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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 × (BC), A × (BC), (A × B) ∪ (A × C).

Answer:

Given:
A = {2, 3}, B = {4, 5} and C ={5, 6}
Also,
(BC) = {4, 5, 6}
Thus, we have:
A × (BC) = {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3,6)}
And,
(BC) = {5}
Thus, we have:
A × (BC) = {(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 × (BC) = (A × B) ∪ (A × C)
(ii) A × (BC) = (A × B) ∩ (A × C)
(iii) A × (BC) = (A × B) − (A × C)

Answer:

Given:
A = {1, 2, 3}, B = {4} and C = {5}

(i) A × (BC) = (A × B) ∪ (A × C)
We have:
(BC) = {4, 5}
LHS: A × (BC)  = {(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 × (BC) = (A × B) ∩ (A × C)
We have:
(BC)  = ϕ
LHS: A × (BC) = ϕ
And,
(A × B) = {(1, 4), (2, 4), (3, 4)}
(A × C) = {(1, 5), (2, 5), (3, 5)}
RHS: (A × B) ∩ (A × C) = ϕ
∴ LHS = RHS


(iii) A × (BC) = (A × B) − (A × C)
We have:
(BC)  = ϕ
LHS: A × (BC) =  ϕ
Now,
(A × B) = {(1, 4), (2, 4), (3, 4)}
And,
(A × C) = {(1, 5), (2, 5), (3, 5)}
RHS: (A × B) − (A × C) = ϕ
∴ 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 × CB × D
(ii) A × (BC) = (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 × CB × 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 × CB × D


(ii) A × (BC) = (A × B) ∩ (A × C)
We have:
(BC)  = ϕ
LHS: A × (BC) = ϕ
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) = ϕ
∴ 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)  (AB) × C = (A × C) ∪ (B × C) (ii) (A ∩ B) × C = (A × C) ∩ (B×C)

Answer:

(i) (AB) × C = (A × C) ∪ (B × C)
Let (a, b) be an arbitrary element of (AB) × C.
Thus, we have:
(a,b) (AB)×Ca (AB) and bC(aA or aB) and  bC(aA and bC) or (aB and  bC)(a,b) (A×C) or (a,b) (B×C)(a,b) (A×C)(B×C)(AB)×C (A×C)(B×C)      ...(i)

Again, let (x, y) be an arbitrary element of (A × C) ∪ (B × C).
Thus, we have:
(x,y) (A×C)(B×C)(x,y) (A×C) or (x,y) (B×C)(xA & yC)  or (xB & yC)(xA or xB) or  yC(xAB) & yC(x,y) (AB)×C(A×C) (B×C) (AB)×C   ...(ii)

From (i) and (ii), we get:
(AB) × 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:
(a,b) (AB)×Ca (AB) & bC(aA & aB) & bC(aA & bC) &  (aB & bC)(a,b) (A×C) & (a,b) (B×C) (a,b) (A×C)(B×C) (AB)×C(A×C)(B×C)     ...iii

Again, let (x, y) be an arbitrary element of (A × C) ∩ (B × C).
Thus, we have:
(x,y) (A×C)(B×C)(x,y) (A×C) & (x,y) (B×C)(xA & yC) & (xB & yC) (xA & xB) & yCx(AB) & yC(x,y) (AB)×C (A×C)(B×C)(AB)×C      ...iv

From (iii) and (iv), we get:
 (AB) × C = (A × C) ∩ (B × C)

Page No 2.12:

Question 7:

If A × BC × D and A × B ≠ ϕ, prove that AC and BD.

Answer:

Let: (x, y) (A×B) xA, yBNow,(A×B)(C×D) (x,y) (C×D) Or, xC and y DThus, we have: AC & BD



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 ⇔ xy
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 ⇔ xy
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.
x = 12-3y2
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 = 8-2y
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, ab 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, ab is odd}
a are the elements of A and b are the elements of B.

a-b=3-7, 3-11, 5-7, 5-11a-b=-4, -8, -2, -6Here, a-b is always an even number.
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,
n (A×B) =n(A) ×n(B) = 2×2 = 4
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) S=a, b : b=a-1, a  Z and a 3

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) S=a, b : b=a-1, a  Z and a 3
Now,
a -3, -2, -1, 0, 1, 2, 3
b = -3-1 = 4b = -2-1 =3b = -1-1 =2b = 0-1 =1b = 1-1 =0b = 2-1 =1b = 3-1 =2
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 = 2×2=4
∴ Number of relations = 24=16 

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 A and B=3×2=6
Number of relations from A to B26=64

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 222.
∴ (2,2)R
False

(ii) (a, b) ∈ R ⇒ (b, a) ∈ R
∵ 4 = 22
(4, 2) ∈ R, but (2,4)R.
False

(iii) (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R
∵ 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) : 3xy = 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) : 3xy = 0, where x, y ∈ A}
Or,
R = {(x, y) : 3x = y, where x, y ∈ A}
As
3×1 = 33×2 = 63×3 = 93×4 = 12

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) ∵ 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,
4-1 = 3 and 6-1 = 5
y = 4, 6
For x = 2,
9-2 = 7
y = 9
For x = 3,
4-3 = 1 and 6-3 = 3
y = 4, 6
For x = 5,
5-4 =1 and 6-5 =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:
5-2 = 3
6-2 = 4
7-2 =  5
∴ R = {(x,y) : y = x-2, xP, yQ}

(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, ab 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 ab 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 = -12 and c = -4
Now,

1,-12R1 and -12,-4R1, as 1+-12 >0  and 1+-12-4>0.

But 1+1×-4 <0.
∴ (1,-4) R1
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
 a+b = b+a for all a,b N(a,b) R (a,b) for all a,b N

(ii) (a, b) R (c, d) ⇒ (c, d) R (a, b) for all (a, b), (c, d) ∈ N × N
(a,b) R (c,d) a+ d = b+c c+b = d+a (c,d) R (a,b)

(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
(a,b) R (c,d) and (c,d) R (e,f)a+d = b+c and c+f = d+ea+d+c+f = b+c+d+e a+f = b+e (a,b) R (e,f)



Page No 2.24:

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 2.24:

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) = nA×A×nB = 3×3×4 = 36

Page No 2.24:

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.
 Now,32+02 = 9-32+02= 9
x can take values -3, 0 and 3.
∴ Domain (R) = {-3, 0, 3}



Page No 2.25:

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:
-22+02 422+02 4-12+02 412+02 4-12+12 402+02 412+12 4-12+-12 4
∴ Domain (R) = {-2, -1, 0, 1, 2}

Page No 2.25:

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 = 11-3
10 = 13-3
∴ R = {(11, 8), (13, 10)}
Or,
R-1 = {(8, 11), (10, 13)}

Page No 2.25:

Question 6:

 Let A = {1, 2, 3} and R=a, b : a2-b25, a, b A. Then write R as set of ordered pairs.

Answer:

Given:
A = {1, 2, 3}
R=a, b : a2-b25, a, b A

We know that
12-12 5,22-22 5,32-32 5,12-22 5,22-12 5,22-32 5,32-22 5
Thus, R ={(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)}

Page No 2.25:

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

So, -2 = 2(a) -42= 2a a =1

Also, b2 = 24-4b2 =4b = ±2

Thus, a =1 and b = ±2

Page No 2.25:

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
1 = 3×2 - 57 = 3×4 - 5-2 = 3×1 - 5

Thus, the linear relation between the components of the ordered pairs of the relation R is y = 3x - 5.

Page No 2.25:

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 2.25:

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}

As y = 8-2xFor x = 0, y = 8For x  =1, y = 6For x = 2, y = 4For x = 3, y = 2For x = 4, y = 0For x = 5, y <0So, y <0 for all x>5  

∴ Domain (R) = {0,1,2,3,4} and Range (R) = {0,2,4,6,8}

Page No 2.25:

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

(x,1) A×B  xA , 1BSimilarly, yA, 2Band zA, 1B

So, A = {x,y,z} and B = {1,2}

Page No 2.25:

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) : xy is odd}. Write R in roster form.

Answer:

Given:
A = {1, 2, 3, 5} and B = {4, 6, 9}
R = {(x, y) : xy is odd}

Since 1-4  = -3 is odd, we have:
1-6 = -5 is odd
2-9 = -7 is odd
3-4  =-1 is odd
3-6 = -3 is odd
5-4 = 1 is odd
5-6 = -1 is odd

∴ R = {(1,4),(1,6),(2,9),(3,4),(3,6),(5,4),(5,6)}

Page No 2.25:

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 2.25:

Question 2:

If R is a relation on the set A = [1, 2, 3, 4, 5, 6, 7, 8, 9] given by x R yy = 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 yy = 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,
-22+02 422+02 4-12+02 412+02 4-12+12 402+02 412+12 4-12+-12 4)2+02 4(2)2+02 4(1)2+02 2)2+02 4(2)2+02 4(1)2+02 (2)2+02 4(2)2+02 4(1)2+02 4(1)2+02 4(1)2+(1)2 4
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 yx 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 yx 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 2.26:

Question 7:

A relation ϕ from C to R is defined by x ϕ yx = y. Which one is correct?

(a) (2 + 3i) ϕ 13
(b) 3ϕ (−3)
(c) (1 + i) ϕ 2
(d) i ϕ 1

Answer:

(d) i ϕ 1

We have i  =12+02 = 1
Thus, i ϕ 1 satisfies x ϕ yx = y.

Page No 2.26:

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 2.26:

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 2.26:

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 2.26:

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 2.26:

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
nA×B = mn
Then, the number of relations from A to B is 2mn.

Page No 2.26:

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) 2n2
(c) n2
(d) nn

Answer:

(b) 2n2

Given : A finite set with n elements
Its Cartesian product with itself will have n2  elements.
∴ Number of relations on A  = 2n2



Page No 2.8:

Question 1:

(i) If a3+1, b-23=53, 13, find the values of a and b.
(ii) If (x + 1, 1) = (3, y − 2), find the values of x and y.

Answer:

(i)  a3+1, b-23=53, 13
By the definition of equality of ordered pairs, we have:
a3+1, b-23=53, 13
a3+1 = 53 and b-23=13a3= 53-1 and b = 13+23a3=23 and b =  1a = 2 and b = 1

(ii) (x + 1, 1) = (3, y − 2)
By the definition of equality of ordered pairs, we have:
(x+1) = 3 and 1 = (y-2)x = 2  and y = 3

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

Page No 2.8:

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.

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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) : aA, bA, a divides b}. Write R explicitly.

Answer:

Given:
A = {1, 2, 3, 4}
R = {(a, b) : aA, bA, 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)}

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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)}

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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 ∩ ϕ) = ϕ.

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 ∩ ϕ) = ϕ
The Cartesian product of any set and an empty set is an empty set.
∴ A × (B ∩ ϕ) = ϕ

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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)}

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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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