问题1.以下哪项是正确的?给出支持您答案的理由。
(i)对于任何两个A和B集,AB或BA。
(ii)无限集的每个子集都是无限的。
(iii)有限集的每个子集都是有限的。
(iv)每个集合都有一个适当的子集。
(v){a,b,a,b,a,b,…。}是一个无限集。
(vi){a,b,c}和{1,2,3}是等价集合。
(vii)一个集合可以有无限多个子集。
解决方案:
(i) False
It is not mandatory for any two set A and B to be either A B or B A.
(ii) False
Let us consider a set, A = {2,3,4}.
It is finite subset with infinite set N of natural numbers.
(iii) True
A finite set can never have an infinite subset.
Therefore, every subset of a finite set is finite.
(iv) False
Null set, also known as empty set doesn’t have a proper subset.
(v) False
A set can never have duplicate entries.
Removing duplications, the set {a, b} becomes a finite set.
(vi) True
Equivalent sets have the same number of elements. Both the sets have three elements, therefore, are equivalent.
(vii) False
Let us consider, a set A = {1}
The subsets of this set A can be ϕ and {1} which are finite.
问题2。说明以下陈述是对还是错:
(i)1∈{1,2,3}
(ii)⊂{b,c,a}
(iii){a}∈{a,b,c}
(iv){a,b} = {a,a,b,b,a}
(v)集合{x:x + 8 = 8}是空集合。
解决方案:
(i) True
1 is a part of the given set, therefore it belongs to this set {1, 2, 3} .
(ii) False
Since, a is an element and not a subset of a set {b, c, a}, therefore this statement is false.
(iii) False
Since, {a} is not an element but a subset of set {b, c, a}.
(iv) True
A set cannot have duplicate entries. Therefore, removing duplicate entries from RHS, the sets become equivalent.
(v) False
Given, x+8 = 8
Solving we get, x = 0
Therefore, the given set becomes a single ton set with the only element being {0}. Where, it is not a null set.
问题3.在以下几组中做出决定,它们是其中的子集:
A = {x:x满足x 2 – 8x + 12 = 0},B = {2,4,6},C = {2,4,6,8,…。},D = {6}
解决方案:
We have,
A = x2 – 8x + 12=0
Solving we get,
⇒ (x–6) (x–2) =0
Solving for x,
⇒ x = 2 or x = 6
Therefore,
A = {2, 6}
Given,
B = {2, 4, 6}
C = {2, 4, 6, 8}
D = {6}
Therefore,
D ⊂ A ⊂ B ⊂ C
问题4.写下以下哪项是对的?证明你的答案。
(i)所有整数的集合都包含在所有有理数的集合中。
(ii)所有乌鸦都包含在所有鸟类中。
(iii)所有矩形的集合都包含在所有正方形的集合中。
(iv)所有矩形的集合都包含在所有正方形的集合中。
(v)集P = {a}和B = {{a}}是相等的。
(vi)集合A = {x:x是单词“ LITTLE”}的字母,并且b = {x:x是单词“ TITLE”}的字母相等。
解决方案:
(i) True
A rational number is a fractional number which is represented by the form p/q where p and q are integers where q is not equal to 0. Substituting, q = 1, we get p=q, which is an integer.
(ii) True
Crows are also birds, so all the crows are contained in the set of all birds.
(iii) False
Every square can be a rectangle, where the length and breadth of the rectangle are same. But, the reverse is not true, that is, not every rectangle cannot be a square.
(iv) False
Every square can be a rectangle, where the length and breadth of the rectangle are same. But, the reverse is not true, that is, not every rectangle cannot be a square.
(v) False
We have,
P = {a}
B = {{a}}
But, {a} = P
B = {P}
Hence, they are not equivalent.
(vi) True
We have,
A = For “LITTLE”
A = {L, I, T, E} = {E, I, L, T}
B = For “TITLE”
B = {T, I, L, E} = {E, I, L, T}
Therefore,
A = B
问题5.以下哪个陈述是正确的?为每个不正确的语句编写正确的形式。
(i)a⊂{a,b,c}
(ii){a} {a,b,c}
(iii)a {{a},b}
(iv){a}⊂{{a},b}
(v){b,c}⊂{a,{b,c}}
(vi){a,b}⊂{a,{b,c}}
(vii)ϕ {a,b}
(viii)⊂{a,b,c}
(ix){x:x + 3 = 3} = ϕ
解决方案:
(i) False
a is not a subset of given set but belongs to the given set.It is just an element.
Correct form is – a ∈ {a, b, c}
(ii) In this {a} is subset of {a, b, c}
Correct form is – {a} ⊂ {a, b, c}
(iii) False
‘a’ is not the element of the set.
Correct form is – {a} ∈ {{a}, b}
(iv) False
{a} is not a subset of given set.
Correct form is – {a} ∈ {{a}, b}
(v) {b, c} is not a subset of given set. But it belongs to the given set.
Correct form is – {b, c} ∈ {a,{b, c}}
(vi) {a, b} is not a subset of given set.
Correct form is – {a, b}⊄{a,{b, c}}
(vii) ϕ does not belong to the given set but it is subset.
Correct form is – ϕ ⊂ {a, b}
(viii) True
It is the correct form. ϕ is subset of every set.
(ix) x + 3 = 3
Evaluating, we get,
x = 0 = {0}, which is not ϕ
Correct form is – {x: x + 3 = 3} ≠ ϕ
问题6.令A = {a,b,{c,d},e}。下列哪些陈述是错误的,为什么?
(i){c,d}⊂A
(ii){c,d}∈A
(iii){{c,d}}⊂A
(iv)a∈A
(v)⊂A。
(vi){a,b,e}⊂A
(vii){a,b,e}∈A
(viii){a,b,c}⊂A
(ix)ϕ∈ A
(x){ϕ}⊂A
解决方案:
(i) False
{c, d} is not a subset of A but it belongs to the set A.
Therefore,
{c, d} ∈ A
(ii) True
{c, d} ∈ A
(iii) True
{c, d} is a subset of A.
(iv) It is true that a belongs to A.
(v) False
The element a is not a subset of A but it belongs to the set A.
(vi) True
{a, b, e} is a subset of A.
(vii) False
{a, b, e} does not belong to A, {a, b, e} ⊂ A this is the correct form.
(viii) False
{a, b, c} is not a subset of A
(ix) False
ϕ is a subset of A.
ϕ ⊂ A.
(x) False
{ϕ} is not subset of A, ϕ is a subset of A. Therefore, it is false.
问题7。令A = {{1、2、3},{4、5},{6、7、8}}。确定以下哪个是对还是错:
(i)1∈A
(ii){1,2,3}-A
(iii){6,7,8}∈A
(iv){4,5}⊂A
(v)ϕ∈A
(vi)⊂A
解决方案:
(i) False
1 is not an element of the set A.
(ii) True
{1,2,3} ∈ A. this is correct form.
(iii) True.
The set {6, 7, 8} is an element of A, that is {6, 7, 8} ∈ A.
(iv) True
{{4, 5}} is a subset of the specified set A={4, 5}.
(v) False
Φ is a subset of the given set A, not an element of A.
(vi) True
Φ is a subset of every set, so it is also a subset of the given set A.
问题8.让A = {ϕ,{ϕ},1,{1,ϕ},2}。以下哪项是正确的?
(i)ϕ∈A
(ii){ϕ}∈A
(iii){1}∈A
(iv){2,ϕ}⊂A
(v)2分
(vi){2,{1}}⊄A
(vii){{2},{1}}⊄A
(viii){ϕ,{ϕ},{1,ϕ}}⊂A
(ix){{ϕ}}⊂A
解决方案:
(i) True
Φ is an element of set A, therefore it belongs to set A. Hence, the given statement is true.
(ii) True
{Φ} is an element of set A, and not a subset. Hence, the given statement is true.
(iii) False
The subset 1 is not an element of A. Hence, the given statement is false
(iv) True
{2, Φ} is a subset of the given set A. Hence, the given statement is true.
(v) False
2 is not a subset of set A, it is an element of set A. Hence, the given statement is false.
(vi) True
{2, {1}} is not a subset of the given set A. Hence, the given statement is true.
(vii) True
Neither {2} and nor {1} is a subset of set A. Hence, the given statement is true.
(viii) True
All three {ϕ, {ϕ}, {1, ϕ}} are subset of set A. Hence, the given statement is true.
(ix) True
{{ϕ}} is a subset of set A. Hence, the given statement is true.