课程： 7.8 上岗和不平等

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上岗和不平等

This is the third in a series of lessons on mathematical proofs. In this lesson we continue to focus mainly on , this time of inequalities, and other kinds of proofs such as proof by geometry.
::这是一系列数学证明课程中的第三门课程。在这一课程中,我们继续主要关注这个不平等的时代,以及其它类型的证据,例如几何证明。

Induction and Inequalities
::上岗和不平等

The Transitive Property of Inequality
::不平等的过境财产

Below, we will prove several statements about inequalities that rely on the transitive property of inequality:
::下面我们将证明若干关于不平等的声明,

If a < b and b < c , then a < c .
::如果a < b和b < c > ,则a < c。

Note that we could also make such a statement by turning around the relationships (i.e., using “greater than” statements) or by making inclusive statements, such as a ≥ b .
::请注意,我们也可以通过扭转关系(即使用“大于”的发言)或作出包容性发言,如_________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

It is also important to note that this property of integers is a postulate , or a statement that we assume to be true. This means that we need not prove the transitive property of inequality.
::还必须指出,这种整数属性是一个假设,或我们假定是真实的陈述,这意味着我们不需要证明不平等的过渡性属性。

You encountered other useful properties of inequalities in earlier algebra courses:
::您在早期代数课程中遇到过不平等的其他有用特性:

Addition property: if a > b , then a + c > b + c .
::添加属性:如果 a > b,则a+ c > b+ c。

Multiplication property: if a > b , and c > 0 then ac > bc .
::乘性属性:如果 a > b, 和 c > 0, 那么 ac > bc。

Examples
::实例

Example 1
::例1

Prove that $\begin{aligned} n! \geq 2^{n} \end{aligned}$ for $\begin{aligned} n \geq 4 \end{aligned}$ .
::证明给N! 2n 证明给N. 4.

Step 1) The base case is n = 4: 4! = 24, 2 ⁴ = 16. 24 ≥ 16 so the base case is true.
::步骤1 基数为 n= 4: 4; 4; 24 = 16; 24 = 16, 因此基数是真实的。

Step 2) Assume that k ! ≥ 2 ^k for some value of k such that k ≥ 4
::第2步 2) 假设k! k! k2k 等于 k, k2k 等于 k, k4

Step 3) Show that ( k +1)! ≥ 2 ^k+1
::步骤 3 显示 (k+1) !% 2k+1

( k +1)! = k !( k +1)	Rewrite ( k +1)! in terms of k !
≥ 2 ^k ( k +1)	Use step 2 and the multiplication property.
≥ 2 ^k (2)	k +1 ≥ 5 >2, so we can use the multiplication property again.
= 2 ^k+1

Therefore n ! ≥ 2 ⁿ for n ≥ 4.
::因此,n! 2n = n 4。

Example 2
::例2

For what values of x is the inequality x > x ² true?
::对于 x 的值, 不平等 x > x2 是真实的吗 ?

The inequality is true if x is a number between -1 and 1 but not 0.
::如果 x 是-1 和 1 之间的数字, 但不是 0 , 不平等是真实的。

Example 3
::例3

Prove that 9 ⁿ - 1 is divisible by 8 for all positive integers n .
::证明所有正整数 n 的 9n - 1 可除以 8 。

1. Base case:	If n = 1, 9 ⁿ - 1 = 9-1 = 8 = 8(1)
2. Inductive hypothesis:	Assume that 9 ^k - 1 is divisible by 8.
3. Inductive step:	Show that 9 ^k+1 - 1 is divisible by 8.

9 ^k - 1 divisible by 8 $\begin{aligned} \Rightarrow \end{aligned}$ 8 W = (9 ^k -1) for some integer W
9 ^k+1 - 1 = 9(9 ^k - 1) + 8 = 9(8W) + 8,which is divisible by 8

Example 4
::例4

Prove that $\begin{aligned} 2^{n} < n! \end{aligned}$ for all positive integers n where $\begin{aligned} n \geq 4 \end{aligned}$ .
::证明 2n <n! 所有正整数 n 的 nnn4 。

Use the three steps of proof by induction:
::通过上岗培训使用三个证明步骤:

Step 1) Base Case: $\begin{aligned} 2^{4} < 4! \end{aligned}$
::基本情况:24 < 4!

$\begin{aligned} 2 \cdot 2 \cdot 2 \cdot 2 < 1 \cdot 2 \cdot 3 \cdot 4 \end{aligned}$

$\begin{aligned} 16 < 24 \end{aligned}$ ..... This checks out
::16 < 24... .

Step 2) Assumption: $\begin{aligned} 2^{k} < k! \end{aligned}$
::步骤2,假设: 2k<k!

Step 3) Induction Step: starting with $\begin{aligned} 2^{k} < k! \end{aligned}$ prove $\begin{aligned} 2^{k} (k + 1) < k! (k + 1) \end{aligned}$
::步骤3) 上岗步骤: 以 2k < k! 证明 2k( k+1) < k! (k+1)

$\begin{aligned} 2^{k} (k + 1) < (k + 1)! \end{aligned}$
::2k(k+1) <(k+1)!

$\begin{aligned} 2 < k + 1 \end{aligned}$ ..... If $\begin{aligned} k \geq 4 \end{aligned}$ then this is true
::2 <k+1.... 如果 k+4, 那么这是真的

$\begin{aligned} 2^{k} \cdot 2 < 2^{k} (k + 1) \end{aligned}$ ... Multiply both sides by $\begin{aligned} 2^{k} \end{aligned}$
::2k2 <2k( k+1)... 将两边乘以 2k

$\begin{aligned} 2^{k + 1} < 2^{k} (k + 1) \end{aligned}$
::2k+1 <2k(k+1)

$\begin{aligned} 2^{k + 1} < (k + 1)! \end{aligned}$
::2k+1 <(k+1)!

$\begin{aligned} ∴ 2^{n} < n! \end{aligned}$ for all positive integers n where $\begin{aligned} n \geq 4 \end{aligned}$
::*% 2n < n! 所有正数整数 n 的位置 n 4

Example 5
::例5

Prove that $\begin{aligned} n^{2} < 3^{n} \end{aligned}$ for all integers $\begin{aligned} n > 2 \end{aligned}$ .
::对所有整数 n>2 证明 n2 < 3n。

Use the three steps of proof by induction:
::通过上岗培训使用三个证明步骤:

Step 1) Base Case: (n = 1) $\begin{aligned} 1^{2} < 3^{1} \end{aligned}$ or, if you prefer, (n = 2) $\begin{aligned} 2^{2} < 3^{2} \end{aligned}$
::基本情况n = 1) 12 < 31或(n = 2) 22 < 32

Step 2) Assumption: $\begin{aligned} k^{2} < 3^{k} \end{aligned}$
::步骤2) 假设: k2<3k

Step 3) Induction Step: starting with $\begin{aligned} k^{2} < 3^{k} \end{aligned}$ prove $\begin{aligned} (k + 1)^{2} < 3^{k + 1} \end{aligned}$
::步骤3 步骤3 上岗步骤: 从 k2 < 3k 验证( k+1) 2 < 3k+1 开始

$\begin{aligned} k^{2} \cdot 3 < 3^{k} \cdot 3 \end{aligned}$
::k23<3k3

$\begin{aligned} 2 k < k^{2} \end{aligned}$ and $\begin{aligned} 1 < k^{2} \end{aligned}$ ..... assuming $\begin{aligned} 2 < k \end{aligned}$ as specified in the question
::2 k < k2 和 1 k2.... 假设2 kk 如问题中指定的那样

$\begin{aligned} 2 k + 1 < 2 k^{2} \end{aligned}$ ..... combine the two statements above
::2k+1 <2k2.... 合并上述两个语句

$\begin{aligned} k^{2} + 2 k + 1 < 3 k^{2} \end{aligned}$ ..... add $\begin{aligned} k^{2} \end{aligned}$ to both sides
::k2+2k+1 <3k2..... 在两侧添加 k2

$\begin{aligned} (k + 1)^{2} < 3 k^{2} \end{aligned}$
:k+1)2 <3k2

$\begin{aligned} (k + 1)^{2} < 3 \cdot 3^{k} \end{aligned}$ ..... from above
:k+1)2 <3__3k....从上方.

$\begin{aligned} (k + 1)^{2} < 3^{k + 1} \end{aligned}$
:k+1)2 <3k+1

$\begin{aligned} ∴ n^{2} < 3^{n} \end{aligned}$ for all integers $\begin{aligned} n > 2 \end{aligned}$
::n2 <3n 所有整数 n> 2

Example 6
::例6

Prove that $\begin{aligned} 2 n + 1 < 2^{n} \end{aligned}$ for all integers $\begin{aligned} n > 3 \end{aligned}$ .
::证明所有整数 n>3 2n+1<2n 。

Use the three steps of proof by induction:
::通过上岗培训使用三个证明步骤:

Step 1) Base case: If n = 3, 2(3) + 1 = 7, 2 ³ = 8 : 7 < 8, so the base case is true.
::步骤1 基本情况:如果 n = 3, 2(3) + 1 = 7, 23 = 8 : 7 < 8, 因此基本情况属实。

Step 2) Inductive hypothesis: Assume that 2 k + 1 < 2 ^k for k > 3
::步骤(2) 引言假设:假设2k+1 < 2k/k > 3

Step 3) Inductive step: Show that 2( k + 1) + 1 < 2 ^{k + 1}
::步骤3,引导步骤:显示2(k+1)+1 < 2k+1

2( k + 1) + 1 = 2 k + 2 + 1 = (2k + 1) + 2 < 2 ^k + 2 < 2 ^k + 2 ^k = 2 (2 ^k ) = 2 ^{k + 1}
::2(k+1)+1=2k+2+2+1=2k+2+1=2k+1=2k+1)+2 < 2k+2 < 2k+2 < 2k+2k+2k=2(2k)=2k+1

Review
::回顾

Prove the following inequalities.
::证明以下不平等。

$\begin{aligned} 5^{k} < (k + 5)! \end{aligned}$
::5k <(k+5)!

$\begin{aligned} 1^{k} < (k + 1)! \end{aligned}$
::1k<(k+1) !

$\begin{aligned} 4^{k} < (k + 4)! \end{aligned}$
::4k < (k+4) !

$\begin{aligned} 2^{k} < (k + 2)! \end{aligned}$
::2k <(k+2)!

For what values of x is the inequality x > x ² true?
::对于 x 的值, 不平等 x > x2 是真实的吗 ?

Prove that 3 ⁿ > n ² for all positive integers n .
::证明所有正整数 n 的 3n > n2 。

Prove the following inequalities.
::证明以下不平等。

$\begin{aligned} \frac{1}{n + 1} + \frac{1}{n + 2} + \frac{1}{n + 3} + . . . + \frac{1}{2 n} > \frac{13}{24} (n > 1) \end{aligned}$
::1n+1+1n+2+1n+3+...+12n>1324(n>1)

$\begin{aligned} 2^{n} \geq n^{2} \end{aligned}$ for $\begin{aligned} n = 4, 5, 6, . . . \end{aligned}$
::N=4,5,6,... 2nn2,n=4,5,6,...

$\begin{aligned} \frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + . . . + \frac{1}{n^{2}} \end{aligned}$
::112+122+132+...+1n2

Given: $\begin{aligned} x_{1}, . . ., x_{n} \end{aligned}$ are positive numbers, prove the following: $\begin{aligned} \frac{(x_{1} + . . . + x_{n})}{n} \geq (x_{1} \cdot . . . \cdot x_{n})^{\frac{1}{n}} \end{aligned}$
::给定 : x1,...... xn 是正数, 证明如下 : (x1+...+...+xn) n} (x1...xn) 1n

$\begin{aligned} n! \geq 3^{n} \end{aligned}$ for $\begin{aligned} n = 7, 8, 9, . . . . \end{aligned}$
::N$7,8,9,... 3n=7,8,9,9...

Complete the following geometric induction proofs.
::完成以下几何感应证明。

Prove that side length of a quadrilateral is less than the sum of all its other side lengths.
::证明四边形的侧长小于其所有其他侧长的总和。

Prove that side length of a pentagon is less than the sum of all its other side lengths.
::证明五角形的侧长低于其所有其他侧长的总和。

Prove that it is possible to color all regions of a plane divided by several lines with two different colors, so that any two neighbor regions contain a different color.
::证明可以用两种不同颜色将飞机的所有地区除以几条线,使任何两个相邻区域都含有不同的颜色。

Review (Answers)
::回顾(答复)

Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键。

章节大纲

常规

上岗和不平等

Induction and Inequalities ::上岗和不平等

The Transitive Property of Inequality ::不平等的过境财产

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Example 6 ::例6

Review ::回顾

Review (Answers) ::回顾(答复)

Induction and Inequalities
::上岗和不平等

The Transitive Property of Inequality
::不平等的过境财产

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Example 4
::例4

Example 5
::例5

Example 6
::例6

Review
::回顾

Review (Answers)
::回顾(答复)