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。

couryes-lab™ 为您的留学生涯保驾护航 在代写现代代数Modern 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。我们的专家在代写现代代数Modern Algebra代写方面经验极为丰富，各种代写现代代数Modern Algebra相关的作业也就用不着说。

数学代写|现代代数代写Modern Algebra代考|Congruence of Integers

In Example 4 of Section 1.7, we defined the relation “congruence modulo 4 ” on the set $\mathbf{Z}$ of all integers, and we proved this relation to be an equivalence relation on $\mathbf{Z}$. That example is a special case of congruence modulo $\boldsymbol{n}$, as defined next.
Congruence Modulo $n$
Let $n$ be a positive integer, $n>1$. For integers $x$ and $y, x$ is congruent to $y$ modulo $n$ if and only if $x-y$ is a multiple of $n$. We write
$$
x \equiv y(\bmod n)
$$
to indicate that $x$ is congruent to $y$ modulo $n$.

Thus $x \equiv y(\bmod n)$ if and only if $n$ divides $x-y$, and this is equivalent to $x-y=n q$, or $x=y+n q$. Another way to describe this relation is to say that $x$ and $y$ yield the same remainder when each is divided by $n$. To see that this is true, let
$$
x=n q_1+r_1 \text { with } 0 \leq r_1<n
$$
and
$$
y=n q_2+r_2 \text { with } 0 \leq r_2<n .
$$
Then
$$
x-y=n\left(q_1-q_2\right)+\left(r_1-r_2\right) \text { with } 0 \leq\left|r_1-r_2\right|<n .
$$
Thus $x-y$ is a multiple of $n$ if and only if $r_1-r_2=0$-that is, if and only if $r_1=r_2$. In particular, any integer $x$ is congruent to its remainder when divided by $n$. This means that any $x$ is congruent to one of
$$
0,1,2, \ldots, n-1 .
$$
Congruence modulo $n$ is an equivalence relation on $\mathbf{Z}$, and this fact is important enough to be stated as a theorem.

数学代写|现代代数代写Modern Algebra代考|Equivalence Relation

The relation of congruence modulo $n$ is an equivalence relation on $\mathbf{Z}$.
Proof We shall show that congruence modulo $n$ is (1) reflexive, (2) symmetric, and (3) transitive. Let $n>1$, and let $x, y$, and $z$ be arbitrary in $\mathbf{Z}$.
Reflexive
Symmetric

$x \equiv x(\bmod n)$ since $x-x=(n)(0)$.

$x \equiv y(\bmod n) \Rightarrow x-y=n q$ for some $q \in \mathbf{Z}$
$$
\begin{aligned}
& \Rightarrow y-x=n(-q) \text { and }-q \in \mathbf{Z} \
& \Rightarrow y \equiv x(\bmod n) .
\end{aligned}
$$
Transitive
$$
\text { 3. } \begin{aligned}
x & \equiv y(\bmod n) \text { and } y \equiv z(\bmod n) \
& \Rightarrow x-y=n q \quad \text { and } y-z=n k \text { and } q, k \in \mathbf{Z} \
& \Rightarrow x-z=x-y+y-z \
& =n(q+k), \text { and } \quad q+k \in \mathbf{Z} \
& \Rightarrow x \equiv z(\bmod n) .
\end{aligned}
$$

As with any equivalence relation, the equivalence classes for congruence modulo $n$ form a partition of $\mathbf{Z}$; that is, they separate $\mathbf{Z}$ into mutually disjoint subsets. These subsets are called congruence classes or residue classes. Referring to Definition 1.39, we have
$$
\begin{aligned}
{[a] } & ={x \in \mathbf{Z} \mid x \equiv a(\bmod n)} \
& ={x \in \mathbf{Z} \mid x-a=n k, k \in \mathbf{Z}} \
& ={x \in \mathbf{Z} \mid x=a+n k, k \in \mathbf{Z}} \
& =[a+n k, k \in \mathbf{Z}],
\end{aligned}
$$

so there are $n$ distinct congruence classes modulo $n$, given by
$$
\begin{aligned}
{[0] } & ={\ldots,-2 n,-n, 0, n, 2 n, \ldots} \
{[1] } & ={\ldots,-2 n+1,-n+1,1, n+1,2 n+1, \ldots} \
{[2] } & ={\ldots,-2 n+2,-n+2,2, n+2,2 n+2, \ldots} \
& \vdots \
{[n-1] } & ={\ldots,-n-1,-1, n-1,2 n-1,3 n-1, \ldots} .
\end{aligned}
$$
When $n=4$, these classes appear as
$$
\begin{aligned}
& {[0]={\ldots,-8,-4,0,4,8, \ldots}} \
& {[1]={\ldots,-7,-3,1,5,9, \ldots}} \
& {[2]={\ldots,-6,-2,2,6,10, \ldots}} \
& {[3]={\ldots,-5,-1,3,7,11, \ldots} .}
\end{aligned}
$$

现代代数代考

数学代写|现代代数代写Modern Algebra代考|Congruence of Integers

在第1.7节的例4中，我们在所有整数的集合$\mathbf{Z}$上定义了关系“同余模4”，并证明了这个关系在$\mathbf{Z}$上是等价关系。这个例子是同余模$\boldsymbol{n}$的一个特例，定义如下。
同余模$n$
设$n$为正整数$n>1$。对于整数$x$和$y, x$等于$y$模$n$当且仅当$x-y$是$n$的倍数。我们写
$$
x \equiv y(\bmod n)
$$
表示$x$等于$y$模$n$。

因此$x \equiv y(\bmod n)$当且仅当$n$除$x-y$，这等价于$x-y=n q$或$x=y+n q$。描述这种关系的另一种方法是，$x$和$y$分别除以$n$得到相同的余数。为了证明这是对的，让
$$
x=n q_1+r_1 \text { with } 0 \leq r_1<n
$$
和
$$
y=n q_2+r_2 \text { with } 0 \leq r_2<n .
$$
然后
$$
x-y=n\left(q_1-q_2\right)+\left(r_1-r_2\right) \text { with } 0 \leq\left|r_1-r_2\right|<n .
$$
因此，$x-y$是$n$的倍数，当且仅当$r_1-r_2=0$，即当且仅当$r_1=r_2$。特别地，任何整数$x$除以$n$都等于它的余数。这意味着任何$x$都等于其中一个
$$
0,1,2, \ldots, n-1 .
$$
同余模$n$是$\mathbf{Z}$上的一个等价关系，这个事实非常重要，可以作为一个定理来表述。

数学代写|现代代数代写Modern Algebra代考|Equivalence Relation

同余模的关系 $n$ 是否存在等价关系 $\mathbf{Z}$．
证明我们要证明同余模 $n$ 是(1)自反的，(2)对称的，(3)传递的。让 $n>1$，让 $x, y$，和 $z$ 武断地 $\mathbf{Z}$．
自反性
对称的

$x \equiv x(\bmod n)$ 自从$x-x=(n)(0)$。

$x \equiv y(\bmod n) \Rightarrow x-y=n q$ 对于一些人$q \in \mathbf{Z}$
$$
\begin{aligned}
& \Rightarrow y-x=n(-q) \text { and }-q \in \mathbf{Z} \
& \Rightarrow y \equiv x(\bmod n) .
\end{aligned}
$$
传递的
$$
\text { 3. } \begin{aligned}
x & \equiv y(\bmod n) \text { and } y \equiv z(\bmod n) \
& \Rightarrow x-y=n q \quad \text { and } y-z=n k \text { and } q, k \in \mathbf{Z} \
& \Rightarrow x-z=x-y+y-z \
& =n(q+k), \text { and } \quad q+k \in \mathbf{Z} \
& \Rightarrow x \equiv z(\bmod n) .
\end{aligned}
$$

与任何等价关系一样，同余模$n$的等价类形成了$\mathbf{Z}$的一个划分;也就是说，它们将$\mathbf{Z}$分离为互不相交的子集。这些子集称为同余类或剩余类。参考定义1.39，我们有
$$
\begin{aligned}
{[a] } & ={x \in \mathbf{Z} \mid x \equiv a(\bmod n)} \
& ={x \in \mathbf{Z} \mid x-a=n k, k \in \mathbf{Z}} \
& ={x \in \mathbf{Z} \mid x=a+n k, k \in \mathbf{Z}} \
& =[a+n k, k \in \mathbf{Z}],
\end{aligned}
$$

所以有$n$不同的同余类以$n$为模，由
$$
\begin{aligned}
{[0] } & ={\ldots,-2 n,-n, 0, n, 2 n, \ldots} \
{[1] } & ={\ldots,-2 n+1,-n+1,1, n+1,2 n+1, \ldots} \
{[2] } & ={\ldots,-2 n+2,-n+2,2, n+2,2 n+2, \ldots} \
& \vdots \
{[n-1] } & ={\ldots,-n-1,-1, n-1,2 n-1,3 n-1, \ldots} .
\end{aligned}
$$
当使用$n=4$时，这些类显示为
$$
\begin{aligned}
& {[0]={\ldots,-8,-4,0,4,8, \ldots}} \
& {[1]={\ldots,-7,-3,1,5,9, \ldots}} \
& {[2]={\ldots,-6,-2,2,6,10, \ldots}} \
& {[3]={\ldots,-5,-1,3,7,11, \ldots} .}
\end{aligned}
$$

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。