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10.1 Sets and Set Notation 集合与集合记号

本页译自 LibreTexts · A First Course in Linear Algebra (Kuttler)「Some Prerequisite Topics」第 10.1 节。公式经 MathJax 渲染，自定义宏已注入。

A set is a collection of things called elements. For example $\left\{1,2,3,8\right\}$ would be a set consisting of the elements $1,2,3$, and $8$.

集合是由称为元素的事物汇成的整体。例如，$\left\{1,2,3,8\right\}$ 就是一个由元素 $1,2,3$ 和 $8$ 组成的集合。

To indicate that $3$ is an element of $\left\{1,2,3,8\right\}$, it is customary to write $3\in\left\{1,2,3,8\right\}$. We can also indicate when an element is not in a set, by writing $9\notin\left\{1,2,3,8\right\}$, which says that $9$ is not an element of $\left\{1,2,3,8\right\}$.

为了表示 $3$ 是 $\left\{1,2,3,8\right\}$ 的元素，习惯上写作 $3\in\left\{1,2,3,8\right\}$。我们也可以表示某个元素不在集合中，即写成 $9\notin\left\{1,2,3,8\right\}$，它表示 $9$ 不是 $\left\{1,2,3,8\right\}$ 的元素。

Sometimes a rule specifies a set. For example you could specify a set as all integers larger than $2$. This would be written as $$S=\left\{x\in\mathbb{Z}:x>2\right\}.$$ This notation says: $S$ is the set of all integers $x$, such that $x>2$.

有时可以用一条规则来刻画一个集合。例如你规定一个集合「所有大于 $2$ 的整数」。这可写成 $$S=\left\{x\in\mathbb{Z}:x>2\right\}.$$ 这个记号表示：$S$ 是所有满足 $x>2$ 的整数 $x$ 构成的集合。

Suppose $A$ and $B$ are sets with the property that every element of $A$ is an element of $B$. Then we say that $A$ is a subset of $B$. For example, $\left\{1,2,3,8\right\}$ is a subset of $\left\{1,2,3,4,5,8\right\}$. In symbols, we write $$\left\{1,2,3,8\right\}\subseteq\left\{1,2,3,4,5,8\right\}.$$

设 $A$、$B$ 是集合，且 $A$ 的每个元素都是 $B$ 的元素，则称 $A$ 是 $B$ 的一个子集。例如，$\left\{1,2,3,8\right\}$ 是 $\left\{1,2,3,4,5,8\right\}$ 的子集。用符号可写成 $$\left\{1,2,3,8\right\}\subseteq\left\{1,2,3,4,5,8\right\}.$$

It is sometimes said that “$A$ is contained in $B$” or even “$B$ contains $A$”. The same statement about the two sets may also be written as $$\left\{1,2,3,4,5,8\right\}\supseteq\left\{1,2,3,8\right\}.$$

有时也说「$A$ 被包含于 $B$」甚至「$B$ 包含 $A$」。关于这两个集合的同一命题也可写成 $$\left\{1,2,3,4,5,8\right\}\supseteq\left\{1,2,3,8\right\}.$$

We can also talk about the union of two sets, which we write as $A\cup B$. This is the set consisting of everything which is an element of at least one of the sets, $A$ or $B$. As an example of the union of two sets, consider $$\left\{1,2,3,8\right\}\cup\left\{3,4,7,8\right\}=\left\{1,2,3,4,7,8\right\}.$$ This set is made up of the numbers which are in at least one of the two sets. In general $$A\cup B=\left\{x:x\in A\text{ or }x\in B\right\}.$$

我们还可以谈论两个集合的并集，记为 $A\cup B$。它由至少属于 $A$ 或 $B$ 中某个集合的元素构成。以两个集合的并集为例，考虑 $$\left\{1,2,3,8\right\}\cup\left\{3,4,7,8\right\}=\left\{1,2,3,4,7,8\right\}.$$ 这个集合由至少属于两个集合之一的数组成。一般地，$$A\cup B=\left\{x:x\in A\text{ 或 }x\in B\right\}.$$

Notice that an element which is in both $A$ and $B$ is also in the union, as well as elements which are in only one of $A$ or $B$.

注意：既属于 $A$ 又属于 $B$ 的元素同样在并集之中，而只属于 $A$ 与只属于 $B$ 中某一个的元素也在其中。

Another important set is the intersection of two sets $A$ and $B$, written $A\cap B$. This set consists of everything which is in both of the sets. Thus $\left\{1,2,3,8\right\}\cap\left\{3,4,7,8\right\}=\left\{3,8\right\}$ because $3$ and $8$ are those elements the two sets have in common. In general, $$A\cap B=\left\{x:x\in A\text{ and }x\in B\right\}.$$

另一个重要的集合是两个集合 $A$ 与 $B$ 的交集，记为 $A\cap B$。它由同时属于两个集合的元素构成。因此，$\left\{1,2,3,8\right\}\cap\left\{3,4,7,8\right\}=\left\{3,8\right\}$，因为 $3$ 和 $8$ 正是两个集合共有的元素。一般地，$$A\cap B=\left\{x:x\in A\text{ 且 }x\in B\right\}.$$

If $A$ and $B$ are two sets, $A\setminus B$ denotes the set of things which are in $A$ but not in $B$. Thus $$A\setminus B=\left\{x\in A:x\notin B\right\}.$$ For example, if $A=\left\{1,2,3,8\right\}$ and $B=\left\{3,4,7,8\right\}$, then $A\setminus B=\left\{1,2,3,8\right\}\setminus\left\{3,4,7,8\right\}=\left\{1,2\right\}$.

若 $A$、$B$ 是两个集合，则 $A\setminus B$ 表示在 $A$ 中但不在 $B$ 中的事物所成之集。因此 $$A\setminus B=\left\{x\in A:x\notin B\right\}.$$ 例如，若 $A=\left\{1,2,3,8\right\}$、$B=\left\{3,4,7,8\right\}$，则 $A\setminus B=\left\{1,2,3,8\right\}\setminus\left\{3,4,7,8\right\}=\left\{1,2\right\}$。

A special set which is very important in mathematics is the empty set denoted by $\emptyset$, which is defined as the set which has no elements in it. It follows that the empty set is a subset of every set. This is true because if it were not so, there would have to exist a set $A$, such that $\emptyset$ has something in it which is not in $A$. However, $\emptyset$ has nothing in it and so it must be that $\emptyset\subseteq A$.

数学中一个非常重要而特殊的集合是空集，记为 $\emptyset$，它被定义为不含任何元素的集合。由此可知空集是任何集合的子集。这是因为若非如此，就必然存在一个集合 $A$，使 $\emptyset$ 中有某个元素不在 $A$ 中；但 $\emptyset$ 空无一物，因此必有 $\emptyset\subseteq A$。

We can also use brackets to denote sets which are intervals of numbers. Let $a$ and $b$ be real numbers. Then:

我们还可以用括号表示作为数轴的区间那样的集合。设 $a$、$b$ 是实数，则：

These sorts of sets of real numbers are called intervals. The two points $a$ and $b$ are called endpoints, or bounds, of the interval. In particular, $a$ is the lower bound while $b$ is the upper bound of the above intervals, where applicable. Other intervals such as $\left(-\infty,b\right)$ are defined by analogy. In general, the curved parenthesis indicates that the end point is not included in the interval, while the square parenthesis indicates that this end point is included. The reason that there will always be a curved parenthesis next to $\infty$ or $-\infty$ is that these are not real numbers and cannot be included in the interval in the way a real number can.

这类实数的集合称为区间。点 $a$ 和 $b$ 称为区间的端点（或界）。特别地，在上述区间中，$a$ 是下界、$b$ 是上界（在适用之处）。其他区间如 $\left(-\infty,b\right)$ 按同样的道理定义。一般地，圆括号表示端点不包含在区间内，方括号表示该端点被包含。之所以 $\infty$ 或 $-\infty$ 旁边总出现圆括号，是因为它们不是实数，不能像实数那样被包含进区间。

To illustrate the use of this notation relative to intervals consider three examples of inequalities. Their solutions will be written in the interval notation just described.

为了说明这种区间记号的使用，考虑三个不等式的例子。它们的解将用刚才介绍的区间记号写出。

Example 1: Solving an Inequality

示例 1：求解一个不等式

Solve the inequality $2x+4\leq x-8$.

解不等式 $2x+4\leq x-8$。

Solution. We need to find $x$ such that $2x+4\leq x-8$. Solving for $x$, we see that $x\leq-12$ is the answer. This is written in terms of an interval as $(-\infty,-12]$.

解。我们需要找到使 $2x+4\leq x-8$ 成立的 $x$。解出 $x$，可知 $x\leq-12$ 即为答案。用区间表示就是 $(-\infty,-12]$。

Consider the following example.

考虑下面的例子。

Example 2: Solving an Inequality

示例 2：求解一个不等式

Solve the inequality $\left(x+1\right)\left(2x-3\right)\geq 0$.

解不等式 $\left(x+1\right)\left(2x-3\right)\geq 0$。

Solution. We need to find $x$ such that $\left(x+1\right)\left(2x-3\right)\geq 0$. The solution is given by $x\leq-1$ or $x\geq\frac{3}{2}$. Therefore, those $x$ which fit into either of these intervals give a solution. In terms of set notation this is denoted by $(-\infty,-1]\cup\left[\frac{3}{2},\infty\right)$.

解。我们需要找到使 $\left(x+1\right)\left(2x-3\right)\geq 0$ 成立的 $x$。解为 $x\leq-1$ 或 $x\geq\frac{3}{2}$。因此，落在其中任一区间的 $x$ 都是解。用集合记号可记为 $(-\infty,-1]\cup\left[\frac{3}{2},\infty\right)$。

Consider one last example.

再考虑最后一个例子。

Example 3: Solving an Inequality

示例 3：求解一个不等式

Solve the inequality $x\left(x+2\right)\geq-4$.

解不等式 $x\left(x+2\right)\geq-4$。

Solution. This inequality is true for any value of $x$ where $x$ is a real number. We can write the solution as $\mathbb{R}$ or $(-\infty,\infty)$.

解。对任何实数 $x$，这个不等式都成立。因此解可写成 $\mathbb{R}$ 或 $(-\infty,\infty)$。

In the next section, we examine another important mathematical concept.

下一节我们将考察另一个重要的数学概念。

10.2 Well Ordering and Induction 良序原理与数学归纳

本页同属「Some Prerequisite Topics」第 10.2 节，公式经 MathJax 渲染。

We begin this section with some important notation. Summation notation, written $\sum_{i=1}^{j} i$, represents a sum. Here, $i$ is called the index of the sum, and we add iterations until $i=j$. For example, $$\sum_{i=1}^{j} i=1+2+\cdots+j.$$ Another example: $$a_{11}+a_{12}+a_{13}=\sum_{i=1}^{3}a_{1i}.$$

本节从一些重要的记号谈起。求和记号 $\sum_{i=1}^{j} i$ 表示一个和。这里 $i$ 称为求和的指标，我们一直累加直到 $i=j$。例如，$$\sum_{i=1}^{j} i=1+2+\cdots+j.$$ 又如：$$a_{11}+a_{12}+a_{13}=\sum_{i=1}^{3}a_{1i}.$$

The following notation is a specific use of summation notation.

下面的记号是求和记号的一种具体用法。

Summation Notation. Let $a_{ij}$ be real numbers, and suppose $1\leq i\leq r$ while $1\leq j\leq s$. These numbers can be listed in a rectangular array as given by $$\begin{array}{cccc}a_{11}&a_{12}&\cdots&a_{1s}\\a_{21}&a_{22}&\cdots&a_{2s}\\\vdots&\vdots&&\vdots\\a_{r1}&a_{r2}&\cdots&a_{rs}\end{array}$$ Then $\sum_{j=1}^{s}\sum_{i=1}^{r} a_{ij}$ means to first sum the numbers in each column (using $i$ as the index) and then to add the sums which result (using $j$ as the index). Similarly, $\sum_{i=1}^{r}\sum_{j=1}^{s} a_{ij}$ means to sum the vectors in each row (using $j$ as the index) and then to add the sums which result (using $i$ as the index).

求和记号。设 $a_{ij}$ 为实数，且 $1\leq i\leq r$、$1\leq j\leq s$。这些数可以按如下矩形阵列列出 $$\begin{array}{cccc}a_{11}&a_{12}&\cdots&a_{1s}\\a_{21}&a_{22}&\cdots&a_{2s}\\\vdots&\vdots&&\vdots\\a_{r1}&a_{r2}&\cdots&a_{rs}\end{array}$$ 那么 $\sum_{j=1}^{s}\sum_{i=1}^{r} a_{ij}$ 表示先对各列求和（以 $i$ 为指标），再把所得的各和相加（以 $j$ 为指标）。类似地，$\sum_{i=1}^{r}\sum_{j=1}^{s} a_{ij}$ 表示先对各行求和（以 $j$ 为指标），再把所得各和相加（以 $i$ 为指标）。

Notice that since addition is commutative, $$\sum_{j=1}^{s}\sum_{i=1}^{r} a_{ij}=\sum_{i=1}^{r}\sum_{j=1}^{s} a_{ij}.$$

注意到由于加法满足交换律，$$\sum_{j=1}^{s}\sum_{i=1}^{r} a_{ij}=\sum_{i=1}^{r}\sum_{j=1}^{s} a_{ij}.$$

We now consider the main concept of this section. Mathematical induction and well ordering are two extremely important principles in math. They are often used to prove significant things which would be hard to prove otherwise.

现在我们进入本节的主要概念。数学归纳法与良序原理是数学中两条极其重要的原理。它们常被用来证明一些用其他方法很难证明的重要结论。

Definition 1: Well Ordered

定义 1：良序的

A set is well ordered if every nonempty subset $S$ contains a smallest element $z$ having the property that $z\leq x$ for all $x\in S$.

若一个集合的每个非空子集 $S$ 都包含一个最小元素 $z$，且对 $S$ 中所有 $x$ 都有 $z\leq x$，则称该集合是良序的。

In particular, the set of natural numbers defined as $$\mathbb{N}=\left\{1,2,\cdots\right\}$$ is well ordered.

特别地，由自然数构成的集合 $$\mathbb{N}=\left\{1,2,\cdots\right\}$$ 是良序的。

Consider the following proposition.

考虑下面的命题。

Proposition 1: Well Ordered Sets

命题 1：良序集

Any set of integers larger than a given number is well ordered.

任意由大于某给定数的整数组成的集合都是良序的。

This proposition claims that if a set has a lower bound which is a real number, then this set is well ordered.

这个命题断言：若一个集合存在一个实数作为下界，则该集合是良序的。

Further, this proposition implies the principle of mathematical induction. The symbol $\mathbb{Z}$ denotes the set of all integers. Note that if $a$ is an integer, then there are no integers between $a$ and $a+1$.

进而，此命题蕴含着数学归纳法原理。符号 $\mathbb{Z}$ 表示全体整数构成的集合。注意：若 $a$ 是整数，则在 $a$ 与 $a+1$ 之间不存在整数。

Theorem 1: Mathematical Induction

定理 1：数学归纳法

A set $S\subseteq\mathbb{Z}$, having the property that $a\in S$ and $n+1\in S$ whenever $n\in S$, contains all integers $x\in\mathbb{Z}$ such that $x\geq a$.

若集合 $S\subseteq\mathbb{Z}$ 具有这样的性质：$a\in S$，且每当 $n\in S$ 时都有 $n+1\in S$，则 $S$ 包含所有满足 $x\geq a$ 的整数 $x\in\mathbb{Z}$。

Proof. Let $T$ consist of all integers larger than or equal to $a$ which are not in $S$. The theorem will be proved if $T=\emptyset$. If $T\neq\emptyset$ then by the well ordering principle, there would have to exist a smallest element of $T$, denoted as $b$. It must be the case that $b>a$ since by definition, $a\notin T$. Thus $b\geq a+1$, and so $b-1\geq a$ and $b-1\notin S$ because if $b-1\in S$, then $b-1+1=b\in S$ by the assumed property of $S$. Therefore, $b-1\in T$ which contradicts the choice of $b$ as the smallest element of $T$ ($b-1$ is smaller). Since a contradiction is obtained by assuming $T\neq\emptyset$, it must be the case that $T=\emptyset$ and this says that every integer at least as large as $a$ is also in $S$.

证明。令 $T$ 表示所有大于或等于 $a$ 但不在 $S$ 中的整数。若能证明 $T=\emptyset$，定理即得证。若 $T\neq\emptyset$，则由良序原理，$T$ 必存在一个最小元素，记为 $b$。必有 $b>a$，因为按定义 $a\notin T$。于是 $b\geq a+1$，从而 $b-1\geq a$ 且 $b-1\notin S$——因为若 $b-1\in S$，则由 $S$ 的假定性质可推出 $b-1+1=b\in S$。因此 $b-1\in T$，这与「$b$ 是 $T$ 的最小元素」相矛盾（因为 $b-1$ 更小）。既然假设 $T\neq\emptyset$ 会推出矛盾，就必然有 $T=\emptyset$，这正说明每个至少与 $a$ 一样大的整数都在 $S$ 中。

Mathematical induction is a very useful device for proving theorems about the integers. The procedure is as follows.

数学归纳法是证明关于整数之定理的一种非常有效的工具。其步骤如下。

Procedure 1: Proof by Mathematical Induction

步骤 1：用数学归纳法证明

Suppose $S_n$ is a statement which is a function of the number $n$, for $n=1,2,\cdots$, and we wish to show that $S_n$ is true for all $n\geq 1$. To do so using mathematical induction, use the following steps.

设 $S_n$ 是关于数 $n$（$n=1,2,\cdots$）的命题，我们希望证明对一切 $n\geq 1$，$S_n$ 成立。用数学归纳法证明时，按如下步骤进行。
  1. Base Case: Show $S_1$ is true.
  2. Assume $S_n$ is true for some $n$, which is the induction hypothesis. Then, using this assumption, show that $S_{n+1}$ is true.
  1. 归纳基础：证明 $S_1$ 成立。
  2. 假设对某个 $n$，$S_n$ 成立（这称为归纳假设），然后利用这一假设证明 $S_{n+1}$ 成立。

Proving these two steps shows that $S_n$ is true for all $n=1,2,\cdots$.

完成这两步，就证明了 $S_n$ 对一切 $n=1,2,\cdots$ 成立。

We can use this procedure to solve the following examples.

我们可用此方法解决下面的例子。

Example 1: Proving by Induction

示例 1：用归纳法证明

Prove by induction that $$\sum_{k=1}^{n}k^{2}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}.$$

用数学归纳法证明 $$\sum_{k=1}^{n}k^{2}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}.$$

Solution. By the Procedure above, we first need to show that this statement is true for $n=1$. When $n=1$, the statement says that $$\sum_{k=1}^{1}k^{2}=\frac{1\left(1+1\right)\left(2(1)+1\right)}{6}=\frac{6}{6}=1.$$ The sum on the left hand side also equals $1$, so this equation is true for $n=1$.

解。按上述步骤，我们先要证明该命题对 $n=1$ 成立。当 $n=1$ 时，命题说的是 $$\sum_{k=1}^{1}k^{2}=\frac{1\left(1+1\right)\left(2(1)+1\right)}{6}=\frac{6}{6}=1.$$ 左边这一和也等于 $1$，所以该等式对 $n=1$ 成立。

Now suppose this formula is valid for some $n\geq 1$ where $n$ is an integer. Hence, the following equation is true. $$\sum_{k=1}^{n}k^{2}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}.$$ We want to show that this is true for $n+1$. Suppose we add $\left(n+1\right)^{2}$ to both sides of the equation above.

现设此公式对某个整数 $n\geq 1$ 成立，于是下面的等式成立：$$\sum_{k=1}^{n}k^{2}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}.$$ 我们要证明它对 $n+1$ 也成立。在上述等式两边都加上 $\left(n+1\right)^{2}$，得到 $$\sum_{k=1}^{n+1}k^{2}=\sum_{k=1}^{n}k^{2}+\left(n+1\right)^{2}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}+\left(n+1\right)^{2}.$$

The step going from the first to the second line is based on the assumption that the formula is true for $n$. Now simplify the expression in the second line, $$\frac{n\left(n+1\right)\left(2n+1\right)}{6}+\left(n+1\right)^{2}.$$ This equals $\left(n+1\right)\left(\frac{n\left(2n+1\right)}{6}+\left(n+1\right)\right)$ and $\frac{n\left(2n+1\right)}{6}+\left(n+1\right)=\frac{6\left(n+1\right)+2n^{2}+n}{6}=\frac{\left(n+2\right)\left(2n+3\right)}{6}$. Therefore, $$\sum_{k=1}^{n+1}k^{2}=\frac{\left(n+1\right)\left(n+2\right)\left(2n+3\right)}{6}=\frac{\left(n+1\right)\left(\left(n+1\right)+1\right)\left(2\left(n+1\right)+1\right)}{6},$$ showing the formula holds for $n+1$ whenever it holds for $n$. This proves the formula by mathematical induction. In other words, this formula is true for all $n=1,2,\cdots$.

由第一步到第二步，依据的是「公式对 $n$ 成立」的假设。现在化简第二行的表达式 $$\frac{n\left(n+1\right)\left(2n+1\right)}{6}+\left(n+1\right)^{2}.$$ 它等于 $\left(n+1\right)\left(\frac{n\left(2n+1\right)}{6}+\left(n+1\right)\right)$，而 $\frac{n\left(2n+1\right)}{6}+\left(n+1\right)=\frac{6\left(n+1\right)+2n^{2}+n}{6}=\frac{\left(n+2\right)\left(2n+3\right)}{6}$。因此，$$\sum_{k=1}^{n+1}k^{2}=\frac{\left(n+1\right)\left(n+2\right)\left(2n+3\right)}{6}=\frac{\left(n+1\right)\left(\left(n+1\right)+1\right)\left(2\left(n+1\right)+1\right)}{6},$$ 这表明只要公式对 $n$ 成立，就对 $n+1$ 成立。这样就通过数学归纳法证明了该公式。换句话说，该公式对一切 $n=1,2,\cdots$ 成立。

Consider another example.

再看一个例子。

Example 2: Proving an Inequality by Induction

示例 2：用归纳法证明一个不等式

Show that for all $n\in\mathbb{N}$, $$\frac{1}{2}\cdot\frac{3}{4}\cdots\frac{2n-1}{2n}<\frac{1}{\sqrt{2n+1}}.$$

证明：对一切 $n\in\mathbb{N}$，$$\frac{1}{2}\cdot\frac{3}{4}\cdots\frac{2n-1}{2n}<\frac{1}{\sqrt{2n+1}}.$$

Solution. Again we will use the Procedure above to prove that this statement is true for all $n$. Suppose $n=1$. Then the statement says $\frac{1}{2}<\frac{1}{\sqrt{3}}$, which is true.

解。我们仍用上述步骤来证明该命题对一切 $n$ 成立。设 $n=1$，则命题说的是 $\frac{1}{2}<\frac{1}{\sqrt{3}}$，这是成立的。

Suppose then that the inequality holds for $n$. In other words, $\frac{1}{2}\cdot\frac{3}{4}\cdots\frac{2n-1}{2n}<\frac{1}{\sqrt{2n+1}}$ is true. Now multiply both sides of this inequality by $\frac{2n+1}{2n+2}$. This yields $$\frac{1}{2}\cdot\frac{3}{4}\cdots\frac{2n-1}{2n}\cdot\frac{2n+1}{2n+2}<\frac{1}{\sqrt{2n+1}}\frac{2n+1}{2n+2}=\frac{\sqrt{2n+1}}{2n+2}.$$

再设该不等式对 $n$ 成立，即 $\frac{1}{2}\cdot\frac{3}{4}\cdots\frac{2n-1}{2n}<\frac{1}{\sqrt{2n+1}}$ 成立。现在两边同乘 $\frac{2n+1}{2n+2}$，得到 $$\frac{1}{2}\cdot\frac{3}{4}\cdots\frac{2n-1}{2n}\cdot\frac{2n+1}{2n+2}<\frac{1}{\sqrt{2n+1}}\frac{2n+1}{2n+2}=\frac{\sqrt{2n+1}}{2n+2}.$$

The theorem will be proved if this last expression is less than $\frac{1}{\sqrt{2n+3}}$. This happens if and only if $$\left(\frac{1}{\sqrt{2n+3}}\right)^{2}=\frac{1}{2n+3}>\frac{2n+1}{\left(2n+2\right)^{2}},$$ which occurs if and only if $\left(2n+2\right)^{2}>\left(2n+3\right)\left(2n+1\right)$, and this is clearly true which may be seen from expanding both sides. This proves the inequality.

若最后一个表达式的值小于 $\frac{1}{\sqrt{2n+3}}$，定理即得证。这只在且仅在 $$\left(\frac{1}{\sqrt{2n+3}}\right)^{2}=\frac{1}{2n+3}>\frac{2n+1}{\left(2n+2\right)^{2}}$$ 时发生；而这又在且仅在 $\left(2n+2\right)^{2}>\left(2n+3\right)\left(2n+1\right)$ 时成立。把两边展开即可看出这是显然成立的。于是不等式得证。

Let's review the process just used. If $S$ is the set of integers at least as large as $1$ for which the formula holds, the first step was to show $1\in S$ and then that whenever $n\in S$, it follows $n+1\in S$. Therefore, by the principle of mathematical induction, $S$ contains $\left[1,\infty\right)\cap\mathbb{Z}$, all positive integers. In doing an inductive proof of this sort, the set $S$ is normally not mentioned. One just verifies the steps above.

回顾一下刚才使用的过程。若 $S$ 是使公式成立且至少不小于 $1$ 的整数集合，第一步是证明 $1\in S$，然后证明每当 $n\in S$ 就有 $n+1\in S$。于是由数学归纳法原理，$S$ 包含 $\left[1,\infty\right)\cap\mathbb{Z}$，即全体正整数。做此类归纳证明时通常不提及集合 $S$，只需验证上述步骤即可。