香港中文大学数学课程-现代数学奠基(MATH1050) 额外练习 (二)
香港中文大学数学课程-现代数学奠基(MATH1050) 习作习作 (七)

MATH1050B/C Assignment7Due date:20-3-2017(1900hrs)Part1.Submit your work on at least Questions(3),(4),(5),(6),(7),(8).1.Let A,B,C,D be sets,and f:A−→B,g:B−→C,h:C−→D be functions.Prove that(h◦g)◦f=h◦(g◦f)as functions.2.Let f,g:R−→R be functions defined by f(x)=x21+x2,g(x)=x−1for any x∈R.(a)Compute the respective‘formulae of definition’of the functions g◦f,f◦g explicitly.(b)Choose some x0∈R so that(g◦f)(x0)=(f◦g)(x0).(c)Is it true that(g◦f)(x)=(f◦g)(x)for any x∈R?Justify your answer.Remark.Hence we have dis-proved the statement below:•Let A be a set,and f,g:A−→A be functions.g◦f=f◦g as functions.3.Let f:R−→R be the function defined by f(x)=x53−1for any x∈R.(a)Verify that f is surjective(directly from the definition of surjectivity).(b)Verify that f is injective(directly from the definition of injectivity).4.Denote the interval(0,+∞)by I.Let f:I−→R be the function defined by f(x)=12 x−1xfor any x∈I.(a)Verify that f is injective(directly from the definition of injectivity).(b)Verify that f is surjective(directly from the definition of surjectivity).Remark.It may help if you start by considering whether,for each b∈R,the equation b=f(u)with unknown u has any solution or not.5.Let f:[0,9]−→R be the function defined by f(x)=−x+6√x−5for any x∈[0,9].(a)Show that f(x)=−(A−√x)2+B for any x∈[0,9].Here A,B are some real constants whose respective values you have to determine.(b)Verify that f is injective,directly from the definition of injectivity.(c)Is f surjective?Justify your answer directly from the definition of surjectivity.6.Let f:(0,+∞)−→R be the function defined by f(x)=x2−1x2+1sin 1√x for any x∈(0,+∞).(a)Verify that f is not injective.(b)i.♦Verify that x2−1x2+1 ≤1for any x∈(0,+∞).Remark.A very simple answer can be obtained without using calculus.ii.Apply the previous part,or otherwise,to verify that f is not surjective.7.Let f:C−→C be the function defined by f(z)=¯z for any z∈C.(a)Verify that f is injective.(b)Verify that f is surjective.8.Let f:C\{0}−→C\{0}be the function defined by f(z)=z¯zfor any z∈C\{0}.(a)Verify that f is not injective.(b)Verify that f is not surjective.9.Let f:C\{0}−→C\{0}be the function defined by f(z)=z2¯zfor any z∈C\{0}.(a)Verify that f(z)=z3|z|2for any z∈C\{0}.(b)♦Is f injective?Justify your answer.(c)♣Is f surjective?Justify your answer.Part2.1.♦We recall/introduce these definitions/notations:•Let A,B,C,D be sets,and f:A−→C,g:B−→D be functions.Suppose f(x)=g(x)for any x∈A∩B.Define the function f∪g:A∪B−→C∪D by(f∪g)(x)= f(x)if x∈Ag(x)if x∈BThe function f∪g is called the union of the functions f,g.Consider each of the statements below.Determine whether it is true or false.Justify your answer with an appropriate argument.(a)Let A,B,C,D be sets,and f:A−→C,g:B−→D be functions.Suppose f(x)=g(x)for any x∈A∩B.Suppose f,g are surjective.Then f∪g:A∪B−→C∪D is surjective.(b)Let A,B,C,D be sets,and f:A−→C,g:B−→D be functions.Suppose f(x)=g(x)for any x∈A∩B.Suppose f,g are injective.Then f∪g:A∪B−→C∪D is injective.2.♥Familiarity with the calculus of one variable is assumed in this question.Let J be an open interval in R.Denote by C(J)the set of all real-valued continuous functions on J.Denote by C1(J) the set of all real-valued differentiables functions on J whosefirst derivatives are continuous functions on J.Define the function D:C1(J)−→C(J)by D(ϕ)=ϕ′for anyϕ∈C1(J).For each a∈J,define the function I a:C(J)−→C1(J)by(I a(ψ))(x)= x aψ(t)dt for anyψ∈C(J)for any x∈J.(a)i.Prove that((I a◦D)(ϕ))(x)=ϕ(x)−ϕ(a)for anyϕ∈C1(J)for any x∈J.ii.Prove that((D◦I a)(ψ))(x)=ψ(x)for anyψ∈C(J)for any x∈J.(b)i.Is the function I a◦D:C1(J)−→C1(J)surjective?Justify your answer.ii.Is the function I a◦D:C1(J)−→C1(J)injective?Justify your answer.3.We introduce/recall these definitions:•Let n∈N.A degree-n polynomial with complex coefficients and with indeterminate z is an expression of the form a n z n+···+a1z+a0in which a0,a1,···,a n∈C and a n=0.•A complex-valued function of one complex variable is called a degree-n polynomial function on C exactly when its‘formula of definition’is given by a degree-n polynomial with complex coefficients.•Letζ∈C,and f(z)≡a n z n+···+a1z+a0be a polynomial with complex coefficients and with indeterminate z.ζis said to be a root of the polynomial f(z)in C if f(ζ)=0.The statement(♯)below,first proved by Gauss,is known as the Fundamental Theorem of Algebra:(♯)Every non-constant polynomial with complex coefficients(and with one indeterminate)has a root in C.(a)♦Prove that the statement(♯)is logically equivalent to the statement(♭)below:(♭)For any n∈N\{0},every degree-n polynomial function on C is surjective.(b)♣Let n∈N\{0,1},a0,a1,···,a n∈C,with a n=0,and f:C−→C be the degree-n polynomial function definedby f(z)=a n z n+···+a1z+a0for any z∈C.Apply the Fundamental Theorem of Algebra,or otherwise,to prove that f is not injective.Remark.Here you may also take for granted the Factor Theorem(whose‘real version’you have already learnt at school and may be carried in verbatim to the‘complex situation’):•Letα∈C,and p(z)be a degree-n polynomial(with complex coefficients).Supposeαis a root of p(z).Then there is a degree-(n−1)polynomial q(z)(with complex coefficients)so that p(z)=(z−α)q(z)aspolynomials.。
数学绘本的阅读与教学

課程主題
幾何:面積與周長 數與量:加法與乘法 幾何:多邊形 幾何:幾何圖形(節奏) 數與量:計算數量 連結:察覺(分類) 數與量:分數 數與量:奇數與偶數 幾何:立體幾何 數與量:金錢 數與量:乘法 數與量:除法 數與量:各種運算的綜合練習
遠流出版社之【魔數小子】系列(續)
書名
嗚,地址不見了 嗯,等我一分鐘 呀!怪物別跟我 哇!40公斤番茄大賽 喔,原來我最棒 嘿,圖表真好用 哼,到底藏哪裡? 哎,貓咪數不完 喂,包裹送到 呵,還有一張票 嘻,就是這一杯 噓,螞蟻搬東西
‧時間量感的建立:一日和一星期
《生氣的瓢蟲,誠品書局代理》(時間篇二,國家教育研究院籌備處出版)
‧世界的一天(時區)
《世界的一天,漢聲出版社》(時間篇三,國家教育研究院籌備處出版)
‧讓兒童的想像力展翅飛翔─以《瘋狂星期二》導讀為例
─(一)引起動機、(二)作品導讀、(三)內容探究 (陳海泓,1999,語文教育通 訊第十八期)
課程主題
認識數字 趣味數字兒歌 數字1-100的點算 分類的概念 空間的概念 量的概念 認識基本圖形及面積.體積 認識數的合成與加法 認識數的分解與減法 認識時間 認識錢幣與錢幣的使用 數概念及邏輯關係綜合練習
遠流出版社之【魔數小子】系列
書名
義大利麵與肉丸子 阿曼達的瘋狂大夢 貪心的三角形 夢想家的披風 國王的超級特派員 啊!別進我房間 噢!披薩 噹!奇數撞偶數 咦,箱子裡是啥? 嗨!可愛的天竺鼠 哈!宇宙無敵湯 喵,別再叫我吃了 嘎,下課變上課?
課程主題
數與量:位值 數與量:時間 幾何:平面幾何 數與量:量與實測 數與量:測量 統計與機率:圖表 連結:解題 數與量:加法與減法 數與量:乘法 數與量:除法 數與量:容量 數與量:倍數與等比級數
GMAT数学讲义

Add,plus Subtract Difference Multiply, times Product divide
有关数学的运算
Add,plus
•加
Subtract
•减
Difference
•差
Multiply,times
•乘
Product
•积
divide
•除
isible Divided evenly Divident Divisor Quotient Remainder
• median
• median
• (中数) 将一堆数排序之后, 正中间的一个数(奇 数个数字), 或者中间两个数的平 均数(偶数个数字)
e.g. median of 1,7,4,9 ,2,2,2,2,2,5,8 is 2 median of 1,7,4,9,2,5 is (5+7)/2=6
• mode
• 1.00001*0.99999+1.00002*0.99998=?
fraction improper fraction proper fraction denominator numerator Mixed number
分数
fraction improper fraction proper fraction denominator numerator Mixed number
• 阶乘 N!=1*2*3*....(N-2)*(N-1)*N
且规定0!=1 例如 8!=1*2*3*4*5*6*7*8
• 1、根号8除以2倍根号2
• If United States imports increased 20 percent and export decreased 10 percent during a certain year,the ratio of imports to exports at the end of the year was how many times the ratio at the beginning of the year?
hku aptitude test 题目汇总

Hku Aptitude Test 题目汇总第一部分:数学能力1. 求解方程1.1 求解一元一次方程组1.2 求解二元二次方程1.3 求解含绝对值的方程2. 几何题2.1 计算三角形的面积2.2 计算圆的周长和面积2.3 计算多边形的内角和外角之和3. 数据分析3.1 理解均值、中位数和众数的概念3.2 解析数据分布的正态分布和偏态分布第二部分:语文能力1. 词汇运用1.1 填空题:根据语境填入正确的词语 1.2 选择题:选择同义词或反义词2. 阅读理解2.1 阅读短文并回答问题2.2 分析文章主旨和作者意图3. 写作能力3.1 选择题:选择合适的词语填入句子3.2 作文题:写一篇关于环保的文章第三部分:思维能力1. 逻辑推理1.1 推理填空:根据条件进行推理并填写空白处1.2 排列组合:计算排列或组合的个数2. 比喻与类比2.1 选择题:根据比喻或类比的关系选择正确的答案 2.2 填空题:根据给定的比喻或类比填写空白处3. 情境分析3.1 阅读情境描述并回答问题3.2 分析情境中的人物关系和冲突第四部分:实践能力1. 实验设计1.1 根据题目要求设计小学科学实验1.2 完善实验步骤和材料准备2. 技能操作2.1 图像处理:使用图片编辑软件对图片进行简单处理2.2 文字排版:使用文字编辑软件进行排版操作3. 项目管理3.1 根据给定需求制定项目计划3.2 管理团队完成项目任务结语Hku Aptitude Test 包含了数学能力、语文能力、思维能力和实践能力四个方面的内容。
通过这些题目的考察,可以全面了解考生的学习能力和综合素质。
希望广大考生能够认真准备,取得满意的成绩。
在Hku Aptitude Test的数学能力考核部分中,数学题目不仅涉及基本的数学运算和几何知识,还包括了数据分析和解题能力。
通过求解方程、几何题和数据分析题目的考核,可以全面评估考生的数学素养和解题能力。
1. 求解方程1.1 求解一元一次方程组在求解一元一次方程组的题目中,考生需要通过列方程组、消元、求解变量的值等步骤来解答问题,考验了考生的代数运算能力和解方程的技巧。
数学专业英语第二版的课文翻译

2-A Why study geometry Many leading institutions of higher learning have recognized that positive benefits can be gained by all who study this branch of mathematics. This is evident from the fact that they require study of geometry as a prerequisite to matriculation in those schools. 许多居于领导地位的学术机构承认,所有学习这个数学分支的人都将得到确实的受益,许多学校把几何的学习作为入学考试的先决条件,从这一点上可以证明。
Geometry had its origin long ago in the measurement by the Babylonians and Egyptians of their lands inundated by the floods of the Nile River. The greek word geometry is derived from geo, meaning “earth” and metron, meaning “measure” . As early as 2000 . we find the land surveyors of these people re-establishing vanishing landmarks and boundaries by utilizing the truths of geometry . 几何学起源于很久以前巴比伦人和埃及人测量他们被尼罗河洪水淹没的土地,希腊语几何来源于geo ,意思是”土地“,和metron 意思是”测量“。
香港中文大学数学本科培养方案

香港中文大学数学本科培养方案本科生入学时就其所属人文、理科、工科及商科的相关学系中选择并提出专业意愿,在第一学年修习所选择主修专业的基础科目,及其它通识教育、语文等课程,利用第一年时间了解自己的专业兴趣,适应新的学习及考试模式。
学生于第一或第二学年确定主修专业。
所有本科课程均采用灵活的学分制度。
学生须至少修毕123学分,始准毕业。
其中包括主修课程规定必修学分、体育科目、通识教育、资讯科技分、副修或选修科目。
教育学士课程及中医学学士课程等另有规定者除外。
灵活学分制度并无升级概念。
为方便行政上的管理,学生将按其在校修业年数分级,惟此年级并非表示学生的学业阶段;学生可按个别情况及兴趣订定其修业进度,修满毕业所需学分,并符合主修课程及大学其它毕业规定,便可获颁学士学位。
除课程另有说明外,学生一般修业4年便可毕业。
学士学位的等级如下:甲等荣誉、乙等一级荣誉、乙等二级荣誉、丙等荣誉及学位。
学位等级得根据学生主修科的平均积点,全部科目的总平均积点,及毕业论文专题研究成绩评定。
中医学学士学位并须根据学生临床学习课程的成绩评定。
为确保学生在主修学科中获得充分训练,全日制学生须修毕个别主修课程规定的最低必修学分,最高不多于72学分,经教务会特别核准的专业课程除外。
如主修课程设有毕业论文/专题研究规定,学生必须于该等论文或专题研究获及格成绩,始可毕业。
为贯彻课程结构的灵活性,大学定副修为自愿选修课程。
学生可根据个人兴趣及志向申请修读最多两项副修课程。
副修课程规定至少须达18学分,但不得多于30学分。
本校其中一项重要的教育目标,是提供均衡教育,亦即着重课程的深与博。
主修课程使学生对一专业范畴有深入认识,达致一定学术水平。
通识教育课程则扩阔学生视野,培养学生抽象与综合思考的能力,并对主修课以外的不同学科有更广博的认识,使学生在瞬息万变的现代社会中,能内省外观、高瞻远瞩。
通识教育课程规定学生须修读21学分。
2012年起,所有学科的学生必须修读的通识教育课程将分为三个部分:九学分的大学通识、六学分的书院通识,以及两科共六学分的新通识教育基础课程。
张广GMAT数学讲义

目录第一章Data Sufficiency题型-------------------------------------------2 第二章数论---------------------------------------------------------------4第三章代数--------------------------------------------------------------17第四章几何--------------------------------------------------------------24第五章文字应用题-------------------------------------------------------32第六章综合练习题-------------------------------------------------------44第七章GMAT数学常见术语---------------------------------------------72第一章 Data Sufficiency 题型【DS 题形式】Information required (Introduction or Background)QuestionTwo statements labeled (1) and (2)Option:(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.(D) EACH statement ALONE is sufficient.(E) Statement (1) and (2) TOGETHER are NOT sufficient.【答题步骤】1. 分析问句类型数值计算——特殊疑问句 (Special Question).判断是非——一般疑问句 (General Question).2. 罗列各种充分与不充分条件特殊疑问句——答案:唯一确定实数解.充分:能得到诸如0, –27, 4.567, 2311, , π等唯一解的Statement .不充分:能得到诸如x = x <1等两个或更多解,及其它一切无法计算出解的Statement .例题1:What is the value of x ?(1) 3x = 15(2) 5x < 30例题2:Tom and Jack are in a line to purchase tickets. How many people are in the line?(1) There are 20 people behind Tom and 20 people in front of Jack.(2) There are 5 people between Tom and Jack.一般疑问句——答案:明确回答”YES”或者”NO”.充分:完全符合或者完全不符合Question提出的内容,即能理直气壮回答”YES”或者”NO”,不留任何余地的Statement.不充分:不完全符合Question提出的内容,即只能心虚回答”Yes, but…”或者”No, but…”,及其它一切无法判断结果的Statement.例题3:Is x equal to 1 ?(1)x2 = 1(2)x2 = 4例题4:体会下列两个Question的区别.There are eight balls in the pocket.Question 1: Are all the balls in the pocket red?充分:”YES”:”NO”:Question 2: Are there any red balls in the pocket?充分:”YES”:”NO”:Statement 1: Three balls are removed; whose colors are brown, green, and red, respectively. Statement 2: Three balls are removed; whose colors are brown, green, and yellow, respectively. Statement 3: Three balls are removed; whose colors are red, red, and red, respectively.3.按照Problem Solving常规题型继续思考牢记:当分析Statement (1) 时,不要预测Statement (2);当分析Statement (2) 时,确信忘记Statement (1);第二章数论【奇数与偶数(Odd and Even Numbers) 】1.奇数个奇数相加减,其结果必为奇数。
港中文数学方向研究生专业

港中文数学方向研究生专业
香港中文大学数学系的研究生专业包括但不限于以下方向:
1. 算法和复杂性理论:研究算法设计与分析、计算复杂性理论、组合优化等问题。
2. 应用数学:研究数学在科学、工程等领域的应用,如数值分析、偏微分方程、优化等。
3. 数学物理:研究数学与物理学的交叉领域,如量子力学、统
计物理、流体力学等。
4. 统计学:研究收集、分析和解释数据的方法和理论,包括实
证研究设计、推断统计等。
5. 金融数学:研究数学在金融领域的应用,如金融衍生品定价、风险管理、投资组合优化等。
6. 概率论和数理统计:研究概率论和数理统计的基本理论和方法,包括随机过程、极限理论等。
7. 代数和几何:研究代数学和几何学的基本理论和方法,包括
群论、代数几何、流形等。
8. 数论和密码学:研究数论和密码学的基本理论和方法,包括
素数分布、密码算法等。
9. 数学教育:研究数学教育的理论和实践,包括教学设计、教
学评估等。
以上只是一些常见的数学方向,实际上还有很多其他具体的研究方向可供选择。
学生可以根据自己的兴趣和背景选择适合自己的方向。
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MATH1050B/C Further Exercise2Due date:20-2-20171.Let P,Q,R be statements.Consider each of the pairs of statements below.Determine whether the statements arelogically equivalent.Justify your answer by drawing an appropriate truth table.(a)∼(P∨Q),(∼P)∧(∼Q).(b)P→(Q∧R),(P→Q)∧(P→R).(c)P→(Q→R),(P∧Q)→R.(d)P→(Q∨R),(P→Q)∨(P→R).(e)(P∨Q)→R,(P→R)∧(Q→R).(f)(P→Q)→R,P→(Q→R).(g)P→(Q∨R),[P∧(∼Q)]→R.2.Let P,Q,R be statements.Consider each of the statements below.Determine whether it is a tautology or acontradiction or a contingent statement.Justify your answer by drawing an appropriate truth table.(a)[P→(P→Q)]→(P→Q)(b)(P→R)→[(P∧Q)→R)](c)[(P→Q)∧(Q→R)]→(P→R)(d)[(P→Q)∧(Q→R)∧(R→P)]→(Q→P)(e)(P→R)→[(P→Q)∨(Q→R)](f)(P→Q)→[(Q→R)∨(P∧R)](g)(P→Q)→[(P→R)∨(Q→R)]3.Let C={0,1,1,2,3,3,4},D={0,1,{1,2,3},{{3},4}}.Consider each of the sets below.List every element ofthe set concerned,each element exactly once.You are not required to justify your answer.(a)C.(b)D.(c)C∩D.(d)C∪D.(e)C\D.(f)D\C.(g)C△D.(h)P(C∩D).4.Let C={{0,1},{1},{1,2,3},{3,4}},D={{0,1,1},{1,2,3},{{3},{4}}}.Consider each of the sets below.List every element of the set concerned,each element exactly once.You are not required to justify your answer.(a)C∩D.(b)C∪D.(c)C\D.(d)C△D.(e)P(C\D).5.Let M={m,a,r,c,u,s},T={t,u,l,l,i,u,s},C={c,i,c,e,r,o}.(a)How many elements are there in the set C?(b)How many elements are there in the set M∪T?(c)How many elements are there in the set(M∪T)\C?(d)How many elements are there in the set{(M∪T)\C}?(e)How many elements are there in the set({M}∪{T})\{C}?(f)How many elements are there in the set{M∪T}\{C}?(g)List every element of the set M∩C,each element exactly once.(h)List every element of the set P(M∩C),each element exactly once.6.Let A={x∈R:x2−2x−3≤0},B={x∈R:−1≤x≤3}.Prove‘fromfirst principles’that A=B.7.Let A={n∈Z:n≡1(mod3)},B={n∈Z:n≡4(mod9)}.(a)Prove‘fromfirst principles’that B⊂A.(b)♦Is it true that A⊂/B?Justify your answer.8.Let A={x∈Z:x=k4for some k∈Z},B={x∈Z:x=k2for some k∈Z}.(a)Prove that A⊂B.(b)♦Is it true that B⊂/A?Justify your answer.9.♦Let A={x∈R:x2−x≥0},B={x∈R:x≤0},C={x∈R:x≥1}.Prove‘fromfirst principles’thatA=B∪C.10.Let A={x∈Q:x=r3for some r∈Q},B={x∈Q:x=r9for some r∈Q},C={x∈Z:x=r3for some r∈Q},D={x∈Q:x=r3for some r∈Z}.(a)Is A a subset of Q?Is Q a subset of A?Justify your answer.(b)Is A a subset of B?Is B a subset of A?Justify your answer.(c)Is A a subset of C?Is C a subset of A?Justify your answer.(d)Is A a subset of D?Is D a subset of A?Justify your answer.11.To handle this question,you may make use of what you have learnt in‘linear algebra’and/or‘coordinate geometry’and/or‘vector geometry’.(a)Let A={p∈R3:There exist some x,y,z∈R such that p=(x,y,z)and4x+2y+z=0},B={q∈R3:There exist some s,t∈R such that q=(s,t,−4s−2t)}.Prove‘fromfirst principles’that A=B.Remark.What are A,B really?(b)Let A={p∈R3:There exist some x,y,z∈R such that p=(x,y,z)and2x+y−z=0and x−2y+z=0},B={q∈R3:There exists some t∈R such that q=(t,3t,5t)}.Prove‘fromfirst principles’that A=B.Remark.What are A,B really?12.♣To handle this question,you may make use of what you have learnt in‘linear algebra’and/or‘coordinate geometry’and/or‘vector geometry’.LetS={p∈R3:There exist some x,y,z∈R such that p=(x,y,z)and x2+y2+z2=1},H={p∈R3:There exist some x,y,z∈R such that p=(x,y,z)and x2+y2−z2=1},C={p∈R3:There exist some x,y,z∈R such that p=(x,y,z)and x2+y2=1},Γ={q∈R3:There exist someθ∈R such that q=(cos(θ),sin(θ),0)}.(a)Prove‘fromfirst principles’thatΓis a subset of each of S,H,C.(b)Prove‘fromfirst principles’that each of S∩H,H∩C,C∩S is a subset ofΓ.(c)Deduce thatΓ=S∩H=H∩C=C∩S.Remark.What are S,H,C,Γreally?Make use of what you have learnt in coordinate geometry.13.♦In this question,you may use of the following statements without proof:(♯)Let a,b be two objects(not necessarily distinct).{a}={b}iffa=b.(♭)Let a,b,c be three objects(not necessarily distinct).{a,b}={c}iffa=b=c.(♮)∅={∅}.Let A={∅},B={{∅}},C={∅,{∅,{∅}}},D={∅,{{∅}}}.For each of the following statements,determine whether it is true or false.Prove your answer in each case.(a)A∈C.(b)B⊂D.(c)B∈C.(d)A∪B∈C.(e)C∩D=∅.14.Prove each of the statements below‘fromfirst principles’,using the definitions of set equality,subset relation,intersection,union,complement,where appropriate.(a)Let A,B be sets.A∩B⊂A.(b)Let A,B be sets.A⊂A∪B.(c)Let A,B,C be sets.A\(B∪C)=(A\B)∩(A\C).(d)Let A,B,C be sets.(A∪B)\C=(A\C)∪(B\C).(e)Let A,B be sets.(A∪B)\A=B\(A∩B).(f)Let A,B,C be sets.A\(B\C)=(A\B)∪(A∩C).15.Prove the following statements:(a)♦Let A,B be sets.A△B=(A∪B)\(A∩B).(b)Let A,B be sets.A△B=B△A.(c)♥Let A,B,C be sets.(A△B)△C=A△(B△C).(d)♣Let A,B,C be sets.A∩(B△C)=(A∩B)△(A∩C).16.♦Let A,B be sets.Prove that the following statements are equivalent:(I)A⊂B.(II)A∩B=A.(III)A∪B=B.17.Dis-prove each of the statements below by giving an appropriate counter-example.(It may help if you draw Venndiagrams to investigate the respective statementsfirst.)(a)Let A,B,C be sets.A\(C\B)⊂A∩B.(b)Let A,B,C be non-empty sets.B\A⊂(C\A)\(C\B).(c)Let A,B,C be non-empty sets.A∪(B∩C)⊂(A∪B)∩C.(d)♦Let A,B,C are non-empty sets.B∩C⊂[A\(B\C)]∪[B\(C\A)].(e)♦Let A,B,C be sets.Suppose A∩B⊂C.Then C⊂(A∩C)∪(B∩C).18.♦Consider each of the statements below.In each case,determine whether it is true or false.Justify your answerby giving a proof or constructing a counter-example where appropriate.(a)Let A,B,C be sets.Suppose A∪(B∩C)=(A∪B)∩C.Then A⊂C.(b)Let A,B,C be sets.A\(B\C)=(A\B)\C.(c)Let A,B,C be sets.If A⊂B then C\A⊂C\B.(d)Let A,B,C be sets.Suppose A⊂B and A⊂/C.Then B⊂/C.(e)Let A,B,C be non-empty sets.Suppose A⊂B and B⊂/C.Then A⊂/C.(f)Let A,B,C be non-empty sets.Suppose A⊂B and B⊂/C.Then A⊂/C.(g)Let A,B,C be sets.Then A∪(B△C)=(A△B)∪(A△C).(h)Let A,B,C be sets.Then A∩(B△C)=(A△B)∩(A△C).19.(a)Prove the statements below‘fromfirst principles’,using the definitions of set equality,subset relation,inter-section,union,complement,where appropriate.i.Let E be a set,and A,B be subsets of E.Suppose A⊂B.Then E\B⊂E\A.ii.Let E be a set,and A,B be subsets of E.Suppose A B.Then E\B E\A.(b)Consider each of the statements below.For each of them,determine whether it is true or false.Justify youranswer by giving a proof or constructing a counter-example where appropriate.i.Let A,B,E be a set.Suppose A⊂B.Then E\B⊂E\A.ii.Let A,B,E be a set.Suppose A B.Then E\B E\A.20.(a)♦Consider each of the statements below.For each of them,determine whether it is true or false.Justify youranswer by giving a proof or constructing a counter-example where appropriate.i.Let A,B be sets.B\(B\A)⊂A.ii.Let A,B be sets.A⊂B\(B\A).(b)♣Prove the statements below:i.Let A,B be sets.A⊂B\(B\A)iffA⊂B.ii.Let A,B be sets.B\(B\A)=A iffA⊂B.iii.Let A,B be sets.B\(B\A) A iffA⊂/B.21.Prove the following statements:(a)♠Let A,B be sets.P(A\B)⊂(P(A)\P(B))∪{∅}.(b)♠Let A,B be sets.Suppose(A⊂B or A∩B=∅).Then P(A)\P(B)⊂P(A\B).(c)♠Let A,B be sets.Suppose P(A)\P(B)⊂P(A\B).Then(A⊂B or A∩B=∅).22.Prove the following statements:(a)Let a,b be two objects(not necessarily distinct).{a}={b}iffa=b.(b)Let a,b,c be three objects(not necessarily distinct).{a,b}={c}iffa=b=c.(c)♣Let a,b,c,d be four objects(not necessarily distinct).{a,b}={c,d}iff((a=c and b=d)or(a=d andb=c)or a=b=c=d).(d)♦Let A,B,C,D be sets.Suppose{A,B}={C,D}.Then A∩B=C∩D and A∪B=C∪D.23.Consider the predicate‘x/∈x’,which we denote by P(x).(a)Denote by R the object{x|P(x)},obtained from the Method of Specification.(R is called the Russell set.)Suppose it were true that R was a set.(Hence it makes sense to discuss whether an arbitrarily given object is an element of R or not.)i.Can it happen that the object R is an element of the set R?Why?ii.Can it happen that the object R is an element of the set R?Why?Remark.From the answers to the above questions,you would have to conclude that R is not a set in the first place.(Why?)This tells us the construction{x|P(x)}fails to give a set.(b)Let A be a set.Denote by B the object{x∈A:P(x)},obtained from the Method of Specification.(Thistime it is guaranteed that B is a set,because we are constructing a subset from the given set A.)Prove that B is not an element of A.(Apply the proof-by-contradiction method.)Remark.This shows that given any set,there is always some object which does not belong to it as an element.In other words,no set contains every conceivable object as its element.There is no such thing as ‘universal set’.。