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# 数学代写|数论代写Number Theory代考|MATH2088 Residue classes

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## 数学代写|数论代写Number Theory代考|Residue classes

As we already observed in Theorem $2.2$, for any fixed positive integer $n$, the binary relation “. $\equiv \cdot(\bmod n) “$ is an equivalence relation on the set $\mathbb{Z}$. As such, this relation partitions the set $\mathbb{Z}$ into equivalence classes. We denote the equivalence class containing the integer $a$ by $[a]_n$, or when $n$ is clear from context, we may simply write $[a]$. Historically, these equivalence classes are called residue classes modulo $n$, and we shall adopt this terminology here as well.

It is easy to see from the definitions that
$$[a]_n=a+n \mathbb{Z}:={a+n z: z \in \mathbb{Z}} .$$
Note that a given residue class modulo $n$ has many different “names”; for example, the residue class $[1]_n$ is the same as the residue class $[1+n]_n$. For any integer $a$ in a residue class, we call $a$ a representative of that class.
The following is simply a restatement of Theorem 2.1:
Theorem 2.9. For a positive integer $n$, there are precisely $n$ distinct residue classes modulo $n$, namely, $[a]_n$ for $a=0, \ldots, n-1$.

Fix a positive integer $n$. Let us define $\mathbb{Z}_n$ as the set of residue classes modulo $n$. We can “equip” $\mathbb{Z}_n$ with binary operations defining addition and multiplication in a natural way as follows: for $a, b \in \mathbb{Z}$, we define
$$[a]_n+[b]_n:=[a+b]_n,$$
and we define
$$[a]_n \cdot[b]_n:=[a \cdot b]_n .$$
Of course, one has to check this definition is unambiguous, in the sense that the sum or product of two residue classes should not depend on which particular representatives of the classes are chosen in the above definitions. More precisely, one must check that if $[a]_n=\left[a^{\prime}\right]_n$ and $[b]_n=\left[b^{\prime}\right]_n$, then $[a \text { op } b]_n=\left[a^{\prime} \text { op } b^{\prime}\right]_n$, for op $\in{+, \cdot}$. However, this property follows immediately from Theorem 2.3.
It is also convenient to define a negation operation on $\mathbb{Z}_n$, defining
$$-[a]_n:=[-1]_n \cdot[a]_n=[-a]_n .$$
Having defined addition and negation operations on $\mathbb{Z}_n$, we naturally define a subtraction operation on $\mathbb{Z}_n$ as follows: for $a, b \in \mathbb{Z}$,
$$[a]_n-[b]_n:=[a]_n+\left(-[b]_n\right)=[a-b]_n .$$

## 数学代写|数论代写Number Theory代考|Fermat’s little theorem

Let $n$ be a positive integer, and let $a \in \mathbb{Z}$ with $\operatorname{gcd}(a, n)=1$. Consider the sequence of powers of $\alpha:=[a]_n \in \mathbb{Z}_n^*$ :
$$[1]_n=\alpha^0, \alpha^1, \alpha^2, \ldots$$

Since each such power is an element of $\mathbb{Z}_n^$, and since $\mathbb{Z}_n^$ is a finite set, this sequence of powers must start to repeat at some point; that is, there must be a positive integer $k$ such that $\alpha^k=\alpha^i$ for some $i=0, \ldots, k-1$. Let us assume that $k$ is chosen to be the smallest such positive integer. We claim that $i=0$, or equivalently, $\alpha^k=[1]_n$. To see this, suppose by way of contradiction that $\alpha^k=\alpha^i$, for some $i=1, \ldots, k-1$. Then we can cancel $\alpha$ from both sides of the equation $\alpha^k=\alpha^i$, obtaining $\alpha^{k-1}=\alpha^{i-1}$, and this contradicts the minimality of $k$.

From the above discussion, we see that the first $k$ powers of $\alpha$, that is, $[1]_n=\alpha^0, \alpha^1, \ldots, \alpha^{k-1}$, are distinct, and subsequent powers of $\alpha$ simply repeat this pattern. More generally, we may consider both positive and negative powers of $\alpha$-it is easy to see (verify) that for all $i, j \in \mathbb{Z}$, we have $\alpha^i=\alpha^j$ if and only if $i \equiv j(\bmod k)$. In particular, we see that for any integer $i$, we have $\alpha^i=[1]_n$ if and only if $k$ divides $i$.

This value $k$ is called the multiplicative order of $\alpha$ or the multiplicative order of $a$ modulo $n$. It can be characterized as the smallest positive integer $k$ such that
$$a^k \equiv 1(\bmod n)$$

## 数学代写|数论代写Number Theory代考|Residue classes

$$[a]n=a+n \mathbb{Z}:=a+n z: z \in \mathbb{Z} .$$ 请注意，给定的剩余类模 $n$ 有许多不同的“名称”；例如，残基类 $[1]_n$ 与剩余类相同 $[1+n]_n$. 对于任何整数 $a$ 在残差类中，我猕 $a$ 该班级的代表。 以下是定理 $2.1$ 的简单重述: 定理 2.9。对于一个正整数 $n$, 正好有 $n$ 不同的残基类模 $n$ ，即， $[a]_n$ 为了 $a=0, \ldots, n-1$. 修正一个正整数 $n$. 让我们定义 $\mathbb{Z}_n$ 作为残基类的集合 $n$. 我们可以“装备” $\mathbb{Z}_n$ 二元运算以自然的方式定义加法和乘㴛如下: $a, b \in \mathbb{Z}$ 我们定义 $$[a]_n+[b]_n:=[a+b]_n,$$ 我们定义 $$[a]_n \cdot[b]_n:=[a \cdot b]_n .$$ 当然，必须检圭此定义是否明确，因为两个残基类别的㥕和或乘积不应取决于在上述定义中选择了哪些类别的特定代表。更准确地 说，必须检囩是否 $[a]_n=\left[a^{\prime}\right]_n$ 和 $[b]_n=\left[b^{\prime}\right]_n$ ，然后 $[a \text { op } b]_n=\left[a^{\prime} \text { op } b^{\prime}\right]_n$ ，冲了上去 $\in+, \cdot$ 然而，这个性质直接来自定理 2.3. 定义一个否定操作也很方便 $\mathbb{Z}_n$, 定义 $$-[a]_n:=[-1]_n \cdot[a]_n=[-a]_n .$$ 定义了加法和否定运算 $\mathbb{Z}_n$ ，我们自然定义了一个減法运算 $\mathbb{Z}_n$ 如下: 对于 $a, b \in \mathbb{Z}{\text {， }}$
$$[a]_n-[b]_n:=[a]_n+\left(-[b]_n\right)=[a-b]_n .$$

## 数学代写|数论代写Number Theory代考|Fermat’s little theorem

$$[1]_n=\alpha^0, \alpha^1, \alpha^2, \ldots$$

$$a^k \equiv 1(\bmod n)$$

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