## 数学代写|数论作业代写number theory代考|Math453

2022年12月23日

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## 数学代写|数论作业代写number theory代考|COUNTABLE AND UNCOUNTABLE SETS

Definition 3.6. A set $S$ is said to be countable if there exists a one-to-one function from $S$ to $\mathbb{N}$, the set of natural numbers, and uncountable if it is not countable.

Lemma 3.28. Any finite set is countable; the set of integers is countable.
Lemma 3.29. Let $f$ be a function from $A$ to $B$ and $g$ a function from $B$ to C. If $f$ and $g$ are both one-to-one, so is the composite function $g \circ f$. If $g \circ f$ is one-to-one, then so is $f$.

Corollary 3.30. If $T$ is countable and there is a one-to-one function from $S$ to $T$, then $S$ is countable.

Exercise. Give an example to show that if $g \circ f$ is one-to-one, $g$ need not be one-to-one.

Theorem 3.31. If $S$ and $T$ are countable, then so are $S \times T$ and $S \cup T$. If $S$ is countable and $R \subseteq S$, then $R$ is countable.
Proof. Let $f: S \rightarrow \mathbb{N}$ and $g: T \rightarrow \mathbb{N}$ be one-to-one functions. Then
$$h: S \times T \rightarrow \mathbb{N}, \quad h(s, t)=2^{f(s)} 3^{g(t)}$$
and
$$k: S \cup T \rightarrow \mathbb{N} \times \mathbb{N}, \quad k(x)= \begin{cases}(0, f(x)) & \text { if } x \in S \ (1, g(x)) & \text { if } x \notin S\end{cases}$$
are one-to-one. If $R \subseteq S$, then the restriction $\left.f\right|_R$ of $f$ to $R$ is one-to-one.
The above constructions can easily be generalised to prove the following.
Theorem 3.32. Suppose that the sets $S_0, S_1, S_2, \ldots$ are countable. Then

• $S_0 \times S_1 \times \cdots \times S_n$ is countable for each natural number $n$;
• $S_0 \cup S_1 \cup S_2 \cup \cdots$ is countable.
Exercise. Explain why we need to restrict the Cartesian product in this result to finitely many terms, but do not need to restrict the union in the same way.
The second of the above properties can also be extended a little further.

## 数学代写|数论作业代写number theory代考|DEFINITION AND BASIC PROPERTIES

Definition 4.1. A finite or infinite expression of the form
$$a_0+\frac{b_1}{a_1+\frac{b_2}{a_2+\frac{b_3}{a_3+\cdots}}}$$
is called a continued fraction. A simple continued fraction is one in which every $b_k$ is 1 , every $a_k$ is an integer, and every $a_k$ except possibly $a_0$ is positive. For a (finite or infinite) simple continued fraction we shall also use the notations
$$a_0+\frac{1}{a_1+} \frac{1}{a_2+} \frac{1}{a_3+} \ldots \quad \text { and } \quad\left[a_0, a_1, a_2, a_3, \ldots\right] .$$
A finite simple continued fraction is said to represent the number obtained by performing the arithmetic in the obvious way; an infinite simple continued fraction $\left[a_0, a_1, a_2, a_3, \ldots\right]$ represents the real number $\alpha$ if
$$\alpha=\lim _{n \rightarrow \infty}\left[a_0, a_1, a_2, a_3, \ldots, a_n\right]$$

Let $k \in \mathbb{N}$. The integer $a_k$ is called the $k$ th partial quotient of the continued fraction $\left[a_0, a_1, a_2, \ldots\right]$, or of the number $\alpha$ it represents; the continued fraction $\alpha_k=\left[a_k, a_{k+1}, a_{k+2}, \ldots\right]$ is the kth complete quotient of $\alpha$; and the continued fraction $\left[a_0, a_1, \ldots, a_k\right]$ is the kth convergent to $\alpha$.

Note that a convergent, being defined as a finite continued fraction, is always a rational number. Henceforth we shall blur the distinction between a continued fraction and the number represented by the continued fraction. We shall use such language as, for example, “the continued fraction $\alpha=\left[a_0, a_1, a_2, \ldots\right]$ ” instead of saying more precisely, “the continued fraction $\left[a_0, a_1, a_2, \ldots\right]$ which represents the number $\alpha “$.

Continued fractions, their convergents, partial quotients and complete quotients have many fascinating properties.

# 数论作业代写

## 数学代写|数论作业代写number theory代考|COUNTABLE AND UNCOUNTABLE SETS

$$h: S \times T \rightarrow \mathbb{N}, \quad h(s, t)=2^{f(s)} 3^{g(t)}$$

$k: S \cup T \rightarrow \mathbb{N} \times \mathbb{N}, \quad k(x)={(0, f(x)) \quad$ if $x \in S(1, g(x))$

• $S_0 \times S_1 \times \cdots \times S_n$ 对于每个自然数都是可数的 $n$;
• $S_0 \cup S_1 \cup S_2 \cup \cdots$ 是可数的。
锻炼。解释为什么我们需要将此结果中的笛卡尔积限制为有限 多项，但不需要以相同的方式限制并集。
上面的第二个属性也可以进一步扩展。

## 数学代写|数论作业代写number theory代考|DEFINITION AND BASIC PROPERTIES

$$a_0+\frac{b_1}{a_1+\frac{b_2}{a_2+\frac{b_3}{a_3+\cdots}}}$$

$$a_0+\frac{1}{a_1+} \frac{1}{a_2+} \frac{1}{a_3+} \ldots \quad \text { and } \quad\left[a_0, a_1, a_2, a_3, \ldots\right]$$

$$\alpha=\lim {n \rightarrow \infty}\left[a_0, a_1, a_2, a_3, \ldots, a_n\right]$$ 让 $k \in \mathbb{N}$. 整数 $a_k$ 被称为 $k$ 连分数的第 th 个偏商 $\left[a_0, a_1, a_2, \ldots\right]$, 或 数 $\alpha$ 它代表; 连分数 $\alpha_k=\left[a_k, a{k+1}, a_{k+2}, \ldots\right.$. 是的第 $\mathrm{k}$ 个完全商 $\alpha$; 和连分数 $\left[a_0, a_1, \ldots, a_k\right]$ 是第 $\mathrm{k}$ 个收敛于 $\alpha$.

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## MATLAB代写

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