RD Sharma XI 2020, 2021 Volume 1 Solutions for Class 11 Commerce Maths Chapter 2 Relations are provided here with simple step-by-step explanations. These solutions for Relations are extremely popular among class 11 Commerce students for Maths Relations Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma XI 2020, 2021 Volume 1 Book of class 11 Commerce Maths Chapter 2 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma XI 2020, 2021 Volume 1 Solutions. All RD Sharma XI 2020, 2021 Volume 1 Solutions for class 11 Commerce Maths are prepared by experts and are 100% accurate.
Page No 2.12:
Question 1:
Given A = {1, 2, 3}, B = {3, 4}, C ={4, 5, 6}, find (A × B) ∩ (B × C).
Answer:
Given:
A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}
Now,
(A × B) = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}
(B × C) = {(3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}
∴ (A × B) ∩ (B × C) = {(3, 4)}
Page No 2.12:
Question 2:
If A = {2, 3}, B = {4, 5}, C ={5, 6}, find A × (B ∪ C), A × (B ∩ C), (A × B) ∪ (A × C).
Answer:
Given:
A = {2, 3}, B = {4, 5} and C ={5, 6}
Also,
(B ∪ C) = {4, 5, 6}
Thus, we have:
A × (B ∪ C) = {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3,6)}
And,
(B ∩ C) = {5}
Thus, we have:
A × (B ∩ C) = {(2, 5), (3, 5)}
Now,
(A × B) = {(2, 4), (2, 5), (3, 4), (3, 5)}
(A × C) = {(2, 5), (2, 6), (3, 5), (3, 6)}
∴ (A × B) ∪ (A × C) = {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
Page No 2.12:
Question 3:
If A = {1, 2, 3}, B = {4}, C = {5}, then verify that:
(i) A × (B ∪ C) = (A × B) ∪ (A × C)
(ii) A × (B ∩ C) = (A × B) ∩ (A × C)
(iii) A × (B − C) = (A × B) − (A × C)
Answer:
Given:
A = {1, 2, 3}, B = {4} and C = {5}
(i) A × (B ∪ C) = (A × B) ∪ (A × C)
We have:
(B ∪ C) = {4, 5}
LHS: A × (B ∪ C) = {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}
Now,
(A × B) = {(1, 4), (2, 4), (3, 4)}
And,
(A × C) = {(1, 5), (2, 5), (3, 5)}
RHS: (A × B) ∪ (A × C) = {(1, 4), (2, 4), (3, 4), (1, 5), (2, 5), (3, 5)}
∴ LHS = RHS
(ii) A × (B ∩ C) = (A × B) ∩ (A × C)
We have:
(B ∩ C) =
LHS: A × (B ∩ C) =
And,
(A × B) = {(1, 4), (2, 4), (3, 4)}
(A × C) = {(1, 5), (2, 5), (3, 5)}
RHS: (A × B) ∩ (A × C) =
∴ LHS = RHS
(iii) A × (B − C) = (A × B) − (A × C)
We have:
(B − C) =
LHS: A × (B − C) =
Now,
(A × B) = {(1, 4), (2, 4), (3, 4)}
And,
(A × C) = {(1, 5), (2, 5), (3, 5)}
RHS: (A × B) − (A × C) =
∴ LHS = RHS
Page No 2.12:
Question 4:
Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that:
(i) A × C ⊂ B × D
(ii) A × (B ∩ C) = (A × B) ∩ (A × C)
Answer:
Given:
A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}
(i) A × C ⊂ B × D
LHS: A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}
RHS: B × D = {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)}
∴ A × C ⊂ B × D
(ii) A × (B ∩ C) = (A × B) ∩ (A × C)
We have:
(B ∩ C) =
LHS: A × (B ∩ C) =
Now,
(A × B) = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}
(A × C) = {(1, 5), (1, 6), (2, 5), (2, 6)}
RHS: (A × B) ∩ (A × C) =
∴ LHS = RHS
Page No 2.12:
Question 5:
If A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}, find
(i) A × (B ∩ C)
(ii) (A × B) ∩ (A × C)
(iii) A × (B ∪ C)
(iv) (A × B) ∪ (A × C)
Answer:
Given:
A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}
(i) A × (B ∩ C)
Now,
(B ∩ C) = {4}
∴ A × (B ∩ C) = {(1, 4), (2, 4), (3, 4)}
(ii) (A × B) ∩ (A × C)
Now,
(A × B) = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}
And,
(A × C) = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
∴ (A × B) ∩ (A × C) = {(1, 4), (2, 4), (3, 4)}
(iii) A × (B ∪ C)
Now,
(B ∪ C) = {3, 4, 5, 6}
∴ A × (B ∪ C) = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)}
(iv) (A × B) ∪ (A × C)
Now,
(A × B) = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}
And,
(A × C) = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
∴ (A × B) ∪ (A × C) = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)}
Page No 2.12:
Question 6:
Prove that: (i) (A ∪ B) × C = (A × C) ∪ (B × C) (ii) (A ∩ B) × C = (A × C) ∩ (B×C)
Answer:
(i) (A ∪ B) × C = (A × C) ∪ (B × C)
Let (a, b) be an arbitrary element of (A ∪ B) × C.
Thus, we have:
Again, let (x, y) be an arbitrary element of (A × C) ∪ (B × C).
Thus, we have:
From (i) and (ii), we get:
(A ∪ B) × C = (A × C) ∪ (B × C)
(ii) (A ∩ B) × C = (A × C) ∩ (B×C)
Let (a, b) be an arbitrary element of (A ∩ B) × C.
Thus, we have:
Again, let (x, y) be an arbitrary element of (A × C) ∩ (B × C).
Thus, we have:
From (iii) and (iv), we get:
(A ∩ B) × C = (A × C) ∩ (B × C)
Page No 2.12:
Question 7:
If A × B ⊆ C × D and A × B ≠ Ï, prove that A ⊆ C and B ⊆ D.
Answer:
Page No 2.20:
Question 1:
If A = [1, 2, 3], B = [4, 5, 6], which of the following are relations from A to B? Give reasons in support of your answer.
(i) [(1, 6), (3, 4), (5, 2)]
(ii) [(1, 5), (2, 6), (3, 4), (3, 6)]
(iii) [(4, 2), (4, 3), (5, 1)]
(iv) A × B.
Answer:
Given:
A = {1, 2, 3} and B = {4, 5, 6}
Thus, we have:
A × B = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
(i) {(1, 6), (3, 4), (5, 2)}
Since it is not a subset of A × B, it is not a relation from A to B.
(ii) {(1, 5), (2, 6), (3, 4), (3, 6)}
Since it is a subset of A × B, it is a relation from A to B.
(iii) {(4, 2), (4, 3), (5, 1)}
Since it is not a subset of A × B, it is not a relation from A to B.
(iv) A × B
Since it is a subset (equal to) of A × B, it is a relation from A to B.
Page No 2.20:
Question 2:
A relation R is defined from a set A = [2, 3, 4, 5] to a set B = [3, 6, 7, 10] as follows:
(x, y) ∈ R ⇔ x is relatively prime to y
Express R as a set of ordered pairs and determine its domain and range.
Answer:
Given:
(x, y) ∈ R ⇔ x is relatively prime to y.
Here,
2 is co-prime to 3 and 7.
3 is co-prime to 7 and 10.
4 is co-prime to 3 and 7.
5 is co-prime to 3, 6 and 7.
Thus, we get:
R = {(2, 3), (2, 7), (3, 7), (3, 10), (4, 3), (4, 7), (5, 3), (5, 6), (5, 7)}
Domain of R = {2, 3, 4, 5}
Range of R = {3, 7, 6, 10}
Page No 2.20:
Question 3:
Let A be the set of first five natural numbers and let R be a relation on A defined as follows:
(x, y) ∈ R ⇔ x ≤ y
Express R and R−1 as sets of ordered pairs. Determine also (i) the domain of R−1 (ii) the range of R.
Answer:
Given:
A is the set of the first five natural numbers.
∴ A = {1, 2, 3, 4, 5}
The relation is defined as:
(x, y) ∈ R ⇔ x ≤ y
Now,
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)}
(i) Domain of R-1 = {1, 2, 3, 4, 5}
(ii) Range of R = {1, 2, 3, 4, 5}
Page No 2.20:
Question 4:
Find the inverse relation R−1 in each of the following cases:
(i) R = {(1, 2), (1, 3), (2, 3), (3, 2), (5, 6)}
(ii) R = {(x, y), : x, y ∈ N, x + 2y = 8}
(iii) R is a relation from {11, 12, 13} to (8, 10, 12] defined by y = x − 3.
Answer:
(i) R = {(1, 2), (1, 3), (2, 3), (3, 2), (5, 6)}
R−1= {(2, 1), (3, 1), (3, 2), (2, 3), (6, 5)}
(ii) R = {(x, y) : x, y ∈ N, x + 2y = 8}
On solving x + 2y = 8, we get:
x = 8 2y
On putting y = 1, we get x = 6.
On putting y = 2, we get x = 4.
On putting y = 3, we get x = 2.
∴ R = {(6, 1), (4, 2), (2, 3)}
Or,
R−1= {(1, 6), (2, 4), (3, 2)}
(iii) R is a relation from {11, 12, 13} to {8, 10, 12} defined by y = x − 3.
x belongs to {11, 12, 13} and y belongs to {8, 10, 12}.
Also, 11 − 3 = 8 and 13 − 3 = 10
∴ R = {(11, 8), (13,10)}
Or,
R−1= {(8, 11), (10,13)}
Page No 2.20:
Question 5:
Write the following relation as the sets of ordered pairs:
(i) A relation R from the set [2, 3, 4, 5, 6] to the set [1, 2, 3] defined by x = 2y.
(ii) A relation R on the set [1, 2, 3, 4, 5, 6, 7] defined by (x, y) ∈ R ⇔ x is relatively prime to y.
(iii) A relation R on the set [0, 1, 2, ....., 10] defined by 2x + 3y = 12.
(iv) A relation R from a set A = [5, 6, 7, 8] to the set B = [10, 12, 15, 16,18] defined by (x, y) ∈ R ⇔ x divides y.
Answer:
(i) A relation R from the set [2, 3, 4, 5, 6] to the set [1, 2, 3] is defined by x = 2y.
Putting y = 1, 2, 3 in x = 2y, we get:
x = 2, 4, 6
∴ R = {(2, 1), (4, 2), (6, 3)}
(ii) A relation R on the set [1, 2, 3, 4, 5, 6, 7] defined by (x, y) ∈ R ⇔ x is relatively prime to y.
Here,
2 is relatively prime to 3, 5 and 7.
3 is relatively prime to 2, 4, 5 and 7.
4 is relatively prime to 3, 5 and 7.
5 is relatively prime to 2, 3, 4, 6 and 7.
6 is relatively prime to 5 and 7.
7 is relatively prime to 2, 3, 4, 5 and 6.
∴ R = {(2, 3), (2, 5), (2, 7), (3, 2), (3, 4), (3, 5), (3, 7), (4, 3), (4, 5), (4, 7), (5, 2), (5, 3), (5, 4), (5, 6), (5, 7), (6, 5), (6, 7), (7, 2), (7, 3), (7,4), (7, 5), (7, 6)}
(iii) A relation R on the set [0, 1, 2,..., 10] is defined by 2x + 3y = 12.
Putting y = 0, 2, 4, we get:
x = 6, 3, 0
∴ R = {(0, 4), (3, 2), (6, 0)}
(iv) A relation R from the set A = [5, 6, 7, 8] to the set B = [10, 12, 15, 16, 18] defined by (x, y) ∈ R ⇔ x divides y.
Here,
5 divides 10 and 15.
6 divides 12 and 18.
8 divides 16.
∴ R = {(5, 10), (5, 15), (6, 12), (6, 18), (8,16)}
Page No 2.20:
Question 6:
Let R be a relation in N defined by (x, y) ∈ R ⇔ x + 2y =8. Express R and R−1 as sets of ordered pairs.
Answer:
Let R be a relation in N defined by (x, y) ∈ R ⇔ x + 2y = 8.
We have:
x = 82y
For y = 3, 2, 1, we have:
x = 2, 4, 6
∴ R = {(2, 3), (4, 2), (6, 1)}
And,
R−1 = {(3, 2), (2, 4), (1, 6)}
Page No 2.21:
Question 7:
Let A = (3, 5) and B = (7, 11). Let R = {(a, b) : a ∈ A, b ∈ B, a − b is odd}. Show that R is an empty relation from A into B.
Answer:
Given:
A = (3, 5) and B = (7, 11)
Also,
R = {(a, b) : a ∈ A, b ∈ B, a − b is odd}
a are the elements of A and b are the elements of B.
So, R is an empty relation from A to B.
Hence proved.
Page No 2.21:
Question 8:
Let A = [1, 2] and B = [3, 4]. Find the total number of relation from A into B.
Answer:
We have:
A = {1, 2} and B = {3, 4}
Now,
There are 2n relations from A to B, where n is the number of elements in their Cartesian product.
∴ Number of relations from A to B is 24 = 16.
Page No 2.21:
Question 9:
Determine the domain and range of the relation R defined by
(i) R = [(x, x + 5): x ∈ (0, 1, 2, 3, 4, 5)]
(ii) R = {(x, x3) : x is a prime number less than 10}
Answer:
(i) R = {(x, x + 5): x ∈ (0, 1, 2, 3, 4, 5)}
We have:
R = {(0, 0 + 5), (1, 1 + 5), (2, 2 + 5), (3, 3 + 5), (4, 4 + 5), (5, 5 + 5)}
Or, R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}
∴ Domain (R) = {0, 1, 2, 3, 4, 5}
Range (R) = {5, 6, 7, 8, 9, 10}
(ii) R = {(x, x3) : x is a prime number less than 10}
We have:
x = 2, 3, 5, 7
x3 = 8, 27, 125, 343
Thus, we get:
R = {(2, 8), (3, 27), (5, 125), (7, 343)}
Domain (R) = {2, 3, 5, 7}
Range (R) = {8, 27, 125, 343}
Page No 2.21:
Question 10:
Determine the domain and range of the following relations:
(i) R = {(a, b) : a ∈ N, a < 5, b = 4}
(ii)
Answer:
(i) R = {(a, b) : a ∈ N, a < 5, b = 4}
We have:
a = 1, 2, 3, 4
b = 4
R = {(1, 4), (2, 4), (3, 4), (4, 4)}
Domain (R) = {1, 2, 3, 4}
Range (R) = {4}
(ii)
Now,
a = 3, 2, 1, 0, 1, 2, 3
Thus, we have:
b = 4, 3, 2, 1, 0, 1, 2
Or,
S = {(3, 4), (2, 3), (1, 2), (0, 1), (1, 0), (2, 1), (3, 2)}
Domain (S) = {3, 2, 1, 0, 1, 2, 3}
Range (S) = {0, 1, 2, 3, 4}
Page No 2.21:
Question 11:
Let A = {a, b}. List all relations on A and find their number.
Answer:
Any relation in A can be written as a set of ordered pairs.
The only ordered pairs that can be included are (a, a), (a, b), (b, a) and (b, b).
There are four ordered pairs in the set, and each subset is a unique combination of them.
Each unique combination makes different relations in A.
{ } [the empty set]
{(a, a)}
{(a, b)}
{(a, a), (a, b)}
{(b, a)}
{(a, a), (b, a)}
{(a, b), (b, a)}
{(a, a), (a, b), (b, a)}
{(b, b)}
{(a, a), (b, b)}
{(a, b), (b, b)}
{(a, a), (a, b), (b, b)}
{(b, a), (b, b)}
{(a, a), (b, a), (b, b)}
{(a, b), (b, a), (b, b)}
{(a ,a), (a, b), (b, a), (b, b)}
Number of elements in the Cartesian product of A and A =
∴ Number of relations =
Page No 2.21:
Question 12:
Let A = (x, y, z) and B = (a, b). Find the total number of relations from A into B.
Answer:
Given:
A = (x, y, z) and B = (a, b)
Now,
Number of elements in the Cartesian product of
Number of relations from A to B =
Page No 2.21:
Question 13:
Let R be a relation from N to N defined by R = [(a, b) : a, b ∈ N and a = b2].
Are the following statements true?
(i) (a, a) ∈ R for all a ∈ N
(ii) (a, b) ∈ R ⇒ (b, a) ∈ R
(iii) (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R
Answer:
Given: R = [(a, b) : a, b ∈ N and a = b2]
(i) (a, a) ∈ R for all a ∈ N.
Here, 2 ∈N, but .
∴ (2,2)R
False
(ii) (a, b) ∈ R ⇒ (b, a) ∈ R
âµ 4 = 22
(4, 2) ∈ R, but (2,4)R.
False
(iii) (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R
âµ 16 = 42 and 4 = 22
∴ (16, 4) ∈ R and (4, 2) ∈ R
Here,
(16,2)R
False
Page No 2.21:
Question 14:
Let A = [1, 2, 3, ......., 14]. Define a relation on a set A by
R = {(x, y) : 3x − y = 0, where x, y ∈ A}.
Depict this relationship using an arrow diagram. Write down its domain, co-domain and range.
Answer:
A = [1, 2, 3,..., 14]
R = {(x, y) : 3x − y = 0, where x, y ∈ A}
Or,
R = {(x, y) : 3x = y, where x, y ∈ A}
As
Or,
R = {(1, 3), (2, 6), (3, 9), (4, 12)}
Domain (R) = {1, 2, 3, 4}
Range (R) = {3, 6, 9, 12}
Co-domain (R) = A
Page No 2.21:
Question 15:
Define a relation R on the set N of natural number by R = {(x, y) : y = x + 5, x is a natural number less than 4, x, y ∈ N}. Depict this relationship using (i) roster form (ii) an arrow diagram. Write down the domain and range or R.
Answer:
R = {(x, y) : y = x + 5, x is a natural number less than 4, x, y ∈ N}
(i) âµ x = 1, 2, 3
∴ y = 1 + 5, 2 + 5, 3 + 5
y = 6, 7, 8
Thus, we have:
R = {(1, 6), (2, 7), (3, 8)}
(ii)
Now,
Domain (R) = {1, 2, 3}
Range (R) = {6, 7, 8}
Page No 2.21:
Question 16:
A = [1, 2, 3, 5] and B = [4, 6, 9]. Define a relation R from A to B by R = {(x, y) : the difference between x and y is odd, x ∈ A, y ∈ B}. Write R in Roster form.
Answer:
A = [1, 2, 3, 5] and B = [4, 6, 9]
R = {(x, y) : the difference between x and y is odd, x ∈ A, y ∈ B}
For x = 1,
41 = 3 and 61 = 5
y = 4, 6
For x = 2,
92 = 7
y = 9
For x = 3,
43 = 1 and 63 = 3
y = 4, 6
For x = 5,
54 =1 and 65 =1
y = 4, 6
Thus, we have:
R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}
Page No 2.21:
Question 17:
Write the relation R = {(x, x3) : x is a prime number less than 10} in roster form.
Answer:
R = {(x, x3) : x is a prime number less than 10}
x = 2, 3, 5, 7
x3 = 8, 27, 125, 343
R = {(2, 8), (3, 27), (5, 125), (7, 343)}
Page No 2.21:
Question 18:
Let A = [1, 2, 3, 4, 5, 6]. Let R be a relation on A defined by
{(a, b) : a, b ∈ A, b is exactly divisible by a}
(i) Writer R in roster form
(ii) Find the domain of R
(ii) Find the range of R.
Answer:
A = [1, 2, 3, 4, 5, 6]
R = {(a, b) : a, b ∈ A, b is exactly divisible by a}
(i) Here,
2 is divisible by 1 and 2.
3 is divisible by 1 and 3.
4 is divisible by 1 and 4.
5 is divisible by 1 and 5.
6 is divisible by 1, 2, 3 and 6.
∴ R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (5, 5), (6, 6)}
(ii) Domain (R) = {1, 2, 3, 4, 5, 6}
(iii) Range (R) = {1, 2, 3, 4, 5, 6}
Page No 2.21:
Question 19:
The adjacent figure shows a relationship between the sets P and Q. Write this relation in (i) set builder form (ii) roster form. What is its domain and range?
Figure
Answer:
(i) We have:
52 = 3
62 = 4
72 = 5
∴ R =
(ii) R = {(5, 3), (6, 4), (7, 5)}
(iii) Domain (R) = {5, 6, 7}
Range (R) = {3, 4, 5}
Page No 2.21:
Question 20:
Let R be the relation on Z defined by
R = {(a, b) : a, b ∈ Z, a − b is an integer}
Find the domain and range of R.
Answer:
R = {(a, b) : a, b ∈ Z, a − b is an integer}
We know:
Difference of any two integers is always an integer.
Thus, for all a, b ∈ Z, we get a − b as an integer.
∴ Domain (R) = Z
And,
Range (R) = Z
Page No 2.21:
Question 21:
For the relation R1 defined on R by the rule (a, b) ∈ R1 ⇔ 1 + ab > 0.
Prove that: (a, b) ∈ R1 and (b , c) ∈ R1 ⇒ (a, c) ∈ R1 is not true for all a, b, c ∈ R.
Answer:
We have:
(a, b) ∈ R1 ⇔ 1 + ab > 0
Let:
a = 1, b = and c = 4
Now,
, as .
But .
∴ (1,4)
And,
(a, b) ∈ R1 and (b , c) ∈ R1
Thus, (a, c) ∈ R1 is not true for all a, b, c ∈ R.
Page No 2.21:
Question 22:
Let R be a relation on N × N defined by
(a, b) R (c, d) ⇔ a + d = b + c for all (a, b), (c, d) ∈ N × N
Show that:
(i) (a, b) R (a, b) for all (a, b) ∈ N × N
(ii) (a, b) R (c, d) ⇒ (c, d) R (a, b) for all (a, b), (c, d) ∈ N × N
(iii) (a, b) R (c, d) and (c, d) R (e, f) ⇒ (a, b) R (e, f) for all (a, b), (c, d), (e, f) ∈ N × N
Answer:
We are given ,
(a, b) R (c, d) ⇔ a + d = b + c for all (a, b), (c, d) ∈ N × N
(i) (a, b) R (a, b) for all (a, b) ∈ N × N
(ii) (a, b) R (c, d) ⇒ (c, d) R (a, b) for all (a, b), (c, d) ∈ N × N
(iii) (a, b) R (c, d) and (c, d) R (e, f) ⇒ (a, b) R (e, f) for all (a, b), (c, d), (e, f) ∈ N × N
Page No 2.24:
Question 1:
If A = {1, 2, 4}, B = {2, 4, 5}, C = {2, 5}, then (A − B) × (B − C) is
(a) {(1, 2), (1, 5), (2, 5)}
(b) [(1, 4)]
(c) (1, 4)
(d) none of these
Answer:
(b) {(1, 4)}
A = {1, 2, 4}, B = {2, 4, 5} and C = {2, 5}
(A − B) = {1}
(B − C) = {4}
So, (A − B) × (B − C) = {(1,4)}
Page No 2.24:
Question 2:
If R is a relation on the set A = [1, 2, 3, 4, 5, 6, 7, 8, 9] given by x R y ⇔ y = 3x, then R =
(a) [(3, 1), (6, 2), (8, 2), (9, 3)]
(b) [(3, 1), (6, 2), (9, 3)]
(c) [(3, 1), (2, 6), (3, 9)]
(d) none of these
Answer:
(d) none of these
A = {1, 2, 3, 4, 5, 6, 7, 8, 9}
x R y ⇔ y = 3x
For x = 1, y = 3
For x = 2, y = 6
For x = 3, y = 9
Thus, R = {(1,3),(2,6),(3,9)}
Page No 2.25:
Question 3:
Let A = [1, 2, 3], B = [1, 3, 5]. If relation R from A to B is given by = {(1, 3), (2, 5), (3, 3)}, Then R−1 is
(a) {(3, 3), (3, 1), (5, 2)}
(b) {(1, 3), (2, 5), (3, 3)}
(c) {(1, 3), (5, 2)}
(d) None of these
Answer:
(a) {(3, 3), (3, 1), (5, 2)}
A = {1, 2, 3}, B ={1, 3, 5}
R = {(1, 3), (2, 5), (3, 3)}
∴ R−1 = {(3,1),(5,2),(3,3)}
Page No 2.25:
Question 4:
If A = [1, 2, 3], B = [1, 4, 6, 9] and R is a relation from A to B defined by 'x' is greater than y. The range of R is
(a) {1, 4, 6, 9}
(b) (4, 6, 9)
(c) [1]
(d) none of these.
Answer:
(c) {1}
A = {1, 2, 3} and B = {1, 4, 6, 9}
R is a relation from A to B defined by: x is greater than y.
Then R = {(2,1),(3,1)}
∴ Range (R) = {1}
Page No 2.25:
Question 5:
If R = {(x, y) : x, y ∈ Z, x2 + y2 ≤ 4} is a relation on Z, then the domain of R is
(a) [0, 1, 2]
(b) [0, −1, −2]
(c) {−2, −1, 0, 1, 2]
(d) None of these
Answer:
(c) {−2, −1, 0, 1, 2}
R = {(x, y) : x, y ∈ Z, x2 + y2 ≤ 4}
We know that,
Hence, domain (R) = {2,1,0,1,2,}
Page No 2.25:
Question 6:
A relation R is defined from [2, 3, 4, 5] to [3, 6, 7, 10] by : x R y ⇔ x is relatively prime to y. Then, domain of R is
(a) [2, 3, 5]
(b) [3, 5]
(c) [2, 3, 4]
(d) [2, 3, 4, 5]
Answer:
(d) {2, 3, 4, 5}
Given:
From {2, 3, 4, 5} to {3, 6, 7, 10}, x R y ⇔ x is relatively prime to y
2 is relatively prime to 3,7
3 is relatively prime to 7,10
4 is relatively prime to 3,7
5 is relatively prime to 3,6,7
So, domain of R is {2,3,4,5}
Page No 2.25:
Question 7:
A relation Ï from C to R is defined by x Ï y ⇔ = y. Which one is correct?
(a) (2 + 3i) Ï 13
(b) 3Ï (−3)
(c) (1 + i) Ï 2
(d) i Ï 1
Answer:
(d) i Ï 1
We have
Thus, i Ï 1 satisfies x Ï y ⇔ = y.
Page No 2.25:
Question 8:
Let R be a relation on N defined by x + 2y = 8. The domain of R is
(a) [2, 4, 8]
(b) [2, 4, 6, 8]
(c) [2, 4, 6]
(d) [1, 2, 3, 4]
Answer:
(c) {2, 4, 6}
x + 2y = 8
⇒ x = 8 2y
For y = 1, x = 6
y = 2, x = 4
y = 3, x = 2
Then R = {(2,3),(4,2),(6,1)}
∴ Domain of R = {2,4,6}
Page No 2.25:
Question 9:
R is a relation from [11, 12, 13] to [8, 10, 12] defined by y = x − 3. Then, R−1 is
(a) [(8, 11), (10, 13)]
(b) [(11, 8), (13, 10)]
(c) [(10, 13), (8, 11), (12, 10)]
(d) none of these
Answer:
(a) [(8, 11), (10, 13)]
R is a relation from [11, 12, 13] to [8, 10, 12], defined by y = x − 3
Now, we have:
11 3 = 8
13 3 = 10
So, R = {(13,10),(11,8)}
∴ R−1 = {(10,13),(8,11)}
Page No 2.25:
Question 10:
If the set A has p elements, B has q elements, then the number of elements in A × B is
(a) p + q
(b) p + q + 1
(c) pq
(d) p2
Answer:
(c) pq
n(A × B) = n(A) × n(B)
= p × q = pq
Page No 2.25:
Question 11:
Let R be a relation from a set A to a set B, then
(a) R = A ∪ B
(b) R = A ∩ B
(c) R ⊆ A × B
(d) R ⊆ B × A
Answer:
(c) R ⊆ A × B
If R is a relation from set A to set B, then R is always a subset of A × B.
Page No 2.25:
Question 12:
If R is a relation from a finite set A having m elements of a finite set B having n elements, then the number of relations from A to B is
(a) 2mn
(b) 2mn − 1
(c) 2mn
(d) mn
Answer:
(a) 2mn
Given: n(A) = m
n(B) = n
∴
Then, the number of relations from A to B is 2mn.
Page No 2.25:
Question 13:
If R is a relation on a finite set having n elements, then the number of relations on A is
(a) 2n
(b)
(c) n2
(d) nn
Answer:
(b)
Given : A finite set with n elements
Its Cartesian product with itself will have n2 elements.
∴ Number of relations on A =
Page No 2.25:
Question 14:
Let n(A) = m and n(B) = n. Then the total number of non-empty relations that can be defined from
A to B is
(a) mn
(b) nm − 1
(c) mn − 1
(d) 2mn − 1
Answer:
Let n(A) = m
n(B) = n
since n (A × B ) = mn
where A × B defines A cartesian B.
Total number of relation from A to B
= number of subsets of A × B
= 2mn
i.e, Total number of non-empty relations is 2mn−1
Hence, the correct answer is option D.
Page No 2.25:
Question 1:
Let n(A) = m and n(B) = n. Then the total number of non-empty relations that can be defined from
A to B is ___________ .
Answer:
The total number of non-empty relations that can be defined from A to B is 2mn –1.
Page No 2.25:
Question 2:
The smallest reflexive relation on a set A is the ___________ .
Answer:
Let us define A = {1, 2, 3..........n}
Then smaller possible reflexive relation on set A is {(1,1), (2, 2),....... (n − 1, n − 1), (n, n)}
Page No 2.26:
Question 3:
If A and B are two sets such that n (A) = 5 and n (B) =7, then the the total number of relations on A × B is ........ .
Answer:
n(A) = 5
n(B) = 7
Then total number of relation on A × B is 2n(A).n(B)
Page No 2.26:
Question 4:
A relation R on a set A is a symmetric relation if __________.
Answer:
A relation R on a set A is symmetric relation
i.e if (x, y) R for x, yA
⇔ (y, x)R
i.e if (x, y)R
⇔ (x, y)R −1
R = R−1
Hence relation R on a set A is symmetric
If R = R−1
Page No 2.26:
Question 5:
If R and S are two equivalence relations on a set A, then R ∩ S is __________.
Answer:
If R and S are two equivalence relations on a
Let A,
→ Since ∀ x∈A
⇒ (x, x) ∈ R â S ∀x∈A
i.e R â S is Reflexive
→ Let (x, y) R â S
⇒ (y, x)∈R and (y, x) ∈S (R and S are symmetric)
⇒ (y, x)∈R â S
⇒ R â S is symmetric
→ Let (x, y) (y, z) ∈ R â S
⇒ (x, z) ∈ R and (x, z) ∈ S
⇒ (x, z)∈ R â S
i.e R â S is transitive
Page No 2.26:
Question 6:
If (1, 3), (2, 5) and (3, 3) are three elements of A × B and n (A × B) = 6, then the remaining
three elements of are __________.
Answer:
Given (1, 3), (2, 5) and (3, 3) ∈ A × B and n (A × B) = 6
According to given collection,
A = {1, 2, 3} (since from (x, y) x∈A and y∈B )
and B = {3, 5}
i.e A × B = {(1, 3), (1, 5), (2, 3), (2, 5), (3, 3), (3, 5)}
Remaining elements of A × B are (1, 5) (2, 3) and (3, 5)
i.e Remaining elements of A × B are (1, 5) (2, 3) and (3, 5)
Page No 2.26:
Question 7:
The total number of reflexive relations on a finite set having n elements is __________.
Answer:
The total number of reflexive relations on a finite set having n elements is _________.
Consider a set A with n elements
Say A ={1, 2, ....... n−1, n }
out of n2 elements n elements are compulsory for relation to be reflexive.
i.e (1, 1) (2, 2) (3, 3) .... (n, n)
and for remaining n2 − n elements, we have choice of filling i.e either they are present or absent.
Hence, Total number of reflexive relation are .
Page No 2.26:
Question 8:
If R = {(x, y) : x, y ∈ Z, x2 + y2 = 25}, then Domain (R) = .................... and Range (R) = __________.
Answer:
If R = { (x, y) : x, y ∈ Z , x2 + y2 = 25}
Since x2 + y2 =25
only possibilities for x are 0, 3, 4 and 5 and similarly for y are 0, 3, 4 and 5
i.e R = {(0, 5) , (3, 4) , (4, 3), (5, 0)}
{ all above satisfy x2 + y2 = 25}
Domain (R) = {0, 3, 4, 5}and Range (R) = {0, 3, 4, 5}
Page No 2.26:
Question 9:
The relation R = {(x, y) : x, y ∈ Z, x2 + y2 = 64} is __________.
Answer:
R = {(x, y) : x, y ∈ Z : x2 + y2 = 64}
Since
x2 + y2 = 64
y2 = 64 − x2
The only possibilities for x are 0, 8 and −8, and for y are 0, −8 and 8 .
i.e R = {(0, 8), (8, 0), (−8, 0) (0, −8)}
Hence R = { (0, 8) (−8, 0) (8, 0) (0, −8)}
Page No 2.26:
Question 10:
If A ={(x : x ∈ W, x < 2}, B = {x : x ∈ N, 1 < x < 5} and C = 3, 5, then A × (B ∩ C) = _________.
Answer:
A = {x : x∈ W ; x < 2y = {0, 1}
B = {x : x∈ N ; 1 < x < 5} = {2, 3, 4}
C = {3, 5}
Then B ∩ C = {3}
A × (B ∩ C) = {(0, 3), (1, 3)}
Page No 2.26:
Question 11:
If R = {(x, y) : where x ∈ R and −5 ≤ x ≤ 5} is a relation, then range (R) = _______ .
Answer:
R = {(x, y) : where x ∈ R and −5 ≤ x ≤ 5}
Range R = ??
(Correction ; NO information as y is given )
Page No 2.26:
Question 12:
If n(A ∩ B′) = 9, n(A' ∩ B) = 10 and n(A ∪ B) = 24, then n(A × B) = ___________ .
Answer:
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n(A) = n(A â B) − n(A' ∩ B)
n(B) = n(A â B) − n(A ∩ B')
i.e n(A) = 24 − 10 =14
n(B) = 24 − 9 = 15
n(A × B) = n(A) . n(B)
n(A × B) = 14 × 15 = 210
Page No 2.26:
Question 13:
If A = {3, 5, 6, 9) and R is a relation in A defined as R = {(x, y) : x + y < 18), then R in roster form is ______ .
Answer:
A = {3, 5, 6, 9)
R = {(x, y) : x + y < 18 }
Then roaster form for R is collection of elements of type {(x, y)} satisfying R.
for x = 3, possible values of y are 3, 5, 6, 9
for x = 5, possible values of y are 3, 5, 6, 9
for x = 6, possible values of y are 3, 5, 6, 9
for x = 9, possible values of y are 3, 5, 6.
R = { (3, 3) (3, 5) (3, 6) (3, 9) (5, 3) (5, 5) (5, 6) (5, 9) (6, 3) (6, 5) (6, 6) (6, 9) (9, 3) (9, 5) (9, 6)} is the roaster form for R.
Page No 2.26:
Question 14:
If n (A × B) = 200 and n(A) = 50, then the number of elements in P(B) is _______ .
Answer:
Given n (A × B) = 200
n(A) = 50
Since n(A × B) = n(A) n(B)
⇒ 200 = n(A) n(B)
i.e n(B) =
i.e number of elements of B are 4
i.e number of elements in P(B) = 24
Page No 2.26:
Question 15:
If A = {1, 2, 3, 4, 5, 6), then the number of sub sets of A containing elements 2, 3 and 5 is _______ .
Answer:
If A = {1, 2, 3, 4, 5, 6)
i.e n(A) = 6
Then number of subsets of A containing 2, 3, 5
Then number of subsets of A
Containing 3 elements = 20
Page No 2.26:
Question 1:
If A = {1, 2, 4}, B = {2, 4, 5} and C = {2, 5}, write (A − C) × (B − C).
Answer:
Given:
A = {1, 2, 4}, B = {2, 4, 5} and C = {2, 5}
Now,
(A − C) = {1, 4}
(B − C) = {4}
Thus, we have:
(A − C) × (B − C) = {(1, 4), (4, 4)}
Page No 2.26:
Question 2:
If n(A) = 3, n(B) = 4, then write n(A × A × B).
Answer:
Given:
n(A) = 3 and n(B) = 4
Now, we have:
n(A × A × B) =
Page No 2.26:
Question 3:
If R is a relation defined on the set Z of integers by the rule (x, y) ∈ R ⇔ x2 + y2 = 9, then write domain of R.
Answer:
We need to find (x, y) ∈ R such that x2 + y2 = 9.
x can take values 3, 0 and 3.
∴ Domain (R) = {3, 0, 3}
Page No 2.26:
Question 4:
If R = {(x, y) : x, y ∈ Z, x2 + y2 ≤ 4} is a relation defined on the set Z of integers, then write domain of R.
Answer:
Given:
R = {(x, y) : x, y ∈ Z, x2 + y2 ≤ 4}
We know:
∴ Domain (R) = {2, 1, 0, 1, 2}
Page No 2.26:
Question 5:
If R is a relation from set A = (11, 12, 13) to set B = (8, 10, 12) defined by y = x − 3, then write R−1.
Answer:
Given:
A = (11, 12, 13) and B = (8, 10, 12)
R is defined by (y = x − 3) from A to B.
We know:
8 = 113
10 = 133
∴ R = {(11, 8), (13, 10)}
Or,
R-1 = {(8, 11), (10, 13)}
Page No 2.27:
Question 6:
Let A = {1, 2, 3} and . Then write R as set of ordered pairs.
Answer:
Given:
A = {1, 2, 3}
We know that
Thus, R ={(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)}
Page No 2.27:
Question 7:
Let R = [(x, y) : x, y ∈ Z, y = 2x − 4]. If (a, -2) and (4, b2) ∈ R, then write the values of a and b.
Answer:
R = [(x, y) : x, y ∈ Z, y = 2x − 4]
(a, 2) and (4, b2) ∈ R
Thus, a =1 and b =
Page No 2.27:
Question 8:
If R = {(2, 1), (4, 7), (1, −2), ...}, then write the linear relation between the components of the ordered pairs of the relation R.
Answer:
Given:
R = {(2, 1), (4, 7), (1, −2), ...}
We can observe that
Thus, the linear relation between the components of the ordered pairs of the relation R is y = 3x 5.
Page No 2.27:
Question 9:
If A = [1, 3, 5] and B = [2, 4], list of elements of R, if
R = {(x, y) : x, y ∈ A × B and x > y}
Answer:
Given:
A = {1, 3, 5} and B = {2, 4}
R = {(x, y) : x, y ∈ A × B and x > y}
A × B = {(1,2),(1,4),(3,2),(3,4),(5,2),(5,4)}
As 3 > 2, 5 > 2 and 5 > 4,
we have R = {(3,2),(5,2),(5,4)}
Page No 2.27:
Question 10:
If R = [(x, y) : x, y ∈ W, 2x + y = 8], then write the domain and range of R.
Answer:
R = {(x, y) : x, y ∈ W, 2x + y = 8}
∴ Domain (R) = {0,1,2,3,4} and Range (R) = {0,2,4,6,8}
Page No 2.27:
Question 11:
Let A and B be two sets such that n(A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) are in A × B, write A and B.
Answer:
Given:
(x, 1), (y, 2), (z, 1) are in A × B
n(A) = 3 and n(B) = 2
So, A = {x,y,z} and B = {1,2}
Page No 2.27:
Question 12:
Let A = [1, 2, 3, 5], B = [4, 6, 9] and R be a relation from A to B defined by R = {(x, y) : x − y is odd}. Write R in roster form.
Answer:
Given:
A = {1, 2, 3, 5} and B = {4, 6, 9}
R = {(x, y) : x − y is odd}
Since 14 = 3 is odd, we have:
16 = 5 is odd
29 = 7 is odd
34 =1 is odd
36 = 3 is odd
54 = 1 is odd
56 = 1 is odd
∴ R = {(1,4),(1,6),(2,9),(3,4),(3,6),(5,4),(5,6)}
Page No 2.8:
Question 1:
(i) If , find the values of a and b.
(ii) If (x + 1, 1) = (3, y − 2), find the values of x and y.
Answer:
(i)
By the definition of equality of ordered pairs, we have:
(ii) (x + 1, 1) = (3, y − 2)
By the definition of equality of ordered pairs, we have:
Page No 2.8:
Question 2:
If the ordered pairs (x, −1) and (5, y) belong to the set {(a, b) : b = 2a − 3}, find the values of x and y.
Answer:
The ordered pairs (x, −1) and (5, y) belong to the set {(a, b) : b = 2a − 3}.
Thus, we have:
x = a and −1 = b such that b = 2a − 3.
∴ −1 = 2x − 3
or, 2x = 3 − 1 = 2
or, x = 1
Also,
5 = a and y = b such that b = 2a − 3.
∴ y = 2(5) − 3
or, y = 10 − 3 = 7
Thus, we get:
x = 1 and y = 7
Page No 2.8:
Question 3:
If a ∈ [−1, 2, 3, 4, 5] and b ∈ [0, 3, 6], write the set of all ordered pairs (a, b) such that a + b = 5.
Answer:
Given:
a ∈ [−1, 2, 3, 4, 5] and b ∈ [0, 3, 6]
We know:
−1 + 6 = 5, 2 + 3 = 5 and 5 + 0 = 5
Thus, possible ordered pairs (a, b) are {(−1, 6), (2, 3), (5, 0)} such that a + b = 5.
Page No 2.8:
Question 4:
If a ∈ [2, 4, 6, 9] and b ∈ [4, 6, 18, 27], then form the set of all ordered pairs (a, b) such that a divides b and a < b.
Answer:
Given:
a ∈ [2, 4, 6, 9] and b ∈ [4, 6, 18, 27]
Here,
2 divides 4, 6 and 18 and 2 is less than all of them.
6 divides 18 and 6 and 6 is less than 18.
9 divides 18 and 27 and 9 is less than 18 and 27.
Now,
Set of all ordered pairs (a, b) such that a divides b and a < b = {(2, 4), (2, 6), (2, 18), (6, 18), (9, 18), (9, 27)}
Page No 2.8:
Question 5:
If A = {1, 2} and B = {1, 3}, find A × B and B × A.
Answer:
Given:
A = {1, 2} and B = {1, 3}
Now,
A × B = {(1, 1), (1, 3), (2, 1), (2, 3)}
B × A = {(1, 1), (1, 2), (3, 1), (3, 2)}
Page No 2.8:
Question 6:
Let A = {1, 2, 3} and B = {3, 4}. Find A × B and show it graphically.
Answer:
Given:
A = {1, 2, 3} and B = {3, 4}
Now,
A × B = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}
To represent A × B graphically, follow the given steps:
(a) Draw two mutually perpendicular lines—one horizontal and one vertical.
(b) On the horizontal line, represent the elements of set A; and on the vertical line, represent the elements of set B.
(c) Draw vertical dotted lines through points representing elements of set A on the horizontal line and horizontal lines through points representing elements of set B on the vertical line.
The points of intersection of these lines will represent A × B graphically.
Page No 2.8:
Question 7:
If A = {1, 2, 3} and B = {2, 4}, what are A × B, B × A, A × A, B × B and (A × B) ∩ (B × A)?
Answer:
Given :
A = {1, 2, 3} and B = {2, 4}
Now,
A × B = {(1, 2), (1, 4), (2, 2), (2, 4), (3, 2), (3, 4)}
B × A = {(2, 1), (2, 2), (2, 3), (4, 1), (4, 2), (4, 3)}
A × A = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}
B × B = {(2, 2), (2, 4), (4, 2), (4, 4)}
We observe:
(A × B) ∩ (B × A) = {(2, 2)}
Page No 2.8:
Question 8:
If A and B are two set having 3 elements in common. If n(A) = 5, n(B) = 4, find n(A × B) and n[(A × B) ∩ (B × A)].
Answer:
Given:
n(A) = 5 and n(B) = 4
Thus, we have:
n(A × B) = 5(4) = 20
A and B are two sets having 3 elements in common.
Now,
Let:
A = (a, a, a, b, c) and B = (a, a, a, d)
Thus, we have:
(A × B) = {(a, a), (a, a), (a, a), (a, d), (a, a), (a, a), (a, a), (a, d), (a, a), (a, a), (a, a), (a, d), (b, a), (b, a), (b, a), (b, d), (c, a), (c, a), (c, a), (c, d)}
(B × A) = {(a, a), (a, a), (a, a), (a, b), (a, c), (a, a), (a, a), (a, a), (a, b), (a, c), (a, a), (a, a), (a, a), (a, b), (a, c), (d, a), (d, a), (d, a), (d, b), (d, c)}
[(A × B) ∩ (B × A)] = {(a, a), (a, a), (a, a), (a, a), (a, a), (a, a), (a, a), (a, a), (a, a)}
∴ n[(A × B) ∩ (B × A)] = 9
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Question 9:
Let A and B be two sets. Show that the sets A × B and B × A have elements in common iff the sets A and B have an elements in common.
Answer:
Case (i): Let:
A = (a, b, c)
B = (e, f)
Now, we have:
A × B = {(a, e}), (a, f), (b,e), (b, f), (c, e), (c, f)}
B × A = {(e, a), (e, b), (e, c), (f, a), (f, b), (f, c)}
Thus, they have no elements in common.
Case (ii): Let:
A = (a, b, c)
B = (a, f)
Thus, we have:
A × B = {(a, a}), (a,f), (b, a), (b, f), (c,a), (c, f)}
B × A = {(a, a), (a, b), (a, c), (f, a), (f, b), (f, c)}
Here, A × B and B × A have two elements in common.
Thus, A × B and B × A will have elements in common iff sets A and B have elements in common.
Page No 2.8:
Question 10:
Let A and B be two sets such that n(A) = 3 and n(B) = 2.
If (x, 1), (y, 2), (z, 1) are in A × B, find A and B, where x, y, z are distinct elements.
Answer:
A is the set of all first entries in ordered pairs in A × B and B is the set of all second entries in ordered pairs in A × B.
Also,
n(A) = 3 and n(B) = 2
∴ A = {x, y, z} and B = {1, 2}
Page No 2.8:
Question 11:
Let A = {1, 2, 3, 4} and R = {(a, b) : a ∈ A, b ∈ A, a divides b}. Write R explicitly.
Answer:
Given:
A = {1, 2, 3, 4}
R = {(a, b) : a ∈ A, b ∈ A, a divides b}
We know:
1 divides 1, 2, 3 and 4.
2 divides 2 and 4.
3 divides 3.
4 divides 4.
∴ R = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 2), (2, 4), (3, 3), (4, 4)}
Page No 2.8:
Question 12:
If A = {−1, 1}, find A × A × A.
Answer:
Given:
A = {−1, 1}
Thus, we have:
A × A = {(−1, −1), (−1, 1), (1, −1), (1, 1)}
And,
A × A × A = {(−1, −1, −1), (−1, −1, 1), (−1, 1, −1), (−1, 1, 1), (1, −1, −1), (1, −1, 1), (1, 1, −1), (1, 1, 1)}
Page No 2.8:
Question 13:
State whether each of the following statements are true or false. If the statements is false, re-write the given statements correctly:
(i) If P = {m, n} and Q = {n, m}, then P × Q = {(m, n), (n, m)}
(ii) If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (x, y) such that x ∈ B and y ∈ A.
(iii) If A = {1, 2}, B = {3, 4}, then A × (B ∩ Ï) = Ï.
Answer:
(i) False
Correct statement:
If P = {m, n} and Q = {n, m}, then P × Q = {(m, n), (m, m), (n, n), (n, m)}.
(ii) False
Correct statement:
If A and B are non-empty sets, then A × B is a non-empty set of an ordered pair (x, y) such that x ∈ A and y ∈ B.
(iii) True
A = {1, 2} and B = {3, 4}
Now,
(B ∩ Ï) = Ï
The Cartesian product of any set and an empty set is an empty set.
∴ A × (B ∩ Ï) = Ï
Page No 2.8:
Question 14:
If A = {1, 2}, from the set A × A × A.
Answer:
Given:
A = {1, 2}
Now,
A × A = {(1, 1), (1, 2), (2, 1), (2, 2)}
∴ A × A × A = {(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2), (2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)}
Page No 2.8:
Question 15:
If A = {1, 2, 4} and B = {1, 2, 3}, represent following sets graphically:
(i) A × B
(ii) B × A
(iii) A × A
(iv) B × B
Answer:
Given:
A = {1, 2, 4} and B = {1, 2, 3}
(i) A × B = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (4, 1), (4, 2), (4, 3)}
(ii) B × A = {(1, 1), (1, 2), (1, 4), (2, 1), (2, 2), (2, 4), (3, 1), (3, 2), (3, 4)}
(iii) A × A = {(1, 1), (1, 2), (1, 4), (2, 1), (2, 2), (2, 4), (4, 1), (4, 2), (4, 4)}
(iv) B × B = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}
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