Explore topic-wise InterviewSolutions in .

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}

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

dom (R) = {2, 3, 4} and range \((R)=\begin{Bmatrix}\frac{1}{2},\frac{1}{3},\frac{1}{4}\end{Bmatrix}\)

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

range (R) = {8 27}

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

Consider as R₁

Given that R₁ = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}

Reflexivity:

Here, (1, 1), (2, 2), (3, 3) ∈ R

Clearly, R₁ is reflexive.

Symmetry:

Here, (2, 1) ∈ R₁,

But (1, 2) ∉ R₁

So, R₁ is not symmetric.

Transitivity:

Here, (2, 1) ∈ R₁ and (1, 3) ∈ R₁,

But (2, 3) ∉ R₁

So, R₁ is not transitive.

Now we consider R₂

Given that R₂ = {(2, 2), (3, 1), (1, 3)}

Reflexivity:

Clearly, (1, 1) and (3, 3) ∉ R₂

So, R₂ is not reflexive.

Symmetry:

Here, (1, 3) ∈ R₂ and (3, 1) ∈ R₂

Clearly, R₂ is symmetric.

Transitivity:

Here, (1, 3) ∈ R₂ and (3, 1) ∈ R₂

But (3, 3) ∉ R₂

So, R₂ is not transitive.

Consider as R₃

Given that R₃ = {(1, 3), (3, 3)}

Reflexivity:

Clearly, (1, 1) ∉ R₃

So, R₃ is not reflexive.

Symmetry:

Here, (1, 3) ∈ R₃, but (3, 1) ∉ R₃

So, R₃ is not symmetric.

Transitivity:

Here, (1, 3) ∈ R₃ and (3, 3) ∈ R₃

Also, (1, 3) ∈ R₃

Clearly, R₃ is transitive.

(i) Consider as R₁

Given that R₁ = {(a, a), (a, b), (a, c), (b, b), (b, c), (c, a), (c, b), (c, c)}

Now let us check R₁ is reflexive, symmetric and transitive

Reflexive:

Given (a, a), (b, b) and (c, c) ∈ R₁

Clearly, R₁ is reflexive.

Symmetric:

We see that the ordered pairs obtained by interchanging the components of R₁ are also in R₁.

So, R₁ is symmetric.

Transitive:

Here, (a, b) ∈ R₁, (b, c) ∈ R₁ and also (a, c) ∈ R₁

So, R₁ is transitive.

(ii) Consider as R₂

Given that R₂ = {(a, a)}

Reflexive:

Clearly (a, a) ∈ R₂.

So, R₂ is reflexive.

Symmetric:

Clearly (a, a) ∈ R ⇒ (a, a) ∈ R.

So, R₂ is symmetric.

Transitive:

R₂ is clearly a transitive relation, since there is only one element in it.

(iii) Consider as R₃

Given that R₃ = {(b, c)}

Reflexive:

Here,(b, b) ∉ R₃ neither (c, c) ∉ R₃

Clearly, R₃ is not reflexive.

Symmetric:

Here, (b, c) ∈ R₃, but (c, b) ∉ R₃

So, R₃ is not symmetric.

Transitive:

Here, R₃ has only two elements.

Hence, R₃ is transitive.

(iv) Consider as R₄

Given that R₄ = {(a, b), (b, c), (c, a)}.

Reflexive:

Here, (a, a) ∉ R₄, (b, b) ∉ R₄ (c, c) ∉ R₄

Clearly, R₄ is not reflexive.

Symmetric:

Here, (a, b) ∈ R₄, but (b, a) ∉ R₄.

So, R₄ is not symmetric

Transitive:

Here, (a, b) ∈ R₄, (b, c) ∈ R₄, but (a, c) ∉ R₄

Clearly, R₄ is not transitive.

First let R be a relation on A

Given as set A of ordered pair of integers defined by (x, y) R (u, v) if xv = yu

Let us check whether the given relation is equivalence or not.

To prove equivalence relation, the given relation should be reflexive, symmetric and transitive.

Reflexivity: 

Let (a, b) be an arbitrary element of the set A. 

Then, (a, b) ∈ A

⇒ ab = ba  

⇒ (a, b) R (a, b)

So, R is reflexive on A.

Symmetry: 

Let (x, y) and (u, v) ∈ A such that (x, y) R (u, v). Then,

 xv = yu

⇒ vx = uy

⇒ uy = vx

⇒ (u, v) R (x, y)

Thus, R is symmetric on A.

Transitivity:  

Let (x, y), (u, v) and (p, q) ∈ R such that (x, y) R (u, v) and (u, v) R (p, q)

⇒ xv = yu and uq = vp

By, multiplying the corresponding sides, we get

xv × uq = yu × vp

⇒ xq = yp

⇒ (x, y) R (p, q)

So, R is transitive on A.

∴ R is reflexive, symmetric and transitive.

Hence, R is an equivalence relation on A.

Given as R = {(a, b): a, b ∈ Z and a + b is even} is a relation defined on R.

Given Z be the set of integers

To prove equivalence relation the relation should be reflexive, symmetric and transitive.

We have to check these properties on R.

Reflexivity:

Let a be an arbitrary element of Z. 

Then, a ∈ R

Clearly, a + a = 2a is even for all a ∈ Z.

⇒ (a, a) ∈ R for all a ∈ Z

Thus, R is reflexive on Z.

Symmetry:

Let (a, b) ∈ R

⇒ a + b is even

⇒ b + a is even

⇒ (b, a) ∈ R for all a, b ∈ Z

Therefore, R is symmetric on Z.

Transitivity:

Let (a, b) and (b, c) ∈ R

⇒ a + b and b + c are even

Now, let a + b = 2x for some x ∈ Z

And b + c = 2y for some y ∈ Z

Adding above two equations, we get

A + 2b + c = 2x + 2y

⇒ a + c = 2 (x + y − b), which is even for all x, y, b ∈ Z

Thus, (a, c) ∈ R

Clearly, R is transitive on Z.

∴ R is reflexive, symmetric and transitive.

Hence, R is an equivalence relation on Z

Given as R on Z defined by (a, b) ∈ R ⇔ a − b is divisible by 5

To prove equivalence relation, the relation should be reflexive, symmetric and transitive.

We have to check these properties on R.

Reflexivity:

Let a be an arbitrary element of R. Then,

⇒ a − a = 0 = 0 × 5

⇒ a − a is divisible by 5

⇒ (a, a) ∈ R for all a ∈ Z

Therefore, R is reflexive on Z.

Symmetry:

Let (a, b) ∈ R

⇒ a − b is divisible by 5

⇒ a − b = 5p for some p ∈ Z

⇒ b − a = 5(−p)

Here, −p ∈ Z [Since p ∈ Z]

⇒ b − a is divisible by 5

⇒ (b, a) ∈ R for all a, b ∈ Z

Thus, R is symmetric on Z.

Transitivity:

Let (a, b) and (b, c) ∈ R

⇒ a − b is divisible by 5

⇒ a − b = 5p for some Z

Also, b − c is divisible by 5

⇒ b − c = 5q for some Z

On adding two equations above, we get

a − b + b − c = 5p + 5q

⇒ a − c = 5(p + q)

⇒ a − c is divisible by 5

Here, p + q ∈ Z

⇒ (a, c) ∈ R for all a, c ∈ Z

Clearly, R is transitive on Z.

∴ R is reflexive, symmetric and transitive.

Hence, R is an equivalence relation on Z.

Given as (a, b) ∈ R ⇔ a − b is divisible by n is a relation R defined on Z.

To prove equivalence relation, the relation should be reflexive, symmetric and transitive.

We have to check these properties on R.

Reflexivity:

Let a ∈ N

Here, a − a = 0 = 0 × n

⇒ a − a is divisible by n

⇒ (a, a) ∈ R

⇒ (a, a) ∈ R for all a ∈ Z

Thus, R is reflexive on Z.

Symmetry:

Let (a, b) ∈ R

Here, a − b is divisible by n

⇒ a − b = np for some p ∈ Z

⇒ b − a = n(−p)

⇒ b − a is divisible by n [p ∈ Z ⇒ − p ∈ Z]

⇒ (b, a) ∈ R 

Clearly, R is symmetric on Z.

Transitivity:

Let (a, b) and (b, c) ∈ R

Here, a − b is divisible by n and b − c is divisible by n.

⇒ a − b = n p for some p ∈ Z

And b − c = nq for some q ∈ Z

a – b + b − c = np + nq

⇒ a − c = n(p + q)

⇒ (a, c) ∈ R for all a, c ∈ Z

Thus, R is transitive on Z.

∴ R is reflexive, symmetric and transitive.

Hence, R is an equivalence relation on Z.

Given as (a, b) ∈ R ⇔ a − b is divisible by n is a relation R defined on Z.

To prove equivalence relation, the given relation should be reflexive, symmetric and transitive.

We have to check these properties on R.

Reflexivity:

Let a ∈ N

Here, a − a = 0 = 0 × n

⇒ a − a is divisible by n

⇒ (a, a) ∈ R

⇒ (a, a) ∈ R for all a ∈ Z

Therefore, R is reflexive on Z.

Symmetry:

Let (a, b) ∈ R

Here, a − b is divisible by n

⇒ a − b = np for some p ∈ Z

⇒ b − a = n(−p)

⇒ b − a is divisible by n [ p ∈ Z⇒ − p ∈ Z]

⇒ (b, a) ∈ R 

So, R is symmetric on Z.

Transitivity:

Let (a, b) and (b, c) ∈ R

Here, a − b is divisible by n and b − c is divisible by n.

⇒ a − b = np for some p ∈ Z

And b − c = nq for some q ∈ Z

a – b + b − c = np + nq

⇒ a − c = n(p + q)

⇒ (a, c) ∈ R for all a, c ∈ Z

So, R is transitive on Z.

∴ R is reflexive, symmetric and transitive.

Hence, R is an equivalence relation on Z.

Correct Answer is (B) 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)

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.

Correct Answer is (C) symmetric and transitive but not reflexive

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)

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.

Correct Answer is (A) reflexive and symmetric but not transitive

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)

Correct Answer is (D) an equivalence relation

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)

Correct Answer is (B) symmetric but neither reflexive nor transitive

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)

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.

Correct Answer is (D) an equivalence relation

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)

Correct Answer is (C) transitive but neither reflexive nor symmetric

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

a R b ⇒ |a| ≤ b

∴ |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)

Correct Answer is (D) an equivalence relation

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)

Correct Answer is (B) * is not commutative

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)

Correct Answer is (D) both commutative and associative

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)

Correct Answer is (D) both commutative and associative

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 = \(\frac{\frac{ab}{2}\times c}{2}\)= abc/4

And , a * (b * c) = a * (bc/2) = \(\frac{a\times\frac{bc}{2}}{2}\)= 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)

Correct Answer is (C) neither commutative nor associative

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 =\((a^b)^c\)

And , a * (b * c) = a * (b^c )=\(a^{(b^c)}\)

Ex a=2 b=3 c=4

(a * b) * c = (2³) * c =\((8)^4\)

a * (b * c) = 2 * (3⁴)=\(2^{(81)}\)

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)

Correct Answer is (D) both commutative and associative

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)

Correct Answer is (C) neither commutative nor associative

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)

Correct Answer is (C) . reflexive and transitive but not symmetric

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)

Correct Answer is (A) reflexive and symmetric but not transitive

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)

Correct Answer is (D) an equivalence relation

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)

Correct Answer is (B) reflexive and transitive but not symmetric

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)

Given as R = {(a, b): a – b is divisible by 3; a, b ∈ Z} is a relation

To prove equivalence relation, the given relation should be reflexive, symmetric and transitive.

We have to check these properties on R.

Reflexivity: 

Let a be an arbitrary element of R. 

Then, a – a = 0 = 0 × 3

⇒ a − a is divisible by 3

⇒ (a, a) ∈ R for all a ∈ Z

Therefore, R is reflexive on Z.

Symmetry:

Let (a, b) ∈ R

⇒ a − b is divisible by 3

⇒ a − b = 3p for some p ∈ Z

⇒ b − a = 3 (−p)

Here, −p ∈ Z

⇒ b − a is divisible by 3

⇒ (b, a) ∈ R for all a, b ∈ Z

Clearly, R is symmetric on Z.

Transitivity:

Let (a, b) and (b, c) ∈ R

⇒ a − b and b − c are divisible by 3

⇒ a – b = 3p for some p ∈ Z

And b − c = 3q for some q ∈ Z

On adding above two equations, we get

a − b + b – c = 3p + 3q

⇒ a − c = 3(p + q)

Here, p + q ∈ Z

⇒ a − c is divisible by 3

⇒ (a, c) ∈ R for all a, c ∈ Z

Thus, R is transitive on Z.

∴ R is reflexive, symmetric and transitive.

Clearly, R is an equivalence relation on Z.

Correct Answer is (D) an equivalence relation

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)

Correct Answer is (A) reflexive and symmetric but not transitive

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)

Correct Answer is (A) symmetric but neither reflexive nor transitive

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)