如果你也在 怎样代写现代代数Modern Algebra 这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。现代代数Modern Algebra有时被称为代数结构或抽象代数，或者仅仅在高等数学的背景下被称为代数。虽然这个名字可能只是暗示了一种新的方式来表示微积分之前的代数，但实际上它比微积分更广泛、更深入。

现代代数Modern Algebra这门学科的思想和方法几乎渗透到现代数学的每一个部分。此外，没有一门学科更适合培养处理抽象概念的能力，即理解和处理问题或学科的基本要素。这包括阅读数学的能力，提出正确的问题，解决问题，运用演绎推理，以及写出正确、切中要害、清晰的数学。

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数学代写|现代代数代写Modern Algebra代考|Properties of Composite Mappings (Optional)

In many cases, we will be dealing with mappings of a set into itself; that is, the domain and codomain of the mappings are the same. In these cases, the mappings $f \circ g$ and $g \circ f$ are both defined, and the question of whether $f \circ g$ and $g \circ f$ are equal arises. That is, is mapping composition commutative when the domain and codomain are equal? The following example shows that the answer is no.

Example 1 Let $\mathbf{Z}$ be the set of all integers, and let the mappings $f: \mathbf{Z} \rightarrow \mathbf{Z}$ and $g: \mathbf{Z} \rightarrow \mathbf{Z}$ be defined for each $n \in \mathbf{Z}$ by
$$
\begin{aligned}
& f(n)=2 n \
& g(n)= \begin{cases}\frac{n}{2} & \text { if } n \text { is even } \
4 & \text { if } n \text { is odd. }\end{cases}
\end{aligned}
$$

In this case, the composition mappings $f \circ g$ and $g \circ f$ are both defined. We have, on the one hand,
$$
\begin{aligned}
(g \circ f)(n) & =g(f(n)) \
& =g(2 n) \
& =n,
\end{aligned}
$$
so $(g \circ f)(n)=n$ for all $n \in \mathbf{Z}$. On the other hand,
$$
\begin{aligned}
(f \circ g)(n) & =f(g(n)) \
& = \begin{cases}f\left(\frac{n}{2}\right)=n & \text { if } n \text { is even } \
f(4)=8 & \text { if } n \text { is odd, }\end{cases}
\end{aligned}
$$
so $f \circ g \neq g \circ f$. Thus mapping composition is not commutative.

数学代写|现代代数代写Modern Algebra代考|Binary Operations

We are familiar with the operations of addition, subtraction, and multiplication on real numbers. These are examples of binary operations. When we speak of a binary operation on a set, we have in mind a process that combines two elements of the set to produce a third element of the set. This third element, the result of the operation on the first two, must be unique. That is, there must be one and only one result from the combination. Also, it must always be possible to combine the two elements, no matter which two are chosen. This discussion is admittedly a bit vague, in that the terms process and combine are somewhat indefinite. To eliminate this vagueness, we make the following formal definition.

It is conventional in mathematics to assume that when a formal definition is made, it is automatically biconditional. That is, it is understood to be an “if and only if” statement, without this being written out explicitly. In Definition 1.18 , for example, it is understood as part of the definition that $f$ is a binary operation on the nonempty set $A$ if and only if $f$ is a mapping from $A \times A$ to $A$. Throughout the remainder of this book, we will adhere to this convention when we make definitions.

We now have a precise definition of the term binary operation, but some of the feel for the concept may have been lost. However, the definition gives us what we want. Suppose $f$ is a mapping from $A \times A$ to $A$. Then $f(x, y)$ is defined for every ordered pair $(x, y)$ of elements of $A$, and the image $f(x, y)$ is unique. In other words, we can combine any two elements $x$ and $y$ of $A$ to obtain a unique third element of $A$ by finding the value $f(x, y)$. The result of performing the binary operation on $x$ and $y$ is $f(x, y)$, and the only thing unfamiliar about this is the notation for the result. We are accustomed to indicating results of binary operations by symbols such as $x+y$ and $x-y$. We can use a similar notation and write $x * y$ in place of $f(x, y)$. Thus $x * y$ represents the result of an arbitrary binary operation $*$ on $A$, just as $f(x, y)$ represents the value of an arbitrary mapping from $A \times A$ to $A$.

现代代数代考

数学代写|现代代数代写Modern Algebra代考|Properties of Composite Mappings (Optional)

在许多情况下，我们将处理集合到自身的映射;也就是说，映射的定义域和上域是相同的。在这些情况下，都定义了映射$f \circ g$和$g \circ f$，并产生了$f \circ g$和$g \circ f$是否相等的问题。也就是说，当定义域和上域相等时，映射组合是否可交换?下面的例子表明答案是否定的。

设$\mathbf{Z}$为所有整数的集合，并为每个$n \in \mathbf{Z}$定义映射$f: \mathbf{Z} \rightarrow \mathbf{Z}$和$g: \mathbf{Z} \rightarrow \mathbf{Z}$
$$
\begin{aligned}
& f(n)=2 n \
& g(n)= \begin{cases}\frac{n}{2} & \text { if } n \text { is even } \
4 & \text { if } n \text { is odd. }\end{cases}
\end{aligned}
$$

在本例中，定义了组合映射$f \circ g$和$g \circ f$。一方面，我们有
$$
\begin{aligned}
(g \circ f)(n) & =g(f(n)) \
& =g(2 n) \
& =n,
\end{aligned}
$$
所以$(g \circ f)(n)=n$对应所有$n \in \mathbf{Z}$。另一方面，
$$
\begin{aligned}
(f \circ g)(n) & =f(g(n)) \
& = \begin{cases}f\left(\frac{n}{2}\right)=n & \text { if } n \text { is even } \
f(4)=8 & \text { if } n \text { is odd, }\end{cases}
\end{aligned}
$$
所以$f \circ g \neq g \circ f$。因此，映射组合是不可交换的。

数学代写|现代代数代写Modern Algebra代考|Binary Operations

我们熟悉实数上的加法、减法和乘法运算。这些是二进制运算的例子。当我们谈到一个集合上的二进制操作时，我们想到的是一个将集合的两个元素组合起来产生第三个元素的过程。第三个元素，即前两个元素的运算结果，必须是唯一的。也就是说，这个组合必须有且只能有一个结果。此外，无论选择哪两个元素，都必须始终能够将这两个元素组合在一起。诚然，这个讨论有点模糊，因为过程和组合这两个术语有些不确定。为了消除这种模糊性，我们做出以下正式定义。

在数学中，通常假设当一个正式定义被定义时，它自动是双条件的。也就是说，它被理解为一个“当且仅当”的语句，而不需要明确地写出来。例如，在定义1.18中，当且仅当$f$是从$A \times A$到$A$的映射时，可以理解为定义的一部分，即$f$是对非空集$A$的二进制操作。在本书的其余部分中，我们将在定义时遵循这一惯例。

现在我们对二进制运算有了一个精确的定义，但是对这个概念的一些感觉可能已经丢失了。然而，定义给了我们想要的。假设$f$是从$A \times A$到$A$的映射。然后为$A$的每个有序元素对$(x, y)$定义$f(x, y)$，并且图像$f(x, y)$是唯一的。换句话说，我们可以组合$A$的任意两个元素$x$和$y$，通过查找值$f(x, y)$来获得$A$的唯一第三个元素。对$x$和$y$执行二进制操作的结果是$f(x, y)$，对此唯一不熟悉的是结果的表示法。我们习惯于用$x+y$和$x-y$等符号表示二进制运算的结果。我们可以使用类似的符号，用$x * y$代替$f(x, y)$。因此，$x * y$表示$A$上任意二进制操作$*$的结果，就像$f(x, y)$表示从$A \times A$到$A$的任意映射的值一样。

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