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

2023年4月10日

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

The idea underlying mathematical induction is very simple. It relies on the fact that the set of natural numbers $\mathbb{N}$ is formed by starting with the initial element 1 and then by repeatedly incrementing to get every element of $\mathbb{N}$.

Now suppose that we have a list of statements $P_1, P_2, P_3, \ldots$ that we want to prove, where there is one statement for each natural number. Induction consists of two steps:

• Initialization Step: Verify that Statement $P_1$ is true.
• Induction Step: Let $n \geq 1$. Assume that you know that Statements $P_1, \ldots, P_n$ are true. Prove that Statement $P_{n+1}$ is true.

Suppose that we’ve done both the Initialization and Induction Steps. Why does that imply that every statement $P_1, P_2, P_3, \ldots$ is true? Well, the Initialization Step says that $P_1$ is true, so that gets us started. Next we use the Induction Step with $n=1$. Since we already know that $P_1$ is true, the Induction Step tells us that $P_2$ is true. Okay, now we know that both $P_1$ and $P_2$ are true, so the Induction Step with $n=2$ tells us that $P_3$ is true, and so on. This informal argument may convince you that mathematical induction is a valid method of proof, but a formal proof that mathematical induction works cannot rely on the phrase “and so on.” Here is a formal proof that uses the well-ordering principle.

Theorem 1.26. Mathematical induction as described in the Initialization Step and the Induction Step is a valid way to prove that Statement $P_n$ is true for every $n$.
Proof. We consider the set
$$\left{n \geq 1: P_n \text { is false }\right}$$
If the set (1.9) is empty, then every $P_n$ statement is true, and we are done. So we suppose that the set (1.9) is not empty, and our goal is to derive a contradiction.

Since we are assuming that the set (1.9) is not empty, the Well-Ordering Principle (Theorem 1.12) tells us that this set has a smallest element, which we denote by $m$. Thus statement $P_m$ is false, and for every $n<m$, the statement $P_n$ is true. If $m=1$, the Initialization Step tells us the $P_1$ is true, which is a contradiction. But if $m \geq 2$, then since we know that $P_1, \ldots, P_{m-1}$ are true, the Induction Step with $n=m-1$ tells us that $P_m$ is true, which is also a contradiction. Hence the set (1.9) is empty, which means that every $P_n$ statement is true.

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

Definition 1.28. We denote the integers by
$$\mathbb{Z}={\ldots,-3,-2,-1,0,1,2,3, \ldots}$$
You already know how to add, subtract, and multiply integers, and you know that these operations have various properties. For example, they satisfy the distributive law
$$a \cdot(b+c)=a \cdot b+a \cdot c$$
As we will see later, the integers are an example of a ring, while if we ignore multiplication and just look at addition, the integers are an example of an (infinite cyclic) group. But that’s all for the future. In this section we give some basic definitions and prove some basic properties about $\mathbb{Z}$ that will be useful later. Much of this material will be generalized in Chapters 3 and 7.
Definition 1.29. Let $a, b \in \mathbb{Z}$ with $b \neq 0$. We say that $b$ divides $a$, and we write $b \mid a$, if there is a $c \in \mathbb{Z}$ satisfying $a=b c$.

In general, if you divide $a$ by $b$, you “know” from grade school that you get a quotient and a remainder, where the remainder is smaller than $b$. Constructing a rigorous proof of this fact requires the well-ordering principle, a task that we leave for you; see Exercise 1.26. Here is an informal proof.

Proposition 1.30 (Division with Remainder). Let $a, b \in \mathbb{Z}$ with $b \neq 0$. Then there are unique integers $q, r \in \mathbb{Z}$ satisfying
$$a=b q+r \quad \text { and } \quad 0 \leq r<|b|$$
We call $q$ the quotient and $r$ the remainder.
Proof. We start with $a$, and we add and subtract multiples of $b$. This gives the list of numbers
$$\ldots, a-2 b, a-b, a, a+b, a+2 b, \ldots .$$
Since $b \neq 0$ and since we allow arbitrarily large multiples of $b$ in both the positive and negative directions, we see that the list (1.10) contains non-negative numbers. By choosing an appropriate multiple of $b$, we can get the difference
$a-q b$ to be between 0 and $|b|-1$.

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

Imathbb ${Z}={\backslash$ dots, $-3,-2,-1,0,1,2,3, \backslash$ dots $}$ 表示整数 $\$ \$$你已经知道如何加、减和乘整数，并且您知道这些操作 具有各种属性。例如，它们满足分配律 \ \$$
$a \backslash c d o t(b+c)=a \backslash c d o t ~ b+a ~ \backslash c d o t ~ c$
$\$ \$$正如我们稍后将看到的，整数是环的一个例子，而如果 我们忽略乘法只看加法，整数是 (无限循环) 群的一个 例子。但这都是为了末来。在本节中，我们给出了一些 基本定义并证明了 \\mathbb {Z} \$$ 的一些基本性质，这 些性质稍后会有用。第 3 章和第 7 章将对这些材料的大 部分内容进行概括。

$$q 商和 r 其余的。 a ，我们加上和减去的倍数 b. 这给出了数字列表$$
\ldots, a-2 b, a-b, a, a+b, a+2 b, \ldots

$b \neq 0$ 并且由于我们允许任意大的倍数 $b$ 在正负方向上， 我们看到列表 (1.10) 包含非负数。通过选择适当的倍数 $b$ ，我们可以得到差异
$a-q b$ 介于 0 和 $|b|-1$.

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