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• 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|Properties of Congruences

Returning now to congruences, we state in the following lemma some of their properties which will be needed as we move forward. We prove several of the properties and leave the remaining proofs for you to do in Problem $3.3$ and Problem 3.11. The basic proof technique is to translate a statement about congruences into a corresponding equation in the integers and proceed from there.
Lemma 3.1. Let $n \geq 2$ be a fixed integer. Assume that $a \equiv b(\bmod n)$ and that $c \equiv d(\bmod n)$. Then
(i) $a+c \equiv b+d(\bmod n)$;
(ii) $a-c \equiv b-d(\bmod n)$;
(iii) $a c \equiv h d(\bmod n)$;

(iv) If $m$ is an integer, then $m a \equiv m b(\bmod n)$;
(v) If $d$ is a divisor of $n$, then $a \equiv b(\bmod d)$.
Proof. We prove Parts (i), (iii), and (v), leaving proofs of the remaining parts to you. From the assumptions of the lemma, we have that $a-b=n k$ and $c-d=n j$ for some integers $k$ and $j$. To prove Part (i), we calculate
$$(a+c)-(b+d)=(a-b)+(c-d)=n k+n j=n(k+j) .$$
Since $k+j$ is an integer, we can conclude that $a+c \equiv b+d$ $(\bmod n)$
For Part (iii) we have that $a c=(b+n k)(d+n j)=b d+b n j+d n k+n^{2} k l=b d+n(b j+d k+n k j)$.
Therefore $a c-b d=n(b j+d k+n k j)$ where $b j+d k+n k$ is an integer, and so $a c \equiv b d(\bmod n)$.

## 数学代写|数论作业代写number theory代考|Doing Division in Zn

To end this chapter, let’s take a closer look at division in the sets $\mathbb{Z}{n}$. In the set of real numbers (i.e., all decimal numbers), we know we can always divide by any non-zero number and get a real number as the answer. For example, if we divide $5.4$ by 2 we get 2.7. We also know, however, that this is not true in the integers $\mathbb{Z}$; for example, 5 divided by 2 is not an integer. Back in the real numbers, another way to describe always being able to divide is to say that every non-zero real number possesses a (unique) multiplicative inverse. For example, in the real numbers the multiplicative inverse of 2 is $1 / 2=0.5$. We note then that dividing by a real number is the same thing as multiplying by its multiplicative inverse; for example, in the real numbers, $5.4$ divided by $2=(5.4)(0.5)=2.7$. We also note that in the integers $\mathbb{Z}$, the only elements possessing multiplicative inverses are 1 and $-1$, so they are the only numbers by which you can always divide in $\mathbb{Z}$. Thinking of division in terms of multiplicative inverses will help us understand what happens in our new sets $\mathbb{Z}{n}$.

We ask then, which elements of $\mathbb{Z}{n}$ possess multiplicative inverses and which do not? Now Lemma $3.2$ tells us the answer in slightly different language: You can always divide by $c$ in $\mathbb{Z}{n}$ provided that $c$ and $n$ are relatively prime. Put in our alternative way, $c$ will possess a multiplicative inverse in $\mathbb{Z}{n}$ if and only if $c$ and $n$ are relatively prime. We observe that in the real numbers it’s easy to write down the inverse of a non-zero element $x$ (it’s just $1 / x$ ), but in $\mathbb{Z}{n}$ it’s not so obvious. We need to look at some examples.

# 数论作业代写

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

\begin{aligned} &\text { (一) } a+c \equiv b+d(\bmod n) \text {; } \ &\text { (二) } a-c \equiv b-d(\bmod n) ; \ &\Leftrightarrow a c \equiv h d(\bmod n) \end{aligned}
(iv) 如果 $m$ 是一个整数，那么 $m a \equiv m b(\bmod n)$;
(v) 如果 $d$ 是一个除数 $n$ ，然后 $a \equiv b(\bmod d)$.

$$(a+c)-(b+d)=(a-b)+(c-d)=n k+n j=n(k+j)$$

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## 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。

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