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  • RS AGGARWAL CLASS 11 MATHS SOLUTION RELATIONS-2

    RS AGGARWAL CLASS 11 MATHS SOLUTION RELATIONS-2

    RS AGGARWAL CLASS 11 MATHS SOLUTION RELATIONS

    RELATIONS Exercise 2D, 2E, 2C

    Exercise 2D

    Q. 1. What do you mean by a binary relation on a set A?
    Define the domain and range of relation on A.

    Answer:  A binary relation on a set A is a subset of the Cartesian product A x A.
    It’s a rule or way of associating elements of a set with other elements within the same set. 
    ▶️ The domain of a relation on A is the set of all first elements in the ordered pairs of the relation,
    ▶️ The range of a relation on A is the set of all second elements in those ordered pairs. 

    Q. 2. Let A = {2, 3, 5} and R = {(2, 3), (2, 5), (3, 3), (3, 5)}. 
    Show that R is a binary relation on A.
    Find its domain and range.

    Solution: Given: A = {2, 3, 5}
    ∴ A×A = {(2, 2), (2, 3), (2, 5), (3, 2), (3, 3), (3, 5), (5, 2), (5, 3), (5, 5)}
    ∵ R = {(2, 3), (2, 5), (3, 3), (3, 5)}.
    Since, R is a subset of A × A, it’s a binary relation on A.
    The domain of R is the set of first co-ordinates of R
    ∴ Dom(R) = {2, 3}
    The range of R is the set of second co-ordinates of R.
    ∴ Range(R) = {3, 5}

    Q. 3. Let A = {0, 1, 2, 3, 4, 5, 6, 7, 8} and let R = {(a, b) : a, b ∈ A and 2a + 3b = 12}.
    Express R as a set of ordered pairs.
    Show that R is a binary relation on A.
    Find its domain and range.

    Solution: A = {0, 1, 2, 3, 4, 5, 6, 7, 8}
    2a + 3b = 12
    ⇒ 3b = 12 – 2a
    ⇒ b = 12 – 2a/3
    If a = 0 then b = 4
    a = 3 then b = 2
    and a = 6 then b = 0
    ∴  R = {(0, 4), (3, 2), (6, 0)}
    Since, R is a subset of A × A, it is a binary relation to A.
    ∴  Dom(R) = {0, 3, 6}
    ∴  Range(R) = {4, 2, 0}

    Q. 4. If R is a binary relation on a set A define R–1 on A.
    Let R = {(a, b) : a, b ∈ W and 3a + 2b = 15} and 3a + 2b = 15}, where W is the set of whole numbers.
    Express R and R–1 as sets of ordered pairs.
    Show that (i) dom (R) = range (R–1)
    (ii) range (R) = dom (R–1)

    Solution: R = {(a, b) : a, b ∈ W and 3a + 2b = 15} and 3a + 2b = 15}
    3a + 2b = 15
    ⇒ 2b = 12 – 3a
    ⇒ b = 12 – 3a/2
    If a = 1 then b = 6
    a = 3 then b = 3
    and a = 5 then b = 0
    ∴  R = {(1, 6), (3, 3), (5, 0)}
    ∴  Dom(R) = {1, 3, 5} and
    Range(R) = {6, 3, 0}
    Now
    R–1 = {(6, 1), (3, 3), (0, 5)}
    ∴  Dom(R–1) = {6, 3, 0} and
    Range(R–1) = {1, 3, 5}
    Thus,
    Dom (R) = {1, 3, 5} = Range (R–1) (Proved)
    Range (R) = {6, 3, 0} = Dom (R–1)  (Proved)

    Q. 5. What is an equivalence relation?
    Show that the relation of ‘similarity’ on the set S of all triangles in a plane is an equivalence relation.

    Solution: An equivalence relation is one which possesses the properties of reflexivity, symmetry and transitivity.
    Let S be a set of all triangles in a plane.
    Reflexivity: A relation R on A is said to be reflexive if (a, a) ∈ R for all a ∈ A.
       Since every triangle is similar to itself, it is reflexive.

    Symmetry: A relation R on A is said to be symmetrical if (a, b) ∈ R ⇒ (b, a) ∈ R for all (a, b) ∈ A.
       If one triangle is similar to another triangle, it implies that the other triangle is also similar to the first triangle.
    Hence, it is symmetric.

    Transitivity: A relation R on A is said to be transitive if (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R for all (a, b, c) ∈ A.
       If one triangle(Δ1) is similar to a triangle(Δ2) and another triangle(Δ3) is also similar to that triangle(Δ2), then all the three triangles are similar.
    So the first triangle(Δ1) is also similar to the last triangle(Δ3).
    Hence, it is transitive.

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    Q. 6. Let R = {(a, b) : a, b ∈ Z and (a – b) is even}.
    Then, show that R is an equivalence relation on Z.

    Solution: R = {(a, b) : a, b ∈ Z and (a – b) is even}
    Reflexivity: Let a ∈ Z,
    ∴ a – a = 0 ∈ Z which is also even.
    Thus, (a, a) ∈ R for all value of a ∈ Z.
    Hence, it is reflexive
    Symmetry: Let (a, b) ∈ R,
    (a, b) ∈ R
    ∴ a – b is even
    ∴ -(b – a) is even
    ⇒ (b – a) is even
    ∴ (b, a) ∈ R
    Thus, (a, b) ∈ R ⇒ (b, a) ∈ R
    So it is symmetric.

    Transitivity: Let (a, b) ∈ R and (b, c) ∈ R
    (a – b) is even and (b – c) is even.
    ∴ [(a – b) + (b – c)] is even
    = (a – c) is even.
    Thus (a, c) ∈ R.
    So it is transitive.
    Since, the given relation possesses the properties of reflexivity, symmetry and transitivity,
    So it is an equivalence relation.

    Q. 7. Let A = {1, 2, 3} and R = {(a, b) : a, b ∈ A and |a2 – b2| ≤ 5.
    Write R as a set of ordered pairs.
    Mention whether R is (i) reflexive (ii) symmetric (iii) transitive.
    Give reason in each case.

    Solution: Given: A = {1, 2, 3} and R = {(a, b) : a, b ∈ A and |a2 – b2| ≤ 5.
    If a = 1, b = 1 then |12 – 12| = 0 < 5
    ∴ (1, 1) ∈ R
    If a = 1, b = 2 then |12 – 22| = 3 < 5
    ∴ (1, 2) ∈ R
    If a = 1, b = 3 then |12 – 32| = 8 > 5
    ∴ (1, 3) ∉ R

    If a = 2, b = 1 then |22 – 12| = 3 < 5
    ∴ (2, 1) ∈ R
    If a = 2, b = 2 then |22 – 22| = 0 < 5
    ∴ (2, 2) ∈ R
    If a = 2, b = 3 then |22 – 32| = 5 = 5
    ∴ (2, 3) ∈ R
    If a = 3, b = 1 then |32 – 12| = 8 > 5
    ∴ (3, 1) ∉ R
    If a = 3, b = 2 then |32 – 22| = 5 = 5
    ∴ (3, 2) ∈ R
    If a = 3, b = 3 then |32 – 32| = 0 < 5
    ∴ (3, 3) ∈ R
    ∴ R = {(1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 2), (3, 3)}

    RS AGGARWAL CLASS 11 MATHS SOLUTION RELATIONS-2

    (i) R is reflexive if (a, a) ∈ R for all value of a ∈ Z.
    |a2 – a2| = 0 ≤ 5.
    Thus, it is reflexive.
    (ii) R is symmetric if (a, b) ∈ R ⇒ (b, a) ∈ R for all value of a, b ∈ Z.
    Let (a, b) ∈ R
    (a, b) ∈ R
    ⇒ |a2 – b2| ≤ 5
    ∴ |b2 – a2| ≤ 5
    ∴ (b, a) ∈ R
    Hence, it is symmetric

    (iii) R is transitive if (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R for all value of a, b, c ∈ Z.
    Putting a = 1 , b = 2 , c = 3.
    |12 – 22| = 3 ≤ 5 and
    |22 – 32| = 5 ≤ 5
    but |12 – 32| = 8 > 5
    Thus, it is not transitive.

    Q. 8. Let R = {(a, b) : a, b ∈ Z and b = 2a – 4}.
    If (a, –2} ∈ R and (4, b2) ∈ R. Then, write the values of a and b.

    Solution: b = 2a – 4……(ii)
    Putting  b = -2 in equation (i) ,
    -2 = 2a – 4
    ⇒ 2a = 3
    ⇒ a = 1
    Putting  a = 4 in equation (i)
    b = 2×4 – 4
    ⇒ b = 8 – 4 = 4
    Ans: Value of a = 1 , b = 4

    Q. 9. Let R be a relation on Z, defined by (x, y) ∈ R ⇔ x2 + y2 = 9.
    Then, write R as a set of ordered pairs.
    What is its domain?

    Solution: x2 + y2 = 9
    (x, y) ∈ R
    If x = 0 , y = ±3 , 02 + (±3)2 = 9
    If x = ±3 , y = 0 , (±3)2 + 02 = 9
    ∴ R = {(0, 3) , (0, -3) , (3, 0) , (-3, 0)}
    ∴ Dom(R) = {-3 , 0 , 3}

    Q. 10. Let A be the set of first five natural numbers and let R be a relation on A,
    defined by (x, y) ∈ R ⇔ x ≤ y.
    Express R and R–1 as sets of ordered pairs.
    Find: dom (R–1) and range (R).

    Solution:  A = {1, 2, 3, 4, 5}
    Since, x ≤ y
    ∴ R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5), (3, 3), (3, 4), (3, 5), (4, 4), (4, 5), (5, 5)}

    R–1 = {(1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (2, 2), (3, 2), (4, 2), (5, 2), (3, 3), (4, 3), (5, 3), (4, 4), (5, 4), (5, 5)}
    ∴ Dom(R–1) = {1, 2, 3, 4, 5}
    ∴ Range(R) = {1, 2, 3, 4, 5}

    Q. 11. Let R = (x, y) : x, y ∈ Z and x2 + y2 = 25}.
    Express R and R–1 as sets of ordered pairs.
    Show that R = R–1.

    Solution: x2 + y2 = 25
    (x, y) ∈ R
    If x = 0 , y = ±5 , 02 + (±5)2 = 25
    If x = ±3 , y = ±4 , (±3)2 + (±5)2 = 25
    x = ±4 , y = ±3 , (±4)2 + (±3)2 = 25
    If x = ±5 , y = 0 , (±5)2 + (0)2 = 25
    ∴ R = {(-5, 0), (-4, -3), (-4, 3), (-3, -4), (-3, 4), (0, -5), (0, 5), (3, -4), (3, 4), (4, -3), (4, 3), (5, 0)}
    ∴ R–1 = {(0, -5), (-3, -4), (3, -4), (-4, -3), (4, -3), (-5, 0), (5, 0), (-4, 3), (4, 3), (-3, 4), (3, 4), (0, 5)}
    Since, all ordered pair of R is in R–1.
    Hence, R = R–1 (Proved)

    RS AGGARWAL CLASS 11 MATHS SOLUTION RELATIONS-2
    Exercise 2D

    Q. 12. (i) Find R–1, when
    R = {(1, 2), (1, 3), (2, 3), (3, 2), (4, 5)}

    Solution: R = {(1, 2), (1, 3), (2, 3), (3, 2), (4, 5)}
    ∴ R–1 = {(2, 1), (3, 1), (3, 2), (2, 3), (5, 4)}

    Q. 12. (ii) Find R–1, when
    R = {(x, y) : x, y ∈ N, x + 2y = 8}.

    Solution: R = {(x, y) : x, y ∈ N, x + 2y = 8}.
    Put x = 2, y = 3
    ∴ x + 2y = 2 + 2.3 = 8
    Put x = 4, y = 2
    ∴ x + 2y = 2 + 2×3 = 8
    Put x = 6, y = 1
    ∴ x + 2y = 6 + 2×1 = 8
    ∴ R = {(2, 3), (4, 2), (6, 1)}
    R–1 = {(3, 2), (2, 4), (1, 6)}

    Q. 13. Let A = {a, b}. List all relation on A and find their number.

    Solution:  Any relation on A is a subset of A×A
    ∴ A×A = {(a, a), (a, b), (b, a), (b, b)}
    Number of relation on A is the number of subsets of A×A
    ∴ All relation on A are:
    {}, {(a, a)}, {(a, b)}, {(b, a)}, {(b, b)}, {(a, a), (a, b)}, {(a, a), (b, a)}, {(a, a), (b, b)}, {(a, b), (b, a)}, {(a, b), (b, b)}, {(b, a), (b, b)}, {(a, a), (a, b), (b, a)}, {(a, a), (b, a), (b, b)}, {(a, b), (b, a), (b, b)}, {(a, a), (a, b), (b, b)}, {(a, a), (a, b), (b, a), (b, b)}
    Thus, there are 16 total relations.

    Q. 14. Let R = {(a, b) : a, b, ε N and a < b}.
    Show that R is a binary relation on N, which is neither reflexive nor symmetric.
    Show that R is transitive.

    Solution: N is the set of all the natural numbers.
    ∴ N = {1, 2, 3, 4, 5, 6, 7…..}
    R = {(a, b) : a, b, ∈ N and a < b}
    R = {(1, 2), (1, 3), (1, 4) …. (2, 3), (2, 4), (2, 5) ……}
    For reflexivity,
    A relation R on N is said to be reflexive if (a, a) ∈ R for all value of a ∈ N.
    Here R = {(a, b) : a, b, ∈ N and a < b}
    Hence a < b,
    So the two co-ordinates of the ordered pair are never equal.
    Thus, the relation is not reflexive.

    For symmetry,
    A relation R on N is said to be symmetrical if (a, b) ∈ R ⇒ (b, a) ∈ R for all value of a, b ∈ N
    Here R = {(a, b) : a, b ∈ N and a < b}
    (a, b) ∈ R
    ⇒ a < b
    ⇒ b > a
    ∴ (b, a) ∉ R 
     Thus, it is not symmetric.
    For transitivity,
    A relation R on A is said to be transitive if (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R for all value of (a, b, c) ∈ N.
    Let’s take three values a, b and c such that a < b < c.
    So, (a, b) ∈ R and (b, c) ∈ R
    ⇒ (a,c) ∈ R.
    Thus, it is transitive.

    Exercise 2E

    Q. 1. (i) Let A and B be two sets such that n(A) = 5, n(B) = 3 and n(A ∩ B) = 2. n(A ∪ B)

    Solution: Given: n(A) = 5, n(B) = 3 and n(A ∩ B) = 2
    We know,
      n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
    ∴ n(A ∪ B) = 5 + 3 – 2 = 6

    Q. 1. (ii) Let A and B be two sets such that n(A) = 5, n(B) = 3 and n(A ∩ B) = 2.  n(A × B)

    Solution: Given: n(A) = 5, n(B) = 3 and n(A ∩ B) = 2
    We know, 
    n(A × B) = n(A) × n(B)
    ∴ n(A × B) = 5 × 3 = 15

    Q. 1. (iii) Let A and B be two sets such that n(A) = 5, n(B) = 3 and n(A ∩ B) = 2. n(A × B) ∩ (B × A)

    Solution: Given: n(A) = 5, n(B) = 3 and n(A ∩ B) = 2
    ∴ n(A × B) ∩ (B × A)
    = 2n(A ∩ B)
    = 22 = 4

    Q. 2 Find a and b when (a – 2b, 13) = (7, 2a – 3b)

    Solution: Since the ordered pairs are equal,
    Comparing corresponding elements,
    ∴ a – 2b = 7 …… (ii)  and
    2a – 3b = 13 …… (ii)
    (i)×2 – (ii),
    2a – 4b – 2a + 3b = 14 – 13
    ⇒ -b = 1
    ⇒ b = -1
    Putting b = -1 in equation (i), 
    a – 2(-1) = 7
    ⇒ a = 7 – 2
    ⇒ a = 5
    Hence a = 5; b = -1

    Q. 3. If A = {1, 2}, find A × A × A

    Solution: Given: A = {1, 2}
    ∴ A × A
    = {1, 2} × {1, 2}
    = {(1, 1), (1, 2), (2, 1), (2, 2)}

    ∴ A × A × A
    = {1, 2} × {(1, 1), (1, 2), (2, 1), (2, 2)}
    = {(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2), (2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)}

    Q. 4. If A = {2, 3, 4} and B = {4, 5}, draw an arrow diagram represent (A × B}.

    Solution: Given: A = {2, 3, 4} and B = {4, 5},
    Arrow Diagram:

    ARROW DIAGRAM:

    B        A> > > > > > 3 2            4 4            5

    Q. 5. If A = {3, 4}, B = {4, 5} and C = {5, 6}, find A × (B × C).

    Solution: Given: A = {3, 4}, B = {4, 5} and C = {5, 6}
    B × C
    = {4, 5} × {5, 6}
    = {(4, 5), (4, 6), (5, 5), (5, 6)}

    ∴ A × (B × C)
    = {3, 4} × {(4, 5), (4, 6), (5, 5), (5, 6)}
    = {(3, 4, 5), (3, 4, 6), (3, 5, 5), (3, 5, 6), (4, 4, 5), (4, 4, 6), (4, 5, 5), (4, 5, 6)}

    Q. 6. If A ⊆ B, prove that A × C = B × C

    Solution: Given: A ⊆ B
    Let any arbitrary element (a, c) ∈ A × C
    (a, c) ∈ A × C
    ⇒ a ∈ A and c ∈ C
    ⇒ a ∈ B and c ∈ C …… [∵ A ⊆ B]
    ∴ (a, c) ∈ B × C
    ∴ A × C = B × C [Proved]

    Q. 7. Prove that A × B = B × A ⇒ A = B.

    Solution: Let A and B be any two sets such thatA × B = {(a, b): a ∈ A, b ∈ B}
    Now, B × A = {(b, a): b ∈ A, a ∈ B}
    Since A × B = B × A
    ∴ (a, b) = (b, a)
    This is possible only when the ordered pairs are equal.
    Therefore,a = b and b = a
    Hence  A = B. (Proved)

    Q. 8. If A = {5} and B = {5, 6}, write down all possible subsets of A × B.

    Solution: Given: A = {5}, B = {5, 6}
    ∴ A × B
    = {(5, 5), (5, 6)}
    ∴ All  possible subsets of A × B are:
    {}, {(5, 5)}, {(5, 6)}, {(5, 6), (5, 6)}

    Q. 9. Let R = {(x, x2): x is a prime number less than 10}.
    (i) Write R in roster form.
    (ii) Find dom (R) and range (R).

    Solution: (i) {(x, x2): x is a prime number less than 10}.
    Prime number less than 10 are 2, 3, 5 and 7
    ∴ Roster form:
    R = {(2, 4), (3, 9), (5, 25), (7, 49)}
    (ii) Domain of R is the set of first coordinates of R
    ∴ Dom(R)
    = {2, 3, 5, 7}
    Range of R is the set of second coordinates of R
    Range(R) = {4, 9, 25, 49}

    Q. 10. Let A = (1, 2, 3} and B = {4}
    How many relations can be defined from A to B.

    Solution: Given: A = (1, 2, 3} and B = {4}
    ∴ n(A) = 3 and n(B) = 1
    ∴ n(A×B)
    = n(A) × n(B)
    = 3 × 1= 3
    The number of relations from set A to set B
    = 2n(A×B)
    = 23 = 8
    Total number of relations = 8

    Q. 11. Let A = {3, 4, 5, 6} and R = {(a, b) : a, b ∈ A and a < b}
    (i) Write R in roster form.
    (ii) Find: dom (R) and range (R)
    (iii) Write R–1 in roster form.

    Solution: (i) Given: A = {3, 4, 5, 6} and R = {(a, b) : a, b ∈ A and a > b}
    ∴ Roster form: R = {(4, 3), (5, 3), (5, 4), (6, 3), (6, 4), (6, 5)}
    (ii) Domain of R is the set of first coordinates of R
    Dom(R) = {4, 5, 6}
    Range of R is the set of second coordinates of R
    Range(R) = {3, 4, 5}
    (iii) R–1 = = {(3, 4), (3, 5), (4, 5), (3, 6), (4, 6), (5, 6)}

    Q. 12. Let R = {(a, b) : a, b ∈ N and a < b}.
    Show that R is a binary relation on N which is neither reflexive nor symmetric.
    Show that R is transitive.

    Solution:  A binary relation R on sets A and B is a subset of the Cartesian product A×B
    Given: R = {(a, b) : a, b ∈ N and a < b}
    Since a, b ∈ N
    So the ordered pair of (a, b) is an element of the Cartesian product N×N
    ∴ R is a subset of the Cartesian product N×N
    Therefore R is a binary relation on N. (Proved)

    For reflexivity,
    A relation R on N is said to be reflexive if (a, a) ∈ R for all value of a ∈ N.
    Here R = {(a, b) : a, b, ∈ N and a < b}
    Hence a < b, so the two co-ordinates of the ordered pair are never equal.
    Thus, the relation is not reflexive.
    For symmetry,
    A relation R on N is said to be symmetrical if (a, b) ∈ R ⇒ (b, a) ∈ R
    Here R = {(a, b) : a, b ∈ N and a < b}
    (a, b) ∈ R
    ∴ a < b
    ⇒ b > a
    ∴ (b, a) ∉ R 
    Thus, it is not symmetric.

    For transitivity,
    A relation R on A is said to be transitive if (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R for all value of (a, b, c) ∈ N.
    Let’s take three values a, b and c such that a < b < c.
    So, (a, b) ∈ R and (b, c) ∈ R
    ⇒ (a,c) ∈ R.
    Thus, it is transitive.

    RELATIONS 2A, 2B, 2C
  • RS AGGARWAL CLASS 11 MATHS SOLUTION RELATIONS-1

    RS AGGARWAL CLASS 11 MATHS SOLUTION RELATIONS-1

    RS AGGARWAL CLASS 11 MATHS SOLUTION RELATIONS

    RELATIONS Exercise 2A, 2B, 2C

    Exercise 2A

    1. (i) Find the values of a and b,
             when: (a + 3, b –2) = (5, 1)

    Answer: Since the ordered pairs are equal,
    Comparing corresponding elements,
     ∴ a + 3 = 5
    ⇒ a = 5 – 3 = 2 and
        b – 2 = 1
    ⇒ b = 1 + 2 = 3
    ∴ a = 2 and b = 3.

    1. (ii) Find the values of a and b,
             when: (a + b, 2b – 3) = (4, –5)

    Answer: Since the ordered pairs are equal.
    Comparing corresponding elements,
     ∴ a + b = 4 …(i) and
    2b – 3 = -5 …(ii)
    From (ii) we get,
        2b – 3 = -5
    ⇒ 2b = -5 + 3
    ⇒ 2b = -2
    ∴ b = -1

    Putting the value of b = – 1 in equation (i),
     we get
      ∴ a – 1 = 4
    ⇒ a – 1 = 4
    ⇒ a = 4 + 1 = 5
    ∴ a = 5 and b =- 1.

    Q. 1. (iii) Find the values of a and b,
             when:  (a/3 + 1, b –  1/3) = (5/3, 2/3)

    Answer: Since the ordered pairs are equal.
    Comparing corresponding elements,
     ∴ a/3+ 1 = 5/3⇒ a/3= 5/3– 1
    ⇒ a/3 = 2/3⇒ a = 2 and
        b –  1/3  = 2/3
    ⇒ b  = 2/3 + 1/3
    ⇒ b = 1
     ∴ a = 2 and b = 1.

    Q. 1. (iv) Find the values of a and b,
             when: (a – 2, 2b + 1 = (b – 1, a + 2)

    Answer: Since the ordered pairs are equal,
    Comparing corresponding elements,
     ∴ a – 2 = b – 1 …(i) and
     2b + 1 = a + 2 …(ii)
    From equation (i) we get
        a – 2 = b – 1
    ⇒ a – b = (-1) + 2
    ⇒ a – b = 1 … (iii)
    From equation (ii) we get
        2b + 1 = a + 2
    ⇒ 2b – a = 2 – 1
    ⇒ -a + 2b = 1 …(iv)

    Add (iii) and (iv),
        a – b + (-a) + 2b = 1 + 1
    ⇒ a – b – a + 2b = 2
    ⇒ b = 2
      Putting b = 2 in equation (iii), we get
        a – 2 = 1
    ⇒ a = 1 + 2
    ⇒ a = 3
    ∴  a = 3 and b = 2.

    Q. 2. If A = {0, 1} and B = {1, 2, 3},
             show that A × B ≠ B × A.

    Answer: Given: A = {0, 1} and B = {1, 2, 3}
    ∵ A × B = {(x, y): x ∈ A, y ∈ B}
    ∵ A × B = {(x, y): x ∈ A, y ∈ B} ∵ A × B = {(x, y): x ∈ A, y ∈ B}

     ∴ A × B
    = {0, 1} × {1, 2, 3}
    = {(0, 1), (0, 2), (0, 3), (1, 1), (1, 2), (1, 3)}
    B × A
    = (1, 2, 3) × (0, 1)
    = {(1, 0), (1, 1), (2, 0), (2, 1), (3, 0), (3, 1)}
    Here (0, 1) ∈ A × B but (0, 1) ∉ B × A
    ∴ A × B ≠ B × A (Proved)

    Q. 3. If P = {a, b} and Q = {x, y, z},
            show that P × Q ≠ Q × P.

    Answer: Given: P = {a, b} and Q = {x, y, z}
    ∵ P × Q = {(p, q) : p ∈ P and q ∈ Q}
     ∴ P × Q
    = (a, b) × (x, y, z)
    = {(a, x), (a, y), (a, z), (b, x), (b, y), (b, z)}
        Q × P
    = (x, y, z) × (a, b)
    = {(x, a), (x, b), (y, a), (y, b), (z, a), (z, b)}
    Here (a, x) ∈ P × Q but (a, x) ∉ Q × P
    ∴ P × Q ≠ Q × P (Proved)

    Q. 4. (i) If A = {2, 3, 5} and B = {5, 7}, find: A × B

    Answer: Given: A = {2, 3, 5} and B = {5, 7}
    ∵ A × B = {(x, y): x ∈ A, y ∈ B}
    ∴ A × B
    = {2, 3, 5} × {5, 7}
    = {(2, 5),  (2, 7), (3, 5), (3, 7), (5, 5), (5, 7)} (Ans)

    Q. 4. (ii) If A = {2, 3, 5} and B = {5, 7}, find: B × A

    Answer: Given: A = {2, 3, 5} and B = {5, 7}
    ∵ A × B = {(x, y): x ∈ A, y ∈ B}
    ∴ B × A
    = {5, 7} × {2, 3, 5}
    = {(5, 2), (5, 3), (5, 5), (7, 2), (7, 3), (7, 5)} (Ans)

    Q. 4. (iii) If A = {2, 3, 5} and B = {5, 7}, find: A × A

    Answer: Given: A = {2, 3, 5}∵ A × B = {(x, y): x ∈ A, y ∈ B}
    ∴  A × A
    = (2, 3, 5) × (2, 3, 5)
    = {(2, 2), (2, 3), (2, 5), (3, 2), (3, 3), (3, 5), (5, 2), (5, 3), (5, 5)} (Ans)

    Q. 4. (iv) If A = {2, 3, 5} and B = {5, 7}, find: B × B

    Answer: Given: B = {5, 7} ∵ A × B = {(x, y): x ∈ A, y ∈ B}
    ∴ B × B
    = (5, 7) × (5, 7)
    = {(5, 5), (5, 7), (7, 5), (7, 7)} (Ans)

    Q. 5. If A = {x ∈ N : x ≤ 3} and {x v W : x < 2}, find (A × B) and (B × A).
             Is (A × B) = (B × A)?

    Answer: Given:
    A = {x ∈ N: x ≤ 3}
     ∴ A = {1, 2, 3} and 
    B = {x ∈ W: x < 2}
     ∴ B = {0, 1}
    ∵ A × B = {(x, y): x ∈ A, y ∈ B}
    ∴ A × B = {1, 2, 3} × {0, 1}
    = {(1, 0), (1, 1), (2, 0), (2, 1), (3, 0), (3, 2)} 
       B × A
    = {0, 1} × {1, 2, 3}
    = {(0, 1), (0, 2), (0, 3), (1, 1), (1, 2), (1, 3)}
    Here (0, 1) ∈ A × B but (0, 1) ∉ B × A
    ∴ A × B ≠ B × A

    Q. 6. (i) If A = {1, 3, 5) B = {3, 4} and C = {2, 3},
             verify that: A × (B ∪ C) = (A × B) ∪ (A × C)

    Answer:     Given: A = {1, 3, 5}, B = {3, 4} and C = {2, 3}
       B ∪ C = {2, 3, 4} 
    ∵ A × B = {(x, y): x ∈ A, y ∈ B}
    L. H. S.
    = A × (B ∪ C)
    ⇒  {1, 3, 5} ×  {2, 3, 4}
    = {(1, 2), (1, 3), (1, 4), (3, 2), (3, 3), (3, 4), (5, 2), (5, 3), (5, 4)}

    A × B 
    = {1, 3, 5} × {3, 4}
    = {(1, 3), (1, 4), (3, 3), (3, 4), (5, 3), (5, 4)} and 
    A × C = {1, 3, 5} × {2, 3}
    = {(1, 2), (1, 3), (3, 2), (3, 3), (5, 2), (5, 3)} 
    R. H. S.
    (A × B) ⋃ (A × C)
    = {(1, 3), (1, 4), (3, 3), (3, 4), (5, 3), (5, 4)} ⋃ {(1, 2), (1, 3), (3, 2), (3, 3), (5, 2), (5, 3)}
    = {(1, 2), (1, 3), (1, 4), (3, 2), (3, 3), (3, 4), (5, 2), (5, 3), (5, 4)}
    ⇒ L. H. S.
    ∴ A × (B ∪ C) = (A × B) ∪ (A × C)   (verified)

    Q. 6. (ii) If A = {1, 3, 5) B = {3, 4} and C = {2, 3},
             verify that: A × (B ∩ C) = (A × B) ∩ (A × C)

    Answer:     Given: A = {1, 3, 5}, B = {3, 4} and C = {2, 3} 
    (B ⋂ C) = {3}
    L. H. S.
    = A × (B ⋂ C)
    ⇒ {1, 3, 5} × {3}
    = {(1, 3), (3, 3), (5, 3)}
    Now, 
    A × B = {1, 3, 5} × {3, 4}
    = {(1, 3), (1, 4), (3, 3), (3, 4), (5, 3), (5, 4)} and 
    A × C = {1, 3, 5} × {2, 3}
    = {(1, 2), (1, 3), (3, 2), (3, 3), (5, 2), (5, 3)}
    R. H. S.
    = (A × B) ⋂ (A × C)
    ⇒ {(1, 3), (3, 3), (5, 3)}
    = L. H. S.
    ∴ A × (B ∩ C) = (A × B) ∩ (A × C) (verified)

    Q. 7. (i) Let A = {x ∈ W : x < 2}, B = {x ∈ N : 1 < x ≤ 4} and C = {3, 5}.
             Verify that: A × (B ∪ C) = (A × B) ∪ (A × C)

    Answer: (i) Given:
    A = {x ∈ W : x < 2}
    ∴ A = {0, 1}
    B = {x ∈ N : 1 < x ≤ 4}
    ∴ B = {2, 3, 4}
    and C = {3, 5}
    B ⋃ C = {2, 3, 4, 5}

    L. H. S.
    = A × (B ⋃ C)
    ⇒ {0, 1} × {2, 3, 4, 5}
    = {(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)}
    Now, A × B
         = {0, 1} × {2, 3, 4}
         = {(0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4)}
    and A × C
         = {0, 1} × {3, 5}
         = {(0, 3), (0, 5), (1, 3), (1, 5)}
    R. H. S.
    = (A × B) ⋃ (A × C)
    = {(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)} = L. H. S.
    ∴ A × (B ∪ C) = (A × B) ∪ (A × C) (verified)

    Q. 7. (ii) Let A = {x ∈ W : x < 2}, B = {x ∈ N : 1 < x ≤ 4} and C = {3, 5}.
             Verify that: A × (B ∩ C) = (A × B) ∩ (A × C)

    Answer: Given:
    A = {x ∈ W : x < 2}
    ∴ A = {0, 1}
    B = {x ∈ N : 1 < x ≤ 4}
    ∴ B = {2, 3, 4}
    and C = {3, 5}
    B ⋂ C = {3}
    L. H. S.
    = A × (B ⋂ C)
    ⇒ {0, 1} × {3}
    = {(0, 3), (1, 3)}
    Now, A × B
          = {0, 1} × {2, 3, 4}
          = {(0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4)}
    and A × C
        = {0, 1} × {3, 5}
        = {(0, 3), (0, 5), (1, 3), (1, 5)}
    R. H. S.
    = (A × B) ⋂ (A × C)
    = {(0, 3), (1, 3)} = L. H. S.
    ∴ A × (B ∩ C) = (A × B) ∩ (A × C) (verified)


    Q. 8. If A × B = {(–2, 3), (–2, 4), (0, 3), (0, 4), (3, 3), (3, 4), find A and B.

    Answer: Here, A × B = {(–2, 3), (–2, 4), (0, 3), (0, 4), (3, 3), (3, 4)} 
    ∵ A × B = {(x, y): x ∈ A, y ∈ B}
    Clearly, A is the set of all first entries in ordered pairs in A × B
    ∴ A = {-2, 0, 3}
    and B is the set of all second entries in ordered pairs in A × B
    ∴ B = {3, 4}

    Q. 9. Let A = {2, 3} and B = {4, 5}. Find (A × B).
             How many subsets will (A × B) have?

    Answer: Given: A = {2, 3} and B = {4, 5}
    ∵ A × B = {(x, y): x ∈ A, y ∈ B}
       A × B
    = {2, 3} × {4, 5}
    = {(2, 4), (2, 5), (3, 4), (3, 5)}
    Number of elements of A× B
     = n(A×B) = 4
    ∴ Number of subsets of A × B
     = 2n = 24 = 16
    ∴ A × B has 16 subsets.

    Q. 10. Let A × B = {(a, b): b = 3a – 2}. if (x, –5) and (2, y) belong to A × B,
               find the values of x and y.

    Answer: Given: A × B = {(a, b): b = 3a – 2} and ((x, -5), (2, y)) ∈ A × B
    ∵ b = 3a – 2
    For (x, -5) ∈ A × B
        -5 = 3.x – 2
    ⇒ 3x =  -5 + 2
    ⇒ 3x = -3
    ∴ x = -1
    For (2, y) ∈ A × B
       2 = a and
        y = b
    ⇒ y = 3a – 2
    ⇒ y = 3.2 – 2
    ∴ y = 4
    ∴ x = -1 and y = 4

    আমাদের YOUTUBE CHANNEL “COMPTECH” দেখার জন্য এখানে ক্লিক করো। 

    Q. 11. Let A and B be two sets such that n(A) = 3 and n(B) = 2. If a ≠ b ≠ c and (a, 0), (b, 1), (c, 0) is in A × B, find A and B.

    Answer:  (a, 0), (b, 1), (c, 0) are the elements of A × B.
    ∴ a, b, c ∈ A and 0, 1 ∈ B
    Given: n(A) = 3 and n(B) = 2 
    ∴ A = {a, b, c} and 
       B = {0, 1}

    Q. 12. (i) Let A = {–2, 2} and B = {0, 3, 5}. Find: A × B
    Represent each of the above (a) graphically and (b) by arrow diagram.

    Answer: Given: A = {-2, 2} and B = {0, 3, 5}
    ∵ A × B = {(x, y): x ∈ A, y ∈ B}
    A × B
    = {–2, 2} × {0, 3, 5}
    = {(-2, 0), (-2, 3), (-2, 5), (2, 0), (2, 3), (2, 5)}

    GRAPH:

    (-2,5)        (2,5) (-2,3)        (2,3) (-2,0)        (2,0) X Y5 4 3 2 1 -1-2   -1   0     1   2   3   4

    ARROW DIAGRAM:

    A        B> > > > > > -2            0 3 2            5

    Q. 12. (ii) Let A = {–2, 2} and B = {0, 3, 5}. Find: B × A
    Represent each of the above (a) graphically and (b) by arrow diagram.

    Answer: Given: A = {-2, 2} and B = {0, 3, 5}
    ∵ A × B = {(x, y): x ∈ A, y ∈ B}
    B × A
    = {0, 3, 5} × {–2, 2}
    = {(0, -2), (0, 2), (3, -2), (3, 2), (5, -2), (5, 2)}

    GRAPH:

    (0,2)        (3,2)     (5,2) (0,-2)        (3,-2)     (5,-2) X Y4 3 2 1 -1 -2-1   0   1     2   3   4   5

    ARROW DIAGRAM:

    B        A> > > > > > 3 0            -2 5            2

    Q. 12. (iii) Let A = {–2, 2} and B = {0, 3, 5}. Find: A × A
    Represent each of the above (a) graphically and (b) by arrow diagram.

    Answer: Given: A = {-2, 2}
    ∵ A × B = {(x, y): x ∈ A, y ∈ B}
    A × A
    = {-2, 2} × {-2, 2}
    = {(-2, -2), (-2, 2), (2, -2), (2, 2)} 

    GRAPH:

    \\2,2 (2,2)\\2,-2 (2,-2)\\-2,2 (-2,2)\\-2-2 (-2,-2)X Y 3    2  1    -1   -2 -3  -2   -1     0   1   2   3

    ARROW DIAGRAM:

    A        A>>> > -2            -2 2            2

    Q. 12. (iv) Let A = {–2, 2} and B = {0, 3, 5}. Find: B × B
    Represent each of the above (a) graphically and (b) by arrow diagram.

    Answer: Given: B = {0, 3, 5} ∵ A × B = {(x, y): x ∈ A, y ∈ B}
    B × B
    = {0, 3, 5} × {0, 3, 5}
    = {(0, 0), (0, 3), (0, 5), (3, 0), (3, 3), (3, 5), (5, 0), (5, 3), (5, 5)}

    GRAPH:

    (0,5)        (3,5)     (5,5) (0,3)        (3,3)     (5,3) (0,0)        (3,0)     (5,0) X Y5 4 3 2 1 -1-1   0   1     2   3   4   5

    ARROW DIAGRAM:

    B        B> > > > > > > > > 3            3 0            0 5            5

    Q. 13. (i) If A = {5, 7), find: A × A

    Answer: Given: A = {5, 7}
    ∵ A × B = {(x, y): x ∈ A, y ∈ B}
    ∴ A × A
    =  {5, 7}× {5, 7}
    = {(5, 5), (5, 7), (7, 5), (7. 7)}

    Q. 13. (ii) If A = {5, 7), find: A × A × A.

    Answer: Given: A = {5, 7}
    ∵ A × B × C = {(x, y, z): x ∈ A, y ∈ B, z ∈ C}
    ∴ A × A × A
    =  {5, 7}× {5, 7}× {5, 7}
    ={(5, 5, 5), (5, 5, 7), (5, 7, 5), (5, 7, 7), (7, 5, 5), (7, 5, 7), (7, 7, 5), (7, 7, 7)}

    RS AGGARWAL CLASS 11 MATHS SOLUTION RELATIONS

    Q. 14. (i) Let A = {–3, –1}, B = {1, 3) and C = {3, 5). Find: A × B

    Answer: Given: A = {-3, -1} and B = {1, 3}
    ∵ A × B = {(x, y): x ∈ A, y ∈ B}
     A × B
    = {-3, -1} × {1, 3}
    = {(-3, 1), (-3, 3), (-1, 1), (-1, 3)}

    Q. 14. (ii) Let A = {–3, –1}, B = {1, 3) and C = {3, 5). Find: (A × B) × C

    Answer: Given: A = {-3, -1} and B = {1, 3} and C = {3, 5}
    ∵ A × B = {(x, y): x ∈ A, y ∈ B}
    (A × B) × C
    = [{-3, -1} × {1, 3}] × {3, 5}
    ⇒ {(-3, 1), (-3, 3), (-1, 1), (-1, 3)} × (3, 5)
    = (-3, 1, 3), (-3, 1 , 5), (-3, 3, 3), (-3, 3, 5), (-1, 1, 3), (-1, 1, 5), (-1, 3, 3), (-1, 3, 5)}

    Q. 14. (iii) Let A = {–3, –1}, B = {1, 3) and C = {3, 5). Find: B × C

    Answer: Given: B = {1, 3} and C = {3, 5}
    ∵ A × B
    = {(x, y): x ∈ A, y ∈ B}
    B × C
    = (1, 3) × (3, 5)
    = {(1, 3), (1, 5), (3, 3), (3, 5)}

    Q. 14. (iv) Let A = {–3, –1}, B = {1, 3) and C = {3, 5). Find: A × (B × C)

    Answer: Given: A = {-3, -1}, B = {1, 3} and C = {3, 5}
    ∵ A × B = {(x, y): x ∈ A, y ∈ B}
     B × C
    =  {1, 3} × {3, 5}
    = {(1, 3), (1, 5), (3, 3), (3, 5)} 
    A × (B × C)
    = {-3, -1} × {(1, 3), (1, 5), (3, 3), (3, 5)}
    = (-3, 1, 3), (-3, 1, 5), (-3, 3, 3), (-3, 3, 5), (-1, 1, 3), (-1, 1, 5), (-1, 3, 3), (-1, 3, 5)}

    RS AGGARWAL CLASS 11 MATHS SOLUTION RELATIONS

    Exercise 2B

    Q. 1 (i) For any sets A, B and C prove that:
    A × (B ∪ C) = (A × B) ∪ (A × C)

    Answer: Let us consider, (x, y) ∈ A × (B ∪ C)
    ⇒ x ∈ A and y ∈ (B ∪ C)
    ⇒ x ∈ A and (y ∈ B or y ∈ C)
    = (x ∈ A and y ∈ B) or (x ∈ A and y ∈ C)
    ⇒ (x, y) ∈ (A × B) or (x, y) ∈ (A × C)
    ⇒ (x, y) ∈ (A × B) ∪ (A × C)
    ∴ A × (B ∪ C) ⊆ (A × B) ∪ (A × C) —- (i)

    Let us consider again (a, b) ∈ (A × B) ∪ (A × C)
    ⇒ (a, b) ∈ (A × B) or (a, b) ∈ (A × C)
    ⇒ (a ∈ A and b ∈ B) or (a ∈ A and b ∈ C)
    = a ∈ A and (b ∈ B or b ∈ C)
    ⇒ a ∈ A and b ∈ (B ∪ C)
    ⇒ (a, b) ∈ A × (B ∪ C)
    ∴ (A × B) ∪ (A × C) ⊆ A × (B ∪ C) —- (ii)
    From (i) and (ii) we can conclude that,
    A × (B ∪ C) = (A × B) ∪ (A × C) (Proved)

    Q. 1 (ii) For any sets A, B and C prove that:
    A × (B ∩ C) = (A × B) ∩ (A × C)

    Answer: Let us consider, (x, y) ∈ A × (B ∩ C)
    ⇒ x ∈ A and y ∈ (B ∩ C)
    ⇒ x ∈ A and (y ∈ B and y ∈ C)
    = (x ∈ A and y ∈ B) and (x ∈ A and y ∈ C)
    ⇒ (x, y) ∈ (A × B) and (x, y) ∈ (A × C)
    ⇒ (x, y) ∈ (A × B) ∩ (A × C)
    ∴ A × (B ∩ C) ⊆ (A × B) ∩ (A × C) —- (i)

    Let us consider again,
    (a, b)∈ (A × B) ∩ (A × C)
    ⇒ (a, b) ∈ (A × B) and (a, b) ∈ (A × C)
    ⇒ (a ∈ A and b ∈ B) and (a ∈ A and b ∈ C)
    = a ∈ A and (b ∈ B and b ∈ C)
    ⇒ a ∈ A and b ∈ (B ∩ C)
    ⇒ (a, b) ∈ A × (B ∩ C)
    ∴ (A × B) ∩ (A × C) ⊆ A × (B ∩ C) —- (ii)
    From (i) and (ii) we can conclude that,
    A × (B ∩ C) = (A × B) ∩ (A × C) (Proved)

    Q. 1 (iii) For any sets A, B and C prove that:
    A × (B – C) = (A × B) – (A × C)

    Answer: Let us consider, (x, y) ∈ A × (B – C)
    ⇒ x ∈ A and y ∈ (B – C )
    ⇒ x ∈ A and (y ∈ B and y ∉ C)
    = (x ∈ A and y ∈ B) and (x ∈ A and y ∉ C)
    ⇒ (x, y) ∈ (A × B) and (x, y) ∉ (A × C)
    ⇒ (x, y) ∈ [(A × B) – (A × C)]
    ∴  A × (B – C) ⊆ (A × B) – (A × C) —- (i)

    Let us consider again, (a, b) ∈ [(A × B) – (A × C)]
    ⇒ (a, b) ∈ (A × B) and (a, b) ∉ (A × C)
    ⇒ (a ∈ A and b ∈ B) and (a ∈ A and b ∉ C)
    = a ∈ A and (b ∈ B and b ∉ C)
    ⇒ a ∈ A and b ∈ (B – C)
    ⇒ (a, b) ∈ A × (B ∪ C)
    ∴ (A × B) – (A × C) ⊆ A × (B – C) —- (ii)
    From (i) and (ii) we can conclude that,
    A × (B – C) = (A × B) – (A × C) (Proved)

    Q. 2. For any sets A and B, prove that
    (A × B) ∩ (B × A) = (A ∩ B) × (B ∩ A).

    Answer: Let us consider, (x, y) ∈ (A × B) ∩ (B × A)
    ⇒ (x, y) ∈ (A × B) and (x, y) ∈ (B × A)
    ⇒ (x ∈ A and y ∈ B) and (x ∈ B and y ∈ A)
    = (x ∈ A and x ∈ B) and (y ∈ B and y ∈ A)
    ⇒ x ∈ (A × B) and y ∈ (B × A)
    ⇒ (x, y) ∈ (A × B) ∩ (B × A) 
    ∴ (A × B) ∩ (B × A) ⊆ (A ∩ B) × (B ∩ A) —- (i)

    Let us consider again, (a, b) ∈ (A ∩ B) × (B ∩ A)
    ⇒ a ∈ (A ∩ B) and b ∈ (B ∩ A)
    ⇒ (a ∈ A and a ∈ B) and (b ∈ B and b ∈ A)
    = (a ∈ A and b ∈ B) and (a ∈ B and b ∈ A)
    ⇒ (a, b) ∈ (A × B) and (a, b) ∈  (B × A)
    ⇒ (a, b) ∈ (A × B) ∩ (B × A)
    ∴ (A ∩ B) × (B ∩ A) ⊆ (A × B) ∩ (B × A) —- (ii)
    From (i) and (ii) we can conclude that,
    (A × B) ∩ (B × A) = (A ∩ B) × (B ∩ A) (Proved)

    Q. 3. If A and B are nonempty sets, prove that
    A × B = B × A ⇔ A = B

    Answer: Given: A = B, where A and B are nonempty sets.
    Let us consider, (x, y) ∈ (A × B)
    ⇒ x ∈ A and y ∈ B
    ⇒ x ∈ B and y ∈ A….. [A = B]
    = (x, y) ∈ (B × A)
    ∴ (A × B) ⊆ (B × A) —- (i)
    Let us consider again,
    (x, y) ∈ (B × A)
    ⇒ x ∈ B and y ∈ A
    ⇒ x ∈ A and y ∈ B….. [A = B]
    = (x, y) ∈ (A × B)
    ⇒ (B × A) ⊆ (A × B) —- (ii)
    From (i) and (ii), we can conclude that, A × B = B × A (Proved)

    Q. 4. (i) If A ⊆ B, prove that A × C ⊆ B × C for any set C.

    Answer: Given: A ⊆ B
    Let us consider,
    (x, y) ∈ (A × C)
    ⇒ x ∈ A and y ∈ C
    = x ∈ B and y ∈ C ….. [A ⊆ B]
    ⇒ (x, y) (B × C)
    ∴ A × C ⊆ B × C (Proved)

    Q. 4. (ii) If A ⊆ B and C ⊆ D then prove that A × C ⊆ B × D.

    Answer: Given: A ⊆ B and C ⊆ D
    Let us consider,
    (x, y) ∈ (A × C)
    ⇒ x ∈ A and y ∈ C
    = x ∈ B and y ∈ D ….. [A ⊆ B and C ⊆ D]
    ⇒ (x, y) ∈ (B × D)
    ∴ A × C ⊆ B × D (Proved)

    Q. 5. If A × B ⊆ C × D and A × B ≠ φ, prove that A ⊆ C and B ⊆ D.

    Answer: Given: A × B ⊆ C × D and A × B ≠ φ
    Let us consider,
    (x, y) ∈ (A × B)
    ⇒ x ∈ A and y ∈ B —- (i)
    Again
     (x, y) ∈ (A × B)
    ⇒ (x, y) ∈ (C × D) …….. [ A × B ⊆ C × D]
    ⇒x ∈ C and y ∈ D —- (ii)
    Comparing (i) and (ii) we can say that,
    ⇒ x ∈ A and x ∈ C
    ⇒ A ⊆ C
    Again,
    ⇒ y ∈ B and y ∈ D
    ⇒ B ⊆ D (Proved)

    RS AGGARWAL CLASS 11 MATHS SOLUTION RELATIONS

    Q. 6. (i) If A and B be two sets such that n(A) = 3, n(B) = 4 and n(A ∩ B) = 2 then find. n(A × B)

    Answer: Given: n(A) = 3, n(B) = 4 and n(A ∩ B) = 2
    ∵  n(A × B) = n(A) × n(B)
    ⇒ n(A × B) = 3 × 4
    ⇒ n(A × B) = 12

    Q. 6. (i) If A and B be two sets such that n(A) = 3, n(B) = 4 and n(A ∩ B) = 2 then find. n(B × A)

    Answer: Given: n(A) = 3, n(B) = 4 and n(A ∩ B) = 2
    ∵ n(B × A) = n(B) × n(A)
    ⇒ n(B × A) = 4 × 3
    ⇒ n(B × A) = 12

    Q. 6. (i) If A and B be two sets such that n(A) = 3, n(B) = 4 and n(A ∩ B) = 2 then find. n(A × B) ∩ (B × A)

    Answer: Given: n(A) = 3, n(B) = 4 and n(A ∩ B) = 2
       n((A × B) ∩ (B × A))
    = n[(A ∩ B) × (B ∩ A)]
    = n(A ∩ B) × n(B ∩ A)
    ⇒ 2×2 = 4

    Q. 7. For any two sets A and B, show that A × B and B × A have an element in common if and only if A and B have an element in common.

    Answer: Let us consider A ∩ B ≠ φ
    Then  (A ∩ B) ∩ (B ∩ A)  ≠ φ
    Let x be an element of A ∩ B
    ⇒ x ∈ A and x ∈ B
    ⇒ (x, x) ∈ A × B
    Similarly
        x ∈ B and x ∈ A
    ⇒ (x, x) ∈ B × A
    Therefore
      (x, x) ∈ A × B ∩ B ∩ A
    ⇒ (A × B) ∩ (B × A)  ≠ φ
    That means A × B and B × A have an element in common if and only if A and B have an element in common. (Proved)

    Q. 8. Let A = {1, 2} and B = {2, 3}. Then, write down all possible subsets of A × B.

    Answer: Given: A = {1, 2} and B = {2, 3}
    (A × B) = {(x, y): x ∈ A and y ∈ B}
    ∴   (A × B)
    = {1, 2} × {2, 3}
    = {(1, 2), (1, 3), (2,2), (2,3)}
    So, all the possible subsets of A × B are:
    φ, {(1, 2)}, {(1, 3)}, {(2,2)}, {(2,3)}, {(1, 2), (1, 3)}, {(1, 2), (2,2)}, {(1, 2), (2,3)}, {(1, 3), (2,2)}, {(1, 3), (2,3)}, {(2,2), (2,3)}, {(1, 2), (1, 3), (2,2)}, {(1, 2), (1, 3), (2,3)}, {(1, 3), (2,2), (2,3)}, {(2,2), (2,3), (1, 2)}, {(1, 2), (1, 3), (2,2), (2,3)}

    Q. 9. (i) Let A = {a, b, c, d}, B = {c, d, e} and C = {d, e, f, g}. Then verify each of the following identities:
    A × (B ∩ C) = (A × B) ∩ (A × C)

    Answer: Given: A = {a, b, c, d}, B = {c, d, e} and C = {d, e, f, g}
       (B ∩ C) = {d, e}
    L.H.S.
    = A × (B ∩ C)
    ⇒ {a, b, c, d} × {d, e}
    = {(a, d), (a, e), (b, d), (b, e), (c, d), (c, e), (d, d), (d, e)}

    Now
       (A × B)
    = {a, b, c, d} × {c, d, e}
    = {(a, c), (a, d), (a, e), (b, c), (b, d), (b, e), (c, c), (c, d), (c, e), (d, c), (d, d), (d, e)}
      (A × C)
    = {a, b, c, d} × {d, e, f, g}
    = {(a, d), (a, e), (a, f), (a, g), (b, d), (b, e), (b, f), (b, g), (c, d), (c, e), (c, f), (c, g), (d, d), (d, e), (d, f), (d, g)}
    R.H.S.
    = (A × B) ∩ (A × C)
    = {(a, d), (a, e), (b, d), (b, e), (c, d), (c, e), (d, d), (d, e)} = L.H.S.
    ∴ A × (B ∩ C) = (A × B) ∩ (A × C) (Proved)

    Q. 9. (ii) Let A = {a, b, c, d}, B = {c, d, e} and C = {d, e, f, g}. Then verify each of the following identities:
    A × (B – C) = (A × B) – (A × C)

    Answer: Given: A = {a, b, c, d}, B = {c, d, e} and C = {d, e, f, g}
       (B – C) = {c}
    L.H.S.
    = A × (B – C)
    ⇒ {a, b, c, d} × {c}
    = {(a, c), (b, c), (c, c), (d, c)}

    Now
       (A × B)
    = {a, b, c, d} × {c, d, e}
    = {(a, c), (a, d), (a, e), (b, c), (b, d), (b, e), (c, c), (c, d), (c, e), (d, c), (d, d), (d, e)}
      (A × C)
    = {a, b, c, d} × {d, e, f, g}
    = {(a, d), (a, e), (a, f), (a, g), (b, d), (b, e), (b, f), (b, g), (c, d), (c, e), (c, f), (c, g), (d, d), (d, e), (d, f), (d, g)}
    R.H.S.
        (A × B) – (A × C)
    = {(a, c), (b, c), (c, c), (d, c)} = L.H.S.
    ∴ A × (B – C) = (A × B) – (A × C) (Proved)

    Q. 9. (iii) Let A = {a, b, c, d}, B = {c, d, e} and C = {d, e, f, g}. Then verify each of the following identities:
    (A × B) ∩ (B × A) = (A ∩ B) × (A ∩ B)

    Answer: Given: A = {a, b, c, d}, B = {c, d, e} and C = {d, e, f, g}
       (A × B)
    = {a, b, c, d} × {c, d, e}
    = {(a, c), (a, d), (a, e), (b, c), (b, d), (b, e), (c, c), (c, d), (c, e), (d, c), (d, d), (d, e)}
       (B × A)
    = {c, d, e} × {a, b, c, d}
    = {(c, a), (c, b), (c, c), (c, d), (d, a), (d, b), (d, c), (d, d), (e, a), (e, b), (e, c), (e, d)}

    L.H.S.
    = (A × B) ∩ (B × A)
    = {(c, c), (c, d), (d, c), (d, d)}
    Now
    (A ∩ B) = {c, d}
    R.H.S.
    = (A ∩ B) × (A ∩ B)
    = {(c, c), (c, d), (d, c), (d, d)} = L.H.S.
    ∴ (A × B) ∩ (B × A) = (A ∩ B) × (A ∩ B) (Proved)

    Exercise 2C

    Q. 1. Let A and B be two nonempty sets.
    (i) What do you mean by a relation from A to B?
    (ii) What do you mean by the domain and range of a relation?

    Answer: (i) If A and B are two nonempty sets, then any subset of the set (A × B) is said to a relation R from set A to set B. 
    Therefore R ⊆ (A × B).

    (ii) Let R be a relation from A to B. Then, the set containing all the first elements of the ordered pairs belonging to R is called Domain. 
    For the relation R,
    Dom(R) = {x: (x, y) ∈R}
    The set containing all the second elements of the ordered pair belonging to R is called Range. 
    For the relation R,
    Range(R) = {y: (x, y) ∈ R}

    Q. 2. (i) Find the domain and range of each of the relations given below:
    R = {(–1, 1), (1, 1), (–2, 4), (2, 4), (2, 4), (3, 9)}

    Solution: Given: R = {(–1, 1), (1, 1), (–2, 4), (2, 4), (2, 4), (3, 9)}
    ∴ Dom(R)
    = {x: (x, y) ∈ R}
    = {-2, -1, 1, 2, 3}
    ∴ Range(R)
    = {y: (x, y) ∈ R}
    = {1, 4, 9}

    Q. 2. (ii) Find the domain and range of each of the relations given below:
    R = {(x, 1/x): x is an integer, 0 < x < 5} 

    Solution: Given:
    R = {(x, 1/x): x is an integer, 0 < x < 5}
    ⇒ R = {(1, 1), (2, 1/2), (3, 1/3), (4, 1/4)
    ∴ Dom(R)
    = {x: (x, y) ∈ R}
    = {1, 2, 3, 4}

    ∴ Range(R)
    = {y: (x, y) ∈ R}
    = {1, 1/2, 1/3, 1/4}

    Q. 2. (iii) Find the domain and range of each of the relations given below:
    R = {(x, y) : x + 2y = 8 and x, y ∈ N}

    Solution: Given: R = {(x, y): x + 2y = 8 and x, y ∈ N}
    ⇒ R = {(2, 3), (4, 2), (6, 1)}
    ∴ Dom(R)
    = {x: (x, y) ∈ R}
    = {2, 4, 6}
    ∴ Range(R)
    = {y: (x, y) ∈ R}
    = {1, 2, 3}

    Q. 2. (iv) Find the domain and range of each of the relations given below:
     R = {(x, y), : y = |x – 1|, x ∈ Z and |x| ≤ 3}

    Solution: Given: R = {(x, y): y = |x – 1|, x ∈ Z and |x| ≤ 3}
    ⇒ R =  {(-3, 4), (-2, 3), (-1, 2), (0, 1), (1, 0), (2, 1),  (3, 2)}
    ∴ Dom(R)
    = {x: (x, y) ∈ R}
    = {-3, -2, -1, 0, 1, 2, 3}
    ∴ Range(R)
    = {y: (x, y) ∈ R}
    = {0, 1, 2, 3, 4}

    Q. 3. Let A = {1, 3, 5, 7} and B = {2, 4, 6, 8}.
    Let R = {(x, y), : x ∈ A, y ∈ B and x > y}.
    (i) Write R in roster form.
    (ii) Find dom (R) and range (R).
    (iii) Depict R by an arrow diagram.

    Solution: Given: A = {1, 3, 5, 7} and B = {2, 4, 6, 8}
    (i) R = {(x, y) : x ∈ A, y ∈ B and x > y}
    ∴ R in Roster Form:
    R = {(3, 2), (5, 2), (5, 4), (7, 2), (7, 4), (7, 6)}
    (ii) Dom(R) = {x: (x, y) ∈ R} = {3, 5, 7}
    Range(R) = {y: (x, y) ∈ R} = {2, 4, 6}
    (iii) Aarrow diagram:

    ARROW DIAGRAM:

    A        B> > > > > > 1            2 3            4 5            6 7            8

    Q. 4. Let A = {2, 4, 5} and b = {1, 2, 3, 4, 6, 8}.
    Let R = {(x, y) x ∈ A, y ∈ B and x divides y}.
    (i)Write R in roster form.
    (ii) Find dom (R) and range (R).??????

    Solution: Given: A = {2, 4, 5} and b = {1, 2, 3, 4, 6, 8}
    (i) R = {(x, y): x ∈ A, y ∈ B and x divides y}
    ∴ R in Roster Form:
    R = {(2, 2), (2, 4), (2, 6), (2, 8), (4, 4), (4, 8)}

    (ii) Dom(R)
    = {x: (x, y) ∈ R}
    = {2, 4,}
    Range(R)
    =  {2, 4, 6, 8}

    Q. 5. Let A = {2, 3, 4, 5} and B = {3, 6, 7, 10}.
    Let R = {(x, y): x ∈ A, y ∈ B and x is relatively prime to y}.
    (i) Write R in roster form.
    (ii) Find dom (R) and range (R).

    Solution: Given: A = {2, 3, 4, 5} and B = {3, 6, 7, 10}
    (i) R = {(x, y), : x ∈ A, y ∈ B and x is relatively prime to y}
    ∴ R in Roster Form:
    R = {(2, 3), (2, 7), (3, 7), (3, 10), (4, 3), (4, 7), (5, 3), (5, 6), (5, 7)}

    (ii) Dom(R)
    = {x: (x, y) ∈ R}
    = {2, 3, 4, 5}
    Range(R)
    = {y: (x, y) ∈ R}
    = {3, 6, 7, 10}

    Q. 6. Let A = {1, 2, 3, 5} And B = {4, 6, 9}.
    Let R = {(x, y): x ∈ A, y ∈ B and (x – y) is odd}.
    Write R in roster form.

    Solution: Given: A = {1, 2, 3, 5} AND B = {4, 6, 9}
    R = {(x, y): x ∈ A, y ∈ B and (x – y) is odd}
    ∴ R in Roster Form:
    R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}

    α  β  γ  Δ δ (delta)  ε (epsilon)  ζ (zêta) η (êta) θ thêta  ι (iota)  κ (kappa)  λ   μ  ν (nu)  ξ (xi)  ο (omikron)  π  ρ (rho)  Σ σ, ς (sigma)  τ (tau)  υ (upsilon)  φ  ψ (psi)Ω ω ± ≠ Δ ≅ × π ∠ Z ℕ ℤ AC  ⊂ ∪ ∩ ⊆ ∈  ∉ A ϕ =>i.e.,  ^ ∨ ⇔ ⇒ <-> ∴  ∵ ∀ ≥ ≤ A 2p θ \left( \right) $$∵ I=\frac{× ×}{100}\\⇒ $$ 

    Q. 7. Let A = {(x, y): x + 3y = 12, x ∈ N and y ∈ N}.
    (i) Write R in roster form.
    (ii) Find dom (R) and range (R).

    Solution: Given: A = {(x, y): x + 3y = 12, x ∈ N and y ∈ N}
    ∵ x + 3y = 12
    ⇒ 3y = 12 – x
    ⇒ y = 12 – x/3
    If x= 1, y = 12 – 1/3 = 11/3 ∉ N
    If x= 2, y = 12 – 2/3 = 10/3 ∉ N
    x= 3, y = 12 – 3/3 = 9/3 = 3 ∈ N
    If x= 6, y = 12 – 6/3 = 6/3 = 2 ∈ N
    If x= 9, y = 12 – 9/3 = 3/3 = 1 ∈ N
    x= 12, y = 12 – 12/3 = 0/3 = 0 ∉ N
    (i) ∴ R in Roster Form: R = {(3, 3), (6, 2), (9, 1)}
    (ii) Dom(R) = {x: (x, y) ∈ R} = {3, 6, 9} Range(R) = {y: (x, y) ∈ R} = {1, 2, 3}

    Q. 8. Let A = {1, 2, 3, 4, 5, 6}.
    Define a relation R from A to A by R = {(x, y): y = x + 1}.
    (i) Write R in roster form.
    (ii) Find dom (R) and range (R).
    (iii) What is its co-domain?
    (iv) Depict R by using arrow diagram.

    Solution: Given: A = {1, 2, 3, 4, 5, 6}
    (i) R = {(x, y): y = x + 1}
    ∴ R in Roster Form:
    R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}

    (ii) Dom(R)
    = {x: (x, y) ∈ R}
    = {1, 2, 3, 4, 5}
    Range(R)
    = {y: (x, y) ∈ R}
    = {2, 3, 4, 5, 6}

    (iii) Here, y = x + 1
    ∴ Co-domain(R)
    = {1, 2, 3, 4, 5, 6}

    (iv) Aarrow diagram:

    ARROW DIAGRAM:

    A         A> >> >> 1             1 2             2 3             3 4             4 5             5 6             6

    Q. 9. Let R = {(x, x + 5): x ∈ {0, 1, 2, 3, 4, 5}}.
    (i) Write R in roster form.
    (ii) Find dom (R) and range (R).

    Solution: Given: R = {(x, x + 5): x ∈ {0, 1, 2, 3, 4, 5}}
    (i) R in Roster Form:
    R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}

    (ii) Dom(R) = {0, 1, 2, 3, 4, 5}
    Range(R) = {5, 6, 7, 8, 9, 10}

    Q. 10. Let A = {1, 2, 3, 4, 6} and R = {(a, b) : a, b ∈ A, and a divides b}.
    (i) Write R in roster form.
    (ii) Find dom (R) and range (R).

    Solution: Given: A = {1, 2, 3, 4, 6}
    (i) R = {(a, b) : a, b ∈ A, and a divides b}
    R in Roster Form:
    R = {(1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)

    (ii) Dom(R)
    = {1, 2, 3, 4, 6}
    Range(R)
    = {2, 3, 4, 6}

    Q. 11. Define a relation R from Z to Z, given by R = {(a, b): a, b ∈ Z and (a – b) is an integer}.
    Find dom (R) and range (R).

    Solution: Given: R = {(a, b): a, b ∈ Z and (a – b) is an integer}
    The difference between any two integers is always an integer.
    The domain of a relation is the set of all first elements of the ordered pairs.
    Here first element is a and a is an integer.
    ∴ Dom (R) = Z
    The range of a relation is the set of all second elements of the ordered pairs.
    Here second element is b and b is an integer.
    ∴ Range (R) = Z

    12. Let R = {(x, y): x, y ∈ Z and x2 + y2 ≤ 4}.
    (i) Write R in roster form.
    (ii) Find dom (R) and range (R).

    Solution: Given: R = {(x, y): x, y ∈ Z and x2 + y2 ≤ 4}
    (i) R in Roster Form:
    R = {(-2, 0), (-1, -1), (-1, 0), (-1, 1), (0, -2), (0, -1), (0, 0), (0, 1), (0, 2), (1, -1), (1, 0), (1, 1), (2, 0)}

    (ii) Dom(R)
    = {-2, -1, 0, 1, 2}
    Range(R)
    = {-2, -1, 0, 1, 2}

    Q. 13. Let A = {2, 3} and B= {3, 5}
    (i) Find (A × B) and n(A × B).
    (ii) How many relations can be defined from A to B?

    Solution: Given: A = {2, 3} and B= {3, 5}
    (i) (A × B) = {(2, 3), (2, 5), (3, 3), (3, 5)}
    Therefore, n(A × B) = 4

    (ii) No. of relation from A to B is a subset of Cartesian product of (A × B).
    Here no. of elements in A = 2 and
    no. of elements in B = 2.
    So, (A × B) = 2 × 2 = 4
    So, the total number of relations can be defined from A to B is
    = 24 = 16

    Q. 14. Let A = {3, 5} and B = {7, 9}. Let R = {(a, b): a ∈ A, b ∈ B and (a – b) is odd}.
    Show that R is an empty relation from A to B.

    Solution: Given: A = {3, 5} and B = {7, 9}
    R = {(a, b): a ∈ A, b ∈ B and (a – b) is odd}
    So, R = {(4, 7), (4, 9)}
    The elements of set A and set B are odd.
    The difference between two odd numbers is always even.
    So (a – b) is always even.
    There is not any elements in Relation R from A to B.
    An empty relation means there is no elements in the relation set.
    Therefore R is an empty relation from A to B.

    SETS 1F, 1G, 1H
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