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

#### Doug I. Jones

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## 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

$$x \equiv y(\bmod n)$$

$$x=n q_1+r_1 \text { with } 0 \leq r_1<n$$

$$y=n q_2+r_2 \text { with } 0 \leq r_2<n .$$

$$x-y=n\left(q_1-q_2\right)+\left(r_1-r_2\right) \text { with } 0 \leq\left|r_1-r_2\right|<n .$$

$$0,1,2, \ldots, n-1 .$$

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

$x \equiv x(\bmod n)$ 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}

\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}

\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}

\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}

\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}

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