In a survey it was found that 21 people liked product A, 26 liked product B and 29 liked product C. If 14 people liked products A and B, 12 people liked products C and A, 14 people liked products B and C and 8 liked all the three products. Find how many liked product C only.

In a survey it was found that 21 people liked product A, 26 liked product B and 29 liked product C. If 14 people liked products A and B, 12 people liked products C and A, 14 people liked products B and C and 8 liked all the three products. Find how many liked product C only.

Answer
Let A, B, and C be the set of people who like product A, B, and C respectively.
n(A) = 21, n(B) = 26, n(C) = 29, n(A ∩ B) = 14, n(C ∩ A) = 12, n(B ∩ C) = 14, n(A ∩ B ∩ C) = 8
People who many liked product C only
= n(C) - n(C ∩ A) - n(B ∩ C) + n(A ∩ B ∩ C)
= 29 -12 – 14 + 8
= 11
Hence, 11 liked product C only.

In a survey of 60 people, it was found that 25 people read newspaper H, 26 read newspaper T, 26 read newspaper I, 9 read both H and I,11 read both H and T, 8 read both T and I, 3 read all three newspapers. Find:

In a survey of 60 people, it was found that 25 people read newspaper H, 26 read newspaper T, 26 read newspaper I, 9 read both H and I,11 read both H and T, 8 read both T and I, 3 read all three newspapers. Find:

(i) the number of people who read at least one of the newspapers.
(ii) the number of people who read exactly one newspaper.
Answer
(i) the number of people who read at least one of the newspapers.
Let H,T and I be the set of people who read newspaper H, T and C respectively.
n(H) = 25, n(T) = 26,n(I) = 26, n(H ∩ I) = 9, n(H ∩ T) = 11, n(T ∩ I) = 8 and n(H ∩ T ∩ I) = 3
Use formula
n(H ∪ T ∪ I) = n(H) + n(T) + n(I) - n(H ∩ T) - n(T ∩ I) - n(I ∩ H) + n(H ∩ T ∩ I)
Plug the values we get
= 25 + 26 + 26 - 11 - 8 - 9 + 3
= 52
The number of people who read at least one of the newspapers = n(H ∪ T ∪ I) = 52
(ii)The number of people who read exactly one newspaper
= n(H ∪ T ∪ I) - n(H ∩ T) - n(T ∩ I) - n(I ∩ H) + 2 × n(H ∩ T ∩ I)
Plug the values we get
= 52 -11 -8 -9 + 2× 3
= 52 -28 + 6
= 58 – 28
= 30
Hence, 30 people read exactly one newspaper.

In a group of students 100 students know Hindi, 50 know English and 25 know both. Each of the students knows either Hindi or English. How many students are there in the group?

In a group of students 100 students know Hindi, 50 know English and 25 know both. Each of the students knows either Hindi or English. How many students are there in the group?

Answer
Let H and E is the set of students who know Hindi and English respectively.
n(H) = 100, n(E) = 50 and n(H ∩ E) = 25
Use formula
n ( H ∪ E ) = n (H ) + n ( E ) – n ( H ∩ E)
Plug the values we get,
= [100 + 50 - 25]
⇒ 125
Therefore, there are 125 students in the group.

In a survey of 600 students in a school, 150 students were found to be taking tea and 225 taking coffee, 100 were taking both tea and coffee. Find how many students were taking neither tea nor coffee?

In a survey of 600 students in a school, 150 students were found to be taking tea and 225 taking coffee, 100 were taking both tea and coffee. Find how many students were taking neither tea nor coffee?

Answer
Let T and C is the set of students who take tea and coffee respectively.
n(T) = 150, n(C) = 225, n(T ∩ C) = 100
Use formula
n ( T ∪ C ) = n ( T ) + n ( C ) – n ( T ∩ C)
Plug the values we get,
= [150 + 225 - 100]
⇒ 275
Students taking neither tea nor coffee
⇒ Total students - student taking tea or coffee
⇒600 – 275
⇒ 325
Hence, 325 students were taking neither tea nor coffee.

Find sets A, B and C such that A ∩ B, B ∩ C and A ∩ C are non-empty sets and A ∩ B ∩ C = Φ.

Find sets A, B and C such that A ∩ B, B ∩ C and A ∩ C are non-empty sets and A ∩ B ∩ C = Φ.

Answer
Let take three set A = {1, 2}, B = {2, 3}, and C = {3, 1}.
A ∩ B, B ∩ C and A ∩ C should be non-empty sets
A ∩ B = {2}, B ∩ C = {3}, and A ∩ C = {1}.
Therefore A ∩ B, B ∩ C, and A ∩ C are non-empty.
Intersection of all three sets is null set, A ∩ B ∩ C = Φ.

Let A and B be sets. If A ∩ X = B ∩ X = Φ and A ∪ X = B ∪ X for some set X, show that A = B

Let A and B be sets. If A ∩ X = B ∩ X = Φ and A ∪ X = B ∪ X for some set X, show that A =  B 

Answer
Let x ∈ A
⇒ x ∈ (A or X)
⇒ x ∈ (A ∪ X)
Given that A ∪ X = B ∪ X
⇒ x ∈ (B ∪ X)
⇒ x ∈ B or x ∈ X
Given that A ∩ X = Φ and x ∈ A. So that x ∉ X.
Therefore, A ⊂ B ... (1)
Similarly take
⇒ y ∈ B
⇒ y ∈ (B or X)
⇒ y ∈ (B ∪ X)
Given that A ∪ X = B ∪ X
⇒ y ∈ (A ∪ X)
⇒ y ∈ A or y ∈ X
Given that B ∩ X = Φ and y ∈ B. So that y ∉ X.
Therefore, B ⊂ A ... (2)
From (1) and (2) we get A = B

Show that A ∩ B = A ∩ C need not imply B = C.

Show that A ∩ B = A ∩ C need not imply B = C. 
Answer
Let A = {0, 1}, B = {0, 2, 3}, and C = {0, 4, 5} Accordingly, A ∩ B = {0} and A ∩ C = {0}
Here, A ∩ B = A ∩ C = {0}
However, B ≠ C [2 ∈ B and 2 ∉ C]

Using properties of sets show that (i) A ∪ (A ∩ B) = A (ii) A ∩ (A ∪ B) = A.

Using properties of sets show that
(i) A ∪ (A ∩ B) = A (ii) A ∩ (A ∪ B) = A.

Answer
(i) A ∪ (A ∩ B) = A
LHS
⇒ A∪ ( A ∩ B)
Use distribution property
⇒ (A ∪ A) ∩ ( A ∪ B)
Use relation A ∪ A = A
⇒ (A) ∩ ( A ∪ B)
⇒ A
RHS
Hence, A ∪ (A ∩ B) = A
(ii) A ∩ (A ∪ B) = A
LHS
⇒A ∩ (A ∪ B)
Use distribution property we get
⇒ (A ∩ A) ∪ (A ∩ B)
Use relation A ∩ A = A
⇒ (A ∩ A) ∪ (A ∩ B)
⇒ A ∪ (A ∩ B)
⇒ A
RHS
Hence, A ∩ (A ∪ B) = A

Show that for any sets A and B, A = (A ∩ B) ∪ (A - B) and A ∪ (B - A) = (A ∪ B)

Show that for any sets A and B,
A = (A ∩ B) ∪ (A - B) and A ∪ (B - A) = (A ∪ B)

Answer
(1) A = (A ∩ B) ∪ (A - B)
RHS
(A ∩ B) ∪ (A - B)
Use relation, A – B = A ∩ B’
⇒ (A ∩ B) ∪ (A ∩ B’)
⇒A ∩ (B ∪ B’)
Use relation B ∪ B’ = U
⇒A ∩ U
⇒ A
LHS
Hence A = (A ∩ B) ∪ (A - B)
(2)A ∪ (B - A) = (A ∪ B)
LHS
A ∪ (B - A)
Use relation, B – A = B ∩ A’
⇒ A ∪ (B ∩ A’)
⇒ (A ∪ B) ∩ (A ∪ A’)
Use relation A ∪ A’ = U.
⇒ (A ∪ B) ∩ U
⇒ (A ∪ B)
RHS
Hence, A ∪ (B - A) = (A ∪ B)

Is it true that for any sets A and B, P(A) ∪ P(B) = P (A ∪ B)? Justify your answer.

Is it true that for any sets A and B, P(A) ∪ P(B) = P (A ∪ B)? Justify your answer.

Answer
Given statement is false.
Let A = {1, 2} and B = {2, 3}
A ∪ B = {1, 2} ∪ {2, 3} = {1, 2, 3}
P(A) = {Φ, {1}, {2}, {1, 2}}
P(B) = {Φ, {2}, {3}, {2, 3}}
P(A ∪ B) = {Φ, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}}
P(A) ∪ P(B) = {Φ, {1}, {2}, {3}, {1, 2}, {2, 3}}
we can observe that P(A) ∪ P(B) ≠ P(A ∪ B).
Hence given statement is false.

Assume that P (A) = P (B). Show that A = B.

Assume that P (A) = P (B). Show that A = B.

Answer
Every set is a member of power set so that, A ∈ P(A)
Given that P (A) = P (B) So that
A ∈ P(B)
A is a element of power set of B so that,
A ⊂ B ... (1)
Similarly we can prove that
B ⊂ A ... (2)
From equation (1) and (2) we get, A = B

Show that if A ⊂ B, then C - B ⊂ C - A

Show that if A ⊂ B, then C - B ⊂ C - A.

Answer
Let x ∈ C - B
x ∈ C and x ∉ B
Given that [A ⊂ B] so that if any element doesn’t lies in set B then it cannot be in set A.
x ∈ C and x ∉ A
x ∈ C - A
Hence, C - B ⊂ C - A.

Show that the following four conditions are equivalent: (i) A ⊂ B (ii) A - B = Φ (iii) A ∪ B = B (iv) A ∩ B = A

Show that the following four conditions are equivalent:
(i) A ⊂ B (ii) A - B = Φ (iii) A ∪ B = B (iv) A ∩ B = A 

Answer
First, we shall try to prove A ⊂ B ⇔ A - B = Φ
Given A ⊂ B
To prove A - B = Φ
A ⊂ B so that A ∩ B = A
LHS
=A - B
= A – (A ∩ B)
= A-A
= Φ
RHS
Given A - B = Φ
To prove A ⊂ B
Let x ∈ A
Given that A - B = Φ so all element of A must be in set B
Therefore, x ∈ B
So that A ⊂ B
Hence proved
Similarly you can solve all other parts.

Let A, B and C be the sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C. show that B = C.

Let A, B and C be the sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C. show that B = C.

Answer
Let x ∈ B
There can be two cases, x ∈ A or x ∉ A.
Case -1 x ∈ A
As x ∈ A and x ∈ B so that x ∈ ( A ∩ B)
Given that A∩ B = A ∩ C, so that
x ∈ ( A ∩ C)
x ∈ A and x ∈ C
x ∈ B then x ∈ C so that B ⊂ C.
Case -2, x ∉ A
We have already assumed that x ∈ B.
Hence, x ∈ (A ∪ B)
Given that A ∪ B = A ∪ C, so that
⇒ x ∈ (A ∪ C)
⇒x ∈ A or x ∈ C
But we assumed that x ∉ A Hence x ∈ C
x ∈ B then x ∈ C so that B ⊂ C.
So in both cases B ⊂ C. ...(1)
Similarly, we can prove C ⊂ B ...(2)
From equation (1) and (2)
B= C

In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example. (i) If x ∈ A and A ∈ B, then x ∈ B (ii) If A ⊂ B and B ∈ C, then A ∈ C

In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example.
(i) If x ∈ A and A ∈ B, then x ∈ B
(ii) If A ⊂ B and B ∈ C, then A ∈ C
(iii) If A ⊂ B and B ⊂ C, then A ⊂ C
(iv) If A ⊄ B and B ⊄ C, then A ⊄ C
(v) If x ∈ A and A ⊄ B, then x ∈ B
(vi) If A ⊂ B and x ∉ B, then x ∉ A

Answer

(i) False
Let A = {2, 3} and B = {6, {2, 3}, 8}
Here 2 and 3 belongs to A and A belongs to B but 2 and 3 are not belongs to C. Hence statement is false.
(ii) False
Let A ={1,3} B = {1 , 3, 5} and C ={{1 , 3, 5}}
In above example
A ⊂ B and B ∈ C but A ∉ C. Hence statement is false.

(iii) True
Let A ⊂ B and B ⊂ C.
Let x ∈ A it is given that A ⊂ B, so that
x ∈ B it is given that B ⊂ C, so that
x ∈ C
Hence A ⊂ C .
(iv) False
Let A ={1, 2, 3} , B = {2, 3, 4} and C = {1, 2, 3, 5}
In above example
A ⊄ B and B ⊄ C, but A ⊂ C
Hence given statement is false.
(v) False
Let A = {1 , 2 , 3}, B = {2, 3, 4}
In above example if x = 1
1 ∈ A and A ⊄ B but 1 ∉ B. Hence given statement is false.
(vi) True
Let x ∉ B and A ⊂ B so that all element of set A will belongs to set B. if set B don’t have element x then A cannot have this element.

Decide, among the following sets, which sets are subsets of one and another: A = {x: x ∈ R and x satisfy x2 - 8x + 12 = 0}, B = {2, 4, 6}, C = {2, 4, 6, 8…}, D = {6}.

Decide, among the following sets, which sets are subsets of one and another: A = {x: x ∈ R and x satisfy x² - 8x + 12 = 0}, B = {2, 4, 6}, C = {2, 4, 6, 8…}, D = {6}.

Answer
A = {x: x ∈ R and x satisfies x² - 8x + 12 = 0}
Factorize
x² - 8x + 12 = 0
x² - 6x -2x + 12 = 0
x(x - 6) -2(x-6) = 0
(x-6)(x-2) =0
x - 6 = 0 or x – 2 = 0
x is equal to 6 and 2
So that, A = {2, 6}
Given that B = {2, 4, 6}, C = {2, 4, 6, 8 …}, D = {6}
All element of set D lies in set A , all element of set of A lies in set B and all element of set B lies in set C.
D ⊂ A ⊂ B ⊂ C
D is subset of A, B and C . D ⊂ A, D ⊂ B, D ⊂ C ,
A is subset of B and C.
A ⊂ B, A ⊂ C
B is subset of C.
B ⊂ C
Hence total subsets are A ⊂ B, A ⊂ C, B ⊂ C, D ⊂ A, D ⊂ B, D ⊂ C.

In a committee, 50 people speak French, 20 speak Spanish and 10 speak both Spanish and French. How many speak at least one of these two languages?

In a committee, 50 people speak French, 20 speak Spanish and 10 speak both Spanish and French. How many speak at least one of these two languages?

Answer
Let F and F represent the set of people in the committee who speak Spanish and French respectively.
Given that n(F) = 50, n(S) = 20 and n(S ∩ F) = 10
Use the formula:
n(S ∪ F) = n(S) + n(F) - n(S ∩ F)
Plug the values, we get
⇒ 20 + 50 - 10
⇒ 70 - 10
⇒ 60
Hence, 60 people in the committee speak at least one of these two languages.

In a group of 65 people, 40 like cricket, 10 like both cricket and tennis. How many like tennis only and not cricket? How many like tennis?

In a group of 65 people, 40 like cricket, 10 like both cricket and tennis. How many like tennis only and not cricket? How many like tennis?

Answer
Let C and T represent the set people who like cricket and tennis respectively.
Number of people like cricket or tennis, n(C ∪ T) = 65.
Number of people likes tennis, n(C) = 40.
Number of people likes both cricket and tennis, n(C ∩ T) = 10
Use the formula
n(C ∪ T) = n(C) + n(T) - n(C ∩ T)
⇒ 65 = 40 + n(T) - 10
⇒ 65 = 30 + n(T)
⇒ 65 - 30 = n(T)
⇒ 35 = n(T)
Hence, 35 people like tennis.
Number of people How many like tennis only and not cricket,
⇒  n(T - C) = n(T) – n(T ∩ C) ⇒ 35 -10 ⇒  25.
Thus, 25 people like only tennis.

In a group of 70 people, 37 like coffee, 52 like tea, and each person likes at least one of the two drinks. How many people like both coffee and tea?

In a group of 70 people, 37 like coffee, 52 like tea, and each person likes at least one of the two drinks. How many people like both coffee and tea?

Answer
Let C and T represent the set of people who like coffee and tea respectively.
Number of person likes at least one of the two drinks, n(C ∪ T) = 70.
Number of person like coffee, n(C) = 37
Number of person likes tea, n(T) = 52
Use the formula
n(C ∪ T) = n(C) + n(T) - n(C ∩ T)
Plug the values, we get
70 = 37 + 52 - n(C ∩ T)
⇒ 70 = 89 - n(C ∩ T)
⇒ n(C ∩ T) = 89 - 70 = 19
Thus, 19 people like both coffee and tea.

If X and Y are two sets such that X has 40 elements, X ∪Y has 60 elements and X ∩Y has 10 elements, how many elements does Y have?

If X and Y are two sets such that X has 40 elements, X ∪Y has 60 elements and X ∩Y has 10 elements, how many elements does Y have?

Answer
It is given that
n(X) = 40, n(X ∪ Y) = 60 and n(X ∩ Y) = 10
Use the formula
n(X ∪ Y) = n(X) + n(Y) - n(X ∩ Y)
Plug the values, we get
⇒ 60 = 40 + n(Y) - 10
⇒60 = 30 + n(Y)
⇒n(Y) = 60 - 30 = 30
Hence, the set Y has 30 elements.

If S and T are two sets such that S has 21 elements, T has 32 elements, and S ∩ T has 11 elements, how many elements does S ∪ T have?

If S and T are two sets such that S has 21 elements, T has 32 elements, and S ∩ T has 11 elements, how many elements does S ∪ T have?

Answer
It is given that
n(S) = 21, n(T) = 32 and n(S ∩ T) = 11
use formula
n (S ∪ T) = n (S) + n (T) - n (S ∩ T)
plug the values we get.
n (S ∪ T) = 21 + 32 - 11 = 42
Hence, the set (S ∪ T) has 42 elements.

In a group of 400 people, 250 can speak Hindi and 200 can speak English. How many people can speak both Hindi and English?

In a group of 400 people, 250 can speak Hindi and 200 can speak English. How many people can speak both Hindi and English?

Answer
Let H and E is the set of people speaking Hindi and English respectively.
It is given that
People speaking Hindi or English, n(H ∪ E) = 400,
Hindi Speaker, n(H) = 250,
English speaker n(E) = 200.
People can speak both Hindi and English, n(H ∩ E) = ?
Use formula
n ( H ∪ E ) = n ( H ) + n ( E ) – n ( H ∩ E )
Plug the values we get
400 = 250 + 200 - n ( H∩ E )
n ( H ∩ E) = 450 – 400 = 50
Answer
People can speak both Hindi and English, n(H ∩ E) = 50.

If X and Y are two sets such that X ∪Y has 18 elements, X has 8 elements and Y has 15 elements; how many elements does X ∩Y have

If X and Y are two sets such that X ∪Y has 18 elements, X has 8 elements and Y has 15 elements; how many elements does X ∩Y have?

Answer
It is given that
n(X) = 8, n(Y) = 15 and n(X ∪ Y) = 18, find n(X ∩Y)
Use formula
n ( X ∪ Y ) = n ( X ) + n ( Y ) – n ( X ∩ Y )
Plug the values we get
18 = 8 + 15 - n ( X ∩ Y )
n ( X ∩ Y ) = 23 – 18 = 5
Answer is n(X ∩Y) = 5

If X and Y are two sets such that n(X) = 17, n(Y) = 23 and n(X ∪ Y) = 38, find n(X ∩Y).

If X and Y are two sets such that n(X) = 17, n(Y) = 23 and n(X ∪ Y) = 38, find n(X ∩Y).

Answer
Use formula
n ( X ∪ Y ) = n ( X ) + n ( Y ) – n ( X ∩ Y )
Plug the values we get
38 = 17 + 23 - n ( X ∩ Y )
n ( X ∩ Y ) = 40 – 38 = 2
Answer is n(X ∩Y) = 5

Fill in the blanks to make each of the following a true statement: (i) A ∪ A′ = . . . (ii) φ′ ∩ A = . . . (iii) A ∩ A′ = . . . (iv) U′ ∩ A = . . .

Fill in the blanks to make each of the following a true statement:
(i) A ∪ A′ = . . . (ii) φ′ ∩ A = . . .
(iii) A ∩ A′ = . . . (iv) U′ ∩ A = . . .

Answer
(i) A ∪ A′ = U (Universal set)
(ii) Φ′ ∩ A = U ∩ A = A
(iii) A ∩ A′ = Φ
(iv) U′ ∩ A = Φ ∩ A = Φ

Let U be the set of all triangles in a plane. If A is the set of all triangles with at least one angle different from 60°, what is A′?

Let U be the set of all triangles in a plane. If A is the set of all triangles with at least one angle different from 60°, what is A′?

Answer
A is the set of all triangles without angle different of 60°.

Draw appropriate Venn diagram for each of the following : (i) (A ∪ B)′, (ii) A′ ∩ B′, (iii) (A ∩ B)′, (iv) A′ ∪ B′

Draw appropriate Venn diagram for each of the following:
(i) (A ∪ B)′, (ii) A′ ∩ B′, (iii) (A ∩ B)′, (iv) A′ ∪ B′

Answer
(i) (A ∪ B)′

(ii) A′ ∩ B′

(iii) (A ∩ B)′

 

(iv) A′ ∪ B′

Verify that (i) (A ∪ B)′ = A′ ∩ B′ (ii) (A ∩ B)′ = A′ ∪ B′

If U = {1, 2, 3, 4, 5, 6, 7, 8, 9 }, A = {2, 4, 6, 8} and B = { 2, 3, 5, 7}. Verify that
(i) (A ∪ B)′ = A′ ∩ B′ (ii) (A ∩ B)′ = A′ ∪ B′

Answer

(i) (A ∪ B)′ = A′ ∩ B′

LHS

(A ∪ B)′

= ({2, 4, 6, 8} ∪ { 2, 3, 5, 7})’
= {2, 3, 4, 5, 6, 7, 8}’
= {1, 2, 3, 4, 5, 6, 7, 8, 9 } - {2, 3, 4, 5, 6, 7, 8}
= {1, 9}
RHS
A′ ∩ B′
={2, 4, 6, 8}’ ∩{ 2, 3, 5, 7}’
= [{1, 2, 3, 4, 5, 6, 7, 8, 9 }-{2, 4, 6, 8}]∩[{1, 2, 3, 4, 5, 6, 7, 8, 9 } - { 2, 3, 5, 7}]
={1, 3, 5, 7, 9}∩{1, 4, 6, 8, 9}
= {1,9}
LHS = RHS
(ii) (A ∩ B)′ = A′ ∪ B′
(A ∩ B)′
= ({2, 4, 6, 8} ∩ { 2, 3, 5, 7})’
= {2}’
= {1, 2, 3, 4, 5, 6, 7, 8, 9 } - {2}
= {1, 3, 4, 5, 6, 7, 8, 9 }
RHS
A′ ∪ B′
={2, 4, 6, 8}’ ∪ { 2, 3, 5, 7}’
= [{1, 2, 3, 4, 5, 6, 7, 8, 9 }-{2, 4, 6, 8}]∪ [{1, 2, 3, 4, 5, 6, 7, 8, 9 } - { 2, 3, 5, 7}]
={1, 3, 5, 7, 9}∪ {1, 4, 6, 8, 9}
= {1, 3, 4, 5, 6, 7, 8, 9}
LHS = RHS

Taking the set of natural numbers as the universal set, write down the complements of the following sets

Taking the set of natural numbers as the universal set, write down the complements of the following sets:
(i) {x: x is an even natural number}
(ii) {x: x is an odd natural number}
(iii) {x: x is a positive multiple of 3}
(iv) {x: x is a prime number}
(v) {x: x is a natural number divisible by 3 and 5}
(vi) {x: x is a perfect square}
(vii) {x: x is perfect cube}
(viii) {x: x + 5 = 8}
(ix) {x: 2x + 5 = 9}
(x) {x: x ≥ 7}
(xi) {x: x ∈ N and 2x + 1 > 10}

Answer

Universal set ⇒ Set of natural numbers ⇒ {1, 2, 3, 4, 5...}
Set of even natural numbers ⇒ {2, 4, 6, 8 ...}
Set of odd natural number ⇒ {1, 3, 5...}
Set of prime number ⇒ {2, 3, 5, 7...}

(i) Complements of {x: x is an even natural number}
⇒ {1, 2, 3, 4, 5...} - {2, 4, 6, 8 ...}
⇒ {1, 3, 5...}
⇒ {x: x is an odd natural number}
(ii) Complements of {x: x is an odd natural number}
⇒ {1, 2, 3, 4, 5...} - {1, 3, 5...}
⇒ {2, 4, 6, 8 ...}
⇒ {x: x is an even natural number}
(iii) Complements of {x: x is a positive multiple of 3} 
⇒ {x: x ∈ N and x is not a multiple of 3}

(iv) Complements of {x: x is a prime number} 
⇒{x: x is a positive composite number and x = 1}
(v) Complements of {x: x is a natural number divisible by 3 and 5} 
⇒ {x: x is a natural number that is not divisible by 3 or 5}
(vi) Complements of {x: x is a perfect square} 
⇒ {x: x ∈ N and x is not a perfect square}
(vii) Complements of {x: x is a perfect cube} 
⇒ {x: x ∈ N and x is not a perfect cube}
(viii){x: x + 5 = 8}
Solving the equation x + 5 ⇒ 8 we get x = 3.
Complement of this set will not have x = 3, Hence complement set can be {x: x ∈ N and x ≠ 3}
(ix){x: 2x + 5 = 9}
Solving the equation 2x + 5 = 9 we get x = 2.
Complement of this set will not have x = 2,
Hence complement set can be {x: x ∈ N and x ≠ 2}
(x) Complements of {x: x ≥ 7} 
={x: x ∈ N and x < 7}
(xi) Complements of {x: x ∈ N and 2x + 1 > 10}
Solve the equation
2x + 1 > 10
2x > 9
X > 9/2
All values of x more than 9/2 cannot be in set of complement.
Hence, complements of {x: x ∈ N and 2x + 1 > 10} ⇒ {x: x ∈ N and x ≤ 9/2}

If U = { a, b, c, d, e, f, g, h}, find the complements of the following sets

If U = { a, b, c, d, e, f, g, h}, find the complements of the following sets :
(i) A = {a, b, c} (ii) B = {d, e, f, g}
(iii) C = {a, c, e, g} (iv) D = { f, g, h, a}

Answer
(i) Complements of set A 
⇒ U – A
⇒ { a, b, c, d, e, f, g, h} - {a, b, c}
⇒ { d, e, f, g, h}
(ii) Complement of set B 
⇒ U – B
⇒ { a, b, c, d, e, f, g, h } - {d, e, f, g}
⇒ { a, b, c, h }
(iii) Complement of set C 
⇒ U – C
⇒ { a, b, c, d, e, f, g, h } - {a, c, e, g}
⇒ { b, d , f, h }
(iv) Complement of set D
⇒ U – D
⇒ { a, b, c, d, e, f, g, h } - { f, g, h, a}
⇒{ b, c, d, e )

Find (i) A′ (ii) B′ (iii) (A ∪ C)′ (iv) (A ∪ B)′ (v) (A′)′ (vi) (B – C)′

Let U = { 1, 2, 3, 4, 5, 6, 7, 8, 9 }, A = { 1, 2, 3, 4}, B = { 2, 4, 6, 8 } and C = { 3, 4, 5, 6 }.
Find (i) A′ (ii) B′ (iii) (A ∪ C)′ (iv) (A ∪ B)′ (v) (A′)′ (vi) (B – C)′

Answer
U ={1, 2, 3, 4, 5, 6, 7, 8, 9}
A = {1, 2, 3, 4}
(i) A’ 
⇒U – A
⇒{1, 2, 3, 4, 5, 6, 7, 8, 9} - {1, 2, 3, 4}
⇒{ 5, 6, 7, 8, 9}
(ii) B’
⇒ U – B
⇒{1, 2, 3, 4, 5, 6, 7, 8, 9} - { 2, 4, 6, 8 }
⇒{1, 3, 5, 7, 9 }

(iii) (A ∪ C)′
Plug the value of A and C we get
⇒ ({1, 2, 3, 4} ∪ { 3, 4, 5, 6 })′
⇒ {1, 2, 3, 4, 5, 6}’
⇒ U - {1, 2, 3, 4, 5, 6}
Plug the value of U we get
⇒{1, 2, 3, 4, 5, 6, 7, 8, 9} - {1, 2, 3, 4, 5, 6}
⇒ {7, 8, 9 }

(iv) (A ∪ B)′
Plug the value of A and B we get
⇒ ({1, 2, 3, 4} ∪ { 2, 4, 6, 8 })′
⇒ {1, 2, 3, 4, 6, 8}’
⇒ U - {1 ,3, 5, 7, 9}
Plug the value of U we get
⇒ {1, 2, 3, 4, 5, 6, 7, 8, 9} - {1, 2, 3, 4, 6, 8}
⇒ { 5, 7, 9 )
(v) (A′)′
⇒( A’)’
⇒(U – A)’
⇒({1, 2, 3, 4, 5, 6, 7, 8, 9} - {1, 2, 3, 4})’
⇒{ 5, 6, 7, 8, 9}’
⇒ U - { 5, 6, 7, 8, 9}
⇒ { 1, 2, 3, 4 }
(vi) (B – C)′ 
= ({ 2, 4, 6, 8 } - { 3, 4, 5, 6 })’
= {2, 8}’
= U - {2, 8}
={1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 8}
={ 1, 3, 4, 5, 6, 7, 9 }

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