## 数学代写|数论作业代写number theory代考|Math676

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 e^r

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.

# 数论作业代写

## 数学代写|数论作业代写number theory代考|IRRATIONALITY OF e^r

$$\int_0^r f(x) e^x d x$$

$$\int_0^r f(x) e^x d x=\left(f(r) e^r-f(0)\right)-\int_0^r f^{\prime}(x) e^x d x,$$

$$\int_0^r f(x) e^x d x=\left(f(r)-f^{\prime}(r)+f^{\prime \prime}(r)-\cdots\right) e^r-\left(f(0)-f^{\prime}(0)+f^{\prime \prime}(0)-\cdots\right) .$$

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

$$\int_0^r f(x) e^x d x=F(r) e^r-F(0)$$

## 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|Math453

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 $\pi^\epsilon$ 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].

# 数论作业代写

## 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|MATH3170

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
$$10^t \equiv 1(\bmod q)$$
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.

# 数论作业代写

## 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

$$\alpha=0 . a_1 a_2 \cdots a_s b_1 b_2 \cdots b_t b_1 b_2 \cdots b_t b_1 b_2 \cdots .$$

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

$$10^t \equiv 1(\bmod q)$$

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

$$\alpha=a+\frac{b}{10^t}+\frac{b}{10^{2 t}}+\cdots=a_1 a_2 \cdots a_s . b_1 b_2 \cdots b_t b_1 b_2 \cdots b_t b_1 b_2 \cdots$$

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

## 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。

## 数学代写|数论作业代写number theory代考|Math676

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 Divisor

If $c$ and $d$ be two arbitrary integers, not simultaneously zero, then the set of common divisors of $c$ and $d$ is a finite set of integers, always containing the integers $+1$ and $-1$ (hence, their set of common divisors is non-null). Now every integer divides zero, so that if $c=d=0$, then every integer serves as a common divisor of $c$ and $d$. In this case, the set of common divisors of $c$ and $d$ turns to be infinite. In this article, we are interested on the greatest integer among the common divisors of two integers.

Definition 2.4.1. The greatest common divisor of two integers $c$ and $d$, that are not both zero, is the greatest integer which divides both $c$ and $d$.
In other words, the above definition can be formulated as
Definition 2.4.2. If $c$ and $d$ be two arbitrary integers, not simultaneously zero, the greatest common divisor of $c$ and $d$ is the common divisor e satisfying the following:

1. $e \mid a$ and $e \mid b$.
2. If $f \mid a$ and $f \mid b$ then $e \geq f$.
The greatest common divisor of $c$ and $d$ is written as $(c, d)$ or $\operatorname{gcd}(c, d)$.

Example 2.4.1. The common divisors of 20 and 80 are $\pm 1, \pm 2, \pm 4, \pm 5, \pm 10$ and $\pm$ 20. Hence $\operatorname{gcd}(20,80)=20$. Similarly, looking at sets of common divisors, we find that $(12,18)=6,(50,5)=5,(19,24)=1,(0,56)=56,(-8,-16)=8$, and $(-19,361)=19$.

We can also define the greatest common divisor of more than two integers.
Definition 2.4.3. Let $c_1, c_2, \ldots, c_n$ be integers, that are not all zero. The greatest common divisor of these integers is the greatest integer which is a common divisor of all of the integers in the set. The greatest common divisor of $c_1, c_2, \ldots, c_n$ is denoted by $\left(c_1, c_2, \ldots, c_n\right)$ or $\operatorname{gcd}\left(c_1, c_2, \ldots, c_n\right)$.
Example 2.4.2. We see that $(12,18,30)=6$ and $(10,15,25)=5$.
The following proposition can be used to find the greatest common divisor of a set of more than two integers.

Proposition 2.4.1. If $c_1, c_2, \ldots, c_n$ are integers, not simultaneously zero, then
$$\operatorname{gcd}\left(c_1, c_2, \ldots, c_n\right)=\operatorname{gcd}\left(c_1, c_2, \ldots,\left(c_{n-1}, c_n\right)\right) \text {. }$$
Before proceeding for proof, let us explain the proposition with an example: To find the greatest common divisor of the three integers 105,140 , and 350 , we see that $\operatorname{gcd}(105,140,350)=\operatorname{gcd}(105,(140,350))=\operatorname{gcd}(105,70)=35$.

## 数学代写|数论作业代写number theory代考|Euclid’s Algorithm

While finding the gcd of two integers (not both 0 ), we can of course list all the common divisors and pick the greatest one amongst those. However, if $a$ and $b$ are very large integers, the process is very much time consuming. However, there is a far more efficient way of obtaining the gcd. That is known as the Euclid’s algorithm. This method essentially follows from the division algorithm for integers.
To prove the Fuclidean algorithm, the following lemma will he helpfull.
Lemma 2.4.1. If $a=q b+r$ then the $\operatorname{gcd}(a, b)=\operatorname{gcd}(b, r)$.
Proof. Let $d=\operatorname{gcd}(a, b)$ and $d_1=\operatorname{gcd}(b, r)$. Then, $d|a, d| b$ implies $d \mid(a-q b)$ 1.e, $d \mid r$. Thus $d$ is a common divisor of $b$ and $r$, hence $d \mid d_1$. Similarly, $d_1\left|b, d_1\right| r$ implies $d_1 \mid(b q+r)$ 1.e., $d_1$ divides both $a$ and $b$. Then, $d_1 \mid d$. Thus, $d=d_1$, as both $d$ and $d_1$ are positive by our definition of gcd.

Theorem 2.4.5. Euclid’s Algorithm: Let $a$ and $b(a>b)$ be any two integers If $r_1$ is the remainder when $a$ is divided by $b, r_2$ is the remainder when $b$ is divided by $r_1, r_3$ is the remainder when $r_1$ is divided by $r_2$ and so on. Thus $r_{n+1}=0$, then the last non zero remainder $r_n$ is the $\operatorname{gcd}(a, b)$.

Proof. Euclid’s algorithm is an efficient way of computing the gcd of two integers by repeated application of the above lemma. At each step the size of the integers concerned gets reduced. Suppose we want to find the gcd of two integers $a$ and $b$, neither of them being 0 . As $\operatorname{gcd}(a, b)=\operatorname{gcd}(a,-b)=\operatorname{gcd}(-a, b)=\operatorname{gcd}(-a,-b)$, we may assume $a>b>0$. By performing division algorithm repeatedly, we obtain
\begin{aligned} a &=b q_1+r_1, & & 0 \leq r_1 \leq b . \ b &=r_1 q_2+r_2, & & 0 \leq r_2 \leq r_1 . \ r_1 &=r_2 q_2+r_3, & & 0 \leq r_3 \leq r_3 . \ \vdots &=\vdots & & \ r_{n-2} &=r_{n-1} q_n+r_n, & & 0 \leq r_n \leq r_{n-1} . \ r_{n-1} &=r_n q_{n+1}+r_{n+1}, & & 0 \leq r_{n+1} \leq r_n . \end{aligned}
As we have a decreasing sequence of non-negative integers $b>r_1>r_2>\ldots>$ $r_n>r_{n+1}$ we must have $r_{n+1}=0$ for some $n$. Then, by applying the previous lemma repeatedly, we find that $\operatorname{gcd}(a, b)=\operatorname{gcd}\left(r_1, b\right)=\operatorname{gcd}\left(r_2, r_1\right)=\ldots=$ $\operatorname{gcd}\left(r_{n-1}, r_{n-2}\right)=\operatorname{gcd}\left(r_n, r_{n-1}\right)=r_n$. Thus, the last non-zero remainder $r_n$ in the above process gives us the $\operatorname{gcd}(a, b)$.

# 数论作业代写

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

1.e $e a$ 和 $e \mid b$.

1. 如果 $f \mid a$ 和 $f \mid b$ 然后 $e \geq f$.
的最大公约数 $c$ 和 $d$ 写成 $(c, d)$ 或者 $\operatorname{gcd}(c, d)$.
示例 2.4.1。20和 80 的公约数是 $\pm 1, \pm 2, \pm 4, \pm 5, \pm 10$ 和 $\pm 20$. 因此 $\operatorname{gcd}(20,80)=20$. 同样，查看公约数 集，我们发现 $(12,18)=6,(50,5)=5,(19,24)=1,(0,56)=56,(-8,-16)=8$ ，和 $(-19,361)=19$.
我们还可以定义两个以上整数的最大公约数。
定义 2.4.3。让 $c_1, c_2, \ldots, c_n$ 是整数，不全为零。这些整数的最大公约数是集合中所有整数的公约数最大的整 数。的最大公约数 $c_1, c_2, \ldots, c_n$ 表示为 $\left(c_1, c_2, \ldots, c_n\right)$ 或者 $\operatorname{gcd}\left(c_1, c_2, \ldots, c_n\right)$.
示例 2.4.2。我们看到 $(12,18,30)=6$ 和 $(10,15,25)=5$.
下面的命题可以用来找出一组多于两个整数的最大公约数。
提案 2.4.1。如果 $c_1, c_2, \ldots, c_n$ 是整数，不能同时为零，那么
$$\operatorname{gcd}\left(c_1, c_2, \ldots, c_n\right)=\operatorname{gcd}\left(c_1, c_2, \ldots,\left(c_{n-1}, c_n\right)\right) .$$
在进行证明之前，让我们用一个例子来解释这个命题: 求三个整数 105,140 和 350 的最大公约数，我们看到 $\operatorname{gcd}(105,140,350)=\operatorname{gcd}(105,(140,350))=\operatorname{gcd}(105,70)=35$

## 数学代写|数论作业代写number theory代考|Euclid’s Algorithm

$\operatorname{gcd}(a, b)=\operatorname{gcd}(a,-b)=\operatorname{gcd}(-a, b)=\operatorname{gcd}(-a,-b)$ ，我们可以假设 $a>b>0$. 通过反复执行除法算 法，我们得到
$$a=b q_1+r_1, \quad 0 \leq r_1 \leq b . b \quad=r_1 q_2+r_2, \quad 0 \leq r_2 \leq r_1 . r_1=r_2 q_2+r_3, \quad 0 \leq r_3 \leq r_3$$

## 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|MATH3170

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

When an integer is divided by a second integer $(\neq 0)$, the quotient may or may not be an integer. For instance, $36 / 6=6$ is an integer, while $18 / 7=2.5$ is not. This observation leads to the following definition.

Definition 2.2.1. If $a$ and $b$ are integers, we say that $b$ is divisible by $a(\neq 0)$ if there exists an integer $c$ such that $b=a c$. Also, we say that $a$ is a divisor or factor of $b$, denoted by $a \mid b$. If a does not divides $b$, then we write $a \nmid b$.

Example 2.2.1. 10 is divisible by 5 because there exist an integer 2 such that $10=5 \times 2$. We say $5 \mid 10$.

Proposition 2.2.1. For any integers $a, b, c, d$ the following statements are true:

1. $a|0,1| a, a \mid a$.
2. $a|b \Rightarrow c a| c b, \forall c \in \mathbb{Z}$.
3. $a \mid b$ and $b|c \Rightarrow a| c$.
4. $a \mid b$ and $b \mid a \Rightarrow a=+b$.
5. $a \mid b$ and $a|c \Rightarrow a|(b x+c y)$ for arbitrary integers $x$ and $y$.
Proof. 1. Obvious.
6. Here,
\begin{aligned} a \mid b & \Rightarrow b=d a \text { for some integer } d \ & \Rightarrow c b=d(c a) \ & \Rightarrow c a \mid c b \end{aligned}
7. Here, $a \mid b \Rightarrow b=a q$ and $c \mid d \Rightarrow d=c p$ for some integers $p$ and $q$. Therefore $c=a(p q)$. Hence $b d=a c(p q)$. Therefore $a c \mid b d$, as $p q$ is an integer.
8. Here, $a \mid b \Rightarrow b=a p$ for some integer $p$. Also, $b \mid c \Rightarrow c=b q$ for some integer $q$. Therefore $c=b q=a(p q)$. Therefore $a \mid c$.
9. Here, $a \mid b \Rightarrow b=a p$ for some integer $p$. Therefore $b=b p q$. Also, $b \mid a \Rightarrow a=$ $b q$ for some integer $q$ implies $p q=1$. As $p, q$ are integers either, $p=q=1$ or $p=q=-1$. Therefore $a=\pm b$.

## 数学代写|数论作业代写number theory代考|Worked out Exercises

Problem 2.3.1. For any two integers $a$ and $b$ with $b>0$, there exists unique integers $q_1$ and $r_1$ such that $a=b q_1+c r_1$ where $0 \leq r_1<\frac{b}{2}, c=\pm 1$.
Solution 2.3.1. By division algorithm we have $a=b q+c r, 0 \leq r<b$.
Case I $r<\frac{b}{2}$, take $q_1=q, c=1, r_1=r$. Therefore $a=b q_1+c r_1, 0 \leq r_1<$ $\frac{b}{2}, c=\pm 1$

Case II $r>\frac{b}{2}$, therefore $0<b-r<\frac{b}{2}$ take $q_1=q_0+1, r_1=b-r$ and $c^2=-1$, therefore, $a=b q_1+c r_1$ where $0 \leq r_1<\frac{b}{2}, c=-1$.

Case III $r=\frac{b}{2}$ then $q_1=q, c=1, r_1=r$. Therefore $a=b q_1+c r_1, r_1=\frac{b}{2}, c=1$ and if $q_1=q+1, r_1=b-r$ and $c=-1$. Therefore $a=b(q+1)-(b-$ $r)=b q_1+c r_1, \frac{b}{2}=r, c=-1$. In this case $q_1$ and $r_1$ is not unique, so $a=b q_1+c r_1, 0 \leq r_1<\frac{b}{2}, c=\pm 1$.

Problem 2.3.2. Show that every square integer is of the form $5 k$ or $5 k \pm 1$ for some $k \in \mathbb{Z}$.

# 数论作业代写

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

1. $a|0,1| a, a \mid a$.
2. $a|b \Rightarrow c a| c b, \forall c \in \mathbb{Z}$.
3. $a \mid b$ 和 $b|c \Rightarrow a| c$.
4. $a \mid b$ 和 $b \mid a \Rightarrow a=+b$.
5. $a \mid b$ 和 $a|c \Rightarrow a|(b x+c y)$ 对于任意整数 $x$ 和 $y$.
证明。1. 显而易见。
6. 这里，
$$a \mid b \Rightarrow b=d a \text { for some integer } d \quad \Rightarrow c b=d(c a) \Rightarrow c a \mid c b$$
7. 这里， $a \mid b \Rightarrow b=a q$ 和 $c \mid d \Rightarrow d=c p$ 对于一些整数 $p$ 和 $q$. 所以 $c=a(p q)$. 因此 $b d=a c(p q)$. 所以 $a c \mid b d$ ，作为 $p q$ 是一个整数。
8. 这里， $a \mid b \Rightarrow b=a p$ 对于某个整数 $p$. 还， $b \mid c \Rightarrow c=b q$ 对于某个整数 $q$. 所以 $c=b q=a(p q)$. 所 以 $a \mid c$.
9. 这里， $a \mid b \Rightarrow b=a p$ 对于某个整数 $p$. 所以 $b=b p q$. 还， $b \mid a \Rightarrow a=b q$ 对于某个整数 $q$ 暗示 $p q=1$. 作为 $p, q$ 都是整数， $p=q=1$ 或者 $p=q=-1$. 所以 $a=\pm b$.

## 数学代写|数论作业代写number theory代考|Worked out Exercises

$q_1=q+1, r_1=b-r$ 和 $c=-1$. 所以 $a=b(q+1)-(b-r)=b q_1+c r_1, \frac{b}{2}=r, c=-1$. 在这种 情况下 $q_1$ 和 $r_1$ 不是唯一的，所以 $a=b q_1+c r_1, 0 \leq r_1<\frac{b}{2}, c=\pm 1$.

## 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|Math453

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|Completeness Properties of R

Next we are going to discuss the completeness property of $\mathbb{R}$. The discussion starts with the following definition.

Definition 1.0.1. Let $K \subset \mathbb{R}$. Then a real number $u$ is said to be an upper bound(lower bound) of $K$ if $x \in K$ and $x \leq u(x \geq u)$. If $K$ has an upper bound(lower bound), then it is called bounded above(bounded below).

Example 1.0.6. Let $K={x \in \mathbb{R}: 3<x<4}$. Note that $K$ is bounded above and 4 is the upper bound. Also $K$ is bounded below and 3 is the lower bound.
The last example raises the question, whether the upper bound 4 and the lower bound 3 is greatest or least respectively. The following definition of least upper bound and greatest lower bound will be the answer to the raised question.
Definition 1.0.2. Let $K \subset \mathbb{R}$. If $K$ is bounded above(bounded below), then an upper bound (lower bound) is said to be the least upper bound(greatest lower bound) or supremum(infimum) if it is less(greater) than every upper(lower) bounds of $K$.

Actually it is a deeper property of $\mathbb{R}$ that for any non-empty bounded above(below) subset of $K$ of $\mathbb{R}$, the least upper bound(greatest lower bound) do exists. This property of $\mathbb{R}$ is called the supremum(infimum) property. Note that we can establish these two properties are equivalent and one implies other. Furthermore, the supremum property of $\mathbb{R}$ can be treated as an axiom, known to be the completeness property of $\mathbb{R}$. The statement is as follows:

Statement 1.0.1. Axiom of least upper bound: If a set $K$ is bounded above, then it has a least upper bound i.e. there exists a unique real number $M$ satisfying

1. $x \leq M, \forall x \in K$.
2. for arbitrary $\epsilon(>0)$, there exists an element $\alpha \in K$ such that $M-\epsilon<\alpha \leq$ $M$.

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

Mathematics is the Universe’s natural tongue. From very beginning of our existence as a species, numbers have deeply mesmerised us. Due to Carl Friedrich Gauss “Number theory is one of the oldest branches of Mathematics which established a relationship between numbers belonging to the set of real numbers”.
The pureness of Number Theory has charmed mathematicians generation after generation – each contributing to the branch that Carl Gauss described as the “Queen of Mathematics.” Today, however, a basic understanding of Number Theory is an absolute precursor to cutting-edge software engineering, specifically security-based software. Number Theory is at the heart of cryptography which is itself experiencing a engrossing period of rapid evolution, ranging from the famous RSA algorithm to the wildly-popular blockchain world.

Two clear-cut moments in history stand out as curvature points in the development of Number Theory. First, in archaic times, Fuclid put forth his GCD (Greatest Common Divisor) algorithm – a splendid set of steps that simplifies fractions to their simplest form using geometrical observations. Then, approximately two-thousand years later, Gauss formalized Euclid’s principles by com-bining Euclid’s informal writings with his own extensive proofs in the timeless Disquistiones Arithmeticae.

# 数论作业代写

## 数学代写|数论作业代写number theory代考|Completeness Properties of R

1. $x \leq M, \forall x \in K$.
2. 对于任意 $\epsilon(>0)$, 存在一个元素 $\alpha \in K$ 这样 $M-\epsilon<\alpha \leq M$.

## 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|Security versus Authenticity

We finish this chapter by discussing the need for authenticity as well as security in modern digital communication. In Example $6.5$ Alice decrypts Bob’s message to meet him at 8 , but how can she be sure that Bob sent the message? Perhaps it was the evil Eve who actually sent it and plans to trick her into giving up her decrypting exponent $d$ when they meet. Alice would like to know that the message from Bob is authentic. Well, it turns out that RSA can also be used to establish authenticity as well as guarantee security, via what’s called a digital signature. This can be done by having as the last packet in a message a “signature” which, unlike the main part of the message, is encoded using the sender’s public modulus and private decoding exponent.

Here’s how it works. Let us now denote Alice’s public keys by $n_A$ and $e_A$ and her private decrypting key by $d_A$. Of course Bob can also have public and private keys which we shall denote by $n_B, e_B$, and $d_B$. Now Bob wants to send the message $m$ and his signature $s$ to Alice. As before he uses her public modulus and encrypting exponent on the $m$, but for the signature part he uses his own public modulus and his private decrypting exponent. Hence Alice receives two numbers $c$ and $t$, say, which are
$$c=m^{e_A}\left(\bmod n_A\right) ; t=s^{d_B}\left(\bmod n_B\right) .$$
Now upon receipt of the pair $(c, t)$, she can decrypt both as follows:
$$m=c^{d_A}\left(\bmod n_A\right) ; s=t^{e_B}\left(\bmod n_B\right) .$$
If the message really was from Bob, the resulting digitized signature $s$, when “undigitized,” should make sense. On the other hand, if the signature was actually from anyone besides Bob, what would come out of her computation for $s$, when undigitized, would definitely not be $s$, but rather something unrecognizable.

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

6.1. Using the exact same linear encryption scheme as in Example 6.1, my broker sends me an encrypted reply to my “SELL” message. It translates as “EKMC.” By decrypting, what is her message?
Solution:
The encoded message EKMC, translated into numbers, is the set ${4,10,12,2}$. We now subtract 12 from each of these and then multiply by the inverse of 5 in $\mathbb{Z}_{26}$, which is 21 . Modulo 26 this gives us
\begin{aligned} 21({4-12,10-12,12&-12,2-12}) \equiv-5({-8,-2,0,-10}) \ & \equiv{14,10,0,24} \end{aligned}
which translates to OKAY.

Primitive Ronts and the Diffie-Hellman Key Fxchange Method
6.2. (a) Find the smallest primitive root in $\mathbb{Z}{13}$. (b) Assume that Alice and Bob are using the Diffie-Hellman key exchange method to create a common secure key and have agreed on 13 for the modulus and the answer of Part (a) as the primitive root. If Alice chooses her secret number to be $a=3$ and Bob chooses his secret number to be $b=5$, determine the common key. Solution: (a) As we saw in Example 6.3, since $\phi(13-1)=4$, there will be four primitive roots in $\mathbb{Z}{13}$. Moreover, if $k$ is the smallest exponent on an element $a$ of $\mathbb{Z}_{13}$ for which $a^k \equiv 1(\bmod 13)$, then $k$ divides $13-1=12$. Hence, testing 2 , we compute that $2^6=64 \equiv 12 \equiv$ $-1(\bmod 13)$, so $2^{12} \equiv 1(\bmod 13)$, and 12 is the smallest such exponent, so 2 is a primitive root modulo 13.

## 数学代写|数论作业代写数论代考|安全与真实性

$$c=m^{e_A}\left(\bmod n_A\right) ; t=s^{d_B}\left(\bmod n_B\right) .$$

$$m=c^{d_A}\left(\bmod n_A\right) ; s=t^{e_B}\left(\bmod n_B\right) .$$

## 数学代写|数论作业代写数论代考|解决的问题

\begin{aligned} 21({4-12,10-12,12&-12,2-12}) \equiv-5({-8,-2,0,-10}) \ & \equiv{14,10,0,24} \end{aligned}

## 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。