Transitive Relations MCQ Quiz - Objective Question with Answer for Transitive Relations - Download Free PDF

Explanation -

Let a = 1 + √2, b = 1 - √2 and c = 2√2. Then,

a² + b² = 6 ∈ Q, b² + c² = 3 ∈ Q but a² + c² = 3 + 4√2 ∉ Q

⇒ (a, b) ∈ R₁ and (b, c) ∈ R₁ but (a, c) ∉ R₁

∴ R₁ is not transitive.

Now, let x = 1 + √2, y = √2 and z = 1 - √2. Then,

x² + y² = 5 + 2√2 ∉ Q, y² + z² = 5 - 2√2 ∉ Q but x² + z² = 6 ∈ Q

⇒ (x, y) ∈ R₂, (y, z) ∈ R₂ but (x, z) ∉ R₂

∴ R₂ is not transitive.

Explanation -

Let a = 1 + √2, b = 1 - √2 and c = 2√2. Then,

a² + b² = 6 ∈ Q, b² + c² = 3 ∈ Q but a² + c² = 3 + 4√2 ∉ Q

⇒ (a, b) ∈ R₁ and (b, c) ∈ R₁ but (a, c) ∉ R₁

∴ R₁ is not transitive.

Now, let x = 1 + √2, y = √2 and z = 1 - √2. Then,

x² + y² = 5 + 2√2 ∉ Q, y² + z² = 5 - 2√2 ∉ Q but x² + z² = 6 ∈ Q

⇒ (x, y) ∈ R₂, (y, z) ∈ R₂ but (x, z) ∉ R₂

∴ R₂ is not transitive.

Given:

If A = {1, 2, 3, 4, 5}, then the relation R = {(2, 3), (3, 4), (2, 4)} on A

Concept: 

IF A is a set and R be a relation defined on A then

R is Reflexive: a is related to a. ∀ a ∈ A
R is symmetric: if a is related to b, then b is related to a. ∀ a, b ∈ A
R is transitive: if a is related to b and b is related to c then a is related to c.  ∀ a, b, c ∈ A

Calculation:

If A = {1, 2, 3, 4, 5}, then the relation R = {(2, 3), (3, 4), (2, 4)} on A

1. Reflexive :

Let 1 ∈ A but (1, 1) ∉ R .

Hence R is not Reflexive.

2. Symmetric :

Let 2, 3 ∈ A  and (2, 3) ∈ R but (3, 2) ∉ R

Hence R is not Symmetric.

3. Transitive :

Let 2 , 3, 4 ∈ A and

( 2, 3) and (3, 4) ∈ R then ( 2, 4 ) ∈ R

Hence R is transitive.

Hence R is transitive only.

Hence the option (1) is correct.

Concept: 

Reflexive relation: Relation is reflexive If (a, a) ∈ R ∀ a ∈ A.

Symmetric relation: Relation is symmetric, If (a, b) ∈ R, then (b, a) ∈ R.

Transitive relation: Relation is transitive, If (a, b) ∈ R & (b, c) ∈ R, then (a, c) ∈ R,

If the relation is reflexive, symmetric, and transitive, it is known as an equivalence relation.

Explanation:

Let A = {1, 2, 3}

The relation R is defined by R = {(1, 2)}

Since, (1, 1) ∉ R

∴ It is not reflexive.

Since, (1, 2) ∈ R but (2, 1) ∉ R

∴ It is not symmetric.

But there is no counter-example to disapprove of transitive condition.

∴ It is transitive.

Concept: 

Reflexive relation: Relation is reflexive If (a, a) ∈ R ∀ a ∈ A.

Symmetric relation: Relation is symmetric, If (a, b) ∈ R, then (b, a) ∈ R.

Transitive relation: Relation is transitive, If (a, b) ∈ R & (b, c) ∈ R, then (a, c) ∈ R,

If the relation is reflexive, symmetric, and transitive, it is known as an equivalence relation.

Explanation:

Let A = {1, 2, 3}

The relation R is defined by R = {(1, 2)}

Since, (1, 1) ∉ R

∴ It is not reflexive.

Since, (1, 2) ∈ R but (2, 1) ∉ R

∴ It is not symmetric.

But there is no counter-example to disapprove of transitive condition.

∴ It is transitive.

Given:

If A = {1, 2, 3, 4, 5}, then the relation R = {(2, 3), (3, 4), (2, 4)} on A

Concept: 

IF A is a set and R be a relation defined on A then

R is Reflexive: a is related to a. ∀ a ∈ A
R is symmetric: if a is related to b, then b is related to a. ∀ a, b ∈ A
R is transitive: if a is related to b and b is related to c then a is related to c.  ∀ a, b, c ∈ A

Calculation:

If A = {1, 2, 3, 4, 5}, then the relation R = {(2, 3), (3, 4), (2, 4)} on A

1. Reflexive :

Let 1 ∈ A but (1, 1) ∉ R .

Hence R is not Reflexive.

2. Symmetric :

Let 2, 3 ∈ A  and (2, 3) ∈ R but (3, 2) ∉ R

Hence R is not Symmetric.

3. Transitive :

Let 2 , 3, 4 ∈ A and

( 2, 3) and (3, 4) ∈ R then ( 2, 4 ) ∈ R

Hence R is transitive.

Hence R is transitive only.

Hence the option (1) is correct.

Concept: 

Reflexive relation: Relation is reflexive If (a, a) ∈ R ∀ a ∈ A.

Symmetric relation: Relation is symmetric, If (a, b) ∈ R, then (b, a) ∈ R.

Transitive relation: Relation is transitive, If (a, b) ∈ R & (b, c) ∈ R, then (a, c) ∈ R,

If the relation is reflexive, symmetric, and transitive, it is known as an equivalence relation.

Explanation:

Let A = {1, 2, 3}

The relation R is defined by R = {(1, 2)}

Since, (1, 1) ∉ R

∴ It is not reflexive.

Since, (1, 2) ∈ R but (2, 1) ∉ R

∴ It is not symmetric.

But there is no counter-example to disapprove of transitive condition.

∴ It is transitive.

Explanation -

Let a = 1 + √2, b = 1 - √2 and c = 2√2. Then,

a² + b² = 6 ∈ Q, b² + c² = 3 ∈ Q but a² + c² = 3 + 4√2 ∉ Q

⇒ (a, b) ∈ R₁ and (b, c) ∈ R₁ but (a, c) ∉ R₁

∴ R₁ is not transitive.

Now, let x = 1 + √2, y = √2 and z = 1 - √2. Then,

x² + y² = 5 + 2√2 ∉ Q, y² + z² = 5 - 2√2 ∉ Q but x² + z² = 6 ∈ Q

⇒ (x, y) ∈ R₂, (y, z) ∈ R₂ but (x, z) ∉ R₂

∴ R₂ is not transitive.

Given:

If A = {1, 2, 3, 4, 5}, then the relation R = {(2, 3), (3, 4), (2, 4)} on A

Concept: 

IF A is a set and R be a relation defined on A then

R is Reflexive: a is related to a. ∀ a ∈ A
R is symmetric: if a is related to b, then b is related to a. ∀ a, b ∈ A
R is transitive: if a is related to b and b is related to c then a is related to c.  ∀ a, b, c ∈ A

Calculation:

If A = {1, 2, 3, 4, 5}, then the relation R = {(2, 3), (3, 4), (2, 4)} on A

1. Reflexive :

Let 1 ∈ A but (1, 1) ∉ R .

Hence R is not Reflexive.

2. Symmetric :

Let 2, 3 ∈ A  and (2, 3) ∈ R but (3, 2) ∉ R

Hence R is not Symmetric.

3. Transitive :

Let 2 , 3, 4 ∈ A and

( 2, 3) and (3, 4) ∈ R then ( 2, 4 ) ∈ R

Hence R is transitive.

Hence R is transitive only.

Hence the option (1) is correct.

Explanation -

Let a = 1 + √2, b = 1 - √2 and c = 2√2. Then,

a² + b² = 6 ∈ Q, b² + c² = 3 ∈ Q but a² + c² = 3 + 4√2 ∉ Q

⇒ (a, b) ∈ R₁ and (b, c) ∈ R₁ but (a, c) ∉ R₁

∴ R₁ is not transitive.

Now, let x = 1 + √2, y = √2 and z = 1 - √2. Then,

x² + y² = 5 + 2√2 ∉ Q, y² + z² = 5 - 2√2 ∉ Q but x² + z² = 6 ∈ Q

⇒ (x, y) ∈ R₂, (y, z) ∈ R₂ but (x, z) ∉ R₂

∴ R₂ is not transitive.