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

Concept:

Reflexive and Symmetric Relations:

• A reflexive relation on a set means that every element of the set is related to itself. For example, for a set {a, b, c}, the reflexive relation would contain (a, a), (b, b), and (c, c).
• A symmetrical relation on a set means that if an element a is related to element b, then b is related to a. For example, if (a, b) ∈ R, then (b, a) ∈ R must also be true.
• To determine the total number of elements in a relation R on a set A, we need to ensure that the relation is reflexive (i.e., it contains at least the pairs (a, a) for each element a) and symmetric (i.e., for every pair (a, b) there is a corresponding pair (b, a)).

Combinatorics:

• For a set with n elements, the total number of unordered pairs (a, b) is given by .
• In a symmetric relation, for each unordered pair, there must be two pairs: (α, β) and (β, α).
• The reflexive part of the relation adds n pairs, (a, a) for each element a in the set.

Calculation:

Given,

• The set A = {a, b, c, d, e, f} has 6 elements.
• The relation R is reflexive and symmetric and contains exactly 10 elements.
• We need to calculate how many unordered pairs are possible for a reflexive and symmetric relation.

The number of unordered pairs for 6 elements is given by:

Out of these 15 pairs, we need to select 2 pairs that are reflexive, i.e., (α, α), (β, β), and so on.

Thus, we need 2 pairs to be added to the relation, leaving us with 15 - 2 = 13 unordered pairs for symmetry.

Now, the total number of possible pairs needed is .

Hence, the total number of elements in the relation R is 105.

∴ The number of elements in the relation S is 105.

Solution:

R= {(f,g): f(0) = g(1) and f(1) = g(0)}

Reflexive: (f, f) € R

⇒ f(0) = f(1) and f(1) = f(0) ⇒ must hold

⇒ But this is not true for all functions

So not reflexive

Symmetric: If (f,g) € R ⇒ (g, f) € R

Now, g(0) = f(1) and g(1) = f(0) ⇒ true

∴ symmetric

Transitive: If (f,g) € R and (g,h) € R

⇒ (f,h) € R

Now (f,g) € R= f(0)= g(1) and f(1)= g(0)

⇒ (g,h) € R ⇒ g(0)= h(1) and g(1) = h(0)

⇒ For (f,h) € R we need f(0) = h(1) and f(1) = h(0)

⇒ Now f(0)= g(1) = h(0) nd f(1) = g(0) = h(1)

Hence transitive

Hence, the correct answer is Option 2

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: 

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 ≥ 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  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:

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 ax is a strictly decreasing function.
• Important property: If log_ax > log_ay, then it implies x
• Hence, the given relation reduces to xRy ⇔ x
• Relation properties:
  • Reflexive: A relation is reflexive if xRx for all x in S. But x
  • Symmetric: If x
  • Transitive: If x

Calculation:

Given,

Relation: xRy ⇔ log_ax > log_ay

Base of logarithm: a = 1/2

⇒ Since 0 ax is a decreasing function

⇒ log_ax > log_ay ⇔ x

⇒ So, xRy ⇔ x

⇒ Check reflexivity: x

⇒ Check symmetry: If x

⇒ Check transitivity: If x

∴ The relation is transitive only.

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) 

⇒ Every element maps to itself.

⇒ R is Reflexive 

Now, 1, 2 , 3 

(1, 2), (2, 3)  ⇒  (1, 3) 

⇒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  (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:

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.

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.