Chapter 1: Sets

Solution:
(i) This collection contains exactly the months January, June and July. Since every member can be identified without any doubt, it is well-defined and therefore, it is a set.

(ii) The idea of “most talented” is subjective — what one person considers talent, another may not. Since there is no fixed, universally agreed-upon criterion, the collection is not well-defined and therefore, not a set.

(iii) The phrase “best cricket batsman” is vague and depends entirely on personal opinion. Different selectors would pick different players. Since the membership cannot be determined objectively, the collection is not well-defined and therefore, not a set.

(iv) Every student is either in your class or not — there is no room for ambiguity. The given collection is well-defined and therefore, it is a set.

(v) This collection contains the first 99 natural numbers, each of which can be clearly listed. It is well-defined and therefore, it is a set.

(vi) All novels written by Munshi Prem Chand can be identified precisely. It is a well-defined collection and therefore, it is a set.

(vii) Any integer can be tested to check whether it is even or not. It is a well-defined collection and therefore, it is a set.

(viii) Every question either belongs to this chapter or it does not. It is a well-defined collection and therefore, it is a set.

(ix) The label “most dangerous” is subjective — different people have different levels of fear and different criteria for danger. Since membership cannot be determined without ambiguity, the given collection is not well-defined and therefore, it is not a set.

Solution:
(i) \(5 \in \mathrm{A}\)
(ii) \(8 \notin \mathrm{A}\)
(iii) \(0 \notin \mathrm{A}\)
(iv) \(4 \in \mathrm{A}\)
(v) \(2 \in \mathrm{A}\)
(vi) \(10 \notin \mathrm{A}\)

Solution:
(i) Listing all integers satisfying \(-3 \leq x < 7\), we get: \(-3,-2,-1,0,1,2,3,4,5,6\) (7 is excluded since the inequality is strict).
\(\therefore\) In roster form, \(\mathrm{A}=\{-3,-2,-1,0,1,2,3,4,5,6\}\)

(ii) Writing out all natural numbers that are strictly less than 6, we get: 1, 2, 3, 4, 5.
\(\therefore\) In roster form, \(\mathrm{B}=\{1,2,3,4,5\}\)

(iii) Checking all two-digit numbers whose digits add up to 8, we get: 17, 26, 35, 44, 53, 62, 71, 80.
\(\therefore\) In roster form, \(\mathrm{C}=\{17,26,35,44,53,62,71,80\}\)

(iv) First, listing all divisors of 60: \(1,2,3,4,5,6,10,12,15,20,30,60\).
Among them, picking out only the prime numbers, we get: 2, 3, 5.
\(\therefore\) In roster form, \(\mathrm{D}=\{2,3,5\}\)

(v) Writing out all letters in TRIGONOMETRY and removing repeated ones (T, R and O each appear more than once):
In roster form \(\mathrm{E}=\{\mathrm{T}, \mathrm{R}, \mathrm{I}, \mathrm{G}, \mathrm{O}, \mathrm{N}, \mathrm{M}, \mathrm{E}, \mathrm{Y}\}\)

(vi) Writing out all letters in BETTER and removing repeated ones (E and T each appear more than once):
In roster form \(\mathrm{F}=\{\mathrm{B}, \mathrm{E}, \mathrm{T}, \mathrm{R}\}\)

Solution:
(i) \(\{3,6,9,12\} = \{3\times1, 3\times2, 3\times3, 3\times4\} = \{x: x=3n,\ n \in \mathbb{N} \text{ and } 1 \leq n \leq 4\}\)

(ii) \(\{2,4,8,16,32\} = \{2^1, 2^2, 2^3, 2^4, 2^5\} = \{x: x=2^n,\ n \in \mathbb{N} \text{ and } 1 \leq n \leq 5\}\)

(iii) \(\{5,25,125,625\} = \{5^1, 5^2, 5^3, 5^4\} = \{x: x=5^n,\ n \in \mathbb{N} \text{ and } 1 \leq n \leq 4\}\)

(iv) \(\{2,4,6,\ldots\} = \{2\times1, 2\times2, 2\times3,\ldots\} = \{x: x=2n,\ n \in \mathbb{N}\}\)
Alternatively, we can write \(\{x: x \text{ is an even natural number}\}\).

(v) \(\{1,4,9,\ldots,100\} = \{1^2, 2^2, 3^2, \ldots, 10^2\} = \{x: x=n^2,\ n \in \mathbb{N} \text{ and } 1 \leq n \leq 10\}\)

Solution:
(i) Listing every odd natural number in order:
\(\therefore \mathrm{A}=\{1,3,5,\ldots\}\)

(ii) Identifying every integer that lies strictly between \(-\frac{1}{2}\) and \(\frac{9}{2}\), we get: \(0, 1, 2, 3, 4\).
\(\therefore \mathrm{B}=\{0,1,2,3,4\}\)

(iii) Finding all integers \(x\) such that \(x^2 \leq 4\) (i.e., \(x\) lies between \(-2\) and \(2\) inclusive), we get: \(-2,-1,0,1,2\).
\(\therefore \mathrm{C}=\{-2,-1,0,1,2\}\)

(iv) Writing out the letters of “LOYAL” and dropping the repeated letter L:
\(\mathrm{D}=\{\mathrm{L},\mathrm{O},\mathrm{Y},\mathrm{A}\}\)

(v) Going through the calendar and picking out months that do not have 31 days:
February, April, June, September, November
\(\therefore \mathrm{E}=\{\text{February, April, June, September, November}\}\)

(vi) Going through the English alphabet and picking only consonants that come before K:
\(b, c, d, f, g, h, j\)
\(\therefore \mathrm{F}=\{b,c,d,f,g,h,j\}\)

Solution:
To match the sets, we first convert each set-builder description on the right into roster form:
(a) \(\{2,3\}\)
(b) \(\{1,3,5,7,9\}\)
(c) \(\{1,2,3,6\}\)
(d) \(\{\mathrm{M},\mathrm{A},\mathrm{T},\mathrm{H},\mathrm{E},\mathrm{I},\mathrm{C},\mathrm{S}\}\)

Comparing each roster on the left with the roster forms obtained above:
(i) \(\{1,2,3,6\}\) matches (c)
(ii) \(\{2,3\}\) matches (a)
(iii) \(\{\mathrm{M},\mathrm{A},\mathrm{T},\mathrm{H},\mathrm{E},\mathrm{I},\mathrm{C},\mathrm{S}\}\) matches (d)
(iv) \(\{1,3,5,7,9\}\) matches (b)

Solution:
(i) There is no odd natural number which is divisible by 2. So, the given set is a null set.

(ii) Set of even prime numbers \(= \{2\} \neq \phi\). So, the given set is not a null set. It is a singleton set.
Note: A natural number > 1 is said to be prime if it has only two divisors 1 and itself. The set of prime numbers is \(\{2,3,5,7,11, \ldots\}\).

(iii) There is no natural number which is both less than 5 and greater than 7. So, the given set is a null set.

(iv) Two parallel lines have no common point. So, the given set is a null set.
Hence (i), (iii) and (iv) are examples of the null set.

Solution:
(i) Since there are 12 months (i.e., a definite number of months) in a year, the given set is finite.

(ii) Since the number of elements in the set is infinite, the given set is infinite.

(iii) Since the number of elements in the set is 100 (i.e., a definite number), the given set is finite.

(iv) Since there are infinitely many numbers greater than 100, the given set is infinite.

(v) Since the number of primes less than 99 is a definite number, the given set is finite.

Solution:
(i) Since there are infinite number of lines parallel to the \(x\)-axis, the given set is infinite.

(ii) Since there are 26 letters, i.e., a definite number of letters, in the English alphabet, the given set is finite.

(iii) Since there are infinitely many multiples of 5, the given set is infinite.

(iv) The process of counting the animals living on the earth is terminating. Thus, a definite number of animals live on the earth and hence the given set is finite.

(v) There is no end to the number of circles passing through the origin \((0,0)\). Hence, the given set is infinite.

Solution:
(i) A and B have exactly same elements, though not in the same order.
\(\therefore \quad \mathrm{A}=\mathrm{B}\)

(ii) \(12 \in \mathrm{A}\) but \(12 \notin \mathrm{B}\)
\(\therefore \quad A \neq B\)

(iii) In roster form \(\mathrm{B}=\{2,4,6,8,10\}\)
Since A and B have exactly same elements, therefore, \(\mathrm{A}=\mathrm{B}\)

(iv) In roster form \(\mathrm{A}=\{10,20,30, \ldots\}\)
Since \(15 \in \mathrm{B}\) but \(15 \notin \mathrm{A}\), therefore \(\mathrm{A} \neq \mathrm{B}\).

Solution:
(i) \(\mathrm{B}=\{x: x \ \text{is solution of} \ (x+2)(x+3)=0\}\)
\(\because x^2+5x+6 = x^2+2x+3x+6 = x(x+2)+3(x+2) = (x+2)(x+3)\)
\(\therefore \mathrm{B} = \{-2,-3\}\)
\(2 \in \mathrm{A}\) but \(2 \notin \mathrm{B} \quad \therefore \mathrm{A} \neq \mathrm{B}\)

(ii) Dropping repetitions: \(\mathrm{A}=\{\mathrm{F}, \mathrm{O}, \mathrm{L}, \mathrm{W}\}\)
\(\mathrm{B}=\{\mathrm{W}, \mathrm{O}, \mathrm{L}, \mathrm{F}\}\)
Sets A and B have exactly the same elements, though not in the same order.
\(\therefore \quad A=B\).

Solution:
\(\mathrm{B}=\mathrm{D}, \quad \mathrm{E}=\mathrm{G}\).

Solution:
(i) Every element of the set \(\{2,3,4\}\) is also an element of the set \(\{1,2,3,4,5\}\)
\(\therefore \{2,3,4\} \subset \{1,2,3,4,5\}\).

(ii) \(a \in \{a, b, c\}\) but \(a \notin \{b, c, d\}\) \(\therefore \{a, b, c\} \not\subset \{b, c, d\}\).

(iii) Every student of class XI of your school is a student of your school.
\(\therefore \{x: x \ \text{is a student of class XI of your school}\} \subset \{x: x \ \text{is a student of your school}\}\).

(iv) Every circle in a plane is not a circle with radius 1 unit, as it can have any radius \(r\), (\(r>0\)).
\(\therefore \{x: x \ \text{is a circle in the plane}\} \not\subset \{x: x \ \text{is a circle in the same plane with radius 1 unit}\}\).

(v) Any triangle is never a rectangle.
\(\therefore \{x: x \ \text{is a triangle in the plane}\} \not\subset \{x: x \ \text{is a rectangle in the plane}\}\).

(vi) Every equilateral triangle in a plane is a triangle in the plane.
\(\therefore \{x: x \ \text{is an equilateral triangle in a plane}\} \subset \{x: x \ \text{is a triangle in the same plane}\}\).

(vii) Every even natural number is an integer.
\(\therefore \{x: x \ \text{is an even natural number}\} \subset \{x: x \ \text{is an integer}\}\).

Solution:
(i) False, since every element of the set \(\{a, b\}\) is also an element of the set \(\{b, c, a\}\), therefore, \(\{a, b\} \subset \{b, c, a\}\).

(ii) True, since every element of the set \(\{a, e\}\) is also an element of the set of vowels \(\{a, e, i, o, u\}\), therefore, \(\{a, e\} \subset \{a, e, i, o, u\}\).

(iii) False, since \(2 \in \{1,2,3\}\) but \(2 \notin \{1,3,5\}\).

(iv) True, since \(a \in \{a, b, c\}\).

(v) False, since \(\{a\}\) is a subset of the set \(\{a, b, c\}\) but not an element of the set \(\{a, b, c\}\).

(vi) True, since \(\{x: x \ \text{is an even natural number less than 6}\} = \{2,4\}\) and \(\{x: x \ \text{is a natural number which divides 36}\} = \{1,2,3,4,6,9,12,18,36\}\).
Clearly, \(\{2,4\} \subset \{1,2,3,4,6,9,12,18,36\}\).

Solution:
(i) False, since \(3 \notin \mathrm{A},\ 4 \notin \mathrm{A}\), therefore, \(\{3,4\} \not\subset \mathrm{A}\).

(ii) True, since \(\{3,4\}\) is an element of A.

(iii) True, since \(\{3,4\} \in \mathrm{A}\), therefore, \(\{\{3,4\}\} \subset \mathrm{A}\).

(iv) True, since 1 is an element of A.

(v) False, since 1 is not a set. Only a set can be a subset of another set.

(vi) True, since 1, 2, 5 are elements of A.

(vii) False, since \(\{1,2,5\}\) is not an element of A.

(viii) False, since \(3 \notin A\).

(ix) False, since \(\phi\) is not an element of A.

(x) True, since \(\phi\) is a subset of every set.

(xi) False, since \(\phi \subset \mathrm{A}\) and hence \(\{\phi\} \in \mathrm{P}(\mathrm{A})\), power set of set A.

Solution:
(i) Let \(\mathrm{A}=\{a\}\), then A has one element.
\(\therefore\) Number of subsets of the set \(\mathrm{A} = 2^n = 2^1 = 2\).
Subset of A having no element is \(\phi\)
Subset of A having one element is \(\{a\}\)
\(\therefore\) The subsets of A are \(\phi, \{a\}\).

(ii) Let \(\mathrm{A}=\{a, b\}\), then A has 2 elements.
\(\therefore\) Number of subsets of \(\mathrm{A} = 2^n = 2^2 = 4\).
Subset of A having no element is \(\phi\)
Subsets of A having one element are \(\{a\}, \{b\}\)
Subset of A having two elements is \(\{a, b\}\)
\(\therefore\) The subsets of A are \(\phi, \{a\}, \{b\}, \{a, b\}\).

(iii) Let \(\mathrm{A}=\{1,2,3\}\), then A has 3 elements.
\(\therefore\) Number of subsets of \(\mathrm{A} = 2^n = 2^3 = 8\).
Subset of A having no element is \(\phi\)
Subsets of A having one element are \(\{1\}, \{2\}, \{3\}\)
Subsets of A having two elements are \(\{1,2\}, \{1,3\}, \{2,3\}\)
Subset of A having three elements is \(\{1,2,3\}\)
\(\therefore\) The subsets of A are \(\phi, \{1\}, \{2\}, \{3\}, \{1,2\}, \{1,3\}, \{2,3\}, \{1,2,3\}\).

(iv) The only subset of the empty set \(\phi\) is \(\phi\) itself.

Solution:
(i) \((-4, 6]\), since \(-4\) is not included while \(6\) is included.

(ii) \((-12, -10)\), since both end points are excluded.

(iii) \([0, 7)\), since \(0\) is included while \(7\) is excluded.

(iv) \([3, 4]\), since both end points are included.

Solution:
(i) \(\{x: x \in \mathrm{R}, -3 < x < 0\}\)

(ii) \(\{x: x \in \mathrm{R}, 6 \leq x \leq 12\}\)

(iii) \(\{x: x \in \mathrm{R}, 6 < x \leq 12\}\)

(iv) \(\{x: x \in \mathrm{R}, -23 \leq x < 5\}\)

Solution:
The set of all triangles.

Solution:
All the three sets A, B and C must be subsets of the universal set.

Since \(8 \notin \{0,1,2,3,4,5,6\}\),
\(\therefore \quad \mathrm{C} \not\subset \{0,1,2,3,4,5,6\}\)

None of the sets A, B and C is a subset of \(\phi\).

Since \(0 \notin \{1,2,3,4,5,6,7,8\}\),
\(\therefore \quad C \not\subset \{1,2,3,4,5,6,7,8\}\)

All the three sets A, B and C are subsets of the set \(\{0,1,2,3,4,5,6,7,8,9,10\}\) which is, therefore, the required universal set.

Solution:
(i) \(\mathrm{X} \cup \mathrm{Y}=\{1,3,5\} \cup \{1,2,3\}=\{1,2,3,5\}\).

(ii) \(\mathrm{A} \cup \mathrm{B}=\{a, e, i, o, u\} \cup \{a, b, c\}=\{a, b, c, e, i, o, u\}\)

(iii) \(\mathrm{A} = \{3,6,9,12,\dots\}\)
\(\mathrm{B} = \{1,2,3,4,5\}\)
\(\mathrm{A} \cup \mathrm{B} = \{3,6,9,12,\dots\} \cup \{1,2,3,4,5\} = \{1,2,3,4,5,6,9,12,\dots\}\)
\(= \{x: x=1,2,4,5 \ \text{or a multiple of } 3\}\)

(iv) \(\mathrm{A} = \{2,3,4,5,6\}\)
\(\mathrm{B} = \{7,8,9\}\)
\(\mathrm{A} \cup \mathrm{B} = \{2,3,4,5,6\} \cup \{7,8,9\} = \{2,3,4,5,6,7,8,9\} = \{x: 1 < x < 10 \ \text{and} \ x \in \mathrm{N}\}\)

(v) \(\mathrm{A} \cup \mathrm{B}=\{1,2,3\} \cup \phi=\{1,2,3\}=\mathrm{A}\)
Note. The result of this part (v) is true in general also. \(A \cup \phi = A\) for every set \(A\).

Solution:
Yes, \(\mathrm{A} \cup \mathrm{B}=\{a, b, c\}=\mathrm{B}\).

Solution:
\(\mathrm{A} \cup \mathrm{B} = \{x: x \in \mathrm{A} \ \text{or} \ x \in \mathrm{B}\}\)
\(= \{x: x \in \mathrm{B}\}\)
\((\because \mathrm{A} \subset \mathrm{B} \quad \therefore x \in \mathrm{A} \Rightarrow x \in \mathrm{B})\)
\(= \mathrm{B}.\)

Solution:
(i) \(\mathrm{A} \cup \mathrm{B}=\{1,2,3,4\} \cup \{3,4,5,6\}=\{1,2,3,4,5,6\}\)

(ii) \(\mathrm{A} \cup \mathrm{C}=\{1,2,3,4\} \cup \{5,6,7,8\}=\{1,2,3,4,5,6,7,8\}\)

(iii) \(\mathrm{B} \cup \mathrm{C}=\{3,4,5,6\} \cup \{5,6,7,8\}=\{3,4,5,6,7,8\}\)

(iv) \(\mathrm{B} \cup \mathrm{D}=\{3,4,5,6\} \cup \{7,8,9,10\}=\{3,4,5,6,7,8,9,10\}\)

(v) \(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}=(\mathrm{A} \cup \mathrm{B}) \cup \mathrm{C}=\{1,2,3,4,5,6\} \cup \{5,6,7,8\}=\{1,2,3,4,5,6,7,8\}\)

(vi) \(\mathrm{A} \cup \mathrm{B} \cup \mathrm{D}=(\mathrm{A} \cup \mathrm{B}) \cup \mathrm{D}=\{1,2,3,4,5,6\} \cup \{7,8,9,10\}=\{1,2,3,4,5,6,7,8,9,10\}\)

(vii) \(\mathrm{B} \cup \mathrm{C} \cup \mathrm{D}=(\mathrm{B} \cup \mathrm{C}) \cup \mathrm{D}=\{3,4,5,6,7,8\} \cup \{7,8,9,10\}=\{3,4,5,6,7,8,9,10\}\)

Solution:
(i) \(\mathrm{X} \cap \mathrm{Y}=\{1,3,5\} \cap \{1,2,3\}=\{1,3\}\).

(ii) \(\mathrm{A} \cap \mathrm{B}=\{a, e, i, o, u\} \cap \{a, b, c\}=\{a\}\).

(iii) \(\mathrm{A} = \{3,6,9,12,\dots\}\)
\(\mathrm{B} = \{1,2,3,4,5\}\)
\(\mathrm{A} \cap \mathrm{B} = \{3,6,9,12,\dots\} \cap \{1,2,3,4,5\}=\{3\}\)

(iv) \(\mathrm{A} = \{2,3,4,5,6\}\)
\(\mathrm{B} = \{7,8,9\}\)
\(\mathrm{A} \cap \mathrm{B} = \{2,3,4,5,6\} \cap \{7,8,9\}=\phi\)

(v) \(\mathrm{A} \cap \mathrm{B}=\{1,2,3\} \cap \phi=\phi\)
Note. The result of this part (v) is true in general also. \(\mathrm{A} \cap \phi = \phi\) for every set A.

Solution:
(i) \(\mathrm{A} \cap \mathrm{B}=\{3,5,7,9,11\} \cap \{7,9,11,13\}=\{7,9,11\}\)

(ii) \(\mathrm{B} \cap \mathrm{C}=\{7,9,11,13\} \cap \{11,13,15\}=\{11,13\}\)

(iii) \(\mathrm{A} \cap \mathrm{C} \cap \mathrm{D}=(\mathrm{A} \cap \mathrm{C}) \cap \mathrm{D}\)
\(=(\{3,5,7,9,11\} \cap \{11,13,15\}) \cap \{15,17\}\)
\(=\{11\} \cap \{15,17\}=\phi\)

(iv) \(\mathrm{A} \cap \mathrm{C}=\{3,5,7,9,11\} \cap \{11,13,15\}=\{11\}\)

(v) \(\mathrm{B} \cap \mathrm{D}=\{7,9,11,13\} \cap \{15,17\}=\phi\)

(vi) \(\mathrm{A} \cap (\mathrm{B} \cup \mathrm{C})=\{3,5,7,9,11\} \cap (\{7,9,11,13\} \cup \{11,13,15\})\)
\(=\{3,5,7,9,11\} \cap \{7,9,11,13,15\}=\{7,9,11\}\)

(vii) \(\mathrm{A} \cap \mathrm{D}=\{3,5,7,9,11\} \cap \{15,17\}=\phi\)

(viii) \(\mathrm{A} \cap (\mathrm{B} \cup \mathrm{D})=\{3,5,7,9,11\} \cap (\{7,9,11,13\} \cup \{15,17\})\)
\(=\{3,5,7,9,11\} \cap \{7,9,11,13,15,17\}=\{7,9,11\}\)

(ix) \((\mathrm{A} \cap \mathrm{B}) \cap (\mathrm{B} \cup \mathrm{C})\)
\(=(\{3,5,7,9,11\} \cap \{7,9,11,13\}) \cap (\{7,9,11,13\} \cup \{11,13,15\})\)
\(=\{7,9,11\} \cap \{7,9,11,13,15\}=\{7,9,11\}\)

(x) \((\mathrm{A} \cup \mathrm{D}) \cap (\mathrm{B} \cup \mathrm{C})\)
\(=(\{3,5,7,9,11\} \cup \{15,17\}) \cap (\{7,9,11,13\} \cup \{11,13,15\})\)
\(=\{3,5,7,9,11,15,17\} \cap \{7,9,11,13,15\}=\{7,9,11,15\}\)

Solution:
(i) \(\mathrm{A} \cap \mathrm{B}=\mathrm{B} \quad (\because \mathrm{B} \subset \mathrm{A})\)
(\(\because\) Every even natural number is a natural number)

(ii) \(\mathrm{A} \cap \mathrm{C}=\mathrm{C} \quad (\because \mathrm{C} \subset \mathrm{A})\)
(\(\because\) Every odd natural number is a natural number)

(iii) \(\mathrm{A} \cap \mathrm{D}=\mathrm{D} \quad (\because \mathrm{D} \subset \mathrm{A})\)
(\(\because\) Every prime number is a natural number)

(iv) \(\mathrm{B} \cap \mathrm{C}=\phi\)
(\(\because\) There is no natural number which is both even and odd)

(v) \(\mathrm{B} \cap \mathrm{D}=\{2\}\), 2 is only even prime number.

(vi) \(\mathrm{C} \cap \mathrm{D}=\{x: x \ \text{is an odd prime number}\}\).

Solution:
(i) Let \(\mathrm{A}=\{1,2,3,4\}\) and \(\mathrm{B}=\{x: x \in \mathrm{N} \ \text{and} \ 4 \leq x \leq 6\}=\{4,5,6\}\)
\(\mathrm{A} \cap \mathrm{B}=\{4\} \neq \phi \quad \therefore\) Sets A and B are not disjoint.

(ii) \(e\) is a common element of the two sets.
\(\therefore\) Sets are not disjoint.

(iii) Given sets are disjoint sets because there is no natural number which is both even and odd.

Solution:
(i) \(\mathrm{A}-\mathrm{B}=\{3,6,9,12,15,18,21\}-\{4,8,12,16,20\}=\{3,6,9,15,18,21\}\)

(ii) \(\mathrm{A}-\mathrm{C}=\{3,6,9,12,15,18,21\}-\{2,4,6,8,10,12,14,16\}=\{3,9,15,18,21\}\)

(iii) \(\mathrm{A}-\mathrm{D}=\{3,6,9,12,15,18,21\}-\{5,10,15,20\}=\{3,6,9,12,18,21\}\)

(iv) \(\mathrm{B}-\mathrm{A}=\{4,8,12,16,20\}-\{3,6,9,12,15,18,21\}=\{4,8,16,20\}\)

(v) \(\mathrm{C}-\mathrm{A}=\{2,4,6,8,10,12,14,16\}-\{3,6,9,12,15,18,21\}=\{2,4,8,10,14,16\}\)

(vi) \(\mathrm{D}-\mathrm{A}=\{5,10,15,20\}-\{3,6,9,12,15,18,21\}=\{5,10,20\}\)

(vii) \(\mathrm{B}-\mathrm{C}=\{4,8,12,16,20\}-\{2,4,6,8,10,12,14,16\}=\{20\}\)

(viii) \(\mathrm{B}-\mathrm{D}=\{4,8,12,16,20\}-\{5,10,15,20\}=\{4,8,12,16\}\)

(ix) \(\mathrm{C}-\mathrm{B}=\{2,4,6,8,10,12,14,16\}-\{4,8,12,16,20\}=\{2,6,10,14\}\)

(x) \(\mathrm{D}-\mathrm{B}=\{5,10,15,20\}-\{4,8,12,16,20\}=\{5,10,15\}\)

(xi) \(\mathrm{C}-\mathrm{D}=\{2,4,6,8,10,12,14,16\}-\{5,10,15,20\}=\{2,4,6,8,12,14,16\}\)

(xii) \(\mathrm{D}-\mathrm{C}=\{5,10,15,20\}-\{2,4,6,8,10,12,14,16\}=\{5,15,20\}\)

Solution:
(i) \(\mathrm{X}-\mathrm{Y}=\{a, b, c, d\}-\{f, b, d, g\}=\{a, c\}\)

(ii) \(\mathrm{Y}-\mathrm{X}=\{f, b, d, g\}-\{a, b, c, d\}=\{f, g\}\)

(iii) \(\mathrm{X} \cap \mathrm{Y}=\{a, b, c, d\} \cap \{f, b, d, g\}=\{b, d\}\)

Solution:
\(\mathrm{R}-\mathrm{Q} = \{x: x \in \mathrm{R} \ \text{and} \ x \notin \mathrm{Q}\}\)
\(= \{x: x \ \text{is a real number and } x \text{ is not a rational number}\}\)
\(= \{x: x \ \text{is an irrational number}\}\)
(\(\because\) Every real number is either rational or irrational but not both)

Solution:
(i) False, because \(\{2,3,4,5\} \cap \{3,6\}=\{3\} \neq \phi\)

(ii) False, because \(\{a, e, i, o, u\} \cap \{a, b, c, d\}=\{a\} \neq \phi\)

(iii) True, because \(\{2,6,10,14\} \cap \{3,7,11,15\}=\phi\)

(iv) True, because \(\{2,6,10\} \cap \{3,7,11\}=\phi\)

Solution:
(i) \(\mathrm{A}^{\prime}=\mathrm{U}-\mathrm{A}=\{1,2,3,4,5,6,7,8,9\}-\{1,2,3,4\}=\{5,6,7,8,9\}\)

(ii) \(\mathrm{B}^{\prime}=\mathrm{U}-\mathrm{B}=\{1,2,3,4,5,6,7,8,9\}-\{2,4,6,8\}=\{1,3,5,7,9\}\)

(iii) \(\mathrm{A} \cup \mathrm{C}=\{1,2,3,4\} \cup \{3,4,5,6\}=\{1,2,3,4,5,6\}\)
\((\mathrm{A} \cup \mathrm{C})^{\prime}=\mathrm{U}-(\mathrm{A} \cup \mathrm{C})=\{1,2,3,4,5,6,7,8,9\}-\{1,2,3,4,5,6\}=\{7,8,9\}\)

(iv) \(\mathrm{A} \cup \mathrm{B}=\{1,2,3,4\} \cup \{2,4,6,8\}=\{1,2,3,4,6,8\}\)
\((\mathrm{A} \cup \mathrm{B})^{\prime}=\mathrm{U}-(\mathrm{A} \cup \mathrm{B})=\{1,2,3,4,5,6,7,8,9\}-\{1,2,3,4,6,8\}=\{5,7,9\}\)

(v) \(\mathrm{A}^{\prime}=\mathrm{U}-\mathrm{A}=\{1,2,3,4,5,6,7,8,9\}-\{1,2,3,4\}=\{5,6,7,8,9\}\)
\(\left(\mathrm{A}^{\prime}\right)^{\prime}=\mathrm{U}-\mathrm{A}^{\prime}=\{1,2,3,4,5,6,7,8,9\}-\{5,6,7,8,9\}=\{1,2,3,4\}=\mathrm{A}\)

(vi) \(\mathrm{B}-\mathrm{C}=\{2,4,6,8\}-\{3,4,5,6\}=\{2,8\}\)
\((\mathrm{B}-\mathrm{C})^{\prime}=\mathrm{U}-(\mathrm{B}-\mathrm{C})=\{1,2,3,4,5,6,7,8,9\}-\{2,8\}=\{1,3,4,5,6,7,9\}\)

Solution:
(i) \(\mathrm{A}^{\prime}=\mathrm{U}-\mathrm{A}=\{a, b, c, d, e, f, g, h\}-\{a, b, c\}=\{d, e, f, g, h\}\)

(ii) \(\mathrm{B}^{\prime}=\mathrm{U}-\mathrm{B}=\{a, b, c, d, e, f, g, h\}-\{d, e, f, g\}=\{a, b, c, h\}\)

(iii) \(\mathrm{C}^{\prime}=\mathrm{U}-\mathrm{C}=\{a, b, c, d, e, f, g, h\}-\{a, c, e, g\}=\{b, d, f, h\}\)

(iv) \(\mathrm{D}^{\prime}=\mathrm{U}-\mathrm{D}=\{a, b, c, d, e, f, g, h\}-\{f, g, h, a\}=\{b, c, d, e\}\)

Solution:
(i) \(\{x: x \ \text{is an odd natural number}\}\)

(ii) \(\{x: x \ \text{is an even natural number}\}\)

(iii) \(\{x: x \in \mathrm{N} \ \text{and} \ x \ \text{is not a multiple of 3}\}\)

(iv) \(\{x: x \ \text{is a positive composite number and} \ x=1\}\)
Def: Composite number: A natural number \(> 1\) is said to be composite if it is not prime, i.e., if it has at least one divisor other than 1 and itself. For example, \(4,6,8,9,\dots\) are composite.
Note: In fact \(\mathrm{N}\) is the union of (set of primes, set of composites and \(\{1\}\)).

(v) \(\{x: x \in \mathrm{N} \ \text{and} \ x \ \text{is not divisible by 3 or not divisible by 5}\}\)

(vi) \(\{x: x \in \mathrm{N} \ \text{and} \ x \ \text{is not a perfect square}\}\)

(vii) \(\{x: x \in \mathrm{N} \ \text{and} \ x \ \text{is not a perfect cube}\}\)

(viii) \(\{x: x \in \mathrm{N} \ \text{and} \ x \neq 3\}\)

(ix) \(\{x: x \in \mathrm{N} \ \text{and} \ x \neq 2\} \quad (\because 2x+5=9 \Rightarrow 2x=4 \Rightarrow x=2)\)

(x) \(\{x: x \in \mathrm{N} \ \text{and} \ x < 7\}=\{1,2,3,4,5,6\}\)

(xi) \(\{x: x \in \mathrm{N} \ \text{and} \ x \leq \frac{9}{2}\}=\{1,2,3,4\}\)

Solution:
Here \(\mathrm{A}^{\prime}=\mathrm{U}-\mathrm{A}=\{1,3,5,7,9\}\) and \(\mathrm{B}^{\prime}=\mathrm{U}-\mathrm{B}=\{1,4,6,8,9\}\)

(i) \(\mathrm{A} \cup \mathrm{B}=\{2,3,4,5,6,7,8\}\)
\((\mathrm{A} \cup \mathrm{B})^{\prime}=\mathrm{U}-(\mathrm{A} \cup \mathrm{B})=\{1,9\}\)
\(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}=\{1,3,5,7,9\} \cap \{1,4,6,8,9\}=\{1,9\}\)
\(\therefore (\mathrm{A} \cup \mathrm{B})^{\prime}=\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\)

(ii) \(\mathrm{A} \cap \mathrm{B}=\{2\}\)
\((\mathrm{A} \cap \mathrm{B})^{\prime}=\mathrm{U}-(\mathrm{A} \cap \mathrm{B})=\{1,3,4,5,6,7,8,9\}\)
\(\mathrm{A}^{\prime} \cup \mathrm{B}^{\prime}=\{1,3,5,7,9\} \cup \{1,4,6,8,9\}=\{1,3,4,5,6,7,8,9\}\)
\(\therefore (\mathrm{A} \cap \mathrm{B})^{\prime}=\mathrm{A}^{\prime} \cup \mathrm{B}^{\prime}\)

Solution:
(i) Venn diagram for \((\mathrm{A} \cup \mathrm{B})^{\prime}\) is:

(ii) Venn diagram for \(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\) is:

(iii) Venn diagram for \((\mathrm{A} \cap \mathrm{B})^{\prime}\) is:

(iv) Venn diagram for \(\mathrm{A}^{\prime} \cup \mathrm{B}^{\prime}\)is:

Solution:
\(\mathrm{A}^{\prime}=\mathrm{U}-\mathrm{A}\)
= Set of all triangles with no angle different from \(60^{\circ}\)
= Set of all triangles with each angle \(60^{\circ}\)
= Set of all equilateral triangles.

Solution:
(i) \(\mathrm{A} \cup \mathrm{A}^{\prime}=\mathrm{U}\)   [Property of complement sets]

(ii) \(\phi \cap \mathrm{A}=\phi\)

(iii) \(\mathrm{A} \cap \mathrm{A}^{\prime}=\phi\)

(iv) \(\mathrm{U}^{\prime} \cap \mathrm{A}=\phi \cap \mathrm{A}=\phi\)

Solution:
\(\mathrm{A}=\{x: x \in \mathrm{R} \ \text{and} \ (x-2)(x-6)=0\}=\{2,6\}\)
\(\mathrm{B}=\{2,4,6\},\ \mathrm{C}=\{2,4,6,8,\ldots\},\ \mathrm{D}=\{6\}\)

Here, \(\mathrm{A} \subset \mathrm{B},\ \mathrm{A} \subset \mathrm{C},\ \mathrm{B} \subset \mathrm{C}\),
\(\mathrm{D} \subset \mathrm{A},\ \mathrm{D} \subset \mathrm{B},\ \mathrm{D} \subset \mathrm{C}\).

Solution:
(i) False. Let \(\mathrm{A}=\{1\}\) and \(\mathrm{B}=\{\{1\},2\}\). Clearly, \(1 \in \mathrm{A}\) and \(\mathrm{A} \in \mathrm{B}\) but \(1 \notin \mathrm{B}\).
Thus, \(x \in \mathrm{A} \ \text{and} \ \mathrm{A} \in \mathrm{B}\) need not imply \(x \in \mathrm{B}\).

(ii) False. Let \(\mathrm{A}=\{1\},\ \mathrm{B}=\{1,2\}\) and \(\mathrm{C}=\{\{1,2\},3\}\). Clearly, \(\mathrm{A} \subset \mathrm{B} \ \text{and} \ \mathrm{B} \in \mathrm{C}\) but \(\mathrm{A} \notin \mathrm{C}\).
Thus, \(\mathrm{A} \subset \mathrm{B} \ \text{and} \ \mathrm{B} \in \mathrm{C}\) need not imply \(\mathrm{A} \in \mathrm{C}\).

(iii) True. Let \(x\) be any element of A. Then
\(x \in \mathrm{A} \Rightarrow x \in \mathrm{B} \quad [\because \mathrm{A} \subset \mathrm{B}]\)
\(\phantom{x \in \mathrm{A}} \Rightarrow x \in \mathrm{C} \quad [\because \mathrm{B} \subset \mathrm{C}]\)
Thus, \(x \in \mathrm{A} \Rightarrow x \in \mathrm{C}\) for all \(x \in \mathrm{A}\), therefore, \(\mathrm{A} \subset \mathrm{C}\).
Hence, \(\mathrm{A} \subset \mathrm{B}\) and \(\mathrm{B} \subset \mathrm{C} \Rightarrow \mathrm{A} \subset \mathrm{C}\).

(iv) False. Let \(\mathrm{A}=\{1,2\},\ \mathrm{B}=\{2,3\}\) and \(\mathrm{C}=\{1,2,5\}\). Clearly, \(\mathrm{A} \not\subset \mathrm{B}\) since \(1 \in \mathrm{A}\) and \(1 \notin \mathrm{B}\). Also \(\mathrm{B} \not\subset \mathrm{C}\) since \(3 \in \mathrm{B}\) and \(3 \notin \mathrm{C}\). But \(\mathrm{A} \subset \mathrm{C}\).
Thus, \(\mathrm{A} \not\subset \mathrm{B}\) and \(\mathrm{B} \not\subset \mathrm{C}\) need not imply \(\mathrm{A} \not\subset \mathrm{C}\).

(v) False. Let \(\mathrm{A}=\{1,2\}\) and \(\mathrm{B}=\{2,3\}\). Clearly, \(1 \in \mathrm{A}\) and \(\mathrm{A} \not\subset \mathrm{B}\) but \(1 \notin \mathrm{B}\).
Thus, \(x \in \mathrm{A}\) and \(\mathrm{A} \not\subset \mathrm{B}\) need not imply \(x \in \mathrm{B}\).

(vi) True. Let \(\mathrm{A} \subset \mathrm{B}\) and \(x \notin \mathrm{B}\). If possible, suppose \(x \in \mathrm{A}\).
Now, \(x \in \mathrm{A}\) and \(\mathrm{A} \subset \mathrm{B} \Rightarrow x \in \mathrm{B}\), which is a contradiction to the given.
Therefore, our supposition is wrong. Hence \(x \notin \mathrm{A}\).
Thus, \(\mathrm{A} \subset \mathrm{B}\) and \(x \notin \mathrm{B} \Rightarrow x \notin \mathrm{A}\).

Solution:
Let \(x \in \mathrm{B}\)
\(\Rightarrow x \in \mathrm{A} \cup \mathrm{B} \quad [\because \mathrm{B} \subset \mathrm{A} \cup \mathrm{B} \ \text{always}]\)
\(\Rightarrow x \in \mathrm{A} \cup \mathrm{C} \quad [\because \mathrm{A} \cup \mathrm{B} = \mathrm{A} \cup \mathrm{C} \ \text{(given)}]\)
\(\Rightarrow x \in \mathrm{A} \ \text{or} \ x \in \mathrm{C}\)

Case I: \(x \in \mathrm{A}\)
\(\Rightarrow x \in \mathrm{A} \ \text{and} \ x \in \mathrm{B} \quad [\because \ \text{we started with} \ x \in \mathrm{B}]\)
\(\Rightarrow x \in \mathrm{A} \cap \mathrm{B}\)
\(\Rightarrow x \in \mathrm{A} \cap \mathrm{C} \quad [\because \mathrm{A} \cap \mathrm{B} = \mathrm{A} \cap \mathrm{C} \ \text{(given)}]\)
\(\Rightarrow x \in \mathrm{A} \ \text{and} \ x \in \mathrm{C}\)
\(\Rightarrow x \in \mathrm{C} \ \text{also}\)

Case II: \(x \in \mathrm{C}\)

\(\therefore\) In each case \(x \in \mathrm{B} \Rightarrow x \in \mathrm{C}\), hence \(\mathrm{B} \subset \mathrm{C}\)    …(i)
Similarly, \(\mathrm{C} \subset \mathrm{B}\)    …(ii)

From (i) and (ii), we have \(\mathrm{B}=\mathrm{C}\).

Solution:
Let us consider condition (i), i.e., \(\mathrm{A} \subset \mathrm{B}\) (given).

(ii) To prove \(\mathrm{A}-\mathrm{B}=\phi\):
If possible, let \(\mathrm{A}-\mathrm{B} \neq \phi\).
Let \(x \in \mathrm{A}-\mathrm{B} \Rightarrow x \in \mathrm{A}\) and \(x \notin \mathrm{B} \Rightarrow x \in \mathrm{B}\) and \(x \notin \mathrm{B} \quad (\because \mathrm{A} \subset \mathrm{B})\)
which is absurd. So our supposition is wrong. Hence, \(\mathrm{A}-\mathrm{B}=\phi\).

(iii) Given \(\mathrm{A} \subset \mathrm{B}\). To prove \(\mathrm{A} \cup \mathrm{B}=\mathrm{B}\):
Let \(x \in \mathrm{A} \cup \mathrm{B} \Rightarrow x \in \mathrm{A}\) or \(x \in \mathrm{B} \Rightarrow x \in \mathrm{B}\) or \(x \in \mathrm{B} \quad [\because \mathrm{A} \subset \mathrm{B} \ \text{(given)}]\)
\(\Rightarrow x \in \mathrm{B} \Rightarrow \mathrm{A} \cup \mathrm{B} \subset \mathrm{B}\)
Also we know that \(\mathrm{B} \subset \mathrm{A} \cup \mathrm{B}\) (always).
\(\therefore \mathrm{A} \cup \mathrm{B}=\mathrm{B}\).

(iv) To prove \(\mathrm{A} \cap \mathrm{B}=\mathrm{A}\):
We know that \(\mathrm{A} \cap \mathrm{B} \subset \mathrm{A}\) always.
Now let \(x \in \mathrm{A}\), \(\therefore x \in \mathrm{B} \quad [\because \mathrm{A} \subset \mathrm{B} \ \text{(given)}]\)
\(\therefore x \in \mathrm{A}\) and \(x \in \mathrm{B} \Rightarrow x \in \mathrm{A} \cap \mathrm{B}\)
\(\therefore \mathrm{A} \subset \mathrm{A} \cap \mathrm{B}\). Therefore, \(\mathrm{A} \cap \mathrm{B}=\mathrm{A}\).

Remark. Similarly, if we begin with any of the other three (ii), (iii) and (iv) as given, we can prove the remaining three. For example, starting with (iii): \(\mathrm{A} \cup \mathrm{B}=\mathrm{B}\), then \(\mathrm{A} \subset \mathrm{B}\) \([\because x \in \mathrm{A} \Rightarrow x \in \mathrm{A} \cup \mathrm{B} \Rightarrow x \in \mathrm{B}]\).

Solution:
Given \(\mathrm{A} \subset \mathrm{B}\).
Let \(x \in (\mathrm{C}-\mathrm{B}) \Rightarrow x \in \mathrm{C}\) and \(x \notin \mathrm{B}\)
\(\Rightarrow x \in \mathrm{C}\) and \(x \notin \mathrm{A} \quad [\because \mathrm{A} \subset \mathrm{B} \ \text{(given)}, \ \text{so} \ x \notin \mathrm{B} \Rightarrow x \notin \mathrm{A}]\)
\(\Rightarrow x \in \mathrm{C}-\mathrm{A}\)
Hence \(\mathrm{C}-\mathrm{B} \subset \mathrm{C}-\mathrm{A}\).

Solution:
\((A \cap B) \cup (A-B) = (A \cap B) \cup (A \cap B^{\prime})\)
\(= A \cap (B \cup B^{\prime}) \quad \text{[Distributive Law]}\)
\(= A \cap U = A\)

Also,
\(A \cup (B-A) = A \cup (B \cap A^{\prime})\)
\(= (A \cup B) \cap (A \cup A^{\prime}) \quad \text{[Distributive Law]}\)
\(= (A \cup B) \cap U\)
\(= A \cup B\).

Solution:
(i) We know that \(\mathrm{X} \subset \mathrm{Y} \Rightarrow \mathrm{X} \cup \mathrm{Y}=\mathrm{Y}\), the superset.
Here \(\mathrm{A} \cap \mathrm{B} \subset \mathrm{A}\),
\(\therefore \quad A \cup (A \cap B) = A\)

(ii) We know that \(\mathrm{X} \subset \mathrm{Y} \Rightarrow \mathrm{X} \cap \mathrm{Y}=\mathrm{X}\), the subset.
Here \(A \subset A \cup B\),
\(\therefore \quad A \cap (A \cup B) = A\).

Solution:
Let \(\mathrm{A}=\{1,2,3\},\ \mathrm{B}=\{2,4\},\ \mathrm{C}=\{2,5\}\).
Then \(A \cap B=\{2\}=A \cap C\)
But \(B \neq C\).

Solution:
Let \(x \in \mathrm{A}\)
\(\Rightarrow x \in \mathrm{A} \cup \mathrm{X} \quad [\because \mathrm{A} \subset \mathrm{A} \cup \mathrm{X} \ \text{always}]\)
\(\Rightarrow x \in \mathrm{B} \cup \mathrm{X} \quad [\because \mathrm{A} \cup \mathrm{X}=\mathrm{B} \cup \mathrm{X} \ \text{(given)}]\)
\(\Rightarrow x \in \mathrm{B} \ \text{or} \ x \in \mathrm{X}\)

Case I. \(x \in \mathrm{B}\).

Case II. \(x \in \mathrm{X}\). Also \(x \in \mathrm{A}\),
\(\Rightarrow x \in \mathrm{A} \cap \mathrm{X}\)
\(\Rightarrow x \in \mathrm{B} \cap \mathrm{X} \quad [\because \mathrm{A} \cap \mathrm{X}=\mathrm{B} \cap \mathrm{X} \ \text{(given)}]\)
\(\Rightarrow x \in \mathrm{B} \ \text{and} \ x \in \mathrm{X}\)
\(\Rightarrow x \in \mathrm{B} \ \text{also}\)

\(\therefore\) In each case, \(x \in \mathrm{B}\). Therefore \(\mathrm{A} \subset \mathrm{B}\).
Similarly \(\mathrm{B} \subset \mathrm{A}\).
\(\therefore A=B\).

Second solution (Using properties of sets):
We know that \(A \subset A \cup X\),
\(\therefore A \cap (A \cup X) = A\)
\(\Rightarrow A = A \cap (A \cup X) = A \cap (B \cup X) \quad [\because \mathrm{A} \cup \mathrm{X}=\mathrm{B} \cup \mathrm{X} \ \text{(given)}]\)
\(\Rightarrow A = (A \cap B) \cup (A \cap X) \quad \text{[By Distributive Law]}\)
\(\Rightarrow A = (A \cap B) \cup \emptyset \quad [\because A \cap X=\phi \ \text{(given)}]\)
\(\Rightarrow A = A \cap B \quad \text{…(i)}\)

Interchanging A and B in the above argument,
\(\mathrm{B} = \mathrm{B} \cap \mathrm{A} = \mathrm{A} \cap \mathrm{B} \quad \text{[Commutative Law]} \quad \text{…(ii)}\)

From (i) and (ii), we have \(\mathrm{A}=\mathrm{B}\).

Solution:
Let \(\mathrm{A}=\{1,2\},\ \mathrm{B}=\{1,3\},\ \mathrm{C}=\{2,3\}\).
\(A \cap B=\{1\} \neq \phi,\quad B \cap C=\{3\} \neq \phi,\quad A \cap C=\{2\} \neq \phi\)
and \(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C}=\emptyset\).