## 数学代写|抽象代数作业代写abstract algebra代考|A Smidgeon of Set Theory

statistics-lab™ 为您的留学生涯保驾护航 在代写抽象代数abstract algebra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。我们的专家在代写抽象代数abstract algebra代写方面经验极为丰富，各种代写抽象代数abstract algebra相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (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|抽象代数作业代写abstract algebra代考|A Smidgeon of Mathematical Logic and Some Proof Techniques

Our description of “what is a proof” in Section 1.1 says that “each step is required to follow logically from known facts.” But what does it mean to follow logically? There is an entire branch of mathematics devoted to this question. In this section we briefly indicate some of the basic principles of logic that are used in constructing proofs.

Some of the examples in this section involve sets. We informally view a set as a collection of objects. We write $a \in S$ if the object $a$ is an element of the set $S$, and $a \notin S$ if not. We denote the empty set by $\emptyset$; it is the set that has no elements. We refer you to Section 1.4 for additional material pertaining to the theory of sets.

1.3.1. Basic Logical Operations. Basic logical operations are applied to one or more statements in order to build other statements. For example, there are basic logical operations corresponding to words “not” and “and” and “or.” To illustrate, suppose that
$P$ is the statement “the house is red,”
$Q$ is the statement “the house is new.”
Then we can form new statements such as:
“not $P$ ” is the statement “the house is not red,”
” $P$ and $Q$ ” is the statement “the house is both red and new,”
” $P$ or $Q$ ” is the statement “the house is either red or new (or both).”
Note that we are not asserting that any of these statement are true or false; that’s an entirely different issue. In practice, we start with certain statements, called axioms, that we assume are true, and we use the axioms, plus any other statements that we’ve already proved, to deduce that certain other statements are true or false.

Thus once we know the truth values of $P$ and $Q$, basic rules of logic determine the truth values of various other statements. For example, if $P$ is true, then “not $P$ ” is false, while ” $P$ and $Q$ ” is true precisely when both $P$ and $Q$ are true. These sorts of deductions may be described using a Truth Table, as illustrated in Figure 1. The first two columns of Figure 1 give the four possible truth values for the pair $(P, Q)$, the third column gives the resulting conclusion for “not $P$,” and the fourth column gives the conclusion for ” $P$ and $Q$.” Subsequent columns describe additional logical operations which we now describe, while also giving the symbols that mathematicians use for these operations.

## 数学代写|抽象代数作业代写abstract algebra代考|Logical Equivalence and the Algebra of Logical Operations

1.3.2. Logical Equivalence and the Algebra of Logical Operations. In Section 1.3.1 we described an array of logical operations. We now consider how they are related to one another. As noted earlier, if Expression Ex $_1$ axpression $2_2$ are logical expressions involving statements and logical operations, we write
Expression $_1 \Longleftrightarrow$ Expression $_2$
to mean that the two expressions have the same truth value, and we say that the two expressions are logically equivalent.

This is somewhat abstract, but some examples will help to clarify. Suppose that $P, Q$, and $R$ are statements. Then “and” and “or” satisfy associative laws,
\begin{aligned} & (P \wedge Q) \wedge R \Longleftrightarrow P \wedge(Q \wedge R), \ & (P \vee Q) \vee R \Longleftrightarrow P \vee(Q \vee R) . \end{aligned}
These logical equivalences agree with our intuition, since both expressions in (1.1) are true precisely when all three of $P, Q$, and $R$ are true, and both expressions in (1.2) are true precisely when at least one of $P, Q$, and $R$ is true.

Similarly, we know that the double-negation of a statement is the same as the original statement, a fact that is expressed by the logical equivalence
$$\neg(\neg P) \Longleftrightarrow P .$$

Here are some more complicated examples. There are distributive laws for “and”‘ and “or” expressed by the logical equivalences
\begin{aligned} & P \vee(Q \wedge R) \Longleftrightarrow(P \vee Q) \wedge(P \vee R), \ & P \wedge(Q \vee R) \Longleftrightarrow(P \wedge Q) \vee(P \wedge R) . \end{aligned}
There are also distributive laws for negation over “and” and “or,” but with a switch:
$$\neg(P \vee Q) \Longleftrightarrow(\neg P) \wedge(\neg Q) ; \quad \neg(P \wedge Q) \Longleftrightarrow(\neg P) \vee(\neg Q) .$$
We could justify equivalences such as (1.4), (1.5), and (1.6) by talking through the logic behind them, and you should do that yourself. But as mathematicians, we want to give rigorous proofs. How can we prove that two logical statements are equivalent? The answer is to use a truth table. Figure 2 shows how this is done for the first equivalence in (1.6), where the proof consists of observing that the two boldface columns match.

# 抽象代数代写

## 数学代写|抽象代数作业代写abstract algebra代考|A Smidgeon of Mathematical Logic and Some Proof Techniques

1.3.1. 基本逻辑运算。将基本逻辑操作应用于一个或多个语句以构建其他语句。例如，有与“不”、“和”、“或”相对应的基本逻辑运算。为了说明，假设
P是声明“房子是红色的”，

“不是P”是声明“房子不是红色的”，
“P和问”是声明“房子又红又新”，
”P或者问”是声明“房子是红色的或新的（或两者都是）。”

## 数学代写|抽象代数作业代写abstract algebra代考|Logical Equivalence and the Algebra of Logical Operations

1.3.2. 逻辑等价和逻辑运算代数。在 1.3.1 节中，我们描述了一组逻辑操作。我们现在考虑它们之间的关 系。如前所述，如果表达式 $\mathrm{Ex}_1$ 表达 $2_2$ 是涉及语句和逻辑运算的逻辑表达式，我们写 Expression $_1 \Longleftrightarrow$ 表达 $_2$

$$(P \wedge Q) \wedge R \Longleftrightarrow P \wedge(Q \wedge R), \quad(P \vee Q) \vee R \Longleftrightarrow P \vee(Q \vee R)$$

$$\neg(\neg P) \Longleftrightarrow P$$

$$P \vee(Q \wedge R) \Longleftrightarrow(P \vee Q) \wedge(P \vee R), \quad P \wedge(Q \vee R) \Longleftrightarrow(P \wedge Q) \vee(P \wedge R)$$

$$\neg(P \vee Q) \Longleftrightarrow(\neg P) \wedge(\neg Q) ; \quad \neg(P \wedge Q) \Longleftrightarrow(\neg P) \vee(\neg Q) .$$

## 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|抽象代数作业代写abstract algebra代考|A Potpourri of Preliminary Topics

statistics-lab™ 为您的留学生涯保驾护航 在代写抽象代数abstract algebra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。我们的专家在代写抽象代数abstract algebra代写方面经验极为丰富，各种代写抽象代数abstract algebra相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (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|抽象代数作业代写abstract algebra代考|What Are Definitions, Axioms, and Proofs

The Role of Definitions, Axioms, and Proofs in Higher Mathematics: Since at least the time of Euclid, circa $300 \mathrm{BC}$, the ultimate test of mathematical rigor has resided in the construction of proofs of mathematical statements. Without getting into deep matters of philosophy, a proof is a sequence of steps that starts with a known fact and ends with the desired final statement. Each step is required to follow logically from a combination of one or more of the following:

• Steps in the proof that have already been completed.
• Statements that have previously been proven.
• Axioms, which are statements that are assumed to be true.
• Definitions, which describe the properties possessed by objects.
A Further Discussion Regarding Definitions: There is nothing magical about a definition, and in principle there are no restrictions on what may be defined. For example, I might define a Zyglx to be a purple pig with wings. I could then potentially use that definition to prove that Zyglxes are able to fly, since they have wings. Is this useful? No, since as far as I am aware, there is nothing in the real world to which I could apply “Zyglx Theory.” So although definitions are, to some extent, arbitrary, the usefulness of a definition is determined by its applicability to a range of (realistic) situations. We will see many examples of such definitions, including especially the definitions of groups, rings, fields, and vector spaces. The primary goal of theoretical mathematics, and likewise of this book, is to formulate and prove interesting mathematical statements, which in our case means statements about groups, rings, etc. And the only way to get started is to have a firm understanding of the definitions of the objects that we want to study. This is why understanding and applying definitions is a crucial part of modern mathematics and why you should spend time studying definitions when they’re introduced and using definitions when you’re trying to prove things.
• A Further Discussion Regarding Axioms: In Greek mathematics, axioms were viewed as statement that are so self-evident they must be true. The modern viewpoint is that, in principle, one is free to use any set of axioms that one wants. However, not all axiom systems are created equal. The best and most interesting axiom systems are those that start with very few axioms and allow one to prove a very large number of useful, interesting, unexpected, and beautiful statements. The axioms for geometry that appear in Euclid’s work are an example. But one of those axioms, the parallel postulate, led to a revolution in mathematics. This axiom says that given a line $L$ in the plane and a point $P$ not lying on $L$, there is exactly one line $L^{\prime}$ that contains $P$ and does not intersect $L$. Seems reasonable, but maybe not entirely self-evident, so mathematicians spent centuries trying to prove that the parallel postulate follows from Euclid’s other axioms. All failed, and it was ultimately discovered that there are alternatives to the parallel postulate that yield geometries that are as valid as Euclid’s. These non-Euclidean geometries have many uses in modern mathematics and physics, and indeed it is possible that the universe in which we live is itself non-Euclidean.

## 数学代写|抽象代数作业代写abstract algebra代考|Mathematical Credos to Live By!

We assemble in this brief section some advice to help you on the path to becoming a mathematician. And, even should becoming a mathematician not be your primary goal in life(!), we believe that you will find these precepts to be widely applicable to most intellectual endeavors.

• If you can solve every problem that you try, you’re not challenging yourself enough, and you’re not working on interesting enough problems!
• Corollary: Do not be discouraged if you can’t solve a problem. ${ }^1$ Just keep plugging away at it.
• If you can’t solve a problem:
• try doing an example;
• try solving a special case;
• try solving a similar problem;
• put it aside for a few hours, or a few days, and then come back to it.
• If you solve a problem and think that you are done, then you haven’t properly absorbed the true spirit of mathematics. Every solved interesting problem will suggest new phenomena to study and new problems to test your skill. Thus after solving a problem, you should:
• analyze a more general situation;
• obtain more accurate information about special cases;
• study analogous problems;
• fit the problem that you solved into a broader context using other mathematics that you know.
Finally, we cannot stress enough that in order to learn mathematics, you must be an active participant.

This book features lots of interesting problems for you to learn from and for you to use to test your mettle and to hone your skills. Enjoy your victories, but don’t be afraid to lose sometimes, and try to turn even failures into learning experiences.

# 抽象代数代写

## 数学代写|抽象代数作业代写abstract algebra代考|What Are Definitions, Axioms, and Proofs

• 证明中已经完成的步骤。
• 先前已经证明的陈述。
• 公理，即假定为真的陈述。
• 定义，描述对象所拥有的属性。
关于定义的进一步讨论：定义没有什么神奇之处，原则上对可以定义的内容没有任何限制。例如，我可能将 Zyglx 定义为带翅膀的紫色猪。然后我可能会使用该定义来证明 Zyglxes 能够飞翔，因为它们有翅膀。这有用吗？不，因为据我所知，现实世界中没有任何东西可以应用“Zyglx 理论”。因此，尽管定义在某种程度上是任意的，但定义的有用性取决于它对一系列（现实）情况的适用性。我们将看到许多此类定义的示例，尤其包括群、环、域和向量空间的定义。理论数学的主要目标，也是本书的主要目标，是制定和证明有趣的数学陈述，在我们的例子中是指关于群、环等的陈述。唯一的开始方法是对我们想要研究的对象的定义有一个坚定的理解。这就是为什么理解和应用定义是现代数学的重要组成部分，以及为什么你应该在引入定义时花时间研究定义，并在你试图证明事物时使用定义。
• 关于公理的进一步讨论：在希腊数学中，公理被视为不言自明的陈述，它们必须是真实的。现代的观点是，原则上，人们可以自由地使用自己想要的任何一组公理。然而，并非所有的公理系统都是生而平等的。最好和最有趣的公理系统是那些从很少的公理开始并允许人们证明大量有用的、有趣的、意想不到的和美丽的陈述的系统。欧几里德作品中出现的几何公理就是一个例子。但是其中一个公理，即平行公设，导致了数学的一场革命。这个公理说给定一条线大号在平面和一个点P不躺在大号, 只有一行大号′包含P并且不相交大号. 看似合理，但也许并非完全不证自明，因此数学家花费了数个世纪的时间试图证明平行公设源自欧几里德的其他公理。所有这些都失败了，并且最终发现平行假设还有其他替代方法可以产生与欧几里得几何一样有效的几何。这些非欧几里德几何在现代数学和物理学中有很多用途，而且我们生活的宇宙本身也有可能是非欧几何。

## 数学代写|抽象代数作业代写abstract algebra代考|Mathematical Credos to Live By!

• 如果你能解决你尝试的每一个问题，你就没有足够地挑战自己，你也没有在解决足够有趣的问题！
• 推论：如果你不能解决问题，不要气馁。1继续努力吧。
• 如果您无法解决问题：
• 尝试做一个例子；
• 尝试解决一个特例；
• 尝试解决类似的问题；
• 把它放在一边几个小时，或者几天，然后再回来看。
• 如果你解决了一个问题并认为你已经完成了，那么你就没有正确地吸收数学的真正精神。每一个解决的有趣问题都会提出新的研究现象和新的问题来测试你的技能。因此，在解决问题后，您应该：
• 分析更一般的情况；
• 获得有关特殊情况的更准确信息；
• 研究类似的问题；
• 使用您知道的其他数学将您解决的问题纳入更广泛的背景。
最后，我们要强调的是，为了学习数学，您必须是一个积极的参与者。

## 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。

## MATH355 abstract algebra课程简介

Algebra gives us tools to solve equations. The integers, the rationals, and the real numbers have special properties which make algebra work according to the circumstances. In this course, we generalize algebraic processes and the sets upon which they operate in order to better understand, theoretically, when equations can and cannot be solved. We define and study abstract algebraic structures such as groups, rings, and fields, as well as the concepts of factor group, quotient ring, homomorphism, isomorphism, and various types of field extensions. This course introduces students to abstract rigorous mathematics.

## PREREQUISITES

The course will cover roughly chapters $1-5$ and 7 of the text. A list of topics is below.

• The integers and other familiar sets of numbers
• $\mathrm{N}, \mathrm{Z}, \mathrm{Q}, \mathrm{R}, \mathrm{C}$ (Should we include 0 in $\mathrm{N}$ ?)
• The well-ordering principle and induction
• Divisibility
• The division algorithm
• greatest common divisors. Euclidian algorithm?
• Bezout’s identity
• Least common multiples
• Prime numbers
• Euclid’s lemma
• Fundamental theorem of arithmetic
• Infinitely many primes
• Functions, Domain, Range, etc.
• Equivalence Relations
• Partitions
• Kernels of functions
• Well-definededness

## MATH355 abstract algebra HELP（EXAM HELP， ONLINE TUTOR）

Exercise 1. Consider a rhombus that is not square (i.e., the four sides all have the same length, but the angles between sides is not $90^{\circ}$ ). Describe all the symmetries of the rhombus. Write down the Cayley table for the group of symmetries.

Exercise 2 . Consider the set $\mathbb{Z}$ and the binary operation $\star$ on $\mathbb{Z}$ given by
$$x \star y=x+x y .$$
Does this operation make $\mathbb{Z}$ a group? Justify your answer.

## Textbooks

• An Introduction to Stochastic Modeling, Fourth Edition by Pinsky and Karlin (freely
available through the university library here)
• Essentials of Stochastic Processes, Third Edition by Durrett (freely available through
the university library here)
To reiterate, the textbooks are freely available through the university library. Note that
you must be connected to the university Wi-Fi or VPN to access the ebooks from the library
links. Furthermore, the library links take some time to populate, so do not be alarmed if
the webpage looks bare for a few seconds.

Statistics-lab™可以为您提供luc.edu MATH355 abstract algebra抽象代数课程的代写代考辅导服务！ 请认准Statistics-lab™. Statistics-lab™为您的留学生涯保驾护航。

## 数学代写|抽象代数Abstract Algebra代考

Let $n \geq 1$ be arbitrary. Consider the group $G=\operatorname{Aut}\left(\mathbb{Z}_n\right)$.
(a) Find a group we have seen in this course isomorphic to $G$.
(b) Let $F: G \rightarrow \mathbb{Z}_n$ be the function defined by $F(g)=g(1)$. Explain why $F$ is injective.
(c) Let $H \subseteq \mathbb{Z}_n$ be the image of $F(G)$. Explain why $H$ is not a subgroup of $\mathbb{Z}_n$.
(d) Find an example where $H$ is isomorphic to a subgroup of $\mathbb{Z}_n$.
(d) Find an example where $\operatorname{Aut}\left(\mathbb{Z}_n\right)$ is isomorphic to $\operatorname{Aut}\left(\mathbb{Z}_m\right)$ but $m \neq n$.

(a) The group $\mathbb{Z}_n^{\times}$ is isomorphic to $G$. This is because an automorphism of $\mathbb{Z}_n$ is completely determined by where it sends the generator $1$, which must be mapped to another generator, i.e., an element $a$ coprime to $n$. Conversely, any such $a$ gives rise to an automorphism of $\mathbb{Z}_n$ by mapping $1$ to $a$ and extending the mapping multiplicatively.

(b) Suppose $F(g_1) = F(g_2)$, where $g_1, g_2 \in G$. This means $g_1(1) = g_2(1)$, so $g_1-g_2$ maps $1$ to $0$, which implies that $n$ divides $(g_1-g_2)(k)$ for all $k\in \mathbb{Z}_n$. But $n$ is coprime to $1$, so $n$ must divide $g_1-g_2$. Therefore, $g_1=g_2$, and $F$ is injective.

(c) $H$ is not a subgroup of $\mathbb{Z}_n$ because it is not closed under addition. For example, if $n=4$ and $G\cong \mathbb{Z}_2$, then $H={0,2}$, but $1+1=2\notin H$.

(d) Let $n=p_1^{k_1}p_2^{k_2}\cdots p_r^{k_r}$ be the prime factorization of $n$, where $p_i$ are distinct primes. Let $H=\langle 1+p_1\rangle \times \langle 1+p_2^{k_1}\rangle \times \cdots \times \langle 1+p_r^{k_1}\rangle \subseteq \mathbb{Z}n$, where $\langle a \rangle$ denotes the subgroup generated by $a$. Then $H$ is isomorphic to $\mathbb{Z}{p_1} \times \mathbb{Z}{p_2^{k_2}} \times \cdots \times \mathbb{Z}{p_r^{k_r}}$, which is a subgroup of $\mathbb{Z}_n$, and $F(G) \cong H$.

(e) Let $n=4$ and $m=6$. Then $\operatorname{Aut}(\mathbb{Z}_n) \cong \mathbb{Z}_2$ and $\operatorname{Aut}(\mathbb{Z}_m) \cong S_3$, the symmetric group on $3$ elements. These groups are isomorphic because they both have order $2$, but $m\neq n$.

Consider the group $G=\mathbb{Z}^2$.
(a) Prove that $H={(3 m, 2 n): m, n \in \mathbb{Z}}$ is a subgroup.
(b) Find two representative elements from each of the cosets of $H$.
(c) Plot the elements of $\mathbb{Z}^2$ as points in the plane. Colour each of the cosets of $H$ in a different colour.
(d) Describe a group isomorphism from $G$ to $H$.
(e) Is this an automorphism? Explain.

(a) To show that $H$ is a subgroup of $G$, we need to verify the following:

• The identity $(0,0)$ is in $H$.
• If $(3m_1,2n_1)$ and $(3m_2,2n_2)$ are in $H$, then their sum $(3m_1+3m_2,2n_1+2n_2)=(3(m_1+m_2),2(n_1+n_2))$ is also in $H$.
• If $(3m,2n)$ is in $H$, then its inverse $(-3m,-2n)=(-3(m),-2(n))$ is also in $H$.

All of these are true, so $H$ is a subgroup of $G$.

(b) Let $(3m,2n)$ be an arbitrary element of $G$. Then we have $$(3m,2n)=(3,2)(m,n)+(-3,-2)(m,n)+(0,0).$$ Thus, $(3m,2n)$ is in the same coset as $(0,0)$, $(3,2)$, or $(-3,-2)$. Representative elements from each coset are: \begin{aligned} (0,0)+H &={(3m,2n):m, n \in \mathbb{Z}}, \ (3,2)+H &={(3m+1,2n+1):m, n \in \mathbb{Z}}, \ (-3,-2)+H &={(3m-1,2n-1):m, n \in \mathbb{Z}}. \end{aligned}

(c) Here is a plot of $\mathbb{Z}^2$, with each coset of $H$ colored differently:

(d) Define the function $\varphi: G \rightarrow H$ by $\varphi(m,n)=(3m,2n)$. We claim that $\varphi$ is an isomorphism.

To see that $\varphi$ is a homomorphism, observe that $$\varphi((m_1,n_1)+(m_2,n_2))=\varphi((m_1+m_2,n_1+n_2))=(3(m_1+m_2),2(n_1+n_2))=(3m_1,2n_1)+(3m_2,2n_2)=\varphi(m_1,n_1)+\varphi(m_2,n_2).$$

To see that $\varphi$ is injective, suppose $\varphi(m_1,n_1)=\varphi(m_2,n_2)$. Then $3m_1=3m_2$ and $2n_1=2n_2$, which implies $m_1=m_2$ and $n_1=n_2$. Therefore, $(m_1,n_1)=(m_2,n_2)$, and $\varphi$ is injective.

To see that $\varphi$ is surjective, let $(3m,2n)$ be an arbitrary element of $H$. Then $\varphi(m,n)=(3m,2n)$, so $\varphi$ is surjective.

Therefore, $\varphi$ is an isomorphism from $G$ to $H$.

(e) This is not an automorphism, since $G$ and $H$ have different.

## 数学代写|抽象代数作业代写abstract algebra代考|Math417

statistics-lab™ 为您的留学生涯保驾护航 在代写抽象代数abstract algebra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。我们的专家在代写抽象代数abstract algebra代写方面经验极为丰富，各种代写抽象代数abstract algebra相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (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|抽象代数作业代写abstract algebra代考|Useful CAS Commands

In linear algebra, the theorem that every vector space has a basis gives us a standard method to describe elements in vector spaces, especially finite dimensional ones: (1) find a basis $\mathcal{B}$ of the vector space; (2) express a given vector by its coordinates with respect to $\mathcal{B}$. No corresponding theorem exists in group theory. Hence, one of the initial challenging questions of group theory is how to describe a group and its elements in a standard way. This is particularly important for implementing computational packages that study groups. There exist a few common methods and we will introduce them in parallel with the development of needed theory.

In Maple version 16 or below, the command with(group); accesses the appropriate package. In Maple version 17 or higher, the group package was deprecated in favor of with(GroupTheory);. The help files, whether provided by the program or those available online ${ }^2$ provide a list of commands and capabilities. Doing a search on “GroupTheory” locates the help file for the GroupTheory package. The student might find useful the LinearAlgebra package or, to support Example 1.2.11, the linear algebra package for modular arithmetic.

Consider the following lines of Maple code, in which the left justified text is the code and the centered text is the printed result of the code.

The first line makes active the linear algebra package for modular arithmetic. The next two code lines define matrices $A$ and $B$ respectively, both defined in $\mathbb{Z} / 11 \mathbb{Z}$. The next three lines calculate respective the determinant of $A$, the produce of $A B$, and the power $A^5$, always assuming we work in $\mathbb{Z} / 11 \mathbb{Z}$.
For SAGE, a browser search for “SageMath groups” will bring up references manuals and tutorials for group theory. Perhaps the gentlest introductory tutorial is entitled “Group theory and Sage.” ${ }^3$ We show here below the commands and approximate look for the same calculations in the console for SageMath for those we did above in Maple.

## 数学代写|抽象代数作业代写abstract algebra代考|Cyclic Groups

The process of simply considering the successive powers of an element gives rise to an important class of groups.
Definition 1.3.11
A group $G$ is called cyclic if there exists an element $x \in G$ such that every element of $g \in G$ we have $g=x^k$ for some $k \in \mathbb{Z}$. The element $x$ is called a generator of $G$.

For example, we notice that for all integers $n \geq 2$, the group $\mathbb{Z} / n \mathbb{Z}$ (with addition as the operation) is a cyclic group because all elements of $\mathbb{Z} / n \mathbb{Z}$ are multiples of $\overline{1}$. As we saw in Section A.6, one of the main differences with usual arithmetic is that $n \cdot \overline{1}=\overline{0}$. The intuitive sense that the powers (or multiples) of an element “cycle back” motivate the terminology of cyclic group.

Remark 1.3.12. We point out that a finite group $G$ is cyclic if and only if there exists an element $g \in G$ such that $|g|=|G|$.
$\triangle$
Cyclic groups do not have to be finite though. The group $(\mathbb{Z},+)$ is also cyclic because every element in $\mathbb{Z}$ is obtained by $n \cdot 1$ with $n \in \mathbb{Z}$.
Example 1.3.13. Consider the group $U(14)$. The elements are
$$U(14)={\overline{1}, \overline{3}, \overline{5}, \overline{9}, \overline{11}, \overline{13}} .$$
This group is cyclic because, for example, the powers of $\overline{3}$ gives all the elements of $U(14)$ :
\begin{tabular}{c|cccccc}
$i$ & 1 & 2 & 3 & 4 & 5 & 6 \
\hline$\overline{3}^i$ & $\overline{3}$ & $\overline{9}$ & $\overline{13}$ & $\overline{11}$ & $\overline{5}$ & $\overline{1}$
\end{tabular}
We note that the powers of $\overline{3}$ will then cycle around, because $\overline{3}^7=\overline{3}^6 \cdot \overline{3}=\overline{3}$, then $\overline{3}^8=\overline{9}$, and so on.

Example 1.3.14. At first glance, someone might think that to prove that a group is not cyclic we would need to calculate the order of every element. If no element has the same order as the cardinality of the group, only then we could say that the group is not cyclic. However, by an application of the theorems in this section, we may be able to conclude the group is not cyclic with much less work.

# 抽象代数代写

## 数学代写|抽象代数作业代写abstract algebra代考|Cyclic Groups

A组 $G$ 如果存在一个元素，则称为循环 $x \in G$ 这样每一个元素 $g \in G$ 我们有 $g=x^k$ 对于一些 $k \in \mathbb{Z}$. 元素 $x$ 称为生成器 $G$.

$$U(14)=\overline{1}, \overline{3}, \overline{5}, \overline{9}, \overline{11}, \overline{13} .$$

## 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|抽象代数作业代写abstract algebra代考|MATH355

statistics-lab™ 为您的留学生涯保驾护航 在代写抽象代数abstract algebra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。我们的专家在代写抽象代数abstract algebra代写方面经验极为丰富，各种代写抽象代数abstract algebra相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (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|抽象代数作业代写abstract algebra代考|A Few Examples

It is important to develop a robust list of examples of groups that show the breadth and restriction of the group axioms.

Example 1.2.4. The pairs $(\mathbb{Z},+),(\mathbb{Q},+),(\mathbb{R},+)$, and $(\mathbb{C},+)$ are groups. In each case, addition is associative and has 0 as the identity element. For a given element $a$, the additive inverse is $-a$.

Example 1.2.5. The pairs $\left(\mathbb{Q}^, \times\right),\left(\mathbb{R}^, \times\right)$, and $\left(\mathbb{C}^, \times\right)$ are groups. Recall that $A^$ mean $A-{0}$ when $A$ is a set that includes 0 . In each group, 1 is the multiplicative identity, and, for a given element $a$, the (multiplicative) inverse is $\frac{1}{a}$. Note that $\left(\mathbb{Z}^*, \times\right)$ is not a group because it fails the inverse axiom. For example, there is no nonzero integer $b$ such that $2 b=1$.

On the other hand $\left(\mathbb{Q}^{>0}, \times\right)$ and $\left(\mathbb{R}^{>0}, \times\right)$ are groups. Multiplication is a binary operation on $\mathbb{Q}^{>0}$ and on $\mathbb{R}^{>0}$, and it satisfies all the axioms.

Example 1.2.6. A vector space $V$ is a group under vector addition with $\overrightarrow{0}$ as the identity. The (additive) inverse of a vector $\vec{v}$ is $-\vec{v}$. Note that the scalar multiplication of a vector spaces has no bearing on the group properties of vector addition.

Example 1.2.7. In Section A.6, we introduced modular arithmetic. Recall that $\mathbb{Z} / n \mathbb{Z}$ represents the set of congruence classes modulo $n$ and that $U(n)$ is the subset of $\mathbb{Z} / n \mathbb{Z}$ of elements with multiplicative inverses. Given any integer $n \geq 2$, both $(\mathbb{Z} / n \mathbb{Z},+)$ and $(U(n), \times)$ are groups. The element $\overline{0}$ is the identity in $\mathbb{Z} / n \mathbb{Z}$ and the element $\overline{1}$ is the identity $U(n)$.

The tables for addition in (A.13) and (A.14) are the Cayley tables for $(\mathbb{Z} / 5 \mathbb{Z},+)$ and $(\mathbb{Z} / 6 \mathbb{Z},+)$. By ignoring the column and row for $\overline{0}$ in the multiplication table in Equation (A.13), we obtain the Cayley table for $(U(5), \times)$.

## 数学代写|抽象代数作业代写abstract algebra代考|Notation for Arbitrary Groups

In group theory, we will regularly discuss the properties of an arbitrary group. In this case, instead of writing the operation as $a * b$, where $*$ represents some unspecified binary operation, it is common to write the generic group operation as $a b$. With this convention of notation, it is also common to indicate the identity in an arbitrary group as 1 instead of $e$. In this chapter, however, we will continue to write $e$ for the arbitrary group identity in order to avoid confusion. Finally, with arbitrary groups, we denote the inverse of an element $a$ as $a^{-1}$.

This shorthand of notation should not surprise us too much. We already developed a similar habit with vector spaces. When discussing an arbitrary vector space, we regularly say, “Let $V$ be a vector space.” So though, in a strict sense, $V$ is only the set of the vector space, we implicitly understand that part of the information of a vector space is the addition of vectors (some operation usually denoted $+$ ) and the scalar multiplication of vectors.

By a similar abuse of language, we often refer, for example, to “the dihedral group $D_n$,” as opposed to “the dihedral group $\left(D_n, \circ\right)$.” Similarly, when we talk about “the group $\mathbb{Z} / n \mathbb{Z}$,” we mean $(\mathbb{Z} / n \mathbb{Z},+)$ because $(\mathbb{Z} / n \mathbb{Z}, \times)$ is not a group. And when we refer to “the group $U(n)$,” we mean the group $(U(n), \times)$. We will explicitly list the pair of set and binary operation if there could be confusion as to which binary operation the group refers. Furthermore, as we already saw with $D_n$, even if a group is equipped with a natural operation, we often just write $a b$ to indicate that operation. Following the analogy with multiplication, in a group $G$, if $a \in G$ and $k$ is a positive integer, by $a^k$ we mean
$$a^k \stackrel{\text { def }}{=} \overbrace{a a \cdots a}^{k \text { times }} .$$
We extend the power notation so that $a^0=e$ and $a^{-k}=\left(a^{-1}\right)^k$, for any positive integer $k$.

Groups that involve addition give an exception to the above habit of notation. In that case, we always write $a+b$ for the operation, $-a$ for the inverse, and, if $k$ is a positive integer,
$$k \cdot a \stackrel{\text { def }}{=} \overbrace{a+a+\cdots+a}^{k \text { times }} .$$
We refer to $k \cdot a$ as a multiple of $a$ instead of as a power. Again, we extend the notation to nonpositive “multiples” just as above with powers.

# 抽象代数代写

## 数学代写|抽象代数作业代写abstract algebra代考|A Few Examples

(A.13) 和 (A.14) 中的加法表是 Cayley 表 $(\mathbb{Z} / 5 \mathbb{Z},+)$ 和 $(\mathbb{Z} / 6 \mathbb{Z},+)$. 通过忽略列和行 $\overline{0}$ 在等式
(A.13）的乘法表中，我们得到屾莱表为 $(U(5), \times)$.

## 数学代写|抽象代数作业代写abstract algebra代考|Notation for Arbitrary Groups

$$a^k \stackrel{\text { def }}{=} \overbrace{a a \cdots a}^{k \text { times }} .$$

$$k \cdot a \stackrel{\text { def }}{=} \overbrace{a+a+\cdots+a}^{k \text { times }} .$$

## 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|抽象代数作业代写abstract algebra代考|MATH1014

statistics-lab™ 为您的留学生涯保驾护航 在代写抽象代数abstract algebra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。我们的专家在代写抽象代数abstract algebra代写方面经验极为丰富，各种代写抽象代数abstract algebra相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (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|抽象代数作业代写abstract algebra代考|Abstract Notation

We introduce a notation that is briefer and aligns with the abstract notation that we will regularly use in group theory.

Having fixed an integer $n \geq 3$, denote by $r$ the rotation of angle $2 \pi / n$, by $s$ the reflection through the $x$-axis, and by $\iota$ the identity function. In other words,
$$r=R_{2 \pi / n}, \quad s=F_0, \quad \text { and } \quad \iota=R_0 .$$
In abstract notation, similar to our habit of notation for multiplication of real variables, we write $a b$ to mean $a \circ b$ for two elements $a, b \in D_n$. Borrowing from a theorem in the next section (Proposition 1.2.13), since $\circ$ is associative, an expression such as rrsr is well-defined, regardless of the order in which we pair terms to perform the composition. In this example, with $n=4$,
$$r r s r=R_{\pi / 2} \circ R_{\pi / 2} \circ F_0 \circ R_{\pi / 2}=R_\pi \circ F_0 \circ R_{\pi / 2}=F_{\pi / 2} \circ R_{\pi / 2}=F_{\pi / 4}$$
To simplify notations, if $a \in D_n$ and $k \in \mathbb{N}^*$, then we write $a^k$ to represent
$$a^k=\overbrace{a a a \cdots a}^{k \text { times }} .$$
Hence, we write $r^2 s r$ for $r r s r$. Since composition o is not commutative, $r^3 s$ is not necessarily equal to $r^2 s r$.
From Proposition 1.1.3, it is not hard to see that
$$r^k=R_{2 \pi k / n} \quad \text { and } \quad r^k s=F_{\pi k / n}$$
where $k$ satisfies $0 \leq k \leq n-1$. Consequently, as a set
$$D_n=\left{\iota, r, r^2, \ldots, r^{n-1}, s, r s, r^2 s, \ldots, r^{n-1} s\right} .$$
The symbols $r$ and $s$ have a few interesting properties. First, $r^n=\iota$ and $s^2=\iota$. These are obvious as long as we do not forget the geometric meaning of the functions $r$ and $s$. Less obvious is the equality in the following proposition.

## 数学代写|抽象代数作业代写abstract algebra代考|Introduction to Groups

As we now jump into group theory with both feet, the reader might not immediately see the value in the definition of a group. The plethora of examples we provide subsequent to the definition will begin to showcase the breadth of applications.
Definition 1.2.1
A group is a pair $(G, )$ where $G$ is a set and is a binary operation on G that satisfies the following properties: (1) associativity: (a * b) * c=a *(b * c) for all a, b, c \in G; (2) identity: there exists e \in G such that a * e=e * a=a for all a \in G (3) inverses: for all a \in G, there exists b \in G such that a * b=b * a=e. By Proposition A.2.16, if any binary operation has an identity, then that identity is unique. Similarly, any element in a group has exactly one inverse element. Proof. Let a \in G be arbitrary and suppose that b_1 and b_2 satisfy the properties of the inverse axiom for the element a. Then \begin{aligned} b_1 & =b_1 * e & & \text { by identity axiom } \ & =b_1 *\left(a * b_2\right) & & \text { by inverse axiom } \ & =\left(b_1 * a\right) * b_2 & & \text { by associativity } \ & =e * b_2 & & \text { by definition of } b_1 \ & =b_2 & & \text { by identity axiom. }\end{aligned} Therefore, for all a \in G there exists a unique inverse. Since every group element has a unique inverse, our notation for inverses can reflect this. We denote the inverse element of a by a^{-1}. The defining properties of a group are often called the group axioms. In common language, one often uses the term “axiom” to mean a truth that is self-evident. That is not the sense in which we use the term “axiom.” In algebra, when we say that such and such are the axioms of a given algebraic structure, we mean the defining properties listed as (1)-(3) in Definition 1.2.1. In the group axioms, there is no assumption that the binary operation * is commutative. We say that two particular elements a, b \in G commute (or commute with each other) if a * b=b * a. # 抽象代数代写 ## 数学代写|抽象代数作业代写abstract algebra代考|Abstract Notation 我们引入了一个更简洁的符号，并与我们将在群论中经常使用的抽象符号保持一致。 固定一个整数 n \geq 3, 表示为 r 角度的旋转 2 \pi / n ，经过 s 通过反射 x – 轴，并通过 \iota 身份函数。换 句话说，
$$在抽象符号中，类似于我们对实变量乘法的符号习惯，我们写 a b 意味着 a \circ b 对于两个元素 a, b \in D_n. 借用下一节中的一个定理（命题 1.2.13），因为o是关联的，像 rrsr 这样的表达式 是明确定义的，不管我们将术语配对以执行组合的顺序如何。在这个例子中，与 n=4 ，$$
r r s r=R_{\pi / 2} \circ R_{\pi / 2} \circ F_0 \circ R_{\pi / 2}=R_\pi \circ F_0 \circ R_{\pi / 2}=F_{\pi / 2} \circ R_{\pi / 2}=F_{\pi / 4}
$$为了简化符号，如果 a \in D_n 和 k \in \mathbb{N}^* ，然后我们写 a^k 代表$$
a^k=\overbrace{a a a \cdots a}^{k \text { times }} .
$$因此，我们写 r^2 s r 为了rrsr. 由于组合 o 不可交换， r^3 s 不一定等于 r^2 s r. 从命题1.1.3不难看出$$
r^k=R_{2 \pi k / n} \quad \text { and } \quad r^k s=F_{\pi k / n}
$$在哪里 k 满足 0 \leq k \leq n-1. 因此，作为一个集$$
\text { D_n }=\backslash l e f t{{i o t a, r, r \wedge 2, \backslash \text { dots, } r \wedge{n-1}, s, r s, r \wedge 2 s, \backslash \text { dots, } \wedge \wedge{n-1} \text { s right }} \text { 。 }
$$符号 r 和 s 有一些有趣的属性。第一的， r^n=\iota 和 s^2=\iota. 只要我们不忘记函数的几何意义，这 些都是显而易见的 r 和 s. 不太明显的是以下命题中的等式。 ## 数学代写|抽象代数作业代写abstract algebra代考|Introduction to Groups 当我们现在双脚跳入群论时，读者可能不会立即看到群定义的价值。我们在定义之后提供的大 量示例将开始展示应用程序的广度。 定义 1.2.1 群是一对 (G,) 在哪里 G 是一个集合， \$$ 是一个二元运算 $G$ 满足以下性质:
(1) 结合性: $(a * b) * c=a *(b * c)$ 对全部 $a, b, c \in G$;
(2)身份: 存在 $e \in G$ 这样 $a * e=e * a=a$ 对全部 $a \in G$
(3) 逆: 对所有 $a \in G$ ，那里存在 $b \in G$ 这样 $a * b=b * a=e$.

## 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|抽象代数作业代写abstract algebra代考|Math417

statistics-lab™ 为您的留学生涯保驾护航 在代写抽象代数abstract algebra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。我们的专家在代写抽象代数abstract algebra代写方面经验极为丰富，各种代写抽象代数abstract algebra相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (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|抽象代数作业代写abstract algebra代考|If and only i

In this chapter, we studied these two implications:

• If $n$ is odd, then $n^2$ is odd.
• If $n^2$ is odd, then $n$ is odd.
Each is obtained from the other by swapping the if-part and the then-part. Thus, each is called the converse of the other. (Careful: They’re not contrapositives of each other. Why not?) As a shorthand, we can combine the two and write: $n$ is odd if and only if $n^2$ is odd. So, when you’re asked to prove an “if and only if” statement, you’ll have to prove two implications.
Example 1.11. Here is another pair of statements that are converses of each other:
• If I live in Tokyo, then I live in Japan. (True)
• If I live in Japan, then I live in Tokyo. (False)
As you can see, an implication can be true even though its converse is false.
Example 1.12. Similar to Example 1.11, below is a pair of statements that are converses of each other where one is true and the other is false.
• If $n$ is positive, then $n^2$ is positive. (True)
• If $n^2$ is positive, then $n$ is positive. (False)
The first implication is true, but the second one is false. With $n=-3$, we see that even though $n^2=9$ is positive, $n=-3$ is not positive.

## 数学代写|抽象代数作业代写abstract algebra代考|Counterexample

Consider the statement: If $n$ is prime, then $2^n-1$ is prime. As usual, let’s create some concrete examples by letting $n$ take on the values of the first few prime numbers.

• If $n=2$, then $2^2-1=3$ is prime.
• If $n=3$, then $2^3-1=7$ is prime.
• If $n=5$, then $2^5-1=31$ is prime.
• If $n=7$, then $2^7-1=127$ is prime.
So far, so good. But does this mean that the statement is true? Not necessarily. In order for the statement to be true, the expression $2^n-1$ must be prime for every prime $n$. In other words, if we can find just one counterexample, i.e., an example that invalidates the statement, then we can conclude that the statement is false. Here is a counterexample: $n=11$ is prime (which satisfies the hypothesis), but $2^{11}-1=2,047=23 \cdot 89$ is not prime (which fails the conclusion). Thus, the implication is false.

Proof know-how. To show that an implication is false, we only need to find one counterexample. Thus, disproving an implication (when it’s false) tends to be easier than proving one (when it’s true).

Example 1.13. Consider the statement: If $n$ is an odd prime, then $n+2$ is prime. Many values of $n$ serve as valid examples of this statement, as shown below:

• If $n=3$, then $n+2=5$ is prime.
• If $n=11$, then $n+2=13$ is prime.
• If $n=101$, then $n+2=103$ is prime.
However, the statement is false because we can find a counterexample: $n=7$ is an odd prime (which satisfies the hypothesis), but $n+2=9$ is not prime (which fails the conclusion).

# 抽象代数代写

## 数学代写|抽象代数作业代写abstract algebra代考|If and only i

• 如果 $n$ 是奇数，那么 $n^2$ 很奇怪。
• 如果 $n^2$ 是奇数，那么 $n$ 很奇怪。
每个都是通过交换 if 部分和 then 部分从另一个获得的。因此，每个都称为另一个的逆。（注意：它 们不是彼此的对立面。为什么不呢?）作为速记，我们可以将两者结合起来写成： $n$ 是奇数当且仅当 $n^2$ 很奇怪。所以，当你被要求证明“当且仅当”陈述时，你必须证明两个推论。
示例 1.11。这是另一对彼此相反的陈述:
• 如果我住在东京，那我就住在日本。（真的）
• 如果我住在日本，那我就住在东京。（错误） 如您所见，即使其反义词为假，蕴涵也可能为真。
示例 1.12。与示例 $1.11$ 类似，下面是一对彼此相反的陈述，其中一个为真，另一个为假。
• 如果 $n$ 是正的，那么 $n^2$ 是积极的。（真的)
• 如果 $n^2$ 是正的，那么 $n$ 是积极的。（假)
第一个蕴涵是真的，但第二个是假的。和 $n=-3$, 我们看到即使 $n^2=9$ 是积极的， $n=-3$ 不是 积极的。

## 数学代写|抽象代数作业代写abstract algebra代考|Counterexample

• 如果 $n=2$ ，然后 $2^2-1=3$ 是质数。
• 如果 $n=3$ ，然后 $2^3-1=7$ 是质数。
• 如果 $n=5$ ，然后 $2^5-1=31$ 是质数。
• 如果 $n=7$ ，然后 $2^7-1=127$ 是质数。
到目前为止，一切都很好。但这是否意味着该声明是真实的? 不必要。为了使陈述为真，表达式 $2^n-1$ 对于每个素数都必须是素数 $n$. 换句话说，如果我们只能找到一个反例，即一个使陈述无效的 例子，那么我们就可以断定该陈述是错误的。这是一个反例： $n=11$ 是质数 (满足假设)，但是 $2^{11}-1=2,047=23 \cdot 89$ 不是质数（无法得出结论)。因此，暗示是错误的。
证明诀狖。要证明一个蕴涵是假的，我们只需要找到一个反例。因此，反驳一个蕴浄 (当它为假时) 往往 比证明一个蕴浄 (当它为真时) 更容易。
示例 1.13。考虑以下陈述: 如果 $n$ 是奇素数，那么 $n+2$ 是质数。许多值 $n$ 作为该声明的有效示例，如下 所示:
• 如果 $n=3$ ，然后 $n+2=5$ 是质数。
• 如果 $n=11$ ，然后 $n+2=13$ 是质数。
• 如果 $n=101$ ，然后 $n+2=103$ 是质数。
然而，这个说法是错误的，因为我们可以找到一个反例： $n=7$ 是奇素数（满足假设），但是 $n+2=9$ 不是质数 (无法得出结论)。

## 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|抽象代数作业代写abstract algebra代考|MATH355

statistics-lab™ 为您的留学生涯保驾护航 在代写抽象代数abstract algebra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。我们的专家在代写抽象代数abstract algebra代写方面经验极为丰富，各种代写抽象代数abstract algebra相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (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|抽象代数作业代写abstract algebra代考|Contrapositive

Consider the statement: If $n^2$ is odd, then $n$ is odd. (Here, $n$ is an integer.) To prove it, we might start by assuming the hypothesis, i.e., $n^2$ is odd. Then $n^2=2 k+1$ for some integer $k$. We wish to show that $n$ is odd, but we’re stuck, since solving $n^2=2 k+1$ for $n$ requires us to take the square root of $2 k+1$. Yikes!

We will introduce a new proof technique to handle a statement like the following: If $n^2$ is odd, then $n$ is odd. But first, consider these four implications:
(a) If I live in Tokyo, then I live in Japan.
(b) If I live in Japan, then I live in Tokyo.
(c) If I don’t live in Tokyo, then I don’t live in Japan.
(d) If I don’t live in Japan, then I don’t live in Tokyo.
Statement (a) is true, because Tokyo is a city inside Japan. For the same reason, statement (d) is also true; i.e., if I live outside of Japan, then I can’t possibly live in Tokyo. However, statements (b) and (c) are false, because I could be living in Osaka, for example.

Note how (d) is obtained from (a) by swapping the hypothesis and conclusion and negating both of them; and (a) is obtained from (d) in the same way. Thus, (a) and (d) are said to be contrapositives of each other. Similarly, (b) and (c) are contrapositive pairs. The key fact about contrapositives is that they are equivalent; i.e., proving one ensures that the other must be true also.

Here, to negate a statement means to write down its opposite. Thus, when we negate “I live in Tokyo,” we obtain its negation: “I don’t live in Tokyo.” Observe that if a statement is true, then its negation is false; and if a statement is false, then its negation is true.

Example 1.6. When $n=7$, then the statement ” $n$ is odd” is true, and its negation ” $n$ is not odd” is false. Moreover, when $n=6$, the statement ” $n^2$ is odd” is false, and its negation ” $n^2$ is not odd” is true.

Proof by contradiction is another technique for proving an implication, i.e., an “if …, then …” statement. Here are the steps of this proof method:
(1) Assume that the hypothesis is true (as usual).
(2) Also assume that the conclusion is false, or equivalently, that the negation of the conclusion is true.
(3) Obtain a contradiction, i.e., an absurd outcome. This would indicate that the conclusion couldn’t have been false, and so it must be true.

Consider the statement: If $n^2$ is even, then $n$ is even. Let’s prove this using proof by contradiction. To start, we assume that the hypothesis is true, which means that the first sentence of the proof should be: Assume $n^2$ is even. Next, we must assume that the negation of the conclusion is true: Assume $n$ is not even, i.e., $n$ is odd. To complete the proof, we must obtain a contradiction. Knowing which contradiction to derive is typically the most challenging aspect of proof by contradiction. Here, we will show that $n^2$ is odd (because $n$ is odd), which contradicts our assumption that $n^2$ is even.
Theorem 1.9. Let $n$ be an integer. If $n^2$ is even, then $n$ is even.
ProOf. Assume $n^2$ is even. Also assume for contradiction that $n$ is odd. Since $n$ is odd, Theorem $1.3$ implies that $n^2$ is odd. But this contradicts the fact that $n^2$ is even. Hence, $n$ cannot be odd. Thus, $n$ is even.

Proof know-how. In a proof by contradiction, we make two assumptions: (1) The hypothesis is true and (2) the negation of the conclusion is true. The phrase “for contradiction” (as seen in the above proof) is often used with the negation of the conclusion to differentiate between these two assumptions.Here is another example. Note that a rational number is a fraction of the form $\frac{m}{n}$ where $m$ and $n$ are integers (and $n$ is non-zero, since we cannot divide by zero).

# 抽象代数代写

## 数学代写|抽象代数作业代写abstract algebra代考|Contrapositive

(a) 如果我住在东京，那么我就住在日本。
(b) 如果我住在日本，那么我就住在东京。
(c) 如果我不住在东京，那我就不住在日本。
(d) 如果我不住在日本，那我就不住在东京。

(1) 假设假设为真（像往常一样）。
(2) 还假设结论是错误的，或者等价地，结论的否定是正确的。
(3) 得出一个矛盾，即一个荒谬的结果。这表明结论不可能是假的，所以它一定是真的。

## 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|抽象代数作业代写abstract algebra代考|МАТН1014

statistics-lab™ 为您的留学生涯保驾护航 在代写抽象代数abstract algebra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。我们的专家在代写抽象代数abstract algebra代写方面经验极为丰富，各种代写抽象代数abstract algebra相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (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|抽象代数作业代写abstract algebra代考|Proving an implication

Consider the following statement: If $n$ is an odd integer, then $n^2$ is odd. To better understand this statement, we look at a few examples:

• If $n=7$ is odd, then $n^2=49$ is also odd.
• If $n=213$ is odd, then $n^2=45,369$ is also odd.
• If $n=-1,081$ is odd, then $n^2=1,168,561$ is also odd.
The importance of these examples cannot be overstated. Concrete examples help us make sense of an abstract statement. Sometimes, they provide insight into why the statement is true. But, as we will see shortly, writing a proof is different from creating examples.

Before proceeding, let’s be more precise about what it means for an integer to be odd (and also even). Notice that 7 is odd, because $7=2 \cdot 3+1$; i.e., when we put 7 into groups of two, there is a remainder of 1 . However, 10 is even, because $10=2$ – 5 ; i.e., 10 can be put into groups of two without a remainder.
Definition $1.1$ (Odd and Even). Let $n$ be an integer. Then:

• $n$ is odd when $n=2 k+1$ for some integer $k$.
• $n$ is even when $n=2 k$ for some integer $k$.

Example 1.2. We categorize the following integers as odd or even:

• $213=2 \cdot 106+1$ so that 213 is odd.
$-1,081=2 \cdot(-541)+1$ so that $-1,081$ is odd.
$-314=2 \cdot(-157)$ so that $-314$ is even.
• $0=2 \cdot 0$ so that 0 is even.
Now, back to our statement: If $n$ is an odd integer, then $n^2$ is odd. This is an example of an implication, i.e., an “if …, then …” statement. The if-part is called the hypothesis (” $n$ is an odd integer”) and the then-part is called the conclusion (” $n{ }^2$ is odd”). To prove an implication, we take the following steps:
Proof know-how. To prove an implication:
(1) Assume that the hypothesis is true.
(2) Show that the conclusion is true.

## 数学代写|抽象代数作业代写abstract algebra代考|Proof by cases

Consider the statement: If $n$ is an integer, then $n^2+n$ is even. As before, we begin by creating some concrete examples. Since the only assumption about $n$ is that it is an integer, we consider the cases where (1) $n$ is odd and (2) $n$ is even:

• If $n=7$ (i.e., $n$ is odd), then $n^2+n=56$ is even.
• If $n=213$ (i.e., $n$ is odd), then $n^2+n=45,582$ is even.
• If $n=10$ (i.e., $n$ is even), then $n^2+n=110$ is even.
• If $n=-314$ (i.e., $n$ is even), then $n^2+n=98,282$ is even.
These examples suggest a proof technique called proof by cases. In this method, we split the given scenario into multiple cases and then prove the statement for each case. It is important that the cases considered cover all the possibilities. For instance, if $n$ is an integer, then the cases (1) $n$ is odd and (2) $n$ is even would suffice, since every integer is either odd or even.
Theorem 1.4. If $n$ is an integer, then $n^2+n$ is even.
Proof. Assume $n$ is an integer. We consider the two cases: (1) $n$ is odd and (2) $n$ is even.
Case (1). Suppose $n$ is odd, so that $n=2 k+1$ for some integer $k$. Then
$$n^2+n=(2 k+1)^2+(2 k+1)=4 k^2+6 k+2=2 \cdot\left(2 k^2+3 k+1\right),$$
where $2 k^2+3 k+1$ is an integer. Thus, $n^2+n$ is even.
Case (2). Suppose $n$ is even, so that $n=2 k$ for some integer $k$. Then
$$n^2+n=(2 k)^2+2 k=4 k^2+2 k=2 \cdot\left(2 k^2+k\right),$$
where $2 k^2+k$ is an integer. Thus, $n^2+n$ is even.
Remark. Notice how Theorem $1.4$ is a statement about all integers. We prove it by showing that it is true for an arbitrary integer $n$. In fact, the first sentence of the proof, “Assume $n$ is an integer”, may be considered as a shorthand for “Assume $n$ is an arbitrary integer.”

# 抽象代数代写

## 数学代写|抽象代数作业代写abstract algebra代考|Proving an implication

• 如果 $n=7$ 是奇数，那么 $n^2=49$ 也奇怪。
• 如果 $n=213$ 是奇数，那么 $n^2=45,369$ 也奇怪。
• 如果 $n=-1,081$ 是奇数，那么 $n^2=1,168,561$ 也奇怪。
这些例子的重要性怎么强调都不为过。具体的例子帮助我们理解抽象的陈述。有时，他们会深入了 解为什么该陈述是正确的。但是，正如我们很快就会看到的，编写证明不同于创建示例。
在继续之前，让我们更准确地了解整数为奇数（以及偶数）的含义。注意 7 是奇数，因为 $7=2 \cdot 3+1$ ； 即，当我们将 7 分成两组时，余数为 1 。然而， 10 是偶数，因为 $10=2-5$ ；即，10可以无余数地分成 两个一组。
定义 $1.1$ (奇数和偶数)。让 $n$ 是一个整数。然后：
• $n$ 奇怪的时候 $n=2 k+1$ 对于某个整数 $k$.
• $n$ 甚至当 $n=2 k$ 对于某个整数 $k$.
示例 1.2。我们将以下整数分类为奇数或偶数：
• $213=2 \cdot 106+1$ 所以 213 是奇数。 $-1,081=2 \cdot(-541)+1$ 以便 $-1,081$ 很奇怪。 $-314=2 \cdot(-157)$ 以便 $-314$ 甚至。
• $0=2 \cdot 0$ 所以 0 是偶数。
现在，回到我们的声明: 如果 $n$ 是奇数，那么 $n^2$ 很奇怪。这是一个蕴涵的例子，即“如果…..，那 么…..”陈述。如果部分称为假设 (” $n$ 是一个奇数”) 然后部分称为结论 (“n $n^2$ 是奇数”)。为了证明蕴 涵式，我们采取以下步骤:
证明诀空。要证明一个蕴涵:
（1）假设假设为真。
(2)证明结论正确。

## 数学代写|抽象代数作业代写abstract algebra代考|Proof by cases

• 如果 $n=7$ (IE， $n$ 是奇数)，那么 $n^2+n=56$ 甚至。
• 如果 $n=213$ (IE， $n$ 是奇数)，那么 $n^2+n=45,582$ 甚至。
• 如果 $n=10$ (IE， $n$ 是偶数)，那么 $n^2+n=110$ 甚至。
• 如果 $n=-314$ (IE， $n$ 是偶数)，那么 $n^2+n=98,282$ 甚至。
这些例子提出了一种称为案例证明的证明技术。在这种方法中，我们将给定的场景分成多个案例，
然后为每个案例证明陈述。重要的是所考虑的案例涵盖了所有的可能性。例如，如果 $n$ 是一个整数， 那么情况 (1) $n$ 是奇数且 (2) $n$ 是偶数就足够了，因为每个整数要么是奇数要么是偶数。
定理 1.4。如果 $n$ 是一个整数，那么 $n^2+n$ 甚至。
证明。认为 $n$ 是一个整数。我们考虑两种情况：(1) $n$ 是奇数且 (2) $n$ 甚至。
情况1)。认为 $n$ 是奇数，所以 $n=2 k+1$ 对于某个整数 $k$. 然后
$$n^2+n=(2 k+1)^2+(2 k+1)=4 k^2+6 k+2=2 \cdot\left(2 k^2+3 k+1\right),$$
在哪里 $2 k^2+3 k+1$ 是一个整数。因此， $n^2+n$ 甚至。
案例 (2)。认为 $n$ 是偶数，所以 $n=2 k$ 对于某个整数 $k$. 然后
$$n^2+n=(2 k)^2+2 k=4 k^2+2 k=2 \cdot\left(2 k^2+k\right)$$
在哪里 $2 k^2+k$ 是一个整数。因此， $n^2+n$ 甚至。
评论。注意定理 $1.4$ 是关于所有整数的陈述。我们通过证明它对任意整数为真来证明它 $n$. 事实上， 证明的第一句话，”假设 $n$ 是一个整数”，可以被认为是”假设 $n$ 是任意整数。”

## 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。