State whether A = B or not:
A = {x : x + 2 = 3}, B = {x : x ∈ N and is less than 2}
Answer
For A,
⇒ x + 2 = 3
⇒ x = 1
So, A = {1}.
The natural numbers less than 2 give only x = 1, so B = {1}.
Both the sets have exactly the same element 1.
Hence, A = B.
State whether A = B or not:
A = {x : x ∈ N and 3x − 1 < 2}, B = {x : x ∈ W and 3x − 1 < 2}
Answer
For A, x ∈ N and 3x − 1 < 2
⇒ 3x < 2 + 1
⇒ 3x < 3
⇒ x < 1
There is no natural number less than 1, so A = φ.
For B, x ∈ W and x < 1. The only whole number less than 1 is 0, so B = {0}.
Here A has no element while B has the element 0.
Hence, A ≠ B.
State whether A = B or not:
A = {x : x ∈ N and is a prime factor of 36}, B = {1, 2, 3, 4, 6, 9, 12}
Answer
Resolving 36 into prime factors, 36 = 22 × 32, so the prime factors of 36 are 2 and 3.
∴ A = {2, 3} and B = {1, 2, 3, 4, 6, 9, 12}
Here 1 ∈ B but 1 ∉ A, so the two sets do not have exactly the same elements.
Hence, A ≠ B.
State whether A = B or not:
A = {x : x ∈ I and x2 ≤ 4}, B = {x : x ∈ R and x2 − 3x + 2 = 0}
Answer
For A, x ∈ I and x2 ≤ 4. The squares of 0, ±1, ±2 are 0, 1, 4 and each of these is less than or equal to 4, while (±3)2 = 9 > 4.
∴ A = {−2, −1, 0, 1, 2}
For B, x2 − 3x + 2 = 0
⇒ (x − 1)(x − 2) = 0
⇒ x = 1 or x = 2
∴ B = {1, 2}
Here −2 ∈ A but −2 ∉ B.
Hence, A ≠ B.
State whether A = B or not:
A = {2, 3} and B = {x : x is a solution of x2 + 5x + 6 = 0}
Answer
Given,
A = {2, 3}
For B, x2 + 5x + 6 = 0
⇒ (x + 2)(x + 3) = 0
⇒ x = −2 or x = −3
∴ B = {−2, −3}
The elements of A are positive while those of B are negative.
Hence, A ≠ B.
State whether A = B or not:
A = {x : x is a letter in the word LOYAL} and B = {y : y is a letter in the word ALLOY}
Answer
The letters of the word LOYAL are L, O, Y, A, L, so A = {L, O, Y, A}.
The letters of the word ALLOY are A, L, L, O, Y, so B = {A, L, O, Y}.
Both the sets consist of exactly the same four letters A, L, O and Y, and the order of listing does not matter.
Hence, A = B.
State whether A = B or not:
A = {x : x is a letter in the word WOLF} and B = {y : y is a letter in the word FOLLOW}
Answer
The letters of the word WOLF are W, O, L, F, so A = {W, O, L, F}.
The letters of the word FOLLOW are F, O, L, L, O, W, so B = {F, O, L, W}.
Both the sets consist of exactly the same four letters F, O, L and W.
Hence, A = B.
State whether A = B or not:
A = {10, 20, 30, 40, 50, ...} and B = {x : x ∈ N and x is a multiple of 10}.
Answer
Given,
A = {10, 20, 30, 40, 50, .....}
The natural numbers which are multiples of 10 are 10, 20, 30, 40, 50, ....., so B = {10, 20, 30, 40, 50, .....}.
Both the sets consist of exactly the same elements.
Hence, A = B.
If A = {1, 2, 3, 4, 5}, B = {2, 3, 4} and C = {2, 4, 5}; state whether the following statements are true or false.
(i) A ⊂ B
(ii) A ⊂ C
(iii) B ⊂ A
(iv) B ⊂ C
(v) C ⊂ A
(vi) C ⊂ B
(vii) B = C
(viii) φ ⊂ B
(ix) A ↔ B
(x) B ↔ C
(xi) A ↔ C.
Answer
Given,
A = {1, 2, 3, 4, 5}, B = {2, 3, 4} and C = {2, 4, 5}
∴ n(A) = 5, n(B) = 3 and n(C) = 3
We know that,
A ⊂ B if every member of A is also a member of B, and A ↔ B if n(A) = n(B).
(i) A ⊂ B
1 ∈ A but 1 ∉ B, so every member of A is not a member of B.
Hence, the statement A ⊂ B is false.
(ii) A ⊂ C
1 ∈ A but 1 ∉ C, so every member of A is not a member of C.
Hence, the statement A ⊂ C is false.
(iii) B ⊂ A
Each of the members 2, 3, 4 of B is also a member of A.
Hence, the statement B ⊂ A is true.
(iv) B ⊂ C
3 ∈ B but 3 ∉ C, so every member of B is not a member of C.
Hence, the statement B ⊂ C is false.
(v) C ⊂ A
Each of the members 2, 4, 5 of C is also a member of A.
Hence, the statement C ⊂ A is true.
(vi) C ⊂ B
5 ∈ C but 5 ∉ B, so every member of C is not a member of B.
Hence, the statement C ⊂ B is false.
(vii) B = C
B = {2, 3, 4} and C = {2, 4, 5}. Here 3 ∈ B but 3 ∉ C, so the two sets do not have exactly the same elements.
Hence, the statement B = C is false.
(viii) φ ⊂ B
The empty set has no element, so it is a subset of every set.
Hence, the statement φ ⊂ B is true.
(ix) A ↔ B
n(A) = 5 and n(B) = 3, so n(A) ≠ n(B).
Hence, the statement A ↔ B is false.
(x) B ↔ C
n(B) = 3 and n(C) = 3, so n(B) = n(C).
Hence, the statement B ↔ C is true.
(xi) A ↔ C
n(A) = 5 and n(C) = 3, so n(A) ≠ n(C).
Hence, the statement A ↔ C is false.
Use appropriate symbol '∈, ∉, ⊂, ⊃, =' to fill in the blanks:
(i) 4 ...... {1, 2, 3, 4}
(ii) −5 ...... {2, 3, 4, 5, 6}
(iii) {2} ...... {2, 3, 4}
(iv) {a, b, c} ...... {b, a, c}
(v) {a, b, c} ...... {a, b, b, a, c}
(vi) {a, i, u} ...... {a, a, i}
(vii) MUMBAI ... {x : x is a capital city of countries in Asia}.
Answer
(i) 4 ...... {1, 2, 3, 4}
The number 4 is one of the elements of the set {1, 2, 3, 4}, so 4 belongs to it.
Hence, 4 ∈ {1, 2, 3, 4}.
(ii) −5 ...... {2, 3, 4, 5, 6}
The elements of {2, 3, 4, 5, 6} are 2, 3, 4, 5, 6 and −5 is not one of them.
Hence, −5 ∉ {2, 3, 4, 5, 6}.
(iii) {2} ...... {2, 3, 4}
Here {2} is a set whose only member 2 is also a member of {2, 3, 4}, so {2} is a subset of {2, 3, 4}.
Hence, {2} ⊂ {2, 3, 4}.
(iv) {a, b, c} ...... {b, a, c}
The order of listing the elements of a set can be changed without altering the set. Both sets consist of exactly the elements a, b and c.
Hence, {a, b, c} = {b, a, c}.
(v) {a, b, c} ...... {a, b, b, a, c}
If one or more elements of a set are repeated, the set remains the same, so {a, b, b, a, c} = {a, b, c}. Both sets consist of exactly the elements a, b and c.
Hence, {a, b, c} = {a, b, b, a, c}.
(vi) {a, i, u} ...... {a, a, i}
On removing the repetition, {a, a, i} = {a, i}. Every member of {a, i} is also a member of {a, i, u}, and u ∈ {a, i, u} but u ∉ {a, i}. So {a, i, u} contains {a, i}.
Hence, {a, i, u} ⊃ {a, a, i}.
(vii) MUMBAI ... {x : x is a capital city of countries in Asia}:
MUMBAI is a city of India but it is not the capital of India; the capital of India is New Delhi. So MUMBAI is not a member of the set of capital cities of countries in Asia.
Hence, MUMBAI ∉ {x : x is a capital city of countries in Asia}.
Examine whether the following statements are true or false:
(i) {a, b} ⊄ {b, c, a}
(ii) {1, 2, 3} ⊂ {1, 3, 5}
(iii) {a} ⊂ {a, b, c}
(iv) {a} ∈ {a, b, c}
(v) {a, e} ⊂ {x : x is a vowel in the English alphabet}
(vi) {x : x is an even natural number less than 8} ⊂ {x : x is a natural number which divides 36}.
Answer
(i) {a, b} ⊄ {b, c, a}
The members of {a, b} are a and b, and both a ∈ {b, c, a} and b ∈ {b, c, a}. So every member of {a, b} is a member of {b, c, a} i.e. {a, b} ⊂ {b, c, a}.
Hence, the statement {a, b} ⊄ {b, c, a} is false.
(ii) {1, 2, 3} ⊂ {1, 3, 5}
Here 2 ∈ {1, 2, 3} but 2 ∉ {1, 3, 5}, so every member of {1, 2, 3} is not a member of {1, 3, 5}.
Hence, the statement {1, 2, 3} ⊂ {1, 3, 5} is false.
(iii) {a} ⊂ {a, b, c}
The only member of {a} is a and a ∈ {a, b, c}.
Hence, the statement {a} ⊂ {a, b, c} is true.
(iv) {a} ∈ {a, b, c}
The elements of {a, b, c} are the letters a, b and c, and not the set {a}. So {a} is not an element of {a, b, c}.
Hence, the statement {a} ∈ {a, b, c} is false.
(v) {a, e} ⊂ {x : x is a vowel in the English alphabet}
The set of vowels in the English alphabet is {a, e, i, o, u}. Both a and e are members of this set.
Hence, the statement {a, e} ⊂ {x : x is a vowel in the English alphabet} is true.
(vi) {x : x is an even natural number less than 8} ⊂ {x : x is a natural number which divides 36}
The even natural numbers less than 8 are 2, 4 and 6, so the first set is {2, 4, 6}.
Since 36 = 22 × 32, the natural numbers which divide 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36, so the second set is {1, 2, 3, 4, 6, 9, 12, 18, 36}.
Each of the members 2, 4 and 6 of the first set is also a member of the second set.
Hence, the given statement is true.
State whether each of the following statements is true or false:
(i) 2 ⊂ {1, 2, 3}
(ii) {2} ∈ {1, 2, 3}
(iii) φ ∈ {1, 2, 3}
(iv) 0 ∈ φ
(v) {1, 2} ⊂ {1, {2, 3}}
(vi) φ ⊂ {φ}.
Answer
(i) 2 ⊂ {1, 2, 3}
The symbol ⊂ is used between two sets, but 2 is a number and not a set.
Hence, the statement 2 ⊂ {1, 2, 3} is false.
(ii) {2} ∈ {1, 2, 3}
The elements of {1, 2, 3} are the numbers 1, 2 and 3, and not the set {2}.
Hence, the statement {2} ∈ {1, 2, 3} is false.
(iii) φ ∈ {1, 2, 3}
The elements of {1, 2, 3} are the numbers 1, 2 and 3, and φ is not one of them.
Hence, the statement φ ∈ {1, 2, 3} is false.
(iv) 0 ∈ φ
The empty set φ contains no element at all, so 0 cannot belong to φ.
Hence, the statement 0 ∈ φ is false.
(v) {1, 2} ⊂ {1, {2, 3}}
The elements of {1, {2, 3}} are the number 1 and the set {2, 3}. Now 2 ∈ {1, 2} but 2 ∉ {1, {2, 3}}, so every member of {1, 2} is not a member of {1, {2, 3}}.
Hence, the statement {1, 2} ⊂ {1, {2, 3}} is false.
(vi) φ ⊂ {φ}
The empty set has no element, so it is a subset of every set, and {φ} is a set.
Hence, the statement φ ⊂ {φ} is true.
Let A = {1, 2, {3, 4}, 5}. Which of the following statements are not true and why?
(i) {3, 4} ⊂ A
(ii) {3, 4} ∈ A
(iii) {{3, 4}} ⊂ A
(iv) {1, 2, 5} ⊂ A
(v) {1, 2, 5} ∈ A
(vi) {1, 2, 3} ⊂ A
(vii) 1 ∈ A
(viii) 1 ⊂ A
(ix) φ ∈ A
(x) φ ⊂ A
(xi) {φ} ⊂ A.
Answer
Given,
A = {1, 2, {3, 4}, 5}
The elements of A are the numbers 1, 2, 5 and the set {3, 4}. Note carefully that 3 ∉ A and 4 ∉ A.
(i) {3, 4} ⊂ A
The members of {3, 4} are 3 and 4, and 3 ∉ A.
Hence, the statement {3, 4} ⊂ A is not true, because 3 is not an element of A.
(ii) {3, 4} ∈ A
The set {3, 4} is itself one of the elements of A.
Hence, the statement {3, 4} ∈ A is true.
(iii) {{3, 4}} ⊂ A
The only member of {{3, 4}} is the set {3, 4}, and {3, 4} ∈ A.
Hence, the statement {{3, 4}} ⊂ A is true.
(iv) {1, 2, 5} ⊂ A
Each of the members 1, 2 and 5 of {1, 2, 5} is also an element of A.
Hence, the statement {1, 2, 5} ⊂ A is true.
(v) {1, 2, 5} ∈ A
The elements of A are 1, 2, {3, 4} and 5; the set {1, 2, 5} is not one of them.
Hence, the statement {1, 2, 5} ∈ A is not true, because {1, 2, 5} is not an element of A.
(vi) {1, 2, 3} ⊂ A
The members of {1, 2, 3} are 1, 2 and 3, and 3 ∉ A.
Hence, the statement {1, 2, 3} ⊂ A is not true, because 3 is not an element of A.
(vii) 1 ∈ A
The number 1 is one of the elements of A.
Hence, the statement 1 ∈ A is true.
(viii) 1 ⊂ A
The symbol ⊂ is used between two sets, but 1 is a number and not a set. The correct statement here is 1 ∈ A.
Hence, the statement 1 ⊂ A is not true, because 1 is an element and not a set.
(ix) φ ∈ A
The elements of A are 1, 2, {3, 4} and 5, and φ is not one of them.
Hence, the statement φ ∈ A is not true, because φ is not an element of A.
(x) φ ⊂ A
The empty set has no element, so it is a subset of every set.
Hence, the statement φ ⊂ A is true.
(xi) {φ} ⊂ A
The only member of {φ} is φ, and φ ∉ A as shown in part (ix).
Hence, the statement {φ} ⊂ A is not true, because φ is not an element of A.
If ξ = {1, 2, 3, ....., 10} and A = {x | x is a prime factor of 66}. List the set A.
Answer
Given,
ξ = {1, 2, 3, ....., 10}
Resolving 66 into prime factors,
66 = 2 × 3 × 11
∴ The prime factors of 66 are 2, 3 and 11.
It is understood that A is a subset of the universal set ξ, so the members of A are to be taken only from ξ. As ξ = {1, 2, 3, ....., 10}, the prime factors 2 and 3 belong to ξ but 11 does not belong to ξ.
Hence, the set A is {2, 3}.
Write down all the subsets of the following sets:
(i) {a}
(ii) {a, b}
(iii) φ
(iv) {φ, 1}
Answer
If A is a set with n(A) = m, then the number of subsets of A is 2m.
(i) {a}
Here n({a}) = 1, so the number of subsets = 21 = 2.
Hence, the subsets of {a} are φ and {a}.
(ii) {a, b}
Here n({a, b}) = 2, so the number of subsets = 22 = 4.
Hence, the subsets of {a, b} are φ, {a}, {b} and {a, b}.
(iii) φ
Here n(φ) = 0, so the number of subsets = 20 = 1. The only subset of the empty set is the empty set itself.
Hence, the only subset of φ is φ.
(iv) {φ, 1}
The set {φ, 1} has two elements, namely φ and 1, so the number of subsets = 22 = 4.
Hence, the subsets of {φ, 1} are φ, {φ}, {1} and {φ, 1}.
Write the power set of A, where A = {−1, 0, 2}.
Answer
Given,
A = {−1, 0, 2}
∴ n(A) = 3
The set formed by all the subsets of a given set A is called the power set of A and is denoted by P(A).
Number of elements in P(A) = 2n(A) = 23 = 8
Writing down all the subsets of A, we have
The subset with no element : φ
The subsets with one element : {−1}, {0}, {2}
The subsets with two elements : {−1, 0}, {−1, 2}, {0, 2}
The subset with three elements : {−1, 0, 2}
Hence, the power set of A is P(A) = {φ, {−1}, {0}, {2}, {−1, 0}, {−1, 2}, {0, 2}, {−1, 0, 2}}.
Write the number of proper subsets of A, where A = {−3, −1, 0, 1, 4}.
Answer
Given,
A = {−3, −1, 0, 1, 4}
∴ n(A) = 5
If A is a set with n(A) = m, then the number of subsets of A is 2m and the number of proper subsets of A is 2m − 1, because the set A itself is a subset of A but not a proper subset of A.
∴ Number of proper subsets of A = 25 − 1
= 32 − 1
= 31.
Hence, the number of proper subsets of A is 31.
If ξ = {1, 2, 3, ..., 40} and A = {x : x is a factor of 42}, then write n(A).
Answer
Given,
ξ = {1, 2, 3, ....., 40}
Resolving 42 into prime factors,
42 = 2 × 3 × 7
∴ The factors of 42 are 1, 2, 3, 6, 7, 14, 21 and 42.
It is understood that A is a subset of the universal set ξ, so the members of A are to be taken only from ξ = {1, 2, 3, ....., 40}. Out of the above factors, 42 does not belong to ξ, so it cannot be taken.
∴ A = {1, 2, 3, 6, 7, 14, 21}, which has 7 elements.
Hence, n(A) = 7.
If ξ = {all digits in our number system} and A = {x : x is a multiple of 3}, then write n(A).
Answer
In our number system every number is written with the help of the ten digits 0, 1, 2, ....., 9.
∴ ξ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
It is understood that A is a subset of ξ, so the members of A are to be taken only from ξ.
The multiples of 3 lying in ξ are 0, 3, 6 and 9, since 0 = 3 × 0, 3 = 3 × 1, 6 = 3 × 2 and 9 = 3 × 3.
∴ A = {0, 3, 6, 9}, which has 4 elements.
Hence, n(A) = 4.
Write the following sets as intervals:
(i) {x : x ∈ R, −3 < x ≤ 5}
(ii) {x : x ∈ R, −5 < x < −1}
(iii) {x : x ∈ R, 2 ≤ x ≤ 7}
(iv) {x : x ∈ R, 0 ≤ x < 3}
(v) {x : x ∈ R, x ≤ 5}
(vi) {x : x ∈ R, x < −3}
(vii) {x : x ∈ R, x ≥ −2}.
Answer
We use the following forms of intervals:
(a, b) = {x : x ∈ R, a < x < b}, [a, b] = {x : x ∈ R, a ≤ x ≤ b},
(a, b] = {x : x ∈ R, a < x ≤ b}, [a, b) = {x : x ∈ R, a ≤ x < b},
(a, ∞) = {x : x ∈ R, x > a}, [a, ∞) = {x : x ∈ R, x ≥ a},
(−∞, a) = {x : x ∈ R, x < a}, (−∞, a] = {x : x ∈ R, x ≤ a}.
(i) {x : x ∈ R, −3 < x ≤ 5}
Here −3 is not included but 5 is included, so the interval is open on the left and closed on the right.
Hence, the given set as an interval is (−3, 5].
(ii) {x : x ∈ R, −5 < x < −1}
Here neither −5 nor −1 is included, so the interval is open at both ends.
Hence, the given set as an interval is (−5, −1).
(iii) {x : x ∈ R, 2 ≤ x ≤ 7}
Here both 2 and 7 are included, so the interval is closed at both ends.
Hence, the given set as an interval is [2, 7].
(iv) {x : x ∈ R, 0 ≤ x < 3}
Here 0 is included but 3 is not included, so the interval is closed on the left and open on the right.
Hence, the given set as an interval is [0, 3).
(v) {x : x ∈ R, x ≤ 5}
Here x can be any real number up to and including 5, so the interval extends indefinitely to the left and is closed on the right.
Hence, the given set as an interval is (−∞, 5].
(vi) {x : x ∈ R, x < −3}
Here x can be any real number less than −3 but −3 is not included.
Hence, the given set as an interval is (−∞, −3).
(vii) {x : x ∈ R, x ≥ −2}
Here x can be any real number from −2 onwards, −2 being included.
Hence, the given set as an interval is [−2, ∞).
Write the following intervals in set builder form:
(i) (−2, 0]
(ii) (2, 7)
(iii) [−5, −2]
(iv) [−9, 4)
(v) (3, ∞)
(vi) (−∞, −1]
(vii) (−∞, 4)
Answer
(i) (−2, 0]
The interval (−2, 0] is open on the left and closed on the right, so −2 is excluded and 0 is included.
Hence, the given interval in set builder form is {x : x ∈ R, −2 < x ≤ 0}.
(ii) (2, 7)
The interval (2, 7) is open at both ends, so both 2 and 7 are excluded.
Hence, the given interval in set builder form is {x : x ∈ R, 2 < x < 7}.
(iii) [−5, −2]
The interval [−5, −2] is closed at both ends, so both −5 and −2 are included.
Hence, the given interval in set builder form is {x : x ∈ R, −5 ≤ x ≤ −2}.
(iv) [−9, 4)
The interval [−9, 4) is closed on the left and open on the right, so −9 is included and 4 is excluded.
Hence, the given interval in set builder form is {x : x ∈ R, −9 ≤ x < 4}.
(v) (3, ∞)
The interval (3, ∞) consists of all real numbers greater than 3, and 3 is excluded.
Hence, the given interval in set builder form is {x : x ∈ R, x > 3}.
(vi) (−∞, −1]
The interval (−∞, −1] consists of all real numbers less than or equal to −1.
Hence, the given interval in set builder form is {x : x ∈ R, x ≤ −1}.
(vii) (−∞, 4)
The interval (−∞, 4) consists of all real numbers less than 4, and 4 is excluded.
Hence, the given interval in set builder form is {x : x ∈ R, x < 4}.
If ξ = {1, 2, 3, ..., 12}, A = {x | 2x + 3 ≤ 18} and B = {x | x2 ≤ 40}; write A and B in the roster form.
Answer
Given,
ξ = {1, 2, 3, ....., 12}
It is understood that A and B are subsets of ξ, so their members are to be taken only from ξ.
For the set A,
⇒ 2x + 3 ≤ 18
⇒ 2x ≤ 18 − 3
⇒ 2x ≤ 15
⇒ x ≤
⇒ x ≤ 7.5
The members of ξ which are not greater than 7.5 are 1, 2, 3, 4, 5, 6, 7.
∴ A = {1, 2, 3, 4, 5, 6, 7}
For the set B,
x2 ≤ 40
The squares of 1, 2, 3, 4, 5, 6 are 1, 4, 9, 16, 25, 36 and each of these is not greater than 40, while 72 = 49 > 40.
∴ B = {1, 2, 3, 4, 5, 6}
Hence, A = {1, 2, 3, 4, 5, 6, 7} and B = {1, 2, 3, 4, 5, 6}.
Let ξ = {0, 1, 2, 3, ..., 50}, A = {x : x = 6n, n ∈ W} B = {x : x = 7n, n ∈ W} and C = {x : x is a factor of 72}, then
(i) write the sets A, B and C in roster form
(ii) state n(A), n(B) and n(C).
Answer
Given,
ξ = {0, 1, 2, 3, ....., 50}
It is understood that A, B and C are subsets of ξ, so their members are to be taken only from ξ.
(i) For the set A, x = 6n where n ∈ W.
Putting n = 0, 1, 2, ....., we get 0, 6, 12, 18, 24, 30, 36, 42, 48, 54, ..... Out of these, only those up to 50 belong to ξ.
∴ A = {0, 6, 12, 18, 24, 30, 36, 42, 48}
For the set B, x = 7n where n ∈ W.
Putting n = 0, 1, 2, ....., we get 0, 7, 14, 21, 28, 35, 42, 49, 56, ..... Out of these, only those up to 50 belong to ξ.
∴ B = {0, 7, 14, 21, 28, 35, 42, 49}
For the set C, resolving 72 into prime factors, 72 = 23 × 32, so the factors of 72 are
1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72.
Out of these, 72 does not belong to ξ, so it cannot be taken.
∴ C = {1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36}
Hence, A = {0, 6, 12, 18, 24, 30, 36, 42, 48}, B = {0, 7, 14, 21, 28, 35, 42, 49} and C = {1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36}.
(ii) Counting the different elements of each set,
A has 9 elements, B has 8 elements and C has 11 elements.
Hence, n(A) = 9, n(B) = 8 and n(C) = 11.
Given A = {x : x is a letter in the word ACCUMULATOR}
(i) Express A in roster form.
(ii) Find the cardinal number of the set of vowels in A.
(iii) Write down the power set of the set of vowels in A.
Answer
(i) The letters of the word ACCUMULATOR are A, C, C, U, M, U, L, A, T, O, R. While listing the elements of a set, each member is written only once, so the repeated letters C, U and A are written only once.
Hence, A in roster form is {A, C, U, M, L, T, O, R}.
(ii) The vowels in the English alphabet are A, E, I, O, U. Out of the elements of A, the vowels are A, U and O.
∴ The set of vowels in A is V = {A, U, O}, which has 3 elements.
Hence, the cardinal number of the set of vowels in A is 3.
(iii) Let V = {A, U, O} be the set of vowels in A, so n(V) = 3.
Number of elements in P(V) = 2n(V) = 23 = 8
Writing down all the subsets of V, we have
The subset with no element : φ
The subsets with one element : {A}, {U}, {O}
The subsets with two elements : {A, U}, {A, O}, {U, O}
The subset with three elements : {A, U, O}
Hence, the power set of the set of vowels in A is P(V) = {φ, {A}, {U}, {O}, {A, U}, {A, O}, {U, O}, {A, U, O}}.
List all the proper subsets of {0, 1, 2, 3}.
Answer
Let A = {0, 1, 2, 3}, so n(A) = 4.
Number of subsets of A = 24 = 16
Number of proper subsets of A = 24 − 1 = 16 − 1 = 15
(The set A itself is a subset of A but not a proper subset of A, so it is left out.)
Writing down all the subsets of A other than A itself, we have
The subset with no element : φ
The subsets with one element : {0}, {1}, {2}, {3}
The subsets with two elements : {0, 1}, {0, 2}, {0, 3}, {1, 2}, {1, 3}, {2, 3}
The subsets with three elements : {0, 1, 2}, {0, 1, 3}, {0, 2, 3}, {1, 2, 3}
Hence, the proper subsets of {0, 1, 2, 3} are φ, {0}, {1}, {2}, {3}, {0, 1}, {0, 2}, {0, 3}, {1, 2}, {1, 3}, {2, 3}, {0, 1, 2}, {0, 1, 3}, {0, 2, 3} and {1, 2, 3}.
If the set A has five elements, how many subsets will A have ? If A has six elements, how many proper subsets will A have?
Answer
If A is a set with n(A) = m, then the number of subsets of A is 2m and the number of proper subsets of A is 2m − 1.
When A has five elements, n(A) = 5.
∴ Number of subsets of A = 25 = 32
When A has six elements, n(A) = 6.
∴ Number of proper subsets of A = 26 − 1 = 64 − 1 = 63
Hence, A will have 32 subsets when it has five elements, and 63 proper subsets when it has six elements.
Two finite sets have m and k elements. If the number of subsets of the first set is 112 more than the number of subsets of the second set, then find the values of m and k.
Answer
Let the two sets be A and B, then n(A) = m and n(B) = k.
As the first set has more subsets than the second set, m > k.
Now, number of subsets of A = 2m and number of subsets of B = 2k.
Given,
Number of subsets of the first set is 112 more than the number of subsets of the second set.
⇒ 2m − 2k = 112
⇒ 2k(2m − k − 1) = 112
⇒ 2k(2m − k − 1) = 16 × 7
⇒ 2k(2m − k − 1) = 24 × 7
As 2m − k − 1 is an odd number and 2k is a power of 2, comparing both sides we get
2k = 24 and 2m − k − 1 = 7
From,
⇒ 2k = 24
⇒ k = 4
From,
⇒ 2m − k − 1 = 7
⇒ 2m − k = 8
⇒ 2m − k = 23
⇒ m − k = 3
⇒ m − 4 = 3
⇒ m = 3 + 4
⇒ m = 7
Hence, m = 7 and k = 4.