问题1.通过在空格中填写符号⊂或⊄来做出正确的陈述:
(i){2,3,4}。 。 。 {1,2,3,4,5}
(ii){a,b,c}。 。 。 {b,c,d}
(iii){x:x是您学校的XI类的学生}。 。 。{x:您学校的x名学生}
(iv){x:x是平面上的一个圆}。 。 。{x:x是同一平面上的一个半径为1单位的圆}
(v){x:x是平面上的三角形}。 。 。 {x:x是平面中的矩形}
(vi){x:x是平面上的等边三角形}。 。 。 {x:x是同一平面上的三角形}
(vii){x:x是一个偶数}。 。 。 {x:x是整数}
解决方案:
(i) {2, 3, 4} ⊂ {1, 2, 3, 4,5}
(ii) {a, b, c} ⊄ {b, c, d}
(iii) {x : x is a student of Class XI of your school} ⊂ {x : x student of your school}
(iv) {x : x is a circle in the plane} ⊄ {x : x is a circle in the same plane with radius 1 unit}
(v) {x : x is a triangle in a plane} ⊄ {x : x is a rectangle in the plane}
(vi) {x : x is an equilateral triangle in a plane} ⊂ {x : x is a triangle in the same plane}
(vii) {x : x is an even natural number} ⊂ {x : x is an integer}
问题2.检查以下陈述是对还是错:
(i){a,b}⊄{b,c,a}
(ii){a,e}⊂{x:x是英语字母中的元音}
(iii){1、2、3}⊂{1、3、5}
(iv){a}⊂{a,b,c}
(v){a}∈{a,b,c}
(vi){x:x是小于6的偶数自然数}⊂{x:x是除以36的自然数}
解决方案:
(i) False. Each element of {a, b} is an element of {b, c, a}.
(ii) True. Since a, e are two vowels of the English alphabet.
(iii) False. 2 is subset of {1, 2, 3} but not subset of {1, 3, 5}
(iv) True. Each element of {a} is also an element of {a, b. c} .
(v) False. Elements of {a, b, c} are a, b, c. Hence, {a} ⊂ {a, b, c}
(vi) True
{x : x is an even natural number less than 6} = {2, 4}
{x: x is a natural number which divides 36} = {1, 2, 3, 4, 6, 9, 12, 18, 36}
{2, 4} ⊂ {1, 2, 3, 4, 6, 9, 12, 18, 36}
问题3。令A = {1,2,3,4},5}。下列哪些陈述是不正确的,为什么?
(i){3,4}⊂A(ii){3,4}∈A(iii){{3,4}}⊂A(iv)1∈A(v)1⊂A(vi){1, 2,5}⊂A
(vii){1,2,5}∈A(viii){1,2,3}⊂A(ix)∅∈A(x)⊂A(xi){∅}⊂A
解决方案:
Given A= {1, 2, {3, 4}, 5}
(i) {3, 4} ⊂ A is incorrect. Here 3 ∈ {3, 4}, where 3 ∉ A.
(ii) {3, 4} ∈ A is correct. {3, 4} is an element of A.
(iii) {{3, 4}} ⊂ A is correct. {3, 4} ∈ {{3, 4}} and {3, 4} ∈ A.
(iv) 1 ∈ A is correct. 1 is an element of A.
(v) 1 ⊂ A is incorrect. An element of a set can never be a subset of itself.
(vi) {1, 2, 5} ⊂ A is correct. Each element of {1, 2, 5} is also an element of A.
(vii) {1, 2, 5} ∈ A is incorrect. { 1, 2, 5 } is not an element of A.
(viii) {1, 2, 3} ⊂ A is incorrect. 3 ∈ {1, 2, 3}; where, 3 ∉ A.
(ix) ∅ ∈ A is incorrect. ∅ is not an element of A.
(x) ∅ ⊂ A is correct. ∅ is a subset of every set.
(xi) {∅} ⊂ A is incorrect. {∅} is not present in A.
问题4.写下以下集合的所有子集
(i){a}(ii){a,b}(iii){1,2,3}(iv)∅
解决方案:
(i) Subsets of {a} are ∅ and {a}.
(ii) Subsets of {a, b} are {a}, {b}, and {a, b}.
(iii) Subsets of {1, 2, 3} are ∅, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, and {1, 2, 3}.
(iv) Only subset of ∅ is ∅.
问题5.如果A = has,P(A)有多少个元素?
解决方案:
For a set A with n(A) = m, then it can be shown that
Number of elements of P(A) = n[P(A)] = 2m
If A = ∅, we get n (A) = 0
So, n[P(A)] = 2° = 1
Therefore, P(A) has one element.
问题6.将以下内容写为间隔:
(i){x:x∈R,– 4
(iii){x:x∈R,0≤x <7}(iv){x:x∈R,3≤x≤4}
解决方案:
(i) {x : x ∈ R, – 4 < x ≤ 6} = (-4, 6]
(ii) {x : x ∈ R, – 12 < x < –10} = (-12, -10)
(iii) {x : x ∈ R, 0 ≤ x < 7} = [0, 7)
(iv) {x : x ∈ R, 3 ≤ x ≤ 4} = [3, 4]
问题7.以set-builder形式编写以下间隔:
(i)(– 3,0)(ii)[6,12](iii)(6,12](iv)[–23,5)
解决方案:
(i) (– 3, 0) = {x : x ∈ R, -3 < x < 0}
(ii) [6, 12] = {x : x ∈ R, 6 ≤ x ≤ 12}
(iii) (6, 12] = {x : x ∈ R, 6 < x ≤ 12}
(iv) [–23, 5) = {x : x ∈ R, -23 ≤ x < 5}
问题8.您将为以下各项建议什么通用集:
(i)直角三角形的集合
(ii)等腰三角形的集合。
解决方案:
(i) The universal set for the set of right triangles is the set of triangles or the set of polygons.
(ii) The universal set for the set of isosceles triangles is the set of triangles or the set of polygons or the set of two-dimensional figures.
问题9.给定集合A = {1、3、5},B = {2、4、6}和C = {0、2、4、6、8},以下哪个集合可以被认为是通用集合(s)所有三个集合A,B和C
(i){0、1、2、3、4、5、6}
(ii)∅
(iii){0、1、2、3、4、5、6、7、8、9、10}
(iv){1、2、3、4、5、6、7、8}
解决方案:
(i) A ⊂ {0, 1, 2, 3, 4, 5, 6}
B ⊂ {0, 1, 2, 3, 4, 5, 6}
But, C ⊄ {0, 1, 2, 3, 4, 5, 6}
Hence, the set {0, 1, 2, 3, 4, 5, 6} cannot be the universal set for the sets A, B, and C.
(ii) A ⊄ ∅, B ⊄ ∅, C ⊄ ∅
Hence, ∅ cannot be the universal set for the sets A, B, and C.
(iii) A ⊂ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
B ⊂ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
C ⊂ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Hence, the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} is the universal set for the sets A, B, and C.
(iv) A ⊂ {1, 2, 3, 4, 5, 6, 7, 8}
B ⊂ {1, 2, 3, 4, 5, 6, 7, 8}
But, C ⊄ {1, 2, 3, 4, 5, 6, 7, 8}
Hence, the set {1, 2, 3, 4, 5, 6, 7, 8} cannot be the universal set for the sets A, B, and C.