## Rajasthan Board RBSE Class 11 Maths Chapter 2 Relations and Functions Ex 2.2

Question 1.

Examine the reflexivity, symmetricity, and transitivity of the following relations:

(i) mR_{1}n ⇔ m and n both are odd, ∀ m, n ∈ N

(ii) In the power set P(A) of set AR_{2}B ⇔ A ⊆ B, ∀ A, B ∈ P(A)

(iii) In set B of straight lines situated in three dimensions space, L_{1}R_{3}L_{2} ⇔ L_{1} and L_{2} are coplanar ∀ L_{1}, L_{2} ∈ S

(iv) aR_{4}b ⇒ b is divisible by a, ∀ a, b ∈ N.

Solution:

(i) Given, set N = {1, 2, 3, 4, …}

A relation R_{1} in N is defined as

mR_{1}n ⇔ m and n both are odd ∀ m, n ∈ N

Reflexivity: Let m ∈ N

m ∈ N ⇒ m : even or odd number

⇒ (m, m) ∈ R_{1} If m is odd

But (m, m) ∉ R_{1} If m is even

So, (m, m) ∉ R_{1}

R_{1} is not reflexivity.

Symmetricity: Let m, n ∈ N, mR_{1} is true,

So, mR_{1}n ⇒ m and n are both odd.

⇒ n and m are both odd

⇒ nR_{1}m

R_{1} is a symmetric relation.

Transitivity: Let m, n, r ∈ N

mR_{1}n and nR_{1}r is true, so,

mR_{1}n and nR_{1}r ⇒ m and n are both odd and n, r are both odd.

⇒ m, n and r are odd.

⇒ mR_{1}r

R_{1} is transitive relation.

(ii) Given set is

AR_{2}B ⇔ A ⊆ B ∀ A, B ∈ P(A)

Reflexivity: Let A ∈ P(A)

A ∈ P(A) ⇒ A ⊆ A (∵ every set is a subset of itself)

⇒ AR_{2}A

Hence, R_{2} reflexive relation.

Symmetricity: Let A, B ∈ P(A), AR_{2}B are true, so,

AR_{2}B ⇒ A ⊆ B

⇒ B ⊄ A (until A = B)

⇒ B ⊄ A

R_{2} is not a symmetric relation.

Transitivity: A, B, C ∈ P(A),

AR_{2}B and BR_{2}C are true. So,

AR_{2}B and BR_{2}C

⇒ A ⊆ B and B ⊆ C

⇒ A ⊆ C(A ⊆ B ⊆ C)

⇒ AR_{2}C

R_{2} is transitive relation.

(iii) Given set is

S = {Set of straight lines situated in three dimensional space}

A relation R_{3} in S is defined as

L_{1}R_{3}L_{2} ⇒ L_{1} and L_{2} are co-planar lines ∀ L_{1}, L_{2} ∈ S

Reflexivity: Let L ∈ S

L ∈ S ⇒ L and L are coplanar

⇒ (L, L) ∈ R_{3} ∀ L ∈ S

R_{3} is a reflexive relation.

Symmetricity: Let L_{1}, L_{2} ∈ S

In this way (L_{1}, L_{2}) ∈ R_{3}

(L_{1}, L_{2}) ∈ R_{3}

⇒ L_{1} and L_{2} are coplanar

⇒ L_{2} and L_{1} are coplanar

⇒ (L_{2}, L_{1}) ∈ R_{3}

So, (L_{1}, L_{2}) ∈ R_{3}

⇒ (L_{2}, L_{1}) ∈ R_{3} ∀ L_{1}, L_{2} ∈ S

R_{3} is a symmetric relation.

Transitivity: Let L_{1}, L_{2}, L_{2} ∈ S is given as

(L_{1}, L_{2}) ∈ R_{3} and (L_{2}, L_{3}) ∈ R_{3}

(L_{1}, L_{2}) ∈ R_{3} ⇒ L_{1} and L_{2} are coplanar ……(1)

(L_{2}, L_{3}) ∈ R_{3} ⇒ L_{2} and L_{3} are coplanar …(2)

So, from (1) and (2) we can not say that L_{1} and L_{3} are also coplanar.

From figure L_{1} and L_{2} are situated in a plane (1),

L_{2} and L_{3} are situated in a plane (2)

but L_{1} and L_{3} are not coplanar.

So, (L_{1}, L_{3}) ∉ R_{3}

R_{3} is not a transitive relation.

(iv) Given set is N = {1,2, 3, 4, …}

Relation R_{4} in N is defined as

aR_{4}b ⇔ b is divided by a ∀ a, b ∈ N

⇔ \(\frac { 1 }{ 2 }\) = k ∈ I

where I is the set of integers ∀ a, b ∈ N

⇔ ∀ a is divided by b

∀ a, b ∈ N

Reflexivity: Let a ∈ N

a ∈ N ⇒ a is divided by a

⇒ (a, a) ∈ R_{4} ∀ a ∈ N

R_{4} is a reflexive relation.

Symmetricity: Let a, b ∈ N is in this way (a, b) ∈ R_{4}

(a, b) ∈ R_{4} ⇔ b is divided by a

⇔ a cannot be divided by b until a = b

⇔ (b, a) ∉ R_{4}

(a, b) ∈ R_{4} ⇔ (b, a) ∉ R_{4}

So, R_{4} is not a symmetric relation.

Transitivity: Let a, b, c ∈ N

In this way

(a, b) ∈ R_{4} and (b, c) ∈ R_{4}

(a, b) ∈ R_{4} ⇒ b is divided by a

⇒ \(\frac { b }{ a }\) = k (Let) where k ∈ I …(1)

(b, c) ∈ R_{4} ⇒ c is divided by b

⇒ \(\frac { c }{ b }\) = m (Let) where m ∈ I …(2)

Putting the value of b from equation (1) in equation (2) we have

m ∈ I, k ∈ I

mk ∈ I

\(\frac { c }{ ak }\) = m ⇒ \(\frac { c }{ a }\) = mk ∈ I

⇒ (a, c) ∈ R_{4}

(a, b) ∈ R_{4}, (b, c) ∈ R_{4}

⇒ (a, c) ∈ R_{4} ∀ a, b, c ∈ N

Hence, R_{4} is a transitive relation.

Question 2.

Any relation P is defined in set R0 of non zero real numbers by following ways:

(i) xPy ⇔ x^{2} + y^{2} = 1

(ii) xPy ⇔ xy = 1

(iii) xPy ⇔ (x + y) is a rational number

(iv) xPy ⇔ x/y is a rational number

Test the reflexivity, symmetricity and transitivity of these relations.

Solution:

(i) Given set

R0 = Set of non-zero real number

Relation P in R0 is defined as

xPy = x^{2} + y^{2} = 1 ∀ x, y ∈ R_{0}

Reflexivity: P is not reflexivie because when 2 ∈ R_{0}

Then (2)^{2} + (2)^{2} ≠ 1 so, (2, 2) ∉ R_{0}

Similarly a^{2} + a^{2} ≠ 1 so, (a, a) ∉ R_{0}

Hence, P is not reflexive.

Symmetricity: Let a, b ∈ R_{0}

In this way (a, b) ∈ P

(a, b) ∈ P

⇒ a^{2} + b^{2} = 1

⇒ b^{2} + a^{2} = 1

⇒ (b, a) ∈ P

So, (a, b) ∈ P

⇒ (b, a) ∈ P ∀ a, b ∈ R_{0}

P is symmetric relation.

Transitivity: P is not transitive relation because

P is not transitive relation.

(ii) Given set R_{0} = {Set of real numbers}

Relation P in R_{0} is defined as

xPy ⇔ xy = 1 ∀ x, y ∈ P

Reflexivity: Let 2 ∈ R_{0}

But 2 × 2 = 4 ≠ 1

⇒ (2, 2) ∉ P

Hence, P is not reflexive

Symmetricity: In this way (a, b) ∈ R_{0}

In this way (a, b) ∈ P

(a, b) ∈ P

⇒ ab = 1

⇒ ba = 1 [∵ Multiplication of real number is commutative]

⇒ b . a ∈ P

⇒ (a, b) ∈ P

⇒ (b, a) ∈ P ∀ a, b ∈ R_{0}

P is a symmetric relation.

Transitivity: Let a, b, c ∈ R_{0} is in this way

a, b ∈ P and b, c ∈ P

(a, b) ∈ P ⇒ a.b = 1 ……(1)

(b, c) ∈ P ⇒ b.c = 1 …..(2)

From equation (1) and (2),

\(\frac { ab }{ bc }\) = \(\frac { 1 }{ 1 }\)

⇒ \(\frac { a }{ c }\) = 1

⇒ (a, c) ≠P ……..(3)

So, (a, b) ∈ P, (b, c) ∈ P but (a, c) ∉ P

Hence, P is not transitive relation.

(iii) Given set R_{0} = Set of real number

Relaton P in R_{0} is defined as

xPy ⇔ x + y is a rational number ∀ x, y ∈ R_{0}

Reflexivity : Let x ∈ R_{0}

x ∈ R_{0} ⇒ x + x is need not to be rational

For example √3 ∈ R_{0} ⇒ √3 + √3 = 2√3

is an irrational number

So, (x, x) ∉ P

P is not reflexive relation.

Symmetricity: Let a, b ∈ R_{0} then (a, b) ∈ P

(a, b) ∈ P

⇒ a + b is a rational number

⇒ (b + a) is also a rational number

⇒ (b, a) ∈ P

So, (a, b) ∈P

⇒ (b, a) ∈ P ∀ a, b ∈ R_{0}

P is a symmetric relation.

Transitivity: Let a, b, c ∈ R_{0} are in this way

(a, b) ∈ P ⇒ a + b is a rational number

(b, c) ∈ P ⇒ b + c is a rational number

⇒ a + c is not necessary a rational number

For example: 2 + √3, -√3 + 6, √3 + √7 ∈ Ro and (2 + √3, -√3 + 6) ∈ P because

2 + √3 – √3 + 6 = 8 is a rational number

(-√3 + 6, √3 + 7) ∈ P because

-√3 + 6 + √3 + 7 = 13 is a rational number

But (2 + √3, √3 + 7) ∈ P because

2 + √3 + √3 + 7 = 2√3 + 9 is an irrational number

Hence, P is not a transitive relation.

(iv) Given: Set : R_{0} = Set of real numbers

Relation P in R_{0} is defined as a rational number

xPy = \(\frac { x }{ y }\) is a rational number ∀ x, y ∈ R_{0}

Reflexivity: Let a ∈ R_{0}

a ∈ R_{0} ⇒ \(\frac { a }{ a }\) = 1 is a rational number

⇒ (a, a) ∈ P

P is a reflexive number.

Symmetricity: Let a, b ∈ R_{0} is in this way (a, b) ∈ P

(a, b) ∈ P

⇒ \(\frac { a }{ b }\) is a rational number

⇒ \(\frac { b }{ a }\) is a rational number

(From infination of set of rational numbers Q)

⇒ (b, a) ∈ P

(a, b) ∈ P

⇒ (b, a) ∈ P ∀ a, b ∈ R_{0}

P is a symmetric relation

Transitivity: Let a, b, c ∈ R_{0} is in this way

(a, b) ∈ P and (b, c) ∈ P

(a, b) ∈ P ⇒ \(\frac { a }{ b }\) is a rational number

(b, c) ∈ P ⇒ \(\frac { b }{ c }\) is a rational number

⇒ (\(\frac { a }{ b }\))(\(\frac { b }{ c }\))is also a rational number.

[∵ Multiplication of rational number is also a rational number]

⇒ \(\frac { a }{ c }\) is a rational number

⇒ (a, c) ∈ P

(a, b) ∈ P, (b, c) ∈ P

⇒ (a, c) ∈ P ∀ a, b, c ∈ R_{0}

P is a transitive relation.

Question 3.

A relation R1 is defined in set R if of real numbers in the following way:

(a, b) ∈ R_{1} ⇔ 1 + ab > 0, ∀ a, b ∈ R

Prove that R_{1} is reflexive and symmetric but not transitive.

Solution:

Given: Set R = set of real numbers.

Relation R_{1} in R is defined as

(a, b) ∈ R_{1} ⇒ 1 + ab > 0 ∀ a, b ∈ R

(i) Reflexivity: Let a ∈ R

a ∈ R ⇒ 1 + a.a > 0

⇒ a.a ∈ R_{1} ∀ a ∈ R

R_{1} is a reflexive relation.

(ii) Symmetricity: Let a, b ∈ R in this way

(a, b) ∈ R_{1}

(a, b) ∈ R_{1}

⇒ 1 + ab > 0

⇒ 1 + b.a > 0 [ab = ba]

⇒ (b, a) ∈ R_{1}

So, (a, b) ∈ R_{1}

⇒ (b, a) ∈ R_{1} ∀ a, b ∈ R

R_{1} is a symmetric relation.

(iii) Transitivity: Let a, b, c ∈ R is in this way

(a, b) ∈ R_{1} and (b, c) ∈ R_{1}

(a, b) ∈ R_{1} ⇒ 1 + ab > 0

(b, c) ∈ R_{1} ⇒ 1 + bc > 0

But 1 + ac > 0 is not necessary.

For example: 1, \(\frac { 1 }{ 2 }\), -1 ∈ R

and (1, \(\frac { 1 }{ 2 }\)) ∈ R_{1} because

R_{1} is not a transitive relation.

Hence Proved.

Question 4.

N is a set of natural numbers. If a relation R is defined in set N × N such that (a, b) R(c, d) ⇔ ad = bc ∀ (a, b), (c, d) ∈ N × N, then prove that R is an equivalence relation.

Solution:

Given:

Set N = {1, 2, 3, 4, …} = Set of natural numbers

A relation R in N × N is defined as

(a, b) R(c, d) ⇔ ad = bc where a, b,c, d ∈ N ∀ (a, b), (c, d) ∈ N × N

Here, to prove R is a equivalence relation, we have to show that R is reflexive, symmetric and transitive.

(i) Reflexivity: Let

(a, b) ∈ N × N

{a, b) ∈ N × N

⇒ a.b = ba (Commutative law of multiplication)

⇒ (a, b) R(a, b) ∀ (a, b) ∈ N × N

R is reflexive relation.

(ii) Symmetricity: Let (a, b) (c, d) ∈ N × N is in this way (a, b) R(c, d)

(a, b) R(c, d)

⇒ ad = bc

⇒ bc = ad

⇒ c.b = d.a

⇒ (c, d) R(a, b)

So, (a, b) R(c, d) ⇒ (c, d) R(a, b) ∀ (a, b)(c, d) ∈ N × N

R is a symmetric relation.

(iii) Transitivity: Let

(a, b), (c, d), (e, f) ∈ N × N

is in this way

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

(a, b) R(c, d) ⇒ ad = bc

(c, d) R(e, f) ⇒ cf = de

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

⇒ (ad) (cf) = (bc)(de) (on multiplication)

⇒ af = be

⇒ (a.b) R(e, f)

So, (a, b) and R(c, d) and (c, f) R(e, f)

⇒ (a, b) R(e, f) ∀ (a, b), (c, d), (e, f) ∈ N × N

R is a transitive relation.

Hence, according to (i), (ii) and (iii), the given relation is an equivalence relation.

Hence Proved.

Question 5.

A relation R is defined in a set Q_{0} set of non zero rational numbers such that aRb ⇔ a = 1/b, ∀ a, b ∈ Q_{0}. Is R is equivalence relations.

Solution:

Given: Set Q_{0} = Set of non zero rational numbers

A relation in Q_{0} is defined as

aRb ⇔ a = 1/b ∀ a, b ∈ Q_{0}

If R is reflexive, symmetric and transitive than R is an equivalence relation.

(i) Reflexivity: Let a ∈ Q_{0}

a ∈ Q_{0} ⇒ a ≠ \(\frac { 1 }{ a }\) (a ≠ 1)

⇒ (a, a) ∉ R ∀ a ∉ Q_{0}

So, R is not a reflexive relation

R is also not equivalenced relation.

Question 6.

Let {(a, b) | a, b ∈ R} where I is set of integers. Relations R_{1} on x is defined in the following way

(a, b) R_{1}(c, d) ⇒ b – d = a – c

Prove that R_{1} is an equivalence relation.

Solution:

Given : Set X = {(a, b) : a, b ∈ I}

where I is the set of integers.

A relation R in X is defined as:

(a, b) R(c, d) ⇔ b – d = a – c ∀ (a, b) (c, d) ∈ X

To prove that R is equivalence relation, we have to prove that R is reflexive, symmetric and transitive.

(i) Reflexivity: Let (a, b) ∈ X

(a, b) ∈ X ⇒ (a, b) ∈ I

⇒ b – b = a – a = 0

⇒ (a, b) R(a, b) ∀ (a, b) ∈ X

R is a reflexive relation.

(ii) Symmetricity: Let (a, b), (c, d) ∈ X is in this way

(a, b) R(c, d)

(a, b) R(c, d)

⇒ b – d = a – c

⇒ -(d – b) = -(c – a)

⇒ d – b = c – a

⇒ (c.d) R(a.b)

(a, b) R(c, d) ⇒ (cd) R(ab) ∀ (a, b), (c, d) ∈ X

R is a symmetric relations.

(iii) Transitivity: Let (a, b), (c, d), (e, f) ∈ X is in this way (a, b) R(c, d) and (c, d) R(e, f)

(a, b) R(c, d) ⇒ b – d = a – c …(1)

(c, d) R(e, f) ⇒ d – f = c – e …(2)

Adding equation (i) and (2), we have

b – d + d – f = a – c + c – e

⇒ b – f = a – e

⇒ (a, b) R(e, f)

So, (a, b) R(c, d) and (c, d) R(e, f)

⇒ (a, b) R(e, f) ∀ (a, b), (c, d), (e, f) ∈ X

R is a transitive relation.

Hence, according to (i), (ii) and (iii), the given relation is equivalence relation.

Hence Proved.

Question 7.

A relation R is defined in a set T of triangles situated in a plane such that xRy ⇔ x is similar to y. Prove that R is an equivalence relation.

Solution:

Given : Set T = {Set of similar triangles}

A relation R in T is defined as:

xRy ⇔ x is similarly to y ∀ x, y ∈ T

To prove R is an equivalence relation, we have to prove that R is reflexive, symmetric and transitive.

(i) Symmetricity: Let x ∈ T

x ∈ f ⇒ x is similar to x.

⇒ (x, x) ∈ R ∀ x ∈ T

R is a reflexive relation.

(ii) Symmetricity: Let x, y ∈ T is in this way

(x, y) ∈ R

(x, y) ∈ R ⇒ x is similar to y

⇒ y is similar to x

⇒ (y, x) ∈ R

So, (x, y) ∈ R ⇒ (y, x) ∈ R ∀ x, y ∈ T

R is a symmetric relation.

(iii) Transitivity: Let x, y, z ∈ T is in this way

(x, y) ∈ R, (y, z) ∈ R

(x, y) ∈ R ⇒ x is similar to y.

(y, z) ∈ R ⇒ y is similar to z.

set x is similar to z.

So, (x, y) ∈ R, (y, z) ∈ R ⇒ (x, z) ∈ R ∀ x, y, z ∈ T

R is a transitive relation.

Hence, according to (i), (ii) and (iii), the given relation is an equivalence relation.

Hence Proved.

Question 8.

Let a relation R is defined in a set A = {1, 2, 3} as : R = {(1, 1), (1, 2), (2, 1) (2, 2), (3, 3), (1, 3), (3, 1), (2, 3), (3, 2)} Examine the reflexivity,

symmetricity and transitivity of R.

Solution:

Given set A = {1, 2, 3}

Reiation R In A is defined as:

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

(i) Reflexivity: Here (1, 1), (2, 2), (3, 3) ∈ R

So, ∀ a ∈ A ⇒ (a, a) ∈ R

R is reflexive.

(ii) Symmetricity: Here, R is symmetric, because

(1, 2) ∈ R ⇔ (2, 1) ∈ R

(1, 3) ∈ R ⇔ (3, 1) ∈ R

(3, 2) ∈ R ⇔ (2, 3) ∈ R

So, (a, b) ∈ R ⇔ (b, a) ∈ R ∀ a, b ∈ A

So, R is a transitive relation.

(iii) Transitivity: (1, 2) ∈ R, (2, 1) ∈ R

⇒ (1, 1) ∈ R

(2, 3) ∈ R, (3, 2) ∈ R

⇒ (2, 2) ∈ R

Hence, by definition of Transitivity.

(a, b) ∈ R, (b, c) ∈ R

⇒ (a, c) ∈ R ∀ a, b, c ∈ A

So, R is a transitive relation.

Question 9.

A relation R in a set C_{0} of non zero complex numbers is defined as:

Prove that R is an equivalence relation.

Solution:

Given: Set C_{0} = set of the non-zero complex number and a relation R in C_{0} is defined as:

Hence from (i), (ii) and (iii), the given relation is an equivalence relation.

Hence Proved.

Question 10.

If R is a relation in set X of subsets defined as “A is disjoint to B” then examine the reflexivity, symmetricity and transitivity of R.

Solution:

Set X = set of subsets A relation R in X is defined as

ARB ⇔ A is disjoint to B ∀ A, B ∈ X

⇔ A ∩ B = Φ ∀ A, B ∈ X (where Φ is a null set)

(i) Reflexivity: Let A ∈ X

A ∈ X ⇒ A ∩ A = A

⇒ A ∩ A ≠ Φ (untill A becomes Φ)

⇒ A is not disjoint to A

⇒ (A, A) ∉ R ∀ A ∈ X

R is not reflexive relation.

(ii) Symmetricity: Let A, B ∈ X in this way

(A, B) ∈ R

(A, B) ∈ R

⇒ A is disjoint to B thus A ∩ B = Φ

⇒ B ∩ A thus, B is disjoint to A (By commutative)

⇒ (B, A) ∈ R

So, (A, B) ∈ R ⇒ (B, A) ∈ R ∀ A, B ∈ X

R is a symmetric relation.

(iii) Transitivity: Let A, B, C ∈ X is in this way

(A, B) ∈ R and (B, C) ∈ R

A, B ∈ R ⇒ A ∩ B = Φ

B, C ∈ R ⇒ B ∩ C = Φ

Then A ∩ C = Φ is not necessary.

For example:

A = {1, 2, 3}, B = {4, 5, 6}, C = {1, 2, 7}

Here A, B ∈ R because A ∩ B = Φ

B, C ∈R because B ∩ C ⇒ Φ

But (A, C) ∉ R because A ∩ C = {1, 2} ≠ Φ

R is not transitive relation.

From (i), (ii) and (iii) given relation R is symmetric but relative R is not reflexive and transitive.

Hence Proved.

Question 11.

A relation R is defined in a set N of natural numbers such that aRb if a is a divisor of b, Prove that R is a partially ordered relation but not a total ordered relation.

Solution:

Given set N = {1, 2, 3, 4,…} = Set of natural numbers

Where a relation is define as

aRb ⇔ a is divisor of b ∀ a, b ∈ N

There R is a partial ordered relation if it is reflexive, anti symmetric and transitive.

(i) Reflexivity: Let a ∈ N

a ∈ N ⇒ a is divisor of a

⇒ (a, a) ∈ R ∀ a ∈ N

R is reflexive relation.

(ii) Anti Symmetricity: Let a, b ∈ N is in this way (a, b) ∈ R

(a, b) ∈ R ⇔ a is divisor of b.

⇔ b is disivor of a.

if a = b ∀ a, b ∈ N ⇔ (b, a) ∈ R; a = b, ∀ a, b ∈ N

(a, b) ∈ R ⇔ (b, a) ∈ R ⇒ a = b ∀ a, b ∈ N

So, R is anti-symmetric relation.

(iii) Transitivity: Let a, b, c ∈ N is in this way

(a, b) ∈ R and (b, c) ∈ R

So, (a, b) ∈ R, (b, c) ∈ R ⇒ (a, c) ∈ R ∀ a, b, c ∈ N

So, R is a transitive relation.

Hence, from (i), (ii) and (iii) the given relation is a partially ordered relation.

Hence Proved.

Question 12.

Whether following subsets of N are total ordered set for relaton “x divides y or not” :

(i) {2, 4, 6, 8,……}

(ii) {0, 2, 4, 6,……..}

(iii) (3, 9, 5, 15,…}

(iv) (5, 15, 30}

(v) {1, 2, 3, 4}

(vi) {a, b, ab} ∀ a, b ∈ R.

Solution:

(i) Let subset A = {2, 4, 6, 8, …} ∈ N

A relation given in N

Hence, the second rule is not followed then the given relation is not a total ordered relation. For this it is clear that to prove a relation is a total ordered relative must be reflexive, symmetric and transitive and also every a, b ∈ A, (a, b) ∈ R or (b, a) ∈ R or a = b is true.

Hence, the given relation and their results are:

Relation |
Conclusion |

(i) (2, 4, 6, 8,…} | No |

(ii) {0, 2, 4, 6,…} | No |

(iii) {3, 9, 5, 15,…} | No |

(iv) {5, 15, 30} | Yes |

(v) {1, 2, 3, 4} | No |

(vi) {a, b, ab} ∀ a, b ∈ R | No |

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