Types of Relations MCQ Quiz in मल्याळम - Objective Question with Answer for Types of Relations - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Explanation:

Equivalence Relation: A relation R on a set A is said to be an equivalence relation if and only if the relation R is reflexive, symmetric, and transitive.

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

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

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)}

Concept:

A relation R on a set A is said to be an equivalence relation if and only if the relation R is reflexive, symmetric and transitive. The equivalence relation is a relationship on the set which is generally represented by the symbol “∼”.

• Reflexive: A relation is said to be reflexive, if (a, a) ∈ R, for every a ∈ A.
• Symmetric: A relation is said to be symmetric, if (a, b) ∈ R, then (b, a) ∈ R.
• Transitive: A relation is said to be transitive if (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R.

Calculations:

Given the relation R on the set of all real numbers R = {(a, b): a ≤ b2} 

Let \((\frac{1}{2}, \frac{1}{2}) ∉ R\), because \(\frac{1}{2}>(\frac{1}{2})^2=\frac{1}{4}\)

∴ R is not reflexive.

Now, let (1, 4) ∈ R as 1 ≤ 4²

But 4 is not less than 1²

∴ (4, 1) ∉ R

∴  R is not symmetric

Next, consider (3, 2), (2, 1.5) ∈ R (as 3 < 2² & 2 < 1.5² = 2.25

But 3 > 1.5² = 2.25

∴ (3, 1.5) ∉ R

∴  R is not transitive 

Thus R is neither reflexive, symmetric, nor transitive

Hence, the correct answer is option 2).

Concept:

• A relation R on the set A is said to be reflexive if (a, a) ∈ R for every a ∈ A.
• A relation R on the set A is said to be symmetric if (a, b) ∈ R then (b, a) ∈ R.

Calculation:

Given that (a, b) ∈ R ⇒ 1 + ab > 0 for all (a, b) ∈ A.

Now, (a, a) ∈ R if 1 + a² > 0.

Since a² ≥ 0, ⇒ 1 + a² > 0.

Therefore the relation is reflexive.

Also, (a, b) ∈ R ⇒ 1 + ab > 0.

And (b, a) = 1 + ba = 1 + ab > 0, which means that (b, a) ∈ R.

Therefore the relation is symmetric as well.

Concept:

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:

Here, xRy → x - y = 5 years Where x,y ∈  X (citizen of India)

For Reflexive:

xRx → x - x ≠ 5, So its not reflexive

For symmetric:

xRy → x - y = 5 and yRx → y - x = 5, difference will be 5 years only

So its symmetric

For transitive:

Let, x = 25, y = 20, and z = 15

x - y = 5, y - z = 5, but x - z ≠ 5

If xRy and yRz, then not necessarily, xRz.

So, its not transitive.

Hence, option (2) is correct. 

Concept:

For a relation R in set A

Reflexive

• The relation is reflexive if (a, a) ∈ R for every a ∈ A

Symmetric

• Relation is symmetric, if (a, b) ∈ R, then (b, a) ∈ R

Transitive

• Relation is transitive, if (a, b) ∈ R & (b, c) ∈ R, then (a, c) ∈ R
• If relation is reflexive, symmetric and transitive, it is an equivalence relation.

Calculation:

∣a - al = 0 <1

Therefore, a R a ∀ a ∈ R

Therefore, R is reflexive.

Again a R b, la - b| ≤ 1

|b - a∣ ≤ 1  b R a

Therefore, R is symmetric.

Again 1 R [\(\frac{1}{2}\)] and [\(\frac{1}{2}\)] R1 but [\(\frac{1}{2}\)] ≠ 1.

Therefore, R is not anti-symmetric.

Further, 1 R 2 and 2 R 3, but 1 R 3 is not possible, [Because, |1 - 3| = 2 > 1]

∴ R is not 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:

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

⇒ y ≥ x + 5

• So option 3 is incorrect.
• 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.
• So option 2 is correct.

Concept:

Reflexive Relation: A relation R on a set A is reflexive if for all the elements a ∈ A, (a , a) ∈ R

Symmetric Relation: A relation R on a set A is symmetric if (a , b) ∈ R, then (b , a) ∈ R for all a, b ∈ A

Transitive Relation: A relation R on a set A is transitive if (a ,b) ∈ R and (b , c) ∈ R, then (a , c) ∈ R for all a, b, c ∈ R

Equivalence Relation: A relation R on a set A is an equivalence relation if it is reflexive, symmetric as well as transitive.

Calculation:

Given:

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

Relation R is on set A

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

• The set, A = {1, 2, 3, 4} and (1 , 1) , (2 , 2) , (3 , 3) , (4 , 4) ∈ R. Hence it is a reflexive relation.
• (1 , 2) ∈ R but (2 , 1) ∉ R. Hence it is not a symmetric relation.
• ∀ (a ,b) ∈ R and (b , c) ∈ R, (a , c) ∈ R for all a, b, c ∈ R. Hence it is a transitive relation.  
• Since it is not symmetric, hence it can not be equivalence.

Concept:

A relation R in a set A is called 

• Reflexive, if (a, a) ∈ R, for every a ∈ A.
• Symmetric, if (a, b) ∈ R implies that (b, a) ∈ R, for all a, b ∈A.
• Transitive, if (a, b) ∈ R and (b, c) ∈ R  implies that (a, c) ∈ R, for all a, b, c ∈A.

A relation R on a set A is said to be an equivalence relation if R is reflexive, symmetric and transitive.

Calculation:

Concept:

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:

S = Set of all integers and R = {(a, b), a, b ϵ  S and ab \(≥\) 0}

For reflexive:

aRa = a.a = a² ≥ 0, so it's reflexive.

For symmetric:

aRb = ab ≥ 0 and bRa = ba ≥  0, So relation is symmetric

For transitive:

For all integers, if ab ≥ 0, bc ≥  0, then for all the cases ac ≥ 0 is not true.

For example,

If a = -1, b = 0, and c =1

Then,  ab ≥ 0, bc ≥  0 but for ac = -1 which is not satisfying ac ≥ 0 so R is not transitive.

So, Relation Is reflexive, symmetric but not transitive. 

Hence, option (2) is correct.

Concept:

A relation R in a set A is called

• Reflexive, if (a, a) ∈ R, for every a ∈ A.
• Symmetric, if (a, b) ∈ R implies that (b, a) ∈ R, for all a, b ∈ A.
• Transitive, if (a, b) ∈ R and (b, c) ∈ R  implies that (a, c) ∈ R, for all a, b, c ∈ A.

A relation R in a set A is said to be an equivalence relation if R is reflexive, symmetric and transitive.

Calculation:

Given: R = {(T1,T2): T1 is similar to T₂ where T₁, T₂ ∈ T} and T is the set of all triangles in a plane

Reflexive:

As we know that, every triangle is similar to itself, so (T₁, T₁) ∈ R ∀ T₁ ∈ T

Hence, relation R is reflexive.

Symmetric:

Suppose if (T₁, T₂) ∈ R ⇒T₁ is similar to T₂ ⇒T₂ is also similar to T₁ ⇒ (T₂, T₁) ∈ R.

Hence, relation R is symmetric.

Transitive:

Now suppose, (T₁, T₂), (T₂, T₃) ∈ R ⇒T₁ is similar to T₂ and T₂ is similar to T₃ ⇒T₁ is similar to T₃ . So (T₁, T₃) ∈ R.

Hence, relation R is transitive.

Hence, relation R is an equivalence relation.