Relations

Given:

To find: values of a and b

By the definition of equality of ordered pairs, we have and simultaneously solving for a and b

and

⇒ b = 1

⇒ a = 2

given: (x + 1, 1) = (3y, y - 1)

To find: values of a and b

By the definition of equality of ordered pairs, we have

x + 1 = 3y and 1 = y - 1

⇒ x = 3y - 1 and y = 2

So,

x = 3(2) - 1
= 6 - 1
= 5

⇒ x = 5 and y = 2

given the ordered pairs (x, - 1) and (5, y) belong to the set {(a, b): b = 2a - 3}

To find: values of x and y

solving for first order pair

(x, - 1) = {(a, b): b = 2a - 3}

x = a and b = - 1

If b = - 1 then 2a = - 1 + 3 = 2

So, a = 1

x = 1

Similarly, solving for second order pair

(5, y) = {(a, b): b = 2a - 3}

a = 5 and y = b

If a = 5 then b = 2×5 - 3

So, b = 7

y = 7

given a ∈ { - 1, 2, 3, 4, 5} and b ∈ {0, 3, 6},

To find: the ordered pair (a, b) such that a + b = 5

then the ordered pair (a, b) such that a + b = 5 are as follows

(a, b)∈ {( - 1, 6), (2, 3), (5, 0)}

given that a ∈ {2, 4, 6, 9} and b ∈{4, 6, 18, 27}.

To find: ordered pairs (a, b) such that a divides b and a<b

Here,

2 divides 4, 6, 18 and is also less than all of them

4 divides 4 and is also less than none of them

6 divides 6, 18 and is less than 18 only

9 divides 18, 27 and is less than all of them

Therefore, ordered pairs (a, b) are (2, 4), (2, 6), (2, 18),

(6, 18), (9, 18) and (9, 27)

Given A = {1, 2} and B = {1, 3}

To find: A × B, B × A

A × B = {(1, 1), (1, 3), (2, 1), (2, 3)}

B × A = {(1, 1), (1, 2), (3, 1), (3, 2)}

given: A = {1, 2, 3} and B = {3, 4}

To find: graphical representation of A × B

A x B = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}

To represent A × B graphically, given steps must be followed:

a. One horizontal and one vertical axis should be drawn

b. Element of set A should be represented in horizontal axis and on vertical axis elements of set B should be represented

c. Draw dotted lines perpendicular to horizontal and vertical axes through the elements of set A and B

d. Point of intersection of these perpendicular represents A × B.

given: A = {1, 2, 3} and B = {2, 4}

To find: A × B, B × A, A × A, (A × B) ∩ (B × A)

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)}

Intersection of two sets represents common elements of both the sets

So,

(A × B) ∩ (B × A) = {(2, 2)}

given: (A) = 5 and n(B) = 4

To find: [(A × B) ∩ (B×A)]

n (A × B) = n(A) × n(B) = 5 x 4 = 20

n (A ∩ B) = 3 (given: A and B has 3 elements in common)

In order to calculate n [(A × B) ∩ (B × A)], we will assume

A = (x, x, x, y, z) and B = (x, x, x, p)

So, we have

(A × B) = {(x, x), (x, x), (x, x), (x, p), (x, x), (x, x), (x, x), (x, p), (x, x), (x, x), (x, x), (x, p), (y, x), (y, x), (y, x), (y, p), (z, x), (z, x), (z, x), (z, p)}

(B × A) = {(x, x), (x, x), (x, x), (x, y), (x, z), (x, x), (x, x), (x, x), (x, y), (x, z), (x, x), (x, x), (x, x), (x, y), (x, z), (p, x), (p, x), (p, x), (p, y), (p, z)}

[(A × B) ∩ (B × A)] = {(x, x), (x, x), (x, x), (x, x), (x, x), (x, x), (x, x), (x, x), (x, x)}

∴ We can say that n [(A × B) ∩ (B × A)] = 9

given: n (A) = 3 n (B) = 2

To prove: The sets A x B and B x A have an element in common if the sets A and B be two sets such that n (A) = 3 and n (B) = 2

Proof:

Case 1: No elements are common

Assuming:

A = (a, b, c) and B = (e, f)

So, 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)}

There are no common ordered pair in A × B and B × A.

Case 2: One element is common

Assuming:

A = (a, b, c) and B = (a, f)

So, 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 one ordered pair in common.

Therefore, we can say that A × B and B × A will have elements in common if and only if sets A and B have an element in common.

given: n(A) = 3 and n(B) = 2

To find: distinct elements of set A and B

Also, it is given that {(x, 1), (y, 2), (z, 1)} A × B

Set A has 3 elements whereas Set B has 2 elements.

Also, A × B = {(a, b): a ∈ A and b ∈ B}

Therefore, A ∈ {x, y, z} and B ∈ {1, 2}

given: A = {1, 2, 3, 4} and R = {(a, b): a ∈ A, b ∈ A, a divides b}

To find: set R

Both elements of R, a and b, belongs to set A and relation between and elements a and b is that a divides b

So,

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)}

given: A = {−1, 1}

To find: A × A × A

So, 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)}

(i) False

given: P = {m, n} and Q = {n, m}, then

P × Q = {(m, n), (m, m), (n, n), (n, m)}.

(ii) False

given: 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

given: A = {1, 2} and B = {3, 4}

∅ is represents null setand intersection of any set with null set gives null set as null set has no elements

(B ∩ ∅) = ∅

The Cartesian product of any set and an empty set will be an empty set.

∴ A × (B ∩ ∅) = ∅

given A = {1, 2}

To find A × A × A

Firstly, we will findCartesian product of A with A

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

Now, Cartesian product of A x A with A

∴ 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)}

given: A = {1, 2, 4} B = {1, 2, 3}

(i) To find: A × B

A × B = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (4, 1), (4, 2), (4, 3)}

(ii) To find: B × A

B × A = {(1, 1), (1, 2), (1, 4), (2, 1), (2, 2), (2, 4), (3, 1), (3, 2), (3, 4)}

(iii) To find: A × A

A × A = {(1, 1), (1, 2), (1, 4), (2, 1), (2, 2), (2, 4), (4, 1), (4, 2), (4, 4)}

(iv) To find: B × B

B × B = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}

given: A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}

To find: (A × B) ∩ (B × C)

(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)}

given: A = {2, 3}, B = {4, 5} and C = {5, 6}

To find: (A × B) ∪ (A × C)

Since, (B ∪ C) = {4, 5, 6}

∴A × (B ∪ C) = {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}

(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)}

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

(i) To prove: A × (B ∪ C) = (A × B) ∪ (A × C)

LHS: (B ∪ C) = {4, 5}

therefore A × (B ∪ C) = {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}

RHS:

(A × B) = {(1, 4), (2, 4), (3, 4)}

(A × C) = {(1, 5), (2, 5), (3, 5)}

(A × B) ∪ (A × C) = {(1, 4), (2, 4), (3, 4), (1, 5), (2, 5), (3, 5)}

∴ LHS = RHS

(ii) To prove: A × (B ∩ C) = (A × B) ∩ (A × C)

LHS: (B ∩ C) = ∅ (No common element)

A × (B ∩ C) = ∅

RHS: (A × B) = {(1, 4), (2, 4), (3, 4)}

(A × C) = {(1, 5), (2, 5), (3, 5)}

(A × B) ∩ (A × C) = ∅

∴ LHS = RHS

(iii) To prove: A × (B − C) = (A × B) − (A × C)

LHS: (B − C) = ∅

A × (B − C) = ∅

RHS: (A × B) = {(1, 4), (2, 4), (3, 4)}

(A × C) = {(1, 5), (2, 5), (3, 5)}

(A × B) − (A × C) = ∅

∴ LHS = RHS

given: A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}

(i) To prove: 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)}

Since, all elements of A × C is in B × D.

∴We can A × C ⊂ B × D

(ii) To prove: A × (B ∩ C) = (A × B) ∩ (A × C)

LHS: (B ∩ C) = ∅

A × (B ∩ C) = ∅

RHS: (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)}

Since, there is no common element between A × B and A × C

(A × B) ∩ (A × C) = ∅

∴ LHS = RHS

given: A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}

(i) To find: A × (B ∩ C)

(B ∩ C) = {4}

∴ A × (B ∩ C) = {(1, 4), (2, 4), (3, 4)}

(ii) To find: (A × B) ∩ (A × C)

(A × B) = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}

(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) To find: A × (B ∪ C)

(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) To find: (A × B) ∪ (A × C)

(A × B) = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}

(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)}

To prove: (A ∪ B) × C = (A × C) ∪ (B × C)

Proof:

Let (x, y) be an arbitrary element of (A ∪ B) × C.

(x, y) ∈ (A ∪ B) C

Since, (x, y) are elements of Cartesian product of (A ∪ B) × C

x ∈ (A ∪ B) and y ∈ C

(x ∈ A or x∈B) and y ∈ C

(x ∈ A and y ∈ C) or (x ∈ Band y ∈ C)

(x, y) ∈ A × C or (x, y) ∈ B × C

(x, y) ∈ (A × C) ∪ (B × C) …1

Let (x, y) be an arbitrary element of (A × C) ∪ (B × C).

(x, y) ∈ (A × C) ∪ (B × C)

(x, y) ∈ (A × C) or (x, y) ∈ (B × C)

(x ∈ A and y ∈ C) or (x ∈ Band y ϵ C)

(x ∈ A or x ∈ B) and y ∈ C

x ∈ (A ∪ B) and y ∈ C

(x, y) ∈ (A ∪ B) × C …2

From 1 and 2, we get: (A ∪ B) × C = (A × C) ∪ (B × C)

To prove: (A ∩ B) × C = (A × C) ∩ (B×C)

Proof:

Let (x, y) be an arbitrary element of (A ∩ B) × C.

(x, y) ∈ (A ∩ B) × C

Since, (x, y) are elements of Cartesian product of (A ∩ B)× C

x ∈ (A ∩ B) and y ∈ C

(x ∈ A and x ∈B) and y ∈ C

(x ∈ A and y ∈ C) and (x ∈ Band y ∈ C)

(x, y) ∈ A × C and (x, y) ∈ B × C

(x, y) ∈ (A × C) ∩ (B × C) …1

Let (x, y) be an arbitrary element of (A × C) ∩ (B × C).

(x, y) ∈ (A × C) ∩ (B × C)

(x, y) ∈ (A × C) and (x, y) ∈ (B × C)

(x ∈A and y ∈ C) and (x ϵ Band y ∈ C)

(x ∈A and x ∈ B) and y ∈ C

x ∈ (A ∩ B) and y ∈ C

(x, y) ∈ (A ∩ B) × C …2

From 1 and 2, we get: (A ∩ B) × C = (A × C) ∩ (B × C)

given A × B C x D and A ∩ B ∈ ∅

To prove: A C and B D

A × B C x D denotes A × B is subset of C × D that is every element A × B is in C × D

And A ∩ B ∈ ∅ denotes A and B does not have any common element between them.

A × B = {(a, b): a ∈ A and b ∈ B}

Since,

A × B C x D (Given)

∴We can say (a, b) C × D

a ∈ C and b ∈ D

A ∈ C and B ∈ D (A and B does not have common elements)

Given,

A = {1, 2, 3}, B = {4, 5, 6}

A relation from A to B can be defined as:

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)}

No, it is not a relation from A to B. The given set is not a subset of A × B as (5, 2) is not a part of the relation from A to B.

ii. {(1, 5), (2, 6), (3, 4),(3, 6)}

Yes, it is a relation from A to B. The given set is a subset of A × B.

iii. {(4, 2), (4, 3), (5, 1)}

No, it is not a relation from A to B. The given set is not a subset of A × B as (4, 2), (4, 3), (5, 1) are not a part of the relation from A to B.

iv. A × B

A × B is a relation from A to B can be defined as:

{(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6),(3, 4),(3, 5),(3, 6)}

Relatively prime numbers are also known as co-prime numbers. If there is no integer greater than one that divides both (that is, their greatest common divisor is one). For example, 12 and 13 are relatively prime, but 12 and14 are not as their greatest common divisor is two.

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.

∴ R = {(2, 3), (2, 7), (3, 7), (3, 10), (4, 3), (4, 7), (5, 3), (5, 6), (5, 7)}

Domain of relation R = {2, 3, 4, 5}

Range of relation R = {3, 6, 7, 10}

A is set of first five natural numbers.

Therefore, A= {1, 2, 3, 4, 5}

Given, (x, y) R x ≤ y

1 is less than 2, 3, 4 and 5.

2 is less than 3, 4 and 5.

3 is less than 4 and 5.

4 is less than 5.

5 is not less than any number A

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

An inverse relation is the set of ordered pairs obtained by interchanging the first and second elements of each pair in the original relation. If the graph of a function contains a point (a, b), then the graph of the inverse relation of this function contains the point (b, a).

∴ R^-1 = {(2, 1), (3, 1), (4, 1), (5, 1), (3, 2), (4, 2), (5, 2), (4, 3), (5, 3), (5, 4)}

⇒ R^-1 = {(2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3), (5, 4)}

i. Domain of R^‑1 = {2, 3, 4, 5}

ii. Range of R = {2, 3, 4, 5}

NOTE: You can see that Domain of R^‑1 is same as Range of R. Similarly, Domain of R is same as Range of R^‑1

An inverse relation is the set of ordered pairs obtained by interchanging the first and second elements of each pair in the original relation. If the graph of a function contains a point (a, b), then the graph of the inverse relation of this function contains the point (b, a).

i. Given, R= {(1, 2), (1, 3), (2, 3), (3, 2), (5, 6)}

∴ R^‑1 = {(2, 1), (3, 1), (3, 2), (2, 3), (6, 5)}

⇒ R^‑1 = {(2, 1), (2, 3), (3, 1), (3, 2), (6, 5)}

ii. Given, R= {(x, y) : x, y N; x + 2y = 8}

Here, x + 2y = 8

⇒ x = 8 – 2y

As y N, Put the values of y = 1, 2, 3,…… till x N

On putting y=1, x = 8 – 2(1) = 8 – 2 = 6

On putting y=2, x = 8 – 2(2) = 8 – 4 = 4

On putting y=3, x = 8 – 2(3) = 8 – 6 = 2

On putting y=4, x = 8 – 2(4) = 8 – 8 = 0

Now, y cannot hold value 4 because x = 0 for y = 4 which is not a natural number.

∴ R = {(2, 3), (4, 2), (6, 1)}

R^‑1 = {(3, 2), (2, 4), (1, 6)}

⇒ R^‑1 = {(1, 6), (2, 4), (3, 2)}

iii. Given, R is a relation from {11, 12, 13} to (8, 10, 12} defined by y = x – 3

Here,

x {11, 12, 13} and y (8, 10, 12}

y = x – 3

On putting x = 11, y = 11 – 3 = 8 (8, 10, 12}

On putting x = 12, y = 12 – 3 = 9 ∉ (8, 10, 12}

On putting x = 13, y = 13 – 3 = 10 (8, 10, 12}

∴ R = {(11, 8), (13, 10)}

R^‑1 = {(8, 11), (10, 13)}

Let A = {2, 3, 4, 5, 6} and B = {1, 2, 3}

Given, x = 2y where x {2, 3, 4, 5, 6} and y {1, 2, 3}

On putting y = 1, x = 2(1) = 2 A

On putting y = 2, x = 2(2) = 4 A

On putting y = 3, x = 2(3) = 6 A

∴ R = {(2, 1), (4, 2), (6, 3)}

Relatively prime numbers are also known as co-prime numbers. If there is no integer greater than one that divides both (that is, their greatest common divisor is one). For example, 12 and 13 are relatively prime, but 12 and14 are not as their greatest common divisor is two.

Given, (x, y) R x is relatively prime to y

Here,

2 is co-prime to 3, 5 and 7.

3 is co-prime to 2, 4, 5 and 7.

4 is co-prime to 3, 5 and 7.

5 is co-prime to 2, 3, 4, 6 and 7.

6 is co-prime to 5 and 7.

7 is co-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)}

Given, (x, y) R 2x + 3y = 12

where x and y {0,1,2,…,10}

2x + 3y = 12

⇒ 2x = 12 – 3y

On putting y=0,

On putting y=2,

On putting y=4,

∴ R = {(0, 4), (3, 2), (6, 0)}

Given, (x, y) R x divides y

where x {5, 6, 7, 8} and y {10, 12, 15, 16, 18}

Here,

5 divides 10 and 15.

6 divides 12 and 18.

7 divides none of the value of set B.

8 divides 16.

∴ R = {(5, 10), (5, 15), (6, 12), (6, 18), (8, 16)}

Given, (x, y) R x + 2y = 8 where x N and y N

x = 8 – 2y

As y N, Put the values of y = 1, 2, 3,…… till x N

On putting y=1, x = 8 – 2(1) = 8 – 2 = 6

On putting y=2, x = 8 – 2(2) = 8 – 4 = 4

On putting y=3, x = 8 – 2(3) = 8 – 6 = 2

On putting y=4, x = 8 – 2(4) = 8 – 8 = 0

Now, y cannot hold value 4 because x = 0 for y = 4 which is not a natural number.

∴ R = {(2, 3), (4, 2), (6, 1)}

An inverse relation is the set of ordered pairs obtained by interchanging the first and second elements of each pair in the original relation. If the graph of a function contains a point (a, b), then the graph of the inverse relation of this function contains the point (b, a).

R^‑1 = {(3, 2), (2, 4), (1, 6)}

⇒R^‑1 = {(1, 6), (2, 4), (3, 2)}

Given, A = {3, 5} and B = {7, 11}

R = {(a, b): a A, b B, a-b is odd}

On putting a = 3 and b = 7:

⇒ a – b = 3 – 7 = -4 which is not odd

On putting a = 3 and b = 11:

⇒ a – b = 3 – 11 = -8 which is not odd

On putting a = 5 and b = 7:

⇒ a – b = 5 – 7 = -2 which is not odd

On putting a = 5 and b = 11:

⇒ a – b = 5 – 11 = -6 which is not odd

∴ R = { } = Φ

⇒ R is an empty relation from into B

Given,

A= {1, 2}, B= {3, 4}

n(A) = 2 (Number of elements in set A).

n(B) = 2 (Number of elements in set B).

We know,

n(A × B) = n(A) × n(B) = 2 × 2 = 4

∴ Number of relations from A to B are 4.

NOTE:

Given,

A= {1, 2}, B= {3, 4}

A relation from A to B can be defined as:

A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}

∴ Number of relations from A to B are 4

Given,

R = {(x, x+5): x {0, 1, 2, 3, 4, 5}

∴ R = {(0, 0+5), (1, 1+5), (2, 2+5), (3, 3+5), (4, 4+5), (5, 5+5)}

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

So,

Domain of relation R = {0, 1, 2, 3, 4, 5}

Range of relation R = {5, 6, 7, 8, 9, 10}

Given,

R = {(x, x³): x is a prime number less than 10}

Prime numbers less than 10 are 2, 3, 5 and 7

∴ R = {(2, 2³), (3, 3³), (5, 5³), (7, 7³)}

⇒ R = {(2, 8), (3, 27), (5, 125), (7, 343)}

So,

Domain of relation R = {2, 3, 5, 7}

Range of relation R = {8, 27, 125, 343}

Given,

R= {a, b): a N, a < 5, b = 4}

Natural numbers less than 5 are 1, 2, 3 and 4

Therefore, a {1, 2, 3, 4} and b {4}

⇒ R = {(1, 4), (2, 4), (3, 4), (4, 4)}

So,

Domain of relation R = {1, 2, 3, 4}

Range of relation R = {4}

Given,

S= {a, b): b = |a-1|, a Z and |a| ≤ 3}

Z denotes integer which can be positive as well as negative

Now, |a| ≤ 3 and b = |a-1|

∴ a {-3, -2, -1, 0, 1, 2, 3}

S = {a, b): b = |a-1|, a Z and |a| ≤ 3}

⇒ S = {a, |a-1|): b = |a-1|, a Z and |a| ≤ 3}

⇒ S = {(-3, |-3 – 1|), (-2, |-2 – 1|), (-1, |-1 – 1|), (0, |0 – 1|), (1, |1 – 1|), (2, |2 – 1|), (3, |3 – 1|)}

⇒S = {(-3, |-4|), (-2, |-3|), (-1, |-2|), (0, |-1|), (1, |0|), (2, |1|), (3, |2|)}

⇒S = {(-3, 4), (-2, 3), (-1, 2), (0, 1), (1, 0), (2, 1), (3, 2)}

So,

Domain of relation S = {-3, -2, -1, 0, 1, 2, 3}

Range of relation S = {0, 1, 2, 3, 4}

The total number of relations that can be defined from a set A to a set B is the number of possible subsets of A × B. If n(A)=p and n(B)=q, then n(A × B)= p q. So, the total number of relations is 2^pq.

Now,

A × A = {(a, a), (a, b), (b, a), (b, b)}

Total number of relations are all possible subsets of A × A:

{ Φ, {(a, a)}, {(a, b)}, {(b, a)}, {(b, b)}, {(a, a), (a, b)}, {(a, a), (b, a)}, {(a, a), (b, b)}, {(a, b), (b, a)}, {(a, b), (b, b)}, {(b, a), (b, b)}, {(a, a), (a, b), (b, a)}, {(a, b), (b, a), (b, b)}, {(a, a), (b, a), (b, b)}, {(a, a), (a, b), (b, b)}, {(a, a), (a, b), (b, a), (b, b)}}

n(A) = 2 ⇒ n(A × A) = 2 × 2 = 4

∴ Total number of relations = 2⁴ = 16

The total number of relations that can be defined from a set A to a set B is the number of possible subsets of A × B. If n(A)=p and n(B)=q, then n(A × B)= pq. So, the total number of relations is 2^pq.

n(A) = 3 and n(B) = 2 ⇒ n(A × B) = 3 × 2 = 6

∴ Total number of relations = 2⁶ = 64

Given, R= {(a, b): a, b N and a = b²}

i. (a, a) R for all a N

Here, take b = 2

⇒ a = b² = 2² = 4

∴ (4, 2) R but (2, 2) ∉ R

As, 2² ≠ 2

So,

No, the statement is false.

ii. (a, b) R (b, a) R

Here, take b = 2

⇒ a = b² = 2² = 4

∴ (4, 2) R but (2, 4) ∉ R

As, 4² ≠ 2

So,

No, the statement is false.

iii. (a, b) R and (b, c) R (a, c) R

Here, take b = 4

⇒ a = b² = 4² = 16

⇒ (16, 4) R

Now, b = c²

⇒ 4 = c²

⇒ c = -2 ∉ N or 2 N

⇒ (4, 2) R

But (16, 2) ∉ R

As, 2² ≠ 16

So,

No, the statement is false.

Given, R= {(x, y): 3x – y = 0, where x, y A}

A= {1, 2, 3,…,14}

As, y = 3x

∴ R= {(x, 3x): where x, 3x A}

⇒ R= {(1, 3×1), (2, 3×2), (3, 3×3), (4, 3×4)}

NOTE: We cannot include (5, 3×5) as 15∉A

⇒ R = {(1, 3), (2, 6), (3, 9), (4, 12)}

So,

Domain of relation R= {1, 2, 3, 4}

Co-Domain of relation R= {1, 2, 3,…,14}= A

Range of relation R= {3, 6, 9, 12}

Given,

R = {(x, y): y = x + 5, x is a natural less than 4, x, y N}

Natural numbers less than 4 are 1, 2 and 3.

On putting x= 1, y = 1 + 5 = 6

On putting x= 2, y = 2 + 5 = 7

On putting x= 3, y = 3 + 5 = 8

i. R = {(1, 6), (2, 7), (3, 8)}

So,

Domain of relation R= {1, 2, 3}

Range of relation R= {6, 7, 8}

Given,

Relation R from A to B by R = {(x, y): the difference between x and y is odd, x A, y B}

A = {1, 2, 3, 5} and B = {4, 6, 9}

For x = 1:

y – x = 4 – 1 = 3 which is odd ⇒ 4 y

y – x = 6 – 1 = 5 which is odd ⇒ 6 y

y – x = 9 – 1 = 8 which is even ⇒ 8 ∉ y

For x = 2:

y – x = 4 – 2 = 2 which is even ⇒ 4 ∉ y

y – x = 6 – 2 = 4 which is even ⇒ 6 ∉ y

y – x = 9 – 2 = 7 which is odd ⇒ 8 y

For x = 3:

y – x = 4 – 3 = 1 which is odd ⇒ 4 y

y – x = 6 – 3 = 3 which is odd ⇒ 6 y

y – x = 9 – 3 = 6 which is even ⇒ 8 ∉ y

For x = 5:

x – y = 5 – 4 = 1 which is odd ⇒ 4 y

y – x = 6 – 5 = 4 which is even ⇒ 6 ∉ y

y – x = 9 – 4 = 4 which is even ⇒ 8 ∉ y

∴ R = {(1, 4), (1, 6), (2, 8), (3, 4), (3, 6), (5, 4)}

NOTE:

Domain of relation R= {1, 2, 3, 5}

Range of relation R= {4, 6, 8}

Given,

R = {(x, x³): x is a prime number less than 10}

Prime numbers less than 10 are 2, 3, 5 and 7

∴ R = {(2, 2³), (3, 3³), (5, 5³), (7, 7³)}

⇒ R = {(2, 8), (3, 27), (5, 125), (7, 343)}

So,

Domain of relation R = {2, 3, 5, 7}

Range of relation R = {8, 27, 125, 343}

Given,

R= {(a, b): a, b A, b is exactly divisible by a}

A= {1, 2, 3, 4, 5, 6}

Here,

6 is exactly divisible by 1, 2, 3 and 6

5 is exactly divisible by 1 and 5

4 is exactly divisible by 1, 2 and 4

3 is exactly divisible by 1 and 3

2 is exactly divisible by 1 and 2

1 is exactly divisible by 1

i. R = {(1, 1), (2, 1), (2, 2), (3, 1), (3, 3), (4, 1), (4, 2), (4,4), (5, 1), (5, 5), (6, 1), (6, 2), (6, 3), (6, 6)}

ii. Domain of relation R = {1, 2, 3, 4, 5, 6}

iii. Range of relation R = {1, 2 , 3, 4, 5, 6}

i. Since 5 – 3 = 6 – 4 = 7 – 5 = 2

∴ x – y = 2 where x P and y Q

So,

R = {(x, y): x – y = 2, x P, y Q}

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

iii. Domain of relation R = {5, 6, 7}

Range of relation R = {3, 4, 5}

Given, R = {(a, b) Z, a – b is an integer}

Z denotes integer, and here a and b both are integers

We know that difference of two integers is always an integer

∴ a and b can be any integer in relation R

⇒ The domain of relation R = Z (as a Z)

The range of relation R = Z (as b Z)

To prove: (a, b) R₁ and (b,c) R₁ (a, c) R₁ is not true for all a, b, c R.

Given R₁ = {(a, b): 1 + ab > 0}

Let a = 1, b = -0.5, c = -4

Here, (1, -0.5) R₁ [∵ 1+(1×-0.5) = 0.5 > 0]

And, (-0.5, -4) R₁ [∵ 1+(-0.5×-4) = 3 > 0]

But, (1, -4) ∉ R₁ [∵ 1+(1×-4) = -3 < 0]

∴ (a, b) R₁ and (b,c) R₁ (a, c) R₁ is not true for all a, b, c R

Hence Proved.

NOTE:

Here R₁ is a relation whereas R denotes a real number.

Given,

(a, b) R (c, d) a + d = b + c for all (a, b), (c, d) N x N

i. (a, b) R (a, b)

⇒ a + b = b + a for all (a, b) N x N

∴ (a, b) R (a, b) for all (a, b) N x N

ii. (a, b) R (c, d)

⇒ a + d = b + c ⇒ c + b = d + a

⇒ (c, d) R (a, b) for all (c, d), (a, b) N x N

iii. (a, b) R (c, d) and (c, d) R (e, f)

a + d = b + c and c + f = d + e

⇒ a + d + c + f = b + c + d + e

⇒ a + f = b + c + d + e – c – d

⇒ a + f = b + e

⇒ (a, b) R (e, f) for all (a, b), (c, d), (e, f) N × N

Here, A={1,2,4} ,B={2,4,5} ,C={2,5}

So, A ∖ C={1,2,4} ∖ {2,5}

={1,4}

Also,B ∖ C={2,4,5}∖ {2,5}

={4}

Now, (A ∖ B)×(B ∖ C)={(x,y):xϵ(A∖ C) and yϵ(B∖ C) }

={(1,4), (4,4)}

We know ,n(A×B)=n(A)×n(B)

Similarly, n(A×B×C)=n(A)×n(B)×n(C)

Here, n(A)=3 and n(B)=4

n(A×A×C)=n(A)×n(A)×n(B)

=3×3×4

=36

Here, relation R is defined on the set Z.

And by the given definition of relation, R={(x,y) : x,yϵ Z ,x² +y²=9}

={(-3,0),(0,-3),(3,0),(0,3)}

Now we know,Domain is the set which consist all first elements of ordered pairs in relation R.

So,Domain(R)={-3,0,3}

Here, relation R is defined on the set Z.

And R={(x,y) : x,yϵ Z,x²+y²≤4}

={(-2,0),(-1,0),(0,0),(1,0),(2,0),(0,-2),(0,-1),(0,1),(0,2),(1,1),(-1,-1),(1,-1),(-1,1)}

Now we know,Domain is the set which consist all first elements of ordered pairs in relation R.

So,Domain(R)={-2,-1,0,1,2}

Here,A={11,12,13} and B={8,10,12}

Also,R={(x,y) : y=x-3 ,xϵ A ,yϵ B}

={(11,8),(13,10)}

Now,R^-1 ={(x,y) : x=y+3 ,xϵB ,yϵ A}

={(8,11),(10,13)}

Here,A={1,2,3}

R={(a,b) : |a²-b²|≤5,a,b ϵ A}

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

Here,R={(x,y) : x, yϵZ, y=2x-4}

Now,(a,-2)ϵ R ⇒ -2=2a-4

∴ 2a=-2+4

∴ 2a=2

∴ a=1

Also,(4,b²)ϵ R ⇒ b²=2×4-4

∴ b²=8-4

∴ b²=4

∴ b=±2

Here,R={(2,1),(4,7),(1,-2),…}

It is seen that all the elements of R have a rule that (x,y)ϵ R ⇒ y+5=3x

i.e y=3x-5.

Here, A={1,3,5} and B={2,4}

Also, R={ (x,y) : x,yϵ A×B and x>y}

={(3,2) ,(5,2) ,(5,4)}

Here, R={ (x, y) :x ,yϵ W,2x+y=8}

={(0,8) ,(1,6) ,(2,4) ,(3,2) ,(4,0)}

Now we know, Domain is the set which consist all first elements of ordered pairs in relation R.

So, Domain(R)= {0,1,2,3,4}

Also we know, Range is the set which consist all second elements of ordered pairs in relation R.

So, Range(R)={0,2,4,6,8}

Here, A×B ={(x,1) ,(y,2) ,(z,1)} and n(A)=3 , n(B)=2

We know ,A×B = {(x,y) : xϵA and yϵ B}

∴ A={x,y,z} and B={1,2}

Here, A= {1, 2, 3, 5} and B={4,6,9}

Also, R= {(x, y) : x-y is odd}

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

When we subtract two sets, say (A – B), the result will be a set obtained on removing those elements from A which also exist in B.

Note : We do not consider the elements of the subtracted set(here B) if it is not present in A.

So, we have (A – B) = {1, 2, 4} – {2, 4, 5}

= {1}

Similarly, we have (B - C) = {2, 4, 5} – {2, 5}

= {4}

[When we multiply two sets, each element of first set is paired with every element of other in an ordered pair of form (x, y)

where x belongs to first set and y to the other.]

Now, (A−B) × (B − C) = {1} × {4}

= {(1, 4)}

Since, it is a set so it is written in curly braces.

Therefore, option B is correct.

Here, y = 3x;

If x = 1; then y = 3.

If x = 2; then y = 6.

If x = 3; then y = 9.

Therefore the required relation will be R = {(1, 3), (2, 6), (3, 9)}.

So, correct answer is D.

Inverse of a relation is given by interchanging the element’s position in each pair.

Ex: Inverse of relation P = {(x, y)} is given by P⁻¹ = {(y, x)}.

Therefore, R⁻¹ = {(3, 1), (5, 2), (3, 3)}.

So, option A is correct.

As per condition of the relation, x > y.

So, required relation will be : {(2, 1), (3, 1)}

Since we know that Range is the set of elements written after comma in each ordered pair.

Therefore, Range = {1}

So, option C is correct.

Domain of R is a set constituting all values of x.

Here, possible values for x by equation x² + y²≤ 4 will be 0, 1, -1, 2, -2.

So, Domain of R is : {-2, -1, 0, 1, 2}.

Therefore, option C is correct.

Relatively prime numbers are those numbers which have only 1 as the common factor.

So, according to this definition we get to know that (2, 3), (2, 7), (3, 7), (3, 10), (4, 3), (4, 7), (5, 3), (5, 6) ,(5, 7) are relatively prime.

So, R ={(2, 3), (2, 7), (3, 7), (3, 10), (4, 3), (4, 7), (5, 3), (5, 6) ,(5, 7)}.

Therefore, Domain of R is the values of x or the first elemenyt of the ordered pair.

So, Domain = {2, 3, 4, 5}.

So, option D is correct.

We have xφy |x| = y

By checking the options,

A. (2 + 3i) φ 13

X = 2 + 3i;

= √13

Therefore, |x|≠ y.

So, option A is incorrect.

B. 3 φ (−3)

X = 3;

= 3

3 ≠(-3)

Therefore, option B is incorrect.

C. (1 + i) φ 2

X = 1+ i;

= √2

√2 ≠ 2

Therefore, option C is also incorrect.

D. iφ 1

x = i;

= 1

1 = 1

|x| = y.

Therefore, option D is correct.

We have , x + 2y = 8

y =

since, x and y are Natural numbers, So .. x must be an even number..

if x = 2, y = 3;

if x = 4, y = 2;

if x = 6, y = 1.

So, relation R = {(2, 3), (4, 2), (6, 1)}

Now, domain of R is {2, 4, 6}.

So, option C is correct.

Since, y = x – 3;

Therefore, for x = 11, y = 8.

For x = 12, y = 9.[ But the value y = 9 does not exist in the given set.]

For x = 13, y =10.

So, we have R = {(11, 8), (13, 10)}

Now, R⁻¹ = {(8, 11), (10, 13)}.

Therefore, option A is correct.

Since A has p elements, so each element of A will make a pair with each element of set B , and so other elements of A will do the same..

This is all forms q + q + q+ q +…….{p-times} ordered pairs…

Therefore, we will have a total of pq elements in A ×B.

Since, R is a relation from set A to set B, therefore it will always be a subset of A x B.

So, option C is correct.

Since we know that a relation from A to B consists of mn ordered pairs if they contain m and n elements respectively..

Each subset of those mn pairs will be a relation..so, each pair has two choices, either to be in that particular relation or not.

So, we have a tptal of 2^mn relations.

Therefore, option A is correct.

A is a set of n elements.

A x A will have a total of elements.

Then, the number of relations on A will be .

Therefore, option B is correct.