Introduction to Mathematical Thinking - Week 7
Q: Why did nineteenth century mathematicians devote time to the proof of self-evident results? Select the best answer.
A: To gain mastery of, and confidence in, the methods of abstract proof to apply them in less obvious cases.
(看这个看的想睡觉,可能是没有动手跟上老师的思路,只是被动吸收。)
习题
2. Say whether the following proof is valid or not. [3 points]
Theorem. The square of any odd number is 1 more than a multiple of 8. (For example, 32=9=8+1,52=25=3⋅8+1.)
Proof: By the Division Theorem, any number can be expressed in one of the forms 4q, 4q+1, 4q+2, 4q+3. So any odd number has one of the forms 4q+1,4q+3. Squaring each of these gives:
(4q+1)2(4q+3)2==16q2+8q+116q2+24q+9==8(2q2+q)+18(2q2+3q+1)+1
In both cases the result is one more than a multiple of 8. This proves the theorem.
不是很理解题意。我的理解:任何奇数的平法都可以表示为 8x + 1 (x 是整数)
Say whether the following verification of the method of induction is valid or not. [3 points]
Proof: We have to prove that if:
* A(1)
* (∀n)[A(n)⇒A(n+1)]
then (∀n)A(n).
We argue by contradiction. Suppose the conclusion is false. Then there will be a natural number n such that ¬A(n). Let m be the least such number. By the first condition, m>1, so m=n+1 for some n. Since n<m, A(n). Then by the second condition, A(n+1), i.e., A(m). This is a contradiction, and that proves the result.
Evaluate this purported proof
Evaluate this purported proof
Evaluate this purported proof
Introduction to Mathematical Thinking - Week 7的更多相关文章
- Introduction to Mathematical Thinking - Week 6 - Proofs with Quantifieers
Mthod of proof by cases 证明完所有的条件分支,然后得出结论. 证明任意 使用任意 注意,对于一个任意的东西,你不知道它的具体信息.比如对于任意正数,你不知道它是 1 还是 2等 ...
- Introduction to Mathematical Thinking - Week 9 评论答案2
根据 rubic 打分. 1. 我认为,如果说明 m, n 是自然数,所以最小值是 1 会更清楚.所以 Clarity 我给了 3 分.其他都是 4 分,所以一共是 23 分. 2. 我给出的分数 ...
- Introduction to Mathematical Thinking - Week 9
错题 评分出错 题目要求的是 "any" ,而答案只给出了一个.所以认为回答者没有理解题意,连 any 都没有理解.所以 0 分. 第一,标准的归纳法只能对自然数使用,而题目要求的 ...
- Introduction to Mathematical Thinking - Week 4
否定的逻辑 应该思考符号背后表示的逻辑,而不是像操作算术运算符一样操作逻辑符号. 比如 对于任意的 x,x属于自然数,那么 x 是偶数或者奇数:这是对的 如果使用“乘法分配律”拆分,变成“对于任意的x ...
- Introduction to Mathematical Thinking - Week 3
there exists and all there exists 证明根号2是无理数 all 习题 3. Which of the following formal propositions say ...
- Introduction to Mathematical Thinking - Week 2
基本数学概念 real number(实数):是有理数和无理数的总称 有理数:可以表达为两个整数比的数(a/b, b!=0) 无理数是指除有理数以外的实数 imply -- 推导出 不需要 A 能推导 ...
- Deep Learning and Shallow Learning
Deep Learning and Shallow Learning 由于 Deep Learning 现在如火如荼的势头,在各种领域逐渐占据 state-of-the-art 的地位,上个学期在一门 ...
- Technical Development Guide---for Google
Technical Development Guide This guide provides tips and resources to help you develop your technica ...
- (转)Awesome Courses
Awesome Courses Introduction There is a lot of hidden treasure lying within university pages scatte ...
随机推荐
- 基于MaterialDesign设计风格的妹纸app的简单实现
代码地址如下:http://www.demodashi.com/demo/11644.html *今天看了郭神的第二行代码,深深的被MaterialDesign的设计风格所吸引,然后就照例做了一个小D ...
- SSH——基于BaseDao和BaseAction实现用户登录
基于BaseDao和BaseAction实现用户登录 1. 首先修改login.jsp页面,点击登录按钮,提交表单 <a onclick="document.forms ...
- LoadRunner+Java接口性能测试
想必各位小伙伴们会对LR还可以调用java感到好奇,之前我也这么一直认为LR只支持C语言.其实LR脚本支持的语言有:C.Java.Visual Basic.VbScript.JavaScript,只不 ...
- genymotion启动黑屏的原因及临时解决方法
解决办法 好像是要激活网络,具体原因不知道!!! sudo ifconfig vboxnet0 up 下面好像更好使一点 sudo ip link set dev vboxnet0 up
- RabbitMQ消息队基本概念
RabbitMQ简介 AMQP,即Advanced Message Queuing Protocol,高级消息队列协议,是应用层协议的一个开放标准,为面向消息的中间件设计.消息中间件主要用于组件之间的 ...
- android应用多线程守护让你非常难杀死它
1.android 应用开启后启动一个服务 public class TestserviceActivity extends Activity { /** Called when the activi ...
- C#--I/O流操作文本文件之StreamWrite类和StreamReader类
使用I/O流操作文本文件时主要用到StreamWrite类和StreamRead类. 1.StreamWrite类 (1)StreamWrite类专门用来处理文本文件的类.能够方便地想文本文件里写入字 ...
- Titel Block不显示
在原有原理图上New page时不显示Titel Block 解决办法:两种(1)(2) 方法1.在新建的原理图上右键选择Schematic Page Properties,按下图勾选即可 方法2.进 ...
- CentOS 加载/挂载光驱
1.mkdir /mnt/cdrom 2.mount /dev/cdrom /mnt/cdrom 3.挂载 umount /dev/cdrom /mnt/cdrom
- Redis之ziplist数据结构
0.前言 redis初始创建hash表,有序集合,链表时, 存储结构采用一种ziplist的存储结构, 这种结构内存排列更紧密, 能提高访存性能. 本文介绍ziplist数据结构 1.ziplist存 ...