Relation

dom (R) = {-1, 1, -2, 2} and range (R) = {1, 4}

range (R) = {8 27}

(i) R = {(2, 8),(3, 27),(5, 125), (7, 343)}

(ii) dom (R) = {2, 3, 5, 7}

(iii) range (R) = {8, 27, 125, 343}

{3, 2, 1}

dom (R) = {3, 6, 9} and range (R) = {3, 2, 1}

dom (R) = {-2, -1, 0, 1, 2} and range (R) = {3, 2, 1, 0}

dom (R) = {2, 3, 4} and range

dom (R) = {1, 2, 3} and range (R) = {6, 7, 8}

Let R = {(A, B) : A ⊂ B)}, i.e., A is a proper subset of B, be a relation defined on S.

Now,

Any set is a subset of itself, but not a proper subset.

⇒ (A,A) ∉ R ∀ A ∈ S

⇒ R is not reflexive.

Let (A,B) ∈ R ∀ A, B ∈ S

⇒ A is a proper subset of B

⇒ all elements of A are in B, but B contains at least one element that is not in A.

⇒ B cannot be a proper subset of A

⇒ (B,A) ∉ R

For e.g. , if B = {1,2,5} then A = {1,5} is a proper subset of B . we observe that B is not a proper subset of A.

⇒ R is not symmetric

Let (A,B) ∈ R and (B,C) ∈ R ∀ A, B,C ∈ S

⇒ A is a proper subset of B and B is a proper subset of C

⇒ A is a proper subset of C

⇒ (A,C) ∈ R

For e.g. , if B = {1,2,5} then A = {1,5} is a proper subset of B .

And if C = {1,2,5,7} then B = {1,2,5} is a proper subset of C.

We observe that A = {1,5} is a proper subset of C also.

⇒ R is transitive.

Thus, R is transitive but not reflexive and not symmetric.

In order to show R is an equivalence relation, we need to show R is Reflexive, Symmetric and Transitive.

Given that, A be the set of all points in a plane and O be the origin. Then, R = {(P, Q) : P, Q ∈ A and OP = OQ)}

Now,

R is Reflexive if (P,P)∈R∀P∈A

∀ P ∈ A , we have

OP=OP

⇒ (P,P) ∈ R

Thus, R is reflexive.

R is Symmetric if (P,Q)∈R⇒(Q,P)∈R∀P, Q∈A

Let P, Q ∈ A such that,

(P,Q) ∈ R

⇒ OP = OQ

⇒ OQ = OP

⇒ (Q,P) ∈ R

Thus, R is symmetric.

R is Transitive if (P,Q)∈R and (Q,S)∈R⇒(P,S)∈R∀P, Q, S∈A

Let (P,Q) ∈ R and (Q,S) ∈ R ∀ P, Q, S ∈ A

⇒ OP = OQ and OQ = OS

⇒ OP = OS

⇒ (P,S) ∈ R

Thus, R is transitive.

Since R is reflexive, symmetric and transitive it is an equivalence relation on A.

Let R = {(a, b) : a ≤ b} be a relation defined on S.

Now,

We observe that any element x ∈ S is less than or equal to itself.

⇒ (x,x) ∈ R ∀ x ∈ S

⇒ R is reflexive.

Let (x,y) ∈ R ∀ x, y ∈ S

⇒ x is less than or equal to y

But y cannot be less than or equal to x if x is less than or equal to y.

⇒ (y,x) ∉ R

For e.g. , we observe that (2,5) ∈ R i.e. 2 < 5 but 5 is not less than or equal to 2 ⇒ (5,2) ∉ R

⇒ R is not symmetric

Let (x,y) ∈ R and (y,z) ∈ R ∀ x, y ,z ∈ S

⇒ x ≤ y and y ≤ z

⇒ x ≤ z

⇒ (x,z) ∈ R

For e.g. , we observe that

(4,5) ∈ R ⇒ 4 ≤ 5 and (5,6) ∈ R ⇒ 5 ≤ 6

And we know that 4 ≤ 6 ∴ (4,6) ∈ R

⇒ R is transitive.

Thus, R is reflexive and transitive but not symmetric.

Given that,

A = {1, 2, 3, 4, 5, 6) and R = {(a, b) : a, b ∈ A and b = a + 1}.

∴ R = {(1,2),(2,3),(3,4),(4,5),(5,6)}

Now,

R is Reflexive if (a,a) ∈ R ∀ a ∈ A

Since, (1,1),(2,2),(3,3),(4,4),(5,5),(6,6) ∉ R

Thus, R is not reflexive .

R is Symmetric if (a,b) ∈ R ⇒ (b,a) ∈ R ∀ a,b ∈ A

We observe that (1,2) ∈ R but (2,1) ∉ R .

Thus, R is not symmetric .

R is Transitive if (a,b) ∈ R and (b,c) ∈ R ⇒ (a,c) ∈ R ∀ a,b,c ∈ A

We observe that (1,2) ∈ R and (2,3) ∈ R but (1,3) ∉ R

Thus, R is not transitive.

Relation: Let A and B be two sets. Then a relation R from set A to set B is a subset of A x B. Thus, R is a relation to A to B ⇔ R ⊆ A x B.

If R is a relation from a non-void set B and if (a,b) ∈ R, then we write a R b which is read as ‘a is related to b by the relation R’. if (a,b) ∉ R, then we write a R b, and we say that a is not related to b by the relation R.

Domain: Let R be a relation from a set A to a set B. Then the set of all first components or coordinates of the ordered pairs belonging to R is called the domain of R.

Thus, domain of R ={a : (a,b) ∈ R}. The domain of R ⊆ A.

Range: let R be a relation from a set A to a set B. then the set of all second component or coordinates of the ordered pairs belonging to R is called the range of R.

Example 1: R = {(-1, 1), (1, 1), (-2, 4), (2, 4)}.

dom (R) = {-1, 1, -2, 2} and range (R) = {1, 4}

Example 2: R ={(a, b): a, b ∈ N and a + 3b = 12}

dom (R) = {3, 6, 9} and range (R) = {3, 2, 1}

Let R = {(∆₁, ∆₂) : ∆₁ ~ ∆₂} be a relation defined on A.

Now,

R is Reflexive if (Δ, Δ)∈R∀Δ∈A

We observe that for each Δ ∈ A we have,

Δ ~ Δ since, every triangle is similar to itself.

⇒ (Δ, Δ) ∈ R ∀ Δ ∈ A

⇒ R is reflexive.

R is Symmetric if (∆₁, ∆₂)∈R⇒(∆₂, ∆₁)∈R∀∆₁, ∆₂∈A

Let (∆₁, ∆₂) ∈ R ∀ ∆₁, ∆₂ ∈ A

⇒ ∆₁ ~ ∆₂

⇒ ∆₂ ~ ∆₁

⇒ (∆₂, ∆₁) ∈ R

⇒ R is symmetric

R is Transitive if (∆₁, ∆₂)∈R and (∆₂, ∆₃)∈R⇒(∆₁, ∆₃)∈R∀∆₁, ∆₂,∆₃∈A

Let (∆₁, ∆₂) ∈ R and ((∆₂, ∆₃) ∈ R ∀ ∆₁, ∆₂, ∆₃ ∈ A

⇒ ∆₁ ~ ∆₂ and ∆₂ ~ ∆₃

⇒ ∆₁ ~ ∆₃

⇒ (∆₁, ∆₃) ∈ R

⇒ R is transitive.

Since R is reflexive, symmetric and transitive, it is an equivalence relation on A.

In order to show R is an equivalence relation, we need to show R is Reflexive, Symmetric and Transitive.

Given that, ∀ a, b ∈ Z, R = {(a, b) : (a + b) is even }.

Now,

R is Reflexive if (a,a)∈R∀a∈Z

For any a ∈ A, we have

a+a = 2a, which is even.

⇒ (a,a) ∈ R

Thus, R is reflexive.

R is Symmetric if (a,b)∈R⇒(b,a)∈R∀a,b∈Z

(a,b) ∈ R

⇒ a+b is even.

⇒ b+a is even.

⇒ (b,a) ∈ R

Thus, R is symmetric .

R is Transitive if (a,b)∈R and (b,c)∈R⇒(a,c)∈R∀a,b,c∈Z

Let (a,b) ∈ R and (b,c) ∈ R ∀ a, b,c ∈ Z

⇒ a+b = 2P and b+c = 2Q

Adding both, we get

a+c+2b = 2(P+Q)

⇒ a+c = 2(P+Q)-2b

⇒ a+c is an even number

⇒ (a, c) ∈ R

Thus, R is transitive on Z.

Since R is reflexive, symmetric and transitive it is an equivalence relation on Z.

In order to show R is an equivalence relation, we need to show R is Reflexive, Symmetric and Transitive.

Given that, ∀ a, b ∈Z, aRb if and only if a – b is divisible by 5.

Now,

R is Reflexive if (a,a)∈R∀a∈Z

aRa ⇒ (a-a) is divisible by 5.

a-a = 0 = 0 × 5 [since 0 is multiple of 5 it is divisible by 5]

⇒ a-a is divisible by 5

⇒ (a,a) ∈ R

Thus, R is reflexive on Z.

R is Symmetric if (a,b)∈R⇒(b,a)∈R∀a,b∈Z

(a,b) ∈ R ⇒ (a-b) is divisible by 5

⇒ (a-b) = 5z for some z ∈ Z

⇒ -(b-a) = 5z

⇒ b-a = 5(-z) [∵ z ∈ Z ⇒ -z ∈ Z ]

⇒ (b-a) is divisible by 5

⇒ (b,a) ∈ R

Thus, R is symmetric on Z.

R is Transitive if (a,b)∈R and (b,c)∈R⇒(a,c)∈R∀a,b,c∈Z

(a,b) ∈ R ⇒ (a-b) is divisible by 5

⇒ a-b = 5z₁ for some z₁∈ Z

(b,c) ∈ R ⇒ (b-c) is divisible by 5

⇒ b-c = 5z₂ for some z₂∈ Z

Now,

a-b = 5z₁ and b-c = 5z₂

⇒ (a-b) + (b-c) = 5z₁ + 5z₂

⇒ a-c = 5(z₁ + z₂ ) = 5z₃ where z₁ + z₂ = z₃

⇒ a-c = 5z₃ [∵ z₁,z₂ ∈ Z ⇒ z₃∈ Z]

⇒ (a-c) is divisible by 5.

⇒ (a, c) ∈ R

Thus, R is transitive on Z.

Since R is reflexive, symmetric and transitive it is an equivalence relation on Z.

In order to show R is an equivalence relation we need to show R is Reflexive, Symmetric and Transitive.

Given that, ∀ a, b ∈A, R = {(a, b) : |a – b| is even}.

Now,

R is Reflexive if (a,a)∈R∀a∈A

For any a ∈ A, we have

|a-a| = 0, which is even.

⇒ (a,a) ∈ R

Thus, R is reflexive.

R is Symmetric if (a,b)∈R⇒(b,a)∈R∀a,b∈A

(a,b) ∈ R

⇒ |a-b| is even.

⇒ |b-a| is even.

⇒ (b,a) ∈ R

Thus, R is symmetric .

R is Transitive if (a,b)∈R and (b,c)∈R⇒(a,c)∈R∀a,b,c∈A

Let (a,b) ∈ R and (b,c) ∈ R ∀ a, b,c ∈ A

⇒ |a – b| is even and |b – c| is even

⇒ (a and b both are even or both odd) and (b and c both are even or both odd)

Now two cases arise:

Case 1 : when b is even

Let (a,b) ∈ R and (b,c) ∈ R

⇒ |a – b| is even and |b – c| is even

⇒ a is even and c is even [∵ b is even]

⇒ |a – c| is even [∵ difference of any two even natural numbers is even]

⇒ (a, c) ∈ R

Case 2 : when b is odd

Let (a,b) ∈ R and (b,c) ∈ R

⇒ |a – b| is even and |b – c| is even

⇒ a is odd and c is odd [∵ b is odd]

⇒ |a – c| is even [∵ difference of any two odd

natural numbers is even]

⇒ (a, c) ∈ R

Thus, R is transitive on Z.

Since R is reflexive, symmetric and transitive it is an equivalence relation on Z.

In order to show R is an equivalence relation we need to show R is Reflexive, Symmetric and Transitive.

Given that, R be the relation in N ×N defined by (a, b) R (c, d) if a + d = b + c for (a, b), (c, d) in N ×N.

R is Reflexive if (a, b) R (a, b) for (a, b) in N ×N

Let (a,b) R (a,b)

⇒ a+b = b+a

which is true since addition is commutative on N.

⇒ R is reflexive.

R is Symmetric if (a,b) R (c,d)⇒(c,d) R (a,b) for (a, b), (c, d) in N ×N

Let (a,b) R (c,d)

⇒ a+d = b+c

⇒ b+c = a+d

⇒ c+b = d+a [since addition is commutative on N]

⇒ (c,d) R (a,b)

⇒ R is symmetric.

R is Transitive if (a,b) R (c,d) and (c,d) R (e,f)⇒(a,b) R (e,f) for (a, b), (c, d),(e,f) in N ×N

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

⇒ a+d = b+c and c+f = d+e

⇒ (a+d) – (d+e) = (b+c ) – (c+f)

⇒ a-e= b-f

⇒ a+f = b+e

⇒ (a,b) R (e,f)

⇒ R is transitive.

Hence, R is an equivalence relation.

In order to show R is an equivalence relation we need to show R is Reflexive, Symmetric and Transitive.

Given that, ∀ a, b ∈ S, R = {(a, b) : a = ± b }

Now,

R is Reflexive if (a,a)∈R∀a∈S

For any a ∈ S, we have

a = ±a

⇒ (a,a) ∈ R

Thus, R is reflexive.

R is Symmetric if (a,b)∈R⇒(b,a)∈R∀a,b∈S

(a,b) ∈ R

⇒ a = ± b

⇒ b = ± a

⇒ (b,a) ∈ R

Thus, R is symmetric .

R is Transitive if (a,b)∈Rand (b,c)∈R⇒(a,c)∈R∀a,b,c∈S

Let (a,b) ∈ R and (b,c) ∈ R ∀ a, b,c ∈ S

⇒ a = ± b and b = ± c

⇒ a = ± c

⇒ (a, c) ∈ R

Thus, R is transitive.

Hence, R is an equivalence relation.

Given that, ∀ A, B ∈ S, R = {(A, B) : d(A, B) < 2 units}.

Now,

R is Reflexive if (A,A)∈R∀A∈S

For any A ∈ S, we have

d(A,A) = 0, which is less than 2 units

⇒ (A,A) ∈ R

Thus, R is reflexive.

R is Symmetric if (A, B)∈R⇒(B,A)∈R∀A,B∈S

(A, B) ∈ R

⇒ d(A, B) < 2 units

⇒ d(B, A) < 2 units

⇒ (B,A) ∈ R

Thus, R is symmetric .

R is Transitive if (A, B)∈R and (B,C)∈R⇒(A,C)∈R∀A,B,C∈S

Consider points A(0,0),B(1.5,0) and C(3.2,0).

d(A,B)=1.5 units < 2 units and d(B,C)=1.7 units < 2 units

d(A,C)= 3.2 ≮ 2

⇒ (A, B) ∈ R and (B,C) ∈ R ⇒ (A,C) ∉ R

Thus, R is not transitive.

Thus, R is reflexive, symmetric but not transitive.

Given that, ∀ a, b ∈ S, R = {(a, b) : a² + b² = 1 }

Now,

R is Reflexive if (a,a)∈R∀a∈S

For any a ∈ S, we have

a²+a² = 2 a² ≠ 1

⇒ (a,a) ∉ R

Thus, R is not reflexive.

R is Symmetric if (a,b)∈R⇒(b,a)∈R∀a,b∈S

(a,b) ∈ R

⇒ a² + b² = 1

⇒ b² + a² = 1

⇒ (b,a) ∈ R

Thus, R is symmetric .

R is Transitive if (a,b)∈R and (b,c)∈R⇒(a,c)∈R∀a,b,c∈S

Let (a,b) ∈ R and (b,c) ∈ R ∀ a, b,c ∈ S

⇒ a² + b² = 1 and b² + c² = 1

Adding both, we get

a²+ c²+2b² = 2

⇒ a²+ c²= 2-2b² ≠ 1

⇒ (a, c) ∉ R

Thus, R is not transitive.

Thus, R is symmetric but neither reflexive nor transitive.

We have, R = {(a, b) : a = b²} relation defined on N.

Now,

We observe that, any element a ∈ N cannot be equal to its square except 1.

⇒ (a,a) ∉ R ∀ a ∈ N

For e.g. (2,2) ∉ R ∵ 2 ≠ 2²

⇒ R is not reflexive.

Let (a,b) ∈ R ∀ a, b ∈ N

⇒ a = b²

But b cannot be equal to square of a if a is equal to square of b.

⇒ (b,a) ∉ R

For e.g., we observe that (4,2) ∈ R i.e 4 = 2² but 2 ≠ 4²⇒ (2,4) ∉ R

⇒ R is not symmetric

Let (a,b) ∈ R and (b,c) ∈ R ∀ a, b,c ∈ N

⇒ a = b² and b = c²

⇒ a ≠ c²

⇒ (a,c) ∉ R

For e.g., we observe that

(16,4) ∈ R ⇒ 16 = 4² and (4,2) ∈ R ⇒ 4 = 2²

But 16 ≠ 2²

⇒ (16,2) ∉ R

⇒ R is not transitive.

Thus, R is neither reflexive nor symmetric nor transitive.

We have, R = {(a, b) : a > b} relation defined on N.

Now,

We observe that, any element a ∈ N cannot be greater than itself.

⇒ (a,a) ∉ R ∀ a ∈ N

⇒ R is not reflexive.

Let (a,b) ∈ R ∀ a, b ∈ N

⇒ a is greater than b

But b cannot be greater than a if a is greater than b.

⇒ (b,a) ∉ R

For e.g., we observe that (5,2) ∈ R i.e 5 > 2 but 2 ≯ 5 ⇒ (2,5) ∉ R

⇒ R is not symmetric

Let (a,b) ∈ R and (b,c) ∈ R ∀ a, b,c ∈ N

⇒ a > b and b > c

⇒ a > c

⇒ (a,c) ∈ R

For e.g., we observe that

(5,4) ∈ R ⇒ 5 > 4 and (4,3) ∈ R ⇒ 4 > 3

And we know that 5 > 3 ∴ (5,3) ∈ R

⇒ R is transitive.

Thus, R is transitive but not reflexive not symmetric.

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

Now,

R is reflexive ∵ (1,1),(2,2),(3,3) ∈ R

R is not symmetric ∵ (1,2),(2,3) ∈ R but (2,1),(3,2) ∉ R

R is not transitive ∵ (1,2) ∈ R and (2,3) ∈ R ⇒ (1,3) ∉ R

Thus, R is reflexive but neither symmetric nor transitive.

Given that, A = {1, 2, 3} and R = {1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (1, 3), (3, 2)}.

Now,

R is reflexive ∵ (1,1),(2,2),(3,3),(4,4) ∈ R

R is not symmetric ∵ (1,2),(1,3),(3,2) ∈ R but (2,1),(3,1),(2,3) ∉ R

R is transitive ∵ (1,3) ∈ R and (3,2) ∈ R ⇒ (1,2) ∈ R

Thus, R is reflexive and transitive but not symmetric.

Given set A = {1, 2, 3}

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

Formula

For a relation R in set A

Reflexive

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

Symmetric

The 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

Equivalence

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

Check for reflexive

Since , (1,1) ∈ R , (2,2) ∈ R , (3,3) ∈ R

Therefore , R is reflexive ……. (1)

Check for symmetric

Since (1,3) ∈ R but (3,1) ∉ R

Therefore , R is not symmetric ……. (2)

Check for transitive

Here , (1,3) ∈ R and (3,2) ∈ R and (1,2) ∈ R

Therefore , R is transitive ……. (3)

Now , according to the equations (1) , (2) , (3)

Correct option will be (B)

Given set A = {a, b, c}

And R = {(a, a), (a, b), (b, a)}

Formula

For a relation R in set A

Reflexive

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

Symmetric

The 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

Equivalence

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

Check for reflexive

Since , (b,b) ∉ R and (c,c) ∉ R

Therefore , R is not reflexive ……. (1)

Check for symmetric

Since , (a,b) ∈ R and (b,a) ∈ R

Therefore , R is symmetric ……. (2)

Check for transitive

Here , (a,b) ∈ R and (b,a) ∈ R and (a,a) ∈ R

Therefore , R is transitive ……. (3)

Now , according to the equations (1) , (2) , (3)

Correct option will be (C)

Given set A = {1, 2, 3}

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

Formula

For a relation R in set A

Reflexive

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

Symmetric

The 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

Equivalence

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

Check for reflexive

Since , (1,1) ∈ R , (2,2) ∈ R , (3,3) ∈ R

Therefore , R is reflexive ……. (1)

Check for symmetric

Since , (1,2) ∈ R and (2,1) ∈ R

(2,3) ∈ R and (3,2) ∈ R

Therefore , R is symmetric ……. (2)

Check for transitive

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

Therefore , R is not transitive ……. (3)

Now , according to the equations (1) , (2) , (3)

Correct option will be (A)

According to the question ,

Given set S = {x, y, z}

And R = {(x, y), (y, z), (x, z) , (y, x), (z, y), (z, x)}

Formula

For a relation R in set A

Reflexive

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

Symmetric

The 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

Equivalence

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

Check for reflexive

Since , (x,x) ∉ R , (y,y) ∉ R , (z,z) ∉ R

Therefore , R is not reflexive ……. (1)

Check for symmetric

Since , (x,y) ∈ R and (y,x) ∈ R

(z,y) ∈ R and (y,z) ∈ R

(x,z) ∈ R and (z,x) ∈ R

Therefore , R is symmetric ……. (2)

Check for transitive

Here , (x,y) ∈ R and (y,x) ∈ R but (x,x) ∉ R

Therefore , R is not transitive ……. (3)

Now , according to the equations (1) , (2) , (3)

Correct option will be (B)

According to the question ,

Given set S = {x, y, z}

And R = {(x, x), (y, y), (z, z)}

Formula

For a relation R in set A

Reflexive

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

Symmetric

The 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

Equivalence

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

Check for reflexive

Since , (x,x) ∈ R , (y,y) ∈ R , (z,z) ∈ R

Therefore , R is reflexive ……. (1)

Check for symmetric

Since , (x,x) ∈ R and (x,x) ∈ R

(y,y) ∈ R and (y,y) ∈ R

(z,z) ∈ R and (z,z) ∈ R

Therefore , R is symmetric ……. (2)

Check for transitive

Here , (x,x) ∈ R and (y,y) ∈ R and (z,z) ∈ R

Therefore , R is transitive ……. (3)

Now , according to the equations (1) , (2) , (3)

Correct option will be (D)

According to the question ,

Given set Z = {1, 2, 3 ,4 …..}

And R = {(a, b) : a,b ∈ Z and (a-b) is divisible by 3}

Formula

For a relation R in set A

Reflexive

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

Symmetric

The 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

Equivalence

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

Check for reflexive

Consider , (a,a)

(a - a) = 0 which is divisible by 3

(a,a) ∈ R where a ∈ Z

Therefore , R is reflexive ……. (1)

Check for symmetric

Consider , (a,b) ∈ R

∴ (a - b) which is divisible by 3

- (a - b) which is divisible by 3

(since if 6 is divisible by 3 then -6 will also be divisible by 3)

∴ (b - a) which is divisible by 3 ⇒ (b,a) ∈ R

For any (a,b) ∈ R ; (b,a) ∈ R

Therefore , R is symmetric ……. (2)

Check for transitive

Consider , (a,b) ∈ R and (b,c) ∈ R

∴ (a - b) which is divisible by 3

and (b - c) which is divisible by 3

[ (a-b)+(b-c) ] is divisible by 3 ] (if 6 is divisible by 3 and 9 is divisible by 3 then 6+9 will also be divisible by 3)

∴ (a - c) which is divisible by 3 ⇒ (a,c) ∈ R

Therefore (a,b) ∈ R and (b,c) ∈ R then (a,c) ∈ R

Therefore , R is transitive ……. (3)

Now , according to the equations (1) , (2) , (3)

Correct option will be (D)

According to the question ,

Given set N = {1, 2, 3 ,4 …..}

And R = {(a, b) : a,b ∈ N and a is a factor of b}

Formula

For a relation R in set A

Reflexive

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

Symmetric

The 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

Equivalence

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

Check for reflexive

Consider , (a,a)

a is a factor of a

(2,2) , (3,3)… (a,a) where a ∈ N

Therefore , R is reflexive ……. (1)

Check for symmetric

a R b ⇒ a is factor of b

b R a ⇒ b is factor of a as well

Ex _ (2,6) ∈ R

But (6,2) ∉ R

Therefore , R is not symmetric ……. (2)

Check for transitive

a R b ⇒ a is factor of b

b R c ⇒ b is a factor of c

a R c ⇒ b is a factor of c also

Ex _(2,6) , (6,18)

∴ (2,18) ∈ R

Therefore , R is transitive ……. (3)

Now , according to the equations (1) , (2) , (3)

Correct option will be (B)

According to the question ,

Given set Z = {1, 2, 3 ,4 …..}

And R = {(a, b) : a,b ∈ Z and a ≥ b}

Formula

For a relation R in set A

Reflexive

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

Symmetric

The 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

Equivalence

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

Check for reflexive

Consider , (a,a) (b,b)

∴ a ≥ a and b ≥ b which is always true.

Therefore , R is reflexive ……. (1)

Check for symmetric

a R b ⇒ a ≥ b

b R a ⇒ b ≥ a

Both cannot be true.

Ex _ If a=2 and b=1

∴ 2 ≥ 1 is true but 1 ≥ 2 which is false.

Therefore , R is not symmetric ……. (2)

Check for transitive

a R b ⇒ a ≥ b

b R c ⇒ b ≥ c

∴ a ≥ c

Ex _a=5 , b=4 and c=2

∴ 5≥4 , 4≥2 and hence 5≥2

Therefore , R is transitive ……. (3)

Now , according to the equations (1) , (2) , (3)

Correct option will be (C)

According to the question ,

Given set S = {…….,-2,-1,0,1,2 …..}

And R = {(a, b) : a,b ∈ S and |a| ≤ b }

Formula

For a relation R in set A

Reflexive

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

Symmetric

The 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

Equivalence

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

Check for reflexive

Consider , (a,a)

∴ |a| ≤ a and which is not always true.

Ex_if a=-2

∴ |-2| ≤ -2 ⇒ 2 ≤ -2 which is false.

Therefore , R is not reflexive ……. (1)

Check for symmetric

a R b ⇒ |a| ≤ b

b R a ⇒ |b| ≤ a

Both cannot be true.

Ex _ If a=-2 and b=-1

∴ 2 ≤ -1 is false and 1 ≤ -2 which is also false.

Therefore , R is not symmetric ……. (2)

Check for transitive

a R b ⇒ |a| ≤ b

b R c ⇒ |b| ≤ c

∴ |a| ≤ c

Ex _a=-5 , b= 7 and c=9

∴ 5 ≤ 7 , 7 ≤ 9 and hence 5 ≤ 9

Therefore , R is transitive ……. (3)

Now , according to the equations (1) , (2) , (3)

Correct option will be (C)

According to the question ,

Given set S = {…….,-2,-1,0,1,2 …..}

And R = {(a, b) : a,b ∈ S and |a – b| ≤ 1 }

Formula

For a relation R in set A

Reflexive

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

Symmetric

The 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

Equivalence

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

Check for reflexive

Consider , (a,a)

∴ |a – a| ≤ 1 and which is always true.

Ex_if a=2

∴ |2-2| ≤ 1 ⇒ 0 ≤ 1 which is true.

Therefore , R is reflexive ……. (1)

Check for symmetric

a R b ⇒ |a – b| ≤ 1

b R a ⇒ |b – a| ≤ 1

Both can be true.

Ex _ If a=2 and b=1

∴ |2 – 1| ≤ 1 is true and |1–2| ≤ 1 which is also true.

Therefore , R is symmetric ……. (2)

Check for transitive

a R b ⇒ |a – b| ≤ 1

b R c ⇒ |b – c| ≤ 1

∴|a – c| ≤ 1 will not always be true

Ex _a=-5 , b= -6 and c= -7

∴ |6-5| ≤ 1 , |7 – 6| ≤ 1 are true But |7 – 5| ≤ 1 is false.

Therefore , R is not transitive ……. (3)

Now , according to the equations (1) , (2) , (3)

Correct option will be (A)

According to the question ,

Given set S = {…….,-2,-1,0,1,2 …..}

And R = {(a, b) : a,b ∈ S and (1 + ab) > 0 }

Formula

For a relation R in set A

Reflexive

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

Symmetric

The 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

Equivalence

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

Check for reflexive

Consider , (a,a)

∴ (1 + a×a) > 0 which is always true because a×a will always be positive.

Ex_if a=2

∴ (1 + 4) > 0 ⇒ (5) > 0 which is true.

Therefore , R is reflexive ……. (1)

Check for symmetric

a R b ⇒ (1 + ab) > 0

b R a ⇒ (1 + ba) > 0

Both the equation are the same and therefore will always be true.

Ex _ If a=2 and b=1

∴ (1 + 2×1) > 0 is true and (1+1×2) > which is also true.

Therefore , R is symmetric ……. (2)

Check for transitive

a R b ⇒ (1 + ab) > 0

b R c ⇒ (1 + bc) > 0

∴(1 + ac) > 0 will not always be true

Ex _a=-1 , b= 0 and c= 2

∴ (1 + -1×0) > 0 , (1 + 0×2) > 0 are true

But (1 + -1×2) > 0 is false.

Therefore , R is not transitive ……. (3)

Now , according to the equations (1) , (2) , (3)

Correct option will be (A)

According to the question ,

Given set S = {…All triangles in plane….}

And R = {(∆₁ , ∆₂) : ∆₁ , ∆₂∈ S and ∆₁ ≡ ∆₂}

Formula

For a relation R in set A

Reflexive

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

Symmetric

The 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

Equivalence

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

Check for reflexive

Consider , (∆₁, ∆₁)

∴ We know every triangle is congruent to itself.

(∆₁, ∆₁) ∈ R all ∆₁ ∈ S

Therefore , R is reflexive ……. (1)

Check for symmetric

(∆₁ , ∆₂) ∈ R then ∆₁ is congruent to ∆₂

(∆₂ , ∆₁) ∈ R then ∆₂ is congruent to ∆₁

Both the equation are the same and therefore will always be true.

Therefore , R is symmetric ……. (2)

Check for transitive

Let ∆₁, ∆₂, ∆₃ ∈ S such that (∆₁, ∆₂) ∈ R and (∆₂, ∆₃) ∈ R

Then (∆₁, ∆₂)∈R and (∆₂, ∆₃)∈R

⇒∆₁ is congruent to ∆_2, and ∆₂ is congruent to ∆₃

⇒∆₁ is congruent to ∆₃

∴(∆₁, ∆₃) ∈ R

Therefore , R is transitive ……. (3)

Now , according to the equations (1) , (2) , (3)

Correct option will be (D)

According to the question ,

Given set S = {…….,-2,-1,0,1,2 …..}

And R = {(a, b) : a,b ∈ S and a² + b² = 1 }

Formula

For a relation R in set A

Reflexive

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

Symmetric

The 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

Equivalence

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

Check for reflexive

Consider , (a,a)

∴ a² + a² = 1 which is not always true

Ex_if a=2

∴ 2² + 2² = 1 ⇒ 4 + 4 = 1 which is false.

Therefore , R is not reflexive ……. (1)

Check for symmetric

a R b ⇒ a² + b² = 1

b R a ⇒ b² + a² = 1

Both the equation are the same and therefore will always be true.

Therefore , R is symmetric ……. (2)

Check for transitive

a R b ⇒ a² + b² = 1

b R c ⇒ b² + c² = 1

∴ a² + c² = 1 will not always be true

Ex _a=-1 , b= 0 and c= 1

∴ (-1)² + 0² = 1 , 0² + 1² = 1 are true

But (-1)² + 1² = 1 is false.

Therefore , R is not transitive ……. (3)

Now , according to the equations (1) , (2) , (3)

Correct option will be (A)

According to the question ,

R = {(a, b) , (c, d) : a + d = b + c }

Formula

For a relation R in set A

Reflexive

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

Symmetric

The 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

Equivalence

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

Check for reflexive

Consider , (a, b) R (a, b)

(a, b) R (a, b) ⇔ a + b = a + b

which is always true .

Therefore , R is reflexive ……. (1)

Check for symmetric

(a, b) R (c, d) ⇔ a + d = b + c

(c, d) R (a, b) ⇔ c + b = d + a

Both the equation are the same and therefore will always be true.

Therefore , R is symmetric ……. (2)

Check for transitive

(a, b) R (c, d) ⇔ a + d = b + c

(c, d) R (e, f) ⇔ c + f = d + e

On adding these both equations we get , a + f = b + e

Also,

(a, b) R (e, f) ⇔ a + f = b + e

∴ It will always be true

Therefore , R is transitive ……. (3)

Now , according to the equations (1) , (2) , (3)

Correct option will be (D)

According to the question ,

O is the origin

R = {(P, Q) : OP = OQ }

Formula

For a relation R in set A

Reflexive

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

Symmetric

The 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

Equivalence

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

Check for reflexive

Consider , (P , P) ∈ R ⇔ OP = OP

which is always true .

Therefore , R is reflexive ……. (1)

Check for symmetric

(P , Q) ∈ R ⇔ OP = OQ

(Q , P) ∈ R ⇔ OQ = OP

Both the equation are the same and therefore will always be true.

Therefore , R is symmetric ……. (2)

Check for transitive

(P , Q) ∈ R ⇔ OP = OQ

(Q , R) ∈ R ⇔ OQ = OR

On adding these both equations, we get , OP = OR

Also,

(P , R) ∈ R ⇔ OP = OR

∴ It will always be true

Therefore , R is transitive ……. (3)

Now , according to the equations (1) , (2) , (3)

Correct option will be (D)

According to the question ,

Q is set of all rarional numbers

R = {(a, b) : a * b = a + 2b }

Formula

* is commutative if a * b = b * a

* is associative if (a * b) * c = a * (b * c)

Check for commutative

Consider , a * b = a + 2b

And , b * a = b + 2a

Both equations will not always be true .

Therefore , * is not commutative ……. (1)

Check for associative

Consider , (a * b) * c = (a + 2b) * c = a+2b + 2c

And , a * (b * c) = a * (b+2c) = a+2(b+2c) = a+2b+4c

Both the equation are not the same and therefore will not always be true.

Therefore , * is not associative ……. (2)

Now , according to the equations (1) , (2)

Correct option will be (C)

According to the question ,

Q = { a,b }

R = {(a, b) : a * b = a + ab }

Formula

* is commutative if a * b = b * a

* is associative if (a * b) * c = a * (b * c)

Check for commutative

Consider , a * b = a + ab

And , b * a = b + ba

Both equations will not always be true .

Therefore , * is not commutative ……. (1)

Check for associative

Consider , (a * b) * c = (a + ab) * c = a+ab + (a+ab)c=a+ab+ac+abc

And , a * (b * c) = a * (b+bc) = a+a(b+bc) = a+ab+abc

Both the equation are not the same and therefore will not always be true.

Therefore , * is not associative ……. (2)

Now , according to the equations (1) , (2)

Correct option will be (B)

According to the question ,

Q = { Positive rationals }

R = {(a, b) : a * b = ab/2 }

Formula

* is commutative if a * b = b * a

* is associative if (a * b) * c = a * (b * c)

Check for commutative

Consider , a * b = ab/2

And , b * a = ba/2

Both equations are the same and will always true .

Therefore , * is commutative ……. (1)

Check for associative

Consider , (a * b) * c = (ab/2) * c = = abc/4

And , a * (b * c) = a * (bc/2) = = abc/4

Both the equation are the same and therefore will always be true.

Therefore , * is associative ……. (2)

Now , according to the equations (1) , (2)

Correct option will be (D)

According to the question ,

Q = { All integers }

R = {(a, b) : a * b = a – b + ab }

Formula

* is commutative if a * b = b * a

* is associative if (a * b) * c = a * (b * c)

Check for commutative

Consider , a * b = a – b + ab

And , b * a = b – a + ba

Both equations are not the same and will not always be true .

Therefore , * is not commutative ……. (1)

Check for associative

Consider , (a * b) * c = (a – b + ab) * c

= a – b + ab – c +(a – b + ab)c

=a – b +ab – c +ac – bc + abc

And , a * (b * c) = a * (b – c + bc)

= a - (b – c + bc) + a(b – c + bc)

=a – b + c – bc + ab – ac + abc

Both the equation are not the same and therefore will not always be true.

Therefore , * is not associative ……. (2)

Now , according to the equations (1) , (2)

Correct option will be (C)

According to the question ,

Q = { All integers }

R = {(a, b) : a * b = a + b - ab }

Formula

* is commutative if a * b = b * a

* is associative if (a * b) * c = a * (b * c)

Check for commutative

Consider , a * b = a + b - ab

And , b * a = b + a - ba

Both equations are the same and will always be true .

Therefore , * is commutative ……. (1)

Check for associative

Consider , (a * b) * c = (a + b - ab) * c

= a + b - ab + c -(a + b - ab)c

=a + b – ab + c – ac – bc + abc

And , a * (b * c) = a * (b + c - bc)

= a + (b + c - bc) - a(b + c - bc)

=a + b + c – bc - ab – ac + abc

Both the equation are the same and therefore will always be true.

Therefore , * is associative ……. (2)

Now , according to the equations (1) , (2)

Correct option will be (D)

According to the question ,

Q = { All integers }

R = {(a, b) : a * b = a^b }

Formula

* is commutative if a * b = b * a

* is associative if (a * b) * c = a * (b * c)

Check for commutative

Consider , a * b = a^b

And , b * a = b^a

Both equations are not the same and will not always be true .

Therefore , * is not commutative ……. (1)

Check for associative

Consider , (a * b) * c = (a^b) * c =

And , a * (b * c) = a * (b^c)=

Ex a=2 b=3 c=4

(a * b) * c = (2³) * c =

a * (b * c) = 2 * (3⁴)=

Both the equation are not the same and therefore will not always be true.

Therefore , * is not associative ……. (2)

Now , according to the equations (1) , (2)

Correct option will be (C)

According to the question ,

R = {(a, b) : a * b = a + b + ab }

Formula

* is commutative if a * b = b * a

* is associative if (a * b) * c = a * (b * c)

Check for commutative

Consider , a * b = a + b + ab

And , b * a = b + a + ba

Both equations are same and will always be true .

Therefore , * is commutative ……. (1)

Check for associative

Consider , (a * b) * c = (a + b + ab) * c

= a + b + ab + c + (a + b + ab)c

=a + b + c + ab + ac + bc + abc

And , a * (b * c) = a * (b + c + bc)

= a + b + c + bc + a(b + c + bc)

=a +b + c + ab + bc + ac + abc

Both the equation are same and therefore will always be true.

Therefore , * is associative ……. (2)

Now , according to the equations (1) , (2)

Correct option will be (D)