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

Calculation

Let S = {1, 2, 3, ........, 10}

R = {A, B) ∶ A ∩ B ≠ ϕ; A, B ∈ M}

For Reflexive,

M is set of all the subset of 'S'

So ϕ ∈ M

for ϕ ∩ ϕ = ϕ
⇒ but relation is A ∩ B ≠ ϕ

So it is not reflexive.

For symmetric,

ARB  ⇒  A ∩ B ≠ ϕ,

⇒ B ∩ A ≠ ϕ ⇒ BRA 

So it is symmetric.

For transitive,

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

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

C = {(3, 4), (5, 6)}

ARB & BRC but A does not relate to C

So it not transitive

Hence option 4 is correct

Concept:

Reflexive relation: A relation is said to be reflexive if (a, a) ∈ R, for every a ∈ A.

Symmetric relation: A relation is said to be symmetric, if (a, b) ∈ R, then (b, a) ∈ R.

Transitive relation: A relation is said to be transitive if (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R.

Calculation:

Given, x = {1, 2, 3, ....... 20}

R₁ = {(x, y) : 2x – 3y = 2}

R₂ = {(x, y) : – 5x + 4y = 0}

∴ R₁ = {(4, 2), (7, 4), (10, 6), (13, 8), (16, 10), (19, 12)}

For symmetric, we need to add 6 elements here as (3, 4), (4, 7) and (6, 10), (8, 13), (10, 16), (12, 19)

∴ M = 6

Also, R₂ = {(4, 5), (8, 10), (12, 15), (16, 20)}

For symmetric, we need to add 4 elements (5, 4), (10, 8) (15, 12), (20, 16)

∴ N = 4

⇒ M + N = 10

∴ The value of M + N is 10.

The correct answer is Option 4.

Calculation

R = {(x, y), x + y is even x, y ∈ z}

reflexive x + x = 2x even

symmetric of x + y is even, then (y + x) is also even

transitive of x + y is even & y + z is even then x + z is also even

So, relation is an equivalence relation.

Hence option 3 is correct

Concept Used:

Equivalence relation: Reflexive, Symmetric, Transitive.

• Reflexive: aRa for all a.
• Symmetric: If aRb, then bRa.
• Transitive: If aRb and bRc, then aRc.

Calculation:

1. aR₁b ⇔ |a| = |b|

⇒ Reflexive: |a| = |a|, so aR₁a. (Reflexive)

⇒ Symmetric: If |a| = |b|, then |b| = |a|, so if aR₁b, then bR₁a. (Symmetric)

⇒ Transitive: If |a| = |b| and |b| = |c|, then |a| = |c|, so if aR₁b and bR₁c, then aR₁c. (Transitive)

⇒ aR₁b is an equivalence relation.

2. aR₃b ⇔ a divides b

⇒ Reflexive: a divides a, so aR₃a. (Reflexive)

⇒ Symmetric: 2 divides 4, but 4 does not divide 2. (Not Symmetric)

⇒ Transitive: If a divides b and b divides c, then a divides c. (Transitive)

⇒ aR₃b is not an equivalence relation.

3. aR₂b ⇔ a ≥ b

⇒ Reflexive: a ≥ a, so aR₂a. (Reflexive)

⇒ Symmetric: 5 ≥ 3, but 3 < 5. (Not Symmetric)

⇒ Transitive: If a ≥ b and b ≥ c, then a ≥ c. (Transitive)

⇒ aR₂b is not an equivalence relation.

4. aR₄b ⇔ a < b

⇒ Reflexive: a < a is false. (Not Reflexive)

⇒ Symmetric: If a < b, then b < a is false. (Not Symmetric)

⇒ Transitive: If a < b and b < c, then a < c. (Transitive)

⇒ aR₄b is not an equivalence relation.

Hence option 1 is correct.

Explanation:

(I) Given that P and Q are reflexive relations of A.

⇒ For any a∈A, (a, a) ∈ P and (a, a) ∈ Q 

⇒ (a, a )∈ P∩Q

⇒ P∩Q is reflexive relation.

So statement (I)is true

(II) Given that P and Q are symmetric relation

Let (a, b) ∈ (P∪Q)

⇒ (a, b) ∈ P or (a, b) ∈ Q.

⇒ (b, a) ∈ P or (b, a) ∈ Q

⇒  (b, a) ∈ (P∪Q)

⇒P∪Q is symmetric relation.

So statement (II) is true.

(III) Given that P and Q are transitive

Let (a, b)∈ (P∩Q) and (b, c)∈ (P∩Q)

⇒ {(a, b)∈ P, (b, c) ∈ P} and {(a, b)∈Q, (b, c) ∈Q}.....     {∵ P and Q are transitive}

⇒ (a, c) ∈ P∩Q

Thus P∩Q is a transitive relation.

So statement (III) is also true.

∴ Option (d) 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:

If x ∈  R then we express it by writing xRy and say that " x is related to y with relation R"

Thus, (x, y) ∈ R ⇔ xRy

Calculation:

Given

2x + 3y = 20

The relation R can be written as

R = {(1,6), (4, 4), (7, 2)}

There are 3 elements in the form (x, y) are there in R.

Concept:

Reflexive: Each element is related to itself.

• R is reflexive if for all x ∈ A, xRx.

Symmetric: If any one element is related to any other element, then the second element is related to the first.

• R is symmetric if for all x, y ∈ A, if xRy, then yRx.

Transitive: If any one element is related to a second and that second element is related to a third, then the first element is related to the third.

• R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

R is an equivalence relation if A is nonempty and R is reflexive, symmetric and transitive.

Calculation:

Here,  R = {(a, b): a, b ϵ N and a2 = b}

1. Relation R is not reflexive since, 2 ≠ 2²

2. Since, 2² = 4 so,  (2, 4) belong to R

But,  4 ≠ 2 and so, R is not symmetric

3. Since, 4² = 16, so (4, 16) belong to R

Also, 16² = 256,  so (16, 256) belong to R

But, 4² ≠ 256  so R is not transitive.

So, R satisfies none of the reflexivity, symmetry and transitivity.

Hence, option (4) is correct. 

Concept:

Let R be a binary relation on a set A.

1. Reflexive: Each element is related to itself.

• R is reflexive if for all x ∈ A, xRx.

2. Symmetric: If any one element is related to any other element, then the second element is related to the first.

• R is symmetric if for all x, y ∈ A, if xRy, then yRx.

3. Transitive: If any one element is related to a second and that second element is related to a third, then the first element is related to the third.

• R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

4. R is an equivalence relation if A is nonempty and R is reflexive, symmetric and transitive.

Calculation:

Given: x < y if y at least 5 years older than x

⇒ y ≥ x + 5

For Reflexive: (x, x) should ∈R for all x ∈ X

Now, x cannot be 5 years older than himself. So the relation is not reflexive.

For Symmetric: If (x, y) ∈ R ⇒(y, x) ∈ R

(x, y) ∈ R ⇒ y is at least 5 years older than x.

(y, x) ∈ R ⇒ x is at least 5 years older than y. This contradicts the above statement. Hence the relation is not symmetric

For Transitive: If (x, y) ∈ R and (y, z) ∈ R ⇒ (x, z) ∈ R.

(x, y) ∈ R ⇒ y is at least 5 years older than x.

(y, z) ∈ R ⇒ z is at least 5 years older than y.

Then, (x, z) ∈ R ⇒ z is at least 5 years older than x.

Since, z is at least 10 years older than x. The relation is transitive.

Concept:

A relation R in a set A is called 

(i) Reflexive, if (a, a) ∈ R, for every a ∈ A.

(ii) Symmetric, if (a, b) ∈ R implies that (b, a) ∈ R, for all a, b ∈A.

(iii) Transitive, if (a, b) ∈ R and (b, c) ∈ R  implies that (a, c) ∈ R, for all a, b, c ∈A.

Calculation:

Given: R = {(a, b): a2 ≥ b}

We know that a2 ≥ a

Therefore (a, a) ∈ R, for all a ∈ Z.

Hence, relation R is reflexive.

Let (a, b) ∈ R

⇒ a2 ≥ b but b2 \(\ngeqslant\) a for all a, b ∈ Z.

So, if (a, b) ∈ R then it does not implies that (b, a) also belongs to R.

Hence, relation R is not symmetric.

Now let (a, b) ∈ R and (b, c) ∈ R.

⇒ a2 ≥ b and b2 ≥ c

This does not implies that a2 ≥ c, therefore (a, c) does not belong to R for all a, b, c ∈ Z.

Hence, relation R is not transitive.

Hence, option 4 is the correct answer.

Concept:

Let A be a set in which the relation R defined. 

1.R is said to be a Reflexive Relation  (a, a) ∈ R

2. R is said to be a symmetric relation, if (a, b) ∈ R ⇒ (b, a) ∈ R

Given set is A = {1, 2, 3}

and the relation is R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)} 

Let A be a set in which the relation R defined.

1.R is said to be a Reflexive Relation  (a, a) ∈ R

2. R is said to be a symmetric relation, if (a, b) ∈ R ⇒ (b, a) ∈ R

Since, 1, 2 , 3 ∈ A and (1, 1), (2, 2), (3, 3) \(\rm ∈ R\)

⇒ Every element maps to itself.

⇒ R is Reflexive 

Now, 1, 2 , 3 \(\rm ∈ R\)

(1, 2), (2, 3) \(\rm ∈ R\) ⇒  (1, 3) \(\rm ∈ R\)

⇒R relates 1 to 2 and 2 to 3, then R also relates 1 to 3

⇒ R is Transitive 

Here, R is not symmetric relation, as (a, b) ∈ R \(\neq \) (b, a) ∈ R

Hence, The relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)} on a set A = {1, 2, 3} is reflexive transitive but not symmetric.

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:

Given that, A = {1, 2, 3} and R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1,3)}.

Now,

 (1,1),(2,2),(3,3) ∈ R

⇒ R is reflexive.

(1,2),(2,3),(1,3) ∈ R but (2,1),(3,2),(3,1) ∉ R

⇒ R is not symmetric.

Also, (1,2) ∈ R and (2,3) ∈ R ⇒ (1,3) ∈ R

⇒ R is transitive.

∴ R is reflexive, and transitive but not symmetric.

Concept:

Relation Based on Logarithmic Comparison:

• A relation R is defined on the set S × S by xRy if log_ax > log_ay.
• The base of the logarithm is a = 1/2, which is a value between 0 and 1.
• For 0 < a < 1, the logarithmic function log_ax is a strictly decreasing function.
• Important property: If log_ax > log_ay, then it implies x < y when 0 < a < 1.
• Hence, the given relation reduces to xRy ⇔ x < y.
• Relation properties:
  • Reflexive: A relation is reflexive if xRx for all x in S. But x < x is never true ⇒ Not reflexive.
  • Symmetric: If x < y, then y < x is false ⇒ Not symmetric.
  • Transitive: If x < y and y < z ⇒ x < z ⇒ Transitive.

Calculation:

Given,

Relation: xRy ⇔ log_ax > log_ay

Base of logarithm: a = 1/2

⇒ Since 0 < a < 1, log_ax is a decreasing function

⇒ log_ax > log_ay ⇔ x < y

⇒ So, xRy ⇔ x < y

⇒ Check reflexivity: x < x ⇒ false ⇒ Not reflexive

⇒ Check symmetry: If x < y, then y < x ⇒ false ⇒ Not symmetric

⇒ Check transitivity: If x < y and y < z ⇒ x < z ⇒ Transitive

∴ The relation is transitive only.

Concept:

Let R be a binary relation on a set A.

1. Reflexive: Each element is related to itself.

• R is reflexive if for all x ∈ A, xRx.

2. Symmetric: If any one element is related to any other element, then the second element is related to the first.

• R is symmetric if for all x, y ∈ A, if xRy, then yRx.

3. Transitive: If any one element is related to a second and that second element is related to a third, then the first element is related to the third.

• R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

4. R is an equivalence relation if A is nonempty and R is reflexive, symmetric and transitive.

Calculation:

A relation is defined on Z such that aRb ⇒ (a − b) is divisible by 5,

For Reflexive: (a, a) ∈ R.

Since, (a − a) = 0 is divisible by 5. 

Therefore, the relation is reflexive.

For symmetric: If (a, b) ∈ R ⇒ (b, a) ∈ R.

(a, b)∈ R ⇒ (a − b) is divisible by 5.

Now, (b − a) = − (a − b) is also divisible by 5.

Therefore, (b, a) ∈ R

Hence, the relation is symmetric.

For Transitive: If (a, b) ∈ R and (b, a) ∈ R ⇒ (a, c) ∈ R.

(a, b) ∈ R ⇒ (a − b) is divisible by 5.

(b, c) ∈ R ⇒ (b − c) is divisible by 5.

Then,

(a − c) = (a – b + b −c)

(a − c)  = (a − b) + (b − c)

We know that (a − b) is divisible by 5 and (b − c) is divisible by 5 then (a − c) is also divisible by 5.  Therefore, (a, c) ∈ R.

Hence, the relation is transitive.

∴ the relation is equivalent.

Now, depending upon the remainder obtained when dividing (a−b) by 5 we can divide the set Z into 5 equivalent classes and they are disjoint i.e., there are no common elements between any two classes.

Concept:

Reflexive: A relation is said to be reflexive if (a, a) ∈ R, for every a ∈ A.

Ex. The relation R in the set {1, 2, 3} given by 

R = {(1,1), (2,2), (3,3)} is reflexive.

Symmetric: A relation is said to be symmetric, if (a, b) ∈ R, then (b, a) ∈ R.

Ex. The relation R in the set {a, b, c} given by

R = {(a,b), (b,a), (b,c), (c,b)} is symmetric.

Transitive: A relation is said to be transitive if (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R.

Ex. The relation R in the set {1, 2, 3} given by

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

Explanation:

Given that, A = {l, m, n} and the relation R be {}.

This relation is called a void relation or empty relation on A.

In other words, a relation R on set A is called an empty relation, if no element of A is related to any other element of A.

So, R will be the empty set.

And, R will be the void relation on set A.

So, void relation is not reflexive because it does not contain (a, a) for any a ∈ A.

As we know the definition of symmetric relation is that if A is a set in which the relation R is defined.

Then R is said to be a symmetric relation if (a, b) ∈ R ⇒ (b, a) ∈ R.

Now for void relation R does not contain any element of set A. So, relation R will be trivially symmetric.

As we know the definition of transitive relation is that a relation R over a set A is transitive if for all elements in A.

Whenever R relates a to b and b to c, then R also relates a to c.

So, a void relation has no element. So, it will also be trivially transitive.

So, void relation (or empty relation) is not reflexive but is symmetric and transitive.