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#### Question 1:

Given A = {1, 2, 3}, B = {3, 4}, C ={4, 5, 6}, find (A × B) ∩ (B × C).

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)}

#### Question 2:

If A = {2, 3}, B = {4, 5}, C ={5, 6}, find A × (BC), A × (BC), (A × B) ∪ (A × C).

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)}

#### 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)

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)  = $\varphi$
LHS: A × (BC) = $\varphi$
And,
(A × B) = {(1, 4), (2, 4), (3, 4)}
(A × C) = {(1, 5), (2, 5), (3, 5)}
RHS: (A × B) ∩ (A × C) = $\varphi$
∴ LHS = RHS

(iii) A × (BC) = (A × B) − (A × C)
We have:
(BC)  = $\varphi$
LHS: A × (BC) =  $\varphi$
Now,
(A × B) = {(1, 4), (2, 4), (3, 4)}
And,
(A × C) = {(1, 5), (2, 5), (3, 5)}
RHS: (A × B) − (A × C) = $\varphi$
∴ LHS = RHS

#### 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)

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)  = $\varphi$
LHS: A × (BC) = $\varphi$
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) = $\varphi$
∴ LHS = RHS

#### 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)

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)}

#### Question 6:

Prove that: (i)  (AB) × C = (A × C) ∪ (B × C) (ii) (A ∩ B) × C = (A × C) ∩ (B×C)

(i) (AB) × C = (A × C) ∪ (B × C)
Let (a, b) be an arbitrary element of (AB) × 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:
(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:

Again, let (x, y) be an arbitrary element of (A × C) ∩ (B × C).
Thus, we have:

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

#### Question 7:

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

#### 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.

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.

#### 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.

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}

#### 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.

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}

#### 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.

(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)}

#### 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.

(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)}

#### 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.

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)}

#### 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.

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.

So, R is an empty relation from A to B.
Hence proved.

#### Question 8:

Let A = [1, 2] and B = [3, 4]. Find the total number of relation from A into B.

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.

#### 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}

(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}

#### Question 10:

Determine the domain and range of the following relations:

(i) R = {(a, b) : a ∈ N, a < 5, b = 4}
(ii)

(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}

#### Question 11:

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

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 = $2×2=4\phantom{\rule{0ex}{0ex}}$
∴ Number of relations = ${2}^{4}=16$

#### Question 12:

Let A = (x, y, z) and B = (a, b). Find the total number of relations from A into B.

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${2}^{6}=64$

#### 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

Given: R = [(a, b) : a, b ∈ N and a = b2]

(i) (a, a) ∈ R for all a ∈ N.
Here, 2 ∈N, but $2\ne {2}^{2}$.
∴ (2,2)$\notin$R
False

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

#### 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.

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

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

#### 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.

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}

#### 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.

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)}

#### Question 17:

Write the relation R = {(x, x3) : x is a prime number less than 10} in roster form.

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)}

#### 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.

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}

#### 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

(i) We have:
5$-$2 = 3
6$-$2 = 4
7$-$2 =  5
∴ R =

(ii) R = {(5, 3), (6, 4), (7, 5)}

(iii) Domain (R) = {5, 6, 7}
Range (R) = {3, 4, 5}

#### 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.

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

#### 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.

We have:
(a, b) ∈ R1 ⇔ 1 + ab > 0
Let:
a = 1, b = $-\frac{1}{2}$ and c = $-$4
Now,

, as .

But .
∴ (1,$-$4) $\notin {R}_{1}$
And,
(a, b) ∈ R1 and (b , c) ∈ R1
Thus, (a, c) ∈ R1 is not true for all a, b, c ∈ R.

#### 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

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

#### Question 1:

If A = {1, 2, 4}, B = {2, 4, 5} and C = {2, 5}, write (A − C) × (B − C).

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)}

#### Question 2:

If n(A) = 3, n(B) = 4, then write n(A × A × B).

Given:
n(A) = 3 and n(B) = 4
Now, we have:
n(A × A × B) =

#### 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.

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}

#### 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.

Given:
R = {(x, y) : x, y ∈ Z, x2 + y2 ≤ 4}
We know:

∴ Domain (R) = {$-$2, $-$1, 0, 1, 2}

#### 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.

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)}

#### Question 6:

Let A = {1, 2, 3} and . Then write R as set of ordered pairs.

Given:
A = {1, 2, 3}

We know that

Thus, R ={(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)}

#### 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.

R = [(x, y) : x, y ∈ Z, y = 2x − 4]
(a, $-$2) and (4, b2) ∈ R

Thus, a =1 and b = $±2$

#### 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.

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.

#### 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}

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)}

#### Question 10:

If R = [(x, y) : x, y ∈ W, 2x + y = 8], then write the domain and range of R.

R = {(x, y) : x, y ∈ W, 2x + y = 8}

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

#### 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.

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}

#### 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.

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)}

#### 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

(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)}

#### 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

(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)}

#### 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

(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)}

#### 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.

(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}

#### 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

(c) {−2, −1, 0, 1, 2}
R = {(x, y) : x, y ∈ Z, x2 + y2 ≤ 4}
We know that,
)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,}

#### 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]

(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}

#### Question 7:

A relation ϕ from C to R is defined by x ϕ y$\left|x\right|$ = y. Which one is correct?

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

(d) i ϕ 1

We have
Thus, i ϕ 1 satisfies x ϕ y$\left|x\right|$ = y.

#### 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]

(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}

#### 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

(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)}

#### 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

(c) pq

n
(A × B) = n(A) × n(B)
= p × q = pq

#### 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

(c) R ⊆ A × B

If R is a relation from set A to set B, then R is always a subset of A × B.

#### 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

(a) 2mn

Given: n(A) = m
n(B) = n

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

#### 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) ${2}^{{n}^{2}}$
(c) n2
(d) nn

(b) ${2}^{{n}^{2}}$

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

#### 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.

(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:

#### 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.

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

#### 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.

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.

#### 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.

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)}

#### Question 5:

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

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)}

#### Question 6:

Let A = {1, 2, 3} and B = {3, 4}. Find A × B and show it graphically.

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.

#### 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)?

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)}

#### 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)].

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

#### 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.

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.

#### 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.

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}

#### Question 11:

Let A = {1, 2, 3, 4} and R = {(a, b) : aA, bA, a divides b}. Write R explicitly.

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)}

#### Question 12:

If A = {−1, 1}, find A × A × A.

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)}

#### 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.

(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

#### Question 14:

If A = {1, 2}, from the set A × A × A.

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)}

#### 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

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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