并且在这个意

## 数学代写|数论作业代写number theory代考|IRRATIONALITY OF er

$$\int_{0}^{r} f(x) e^{x} d x$$

$$\int_{0}^{r} f(x) e^{x} d x=\left(f(r) e^{r}-f(0)\right)-\int_{0}^{r} f^{\prime}(x) e^{x} d x$$

$$\int_{0}^{r} f(x) e^{x} d x=\left(f(r)-f^{\prime}(r)+f^{\prime \prime}(r)-\cdots\right) e^{r}-\left(f(0)-f^{\prime}(0)+f^{\prime \prime}(0)-\cdots\right)$$

$$F(x)=f(x)-f^{\prime}(x)+f^{\prime \prime}(x)-f^{\prime \prime \prime}(x)+\cdots,$$

$$\int_{0}^{r} f(x) e^{x} d x=F(r) e^{r}-F(0)$$

$$f(x)=c x^{n}(r-x)^{n}$$

\begin{aligned } f \wedge { ( k ) } ( 0 ) \& = k \text { ! \timesไleft } { \backslash \text { text } { } \mathrm { x } ^ { \wedge } { k } \backslash \text { right } } \text { 的系数 } \backslash \& = k \mathrm { ~ ! ~ c l

$$f(r-x)=f(x) \Rightarrow(-1)^{k} f^{(k)}(r-x)=f^{(k)}(x) \quad \Rightarrow f^{(k)}(r)=(-1)^{k} f^{(k)}(0)$$

$$q \int_{0}^{r} f(x) e^{x} d x=p F(r)-q F(0)$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 数学代写|数论作业代写number theory代考|MATH538101

statistics-lab™ 为您的留学生涯保驾护航 在代写数论number theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写数论number theory代写方面经验极为丰富，各种代写数论number theory相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|数论作业代写number theory代考|IRRATIONALITY OF THE EXPONENTIAL CONSTANT

Once we get beyond radical expressions and decimals, irrationality proofs, for the most part, become significantly harder. A notable exception is the irrationality of the exponential constant $e$. Apart from the intrinsic interest of the result, its proof provides our first glimpse of an idea which will recur again and again in irrationality arguments, and which we shall employ extensively in Chapters 2 and $5 .$

Theorem 1.9. The exponential constant $e$ is irrational.
Proof. Assume that $e=p / q$ is rational. That is,
$$\frac{p}{q}=1+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+\cdots,$$
and for any positive integer $n$, we have
$$\frac{p n !}{q}=n !+\frac{n !}{1 !}+\frac{n !}{2 !}+\cdots+1+R,$$
where $R$ (which depends on $n$ ) is given by
$$R=\frac{n !}{(n+1) !}+\frac{n !}{(n+2) !}+\cdots$$
We can estimate $R$ in terms of a geometric series:
$$R=\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+\cdots<\frac{1}{n+1}+\frac{1}{(n+1)^{2}}+\cdots=\frac{1}{n} .$$
In particular, choose $n=q$. Then
$$R=\frac{p n !}{q}-\left(n !+\frac{n !}{1 !}+\frac{n !}{2 !}+\cdots+1\right)$$
is clearly an integer; but using (1.1), we have $0<R<1$. This is impossible, and so $e$ is irrational.

Observe that this proof relies essentially on an infinite series for $e$, and therefore has to involve concepts of calculus. In some sense this may be surprising, as number theory is usually thought of as studying discrete systems while calculus is the science of the continuous; in another sense there should be no surprise, as it is not even possible to define the number $e$ without recourse to calculus techniques. Whether it is in fact a surprise or not, we shall find that many of our future proofs will be expressed in terms of calculus.

## 数学代写|数论作业代写number theory代考|OTHER RESULTS, AND SOME OPEN QUESTIONS

It is known that $\pi$ is irrational: we shall prove this in the next chapter. It is not hard to see that at least one of the numbers $\pi+e$ and $\pi e$ must be irrational (in fact, at least one must be transcendental – see Chapter 3 ); although, most likely, both are irrational, this has not been proved for either one individually. As a consequence of a difficult result due to Gelfond and Schneider (Theorem $5.18$ ) we know that $e^{\pi}$ is irrational; however it is still unknown whether or not $\pi^{e}$ is irrational. It can also be shown that various numbers such as, for example, $e^{\sqrt{2}}$ and $2^{\sqrt{2}}$ are irrational. However, the irrationality of $\pi^{\sqrt{2}}$ and $2^{e}$, and that of the Euler-Mascheroni constant
$$\gamma=\lim {n \rightarrow \infty}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}-\log n\right)=0.57721 \cdots$$ remain undecided. Another problem which has attracted much attention is to investigate the irrationality of the numbers $\zeta(n)$. Here $n \geq 2$ is an integer and $\zeta$ is the Riemann zeta function defined by $$\zeta(s)=\sum{k=1}^{\infty} \frac{1}{k^{s}}=1+\frac{1}{2^{s}}+\frac{1}{3^{s}}+\frac{1}{4^{s}}+\cdots$$
for $s>1$. By methods of complex integration we can show that if $n$ is even then $\zeta(n)$ is a rational number times $\pi^{n}$, and this is known to be irrational. On the other hand, it is much harder to find out anything of interest about $\zeta(n)$ for odd $n$. In 1978 the French mathematician R. Apéry sensationally proved that $\zeta(3)$ is irrational. His complicated argument had the appearance of being completely unmotivated, and all of the techniques he had used would have been available two centuries earlier: for these reasons, few people believed that the proof could possibly be correct. Nevertheless it was found possible eventually to confirm all of Apéry’s assertions and thereby establish what has been called “a proof that Euler missed”. A brief (but not easy!) account of Apéry’s work is given in [66].

## 数学代写|数论作业代写number theory代考|IRRATIONALITY OF THE EXPONENTIAL CONSTANT

$$\frac{p}{q}=1+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+\cdots$$

$$\frac{p n !}{q}=n !+\frac{n !}{1 !}+\frac{n !}{2 !}+\cdots+1+R$$

$$R=\frac{n !}{(n+1) !}+\frac{n !}{(n+2) !}+\cdots$$

$$R=\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+\cdots<\frac{1}{n+1}+\frac{1}{(n+1)^{2}}+\cdots=\frac{1}{n}$$

$$R=\frac{p n !}{q}-\left(n !+\frac{n !}{1 !}+\frac{n !}{2 !}+\cdots+1\right)$$

## 数学代写|数论作业代写number theory代考|OTHER RESULTS, AND SOME OPEN QUESTIONS

$$\gamma=\lim n \rightarrow \infty\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}-\log n\right)=0.57721 \cdots$$

$$\zeta(s)=\sum k=1^{\infty} \frac{1}{k^{s}}=1+\frac{1}{2^{s}}+\frac{1}{3^{s}}+\frac{1}{4^{s}}+\cdots$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 数学代写|数论作业代写number theory代考|MATH312

statistics-lab™ 为您的留学生涯保驾护航 在代写数论number theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写数论number theory代写方面经验极为丰富，各种代写数论number theory相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|数论作业代写number theory代考|IRRATIONAL SURDS

The following result is well known, and was, essentially, proved by Pythagoras or one of his followers.
Theorem 1.1. $\sqrt{2}$ is irrational.
Proof by contradiction. Suppose that $\sqrt{2}=p / q$, where $p$ and $q$ are integers with no common factor, and with $q \neq 0$. Squaring both sides and multiplying by $q^{2}$, we have $p^{2}=2 q^{2}$. Thus $p^{2}$ is even and so $p$ is even, say $p=2 r$. Substituting for $p$ gives $q^{2}=2 r^{2}$ and so $q$ is even. Thus $p$ and $q$ have a common factor of 2 , and this contradicts our initial assumption. Therefore, $\sqrt{2}$ is irrational.

Plato records that his teacher Theodorus proved the irrationality of $\sqrt{n}$ for $n$ up to 17. Historians of mathematics have wondered why he stopped just here; the question is made harder by the fact that we don’t know exactly how Theodorus’ proof ran. The following proof of the irrationality of $\sqrt{n}$ for certain values of $n$ suggests a possible reason for stopping just before $n=17$.
First, if $n=4 k$, then the irrationality of $\sqrt{n}$ is equivalent to that of $\sqrt{k}$; and if $n=4 k+2$, then the method used above for $n=2$ can be employed with only minor changes. So we concentrate on odd values of $n$. If $n$ is odd and $\sqrt{n}=p / q$, then $n q^{2}=p^{2}$ and $p$ and $q$ must both be odd; substituting $p=2 r+1$ and $q=2 s+1$ and rearranging yields
$$4 n\left(s^{2}+s\right)-4\left(r^{2}+r\right)+n-1=0 .$$
Consider the case $n=4 k+3$. Cancelling 2 from the above equation gives
$$2 n\left(s^{2}+s\right)-2\left(r^{2}+r\right)+2 k+1=0$$
which is clearly impossible as the left-hand side is odd. This method does not work directly for $n=4 k+1$, so we consider as a subsidiary case $n=8 k+5$. Substituting as above and cancelling 4 we obtain
$$n\left(s^{2}+s\right)-\left(r^{2}+r\right)+2 k+1=0$$
but as $r^{2}+r$ and $s^{2}+s$ are both even, this is again impossible.
The remaining possibility is that $n=8 k+1$; but it appears that this case has to be split up into still further subcases, and the proof becomes much more complicated (try it!), so we shall stop here. Therefore, we have proved the following.

## 数学代写|数论作业代写number theory代考|IRRATIONAL DECIMALS

The following well-known result characterises rational numbers in terms of their decimals. Note that the eventually periodic decimal expansions include the finite expansions, for instance, $0.123=0.123000 \cdots=0.122999 \cdots$.

Theorem 1.7. Rationality of decimals. A real number $\alpha$ is rational if and only if it has an eventually periodic decimal expansion.

Proof. Firstly, suppose that $\alpha$ has an eventually periodic expansion. Without loss of generality we may assume that $0<\alpha<1$, say
$$\alpha=0 . a_{1} a_{2} \cdots a_{s} b_{1} b_{2} \cdots b_{t} b_{1} b_{2} \cdots b_{t} b_{1} b_{2} \cdots .$$
Let $a$ and $b$ be the non-negative integers with digits $a_{1} a_{2} \cdots a_{s}$ and $b_{1} b_{2} \cdots b_{t}$ respectively; then
$$\alpha=\frac{a}{10^{s}}+\frac{b}{10^{s+t}}+\frac{b}{10^{s+2 t}}+\cdots=\frac{a}{10^{s}}+\frac{b}{10^{s+t}} \frac{1}{1-10^{-t}},$$
which is rational. Conversely, suppose that $\alpha=p / q$ is rational, and initially assume that neither 2 nor 5 is a factor of $q$. Choose $t=\phi(q)$, where $\phi$ is Euler’s function: see definition $1.6$ in the appendix to this chapter. By Euler’s Theorem we have
$$10^{t} \equiv 1 \quad(\bmod q)$$
and so $q$ is a factor of $10^{t}-1$, say $10^{t}-1=q r$. Hence we can write
$$\alpha=\frac{p r}{10^{t}-1}=a+\frac{b}{10^{t}-1}$$
here we have used the division algorithm to guarantee that $0 \leq b<10^{t}-1$. We can thus write $b$ as a number of $t$ digits, say $b=b_{1} b_{2} \cdots b_{t}$; it is possible that $b_{1}$ is zero. Similarly, write $a=a_{1} a_{2} \cdots a_{s}$. Then
$$\alpha=a+\frac{b}{10^{t}}+\frac{b}{10^{2 t}}+\cdots=a_{1} a_{2} \cdots a_{s}, b_{1} b_{2} \cdots b_{t} b_{1} b_{2} \cdots b_{t} b_{1} b_{2} \cdots$$ and we see that $\alpha$ has an eventually periodic decimal expansion. To complete the proof we must also consider the case when $q$ has 2 or 5 as a factor. Let $q=2^{m} 5^{n} q^{\prime}$, where neither 2 nor 5 is a factor of $q^{\prime}$; then
$$10^{m+n} \alpha=\frac{2^{n} 5^{m} p}{q^{\prime}}=\frac{p^{\prime}}{q^{\prime}},$$
say; by the previous argument, the decimal expansion of $10^{m+n} \alpha$ is eventually periodic. The expansion of $\alpha$ contains exactly the same digits (with the decimal point shifted $m+n$ places), so it too is eventually periodic.

## 数学代写|数论作业代写number theory代考|IRRATIONAL SURDS

$$4 n\left(s^{2}+s\right)-4\left(r^{2}+r\right)+n-1=0 .$$

$$2 n\left(s^{2}+s\right)-2\left(r^{2}+r\right)+2 k+1=0$$

$$n\left(s^{2}+s\right)-\left(r^{2}+r\right)+2 k+1=0$$

## 数学代写|数论作业代写number theory代考|IRRATIONAL DECIMALS

$$0.123=0.123000 \cdots=0.122999 \cdots$$

$$\alpha=0 . a_{1} a_{2} \cdots a_{s} b_{1} b_{2} \cdots b_{t} b_{1} b_{2} \cdots b_{t} b_{1} b_{2} \cdots$$

$$\alpha=\frac{a}{10^{s}}+\frac{b}{10^{s+t}}+\frac{b}{10^{s+2 t}}+\cdots=\frac{a}{10^{s}}+\frac{b}{10^{s+t}} \frac{1}{1-10^{-t}}$$

$$10^{t} \equiv 1 \quad(\bmod q)$$

$$\alpha=\frac{p r}{10^{t}-1}=a+\frac{b}{10^{t}-1}$$

$$\alpha=a+\frac{b}{10^{t}}+\frac{b}{10^{2 t}}+\cdots=a_{1} a_{2} \cdots a_{s}, b_{1} b_{2} \cdots b_{t} b_{1} b_{2} \cdots b_{t} b_{1} b_{2} \cdots$$

$$10^{m+n} \alpha=\frac{2^{n} 5^{m} p}{q^{\prime}}=\frac{p^{\prime}}{q^{\prime}}$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。