Explore topic-wise InterviewSolutions in .

Given a set A = {1, 2, 3, 4, 5} and relation R = {(a, b) : |a² – b²| < 8}

Now according to the question |a² – b²| < 8

⇒ R = {(1,1),(2,2), (3,3), (4,4), (5,5), (1,2), (2,1), (2,3), (3,2), (3,4), (4,3),}

Given R = {(a, b) : 2a + 3b = 30} ∀ (a,b) ∈ N

Now according to the question 2a + 3b = 30:

⇒ R={(3,8),(6,6),(9,4),(12,2)}

NOTE: 0 is a whole number that’s why its not considered in this set, Although if we consider 0 as a natural number then the answer would be:

⇒ R={(0,10),(3,8),(6,6),(9,4),(12,2),(15,0)}

Given R = {(a, b) : a = b – 2, b > 6}

Now according to the question a = b – 2 and b > 6

⇒ R={(5,7),(6,8),(7,9)…….∞ }

A relation is an equivalence relation if and only if it is reflexive, symmetric and transitive:

The smallest equivalence relation on the set A={1,2,3} is :

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

∵ (1,1) ∈ R → Reflexive

(1,3) ∈ R and (3,1) ∈ R → Symmetric

(1,3) ∈ R and (3,1) ∈ R and (1,1) ∈ R → Transitive

C. 6

Let (3,2) ≃ (x,y)

⇒ 3y = 2x

This is possible in the cases:

x = 3, y = 2

x = 6, y = 4

x= 9, y =6

x=12, y = 3

x=15, y = 10

x=18,y=12

Hence total pairs are 6.

The total number of relations that can be defined from a set A to a set B is the number of possible subsets of A × B. 

If n(A) = p and n(B) = q, then n(A × B) = pq. 

So, 

The total number of relations is 2^pq. 

n(A) = 3 and n(B) = 2 

⇒ n(A × B) = 3 × 2 = 6 

∴ Total number of relations = 2⁶ = 64

Given,

R = {(a, b) : a, b N and a = b²} 

i. (a, a) R for all a N 

Here, 

take b = 2 

⇒ a = b² 

= 2² = 4 

∴ (4, 2) R but (2, 2) ∉ R 

As, 

2² ≠ 2 

So, 

No, the statement is false. 

ii. (a, b) R (b, a) R 

Here, 

take b = 2 

⇒ a = b² 

= 2² = 4 

∴ (4, 2) R but (2, 4) ∉ R 

As, 

4² ≠ 2 

So, 

No, the statement is false. 

iii. (a,b) R and (b,c) R (a,c) R 

Here, 

take b = 4 

⇒ a = b² 

= 4² = 16 

⇒ (16,4) R 

Now, 

b = c² 

⇒ 4 = c² 

⇒ c = - 2 ∉ N or 2 N 

⇒ (4,2) R 

But (16,2) ∉ R 

As, 

2² ≠ 16 

So, 

No, the statement is false.

We have been given that,

A is the set of all human beings in a town at a particular time.

Here, R is the binary relation on set A.

So, recall that

R is reflexive if for all x ∈ A, xRx.

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

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

Using these criteria, we can solve these.

We have,

R = {(x, y): x and y work at the same place}

Check for Reflexivity:

Since x & x are the same people then, x & x works at the same place.

Take yourself, for example, if you work at Bloomingdale then you work at Bloomingdale.

Since you can’t work in two places at a particular time,

So, ∀ x ∈ A, then (x, x) ∈ R.

∴ R is Reflexive.

Check for Symmetry:

If x & y works at the same place, then, y and x also work at the same place.

If you & your friend, Chris was working in Bloomindale, then Chris and you are working in Bloomingdale only.

The only difference is in the way of writing, either you write your name and your friend’s name or your friend’s name and your name, it’s the same.

So, if (x, y) ∈ R, then (y, x) ∈ R

∀ x, y ∈ A

∴ R is Symmetric.

Check for Transitivity:

If x & y works at the same place and y & z works at the same place.

Then, x & z also works at the same place.

Say, if she & I was working in Bloomingdale and she & you were also working in Bloomingdale. Then, you and I are working in the same company.

So, if (x, y) ∈ R and (y, z) ∈ R, then (x, z) ∈ R.

∀ x, y, z ∈ A

∴ R is Transitive.

Hence, R is reflexive, symmetric and transitive.

According to the question,

m is related to n if m is a multiple of n.

∀ m, n ∈ I (I being set of integers)

The relation comes out to be:

R = {(m, n): m = kn, k ∈ ℤ}

Recall that for any binary relation R on set A. We have,

R is reflexive if for all x ∈ A, xRx.

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

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

Check for Reflexivity:

∀ m ∈ I

If (m, m) ∈ R

⇒ m = k m, holds.

As an integer is always a multiple of itself, So, ∀ m ∈ I, then (m, m) ∈ R.

⇒ R is reflexive.

∴ R is reflexive.

Check for Symmetry:

∀ m, n ∈ I

If (m, n) ∈ R

⇒ m = k n, holds.

Now, replace m by n and n by m, we get

n = k m, which may or not be true.

Let us check:

If 12 is a multiple of 3, but 3 is not a multiple of 12.

⇒ n = km does not hold.

So, if (m, n) ∈ R, then (n, m) ∉ R.

∀ m, n ∈ I

⇒ R is not symmetric.

∴ R is not symmetric.

Check for Transitivity:

∀ m, n, o ∈ I

If (m, n) ∈ R and (n, o) ∈ R

⇒ m = kn and n = ko

Where k ∈ ℤ

Substitute n = ko in m = kn, we get

m = k(ko)

⇒ m = k²o

If k ∈ ℤ, then k²∈ ℤ.

Let k² = r

⇒ m = ro, holds true.

⇒ (m, o) ∈ R

So, if (m, n) ∈ R and (n, o) ∈ R, then (m, o) ∈ R.

∀ m, n ∈ I

⇒ R is transitive.

∴ R is transitive.

We have the set A = {1, 2, 3, 4, 5, 6}

So, recall that for any binary relation R on set A. We have,

R is reflexive if for all x ∈ A, xRx.

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

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

We have

R = {(a, b): b = a + 1}

∵ Every a, b ∈ A.

And A = {1, 2, 3, 4, 5, 6}

The relation R on set A can be defined as:

Put a = 1

⇒ b = a + 1

⇒ b = 1 + 1

⇒ b = 2

⇒ (a, b) ≡ (1, 2)

Put a = 2

⇒ b = 2 + 1

⇒ b = 3

⇒ (a, b) ≡ (2, 3)

Put a = 3

⇒ b = 3 + 1

⇒ b = 4

⇒ (a, b) ≡ (3, 4)

Put a = 4

⇒ b = 4 + 1

⇒ b = 5

⇒ (a, b) ≡ (4, 5)

Put a = 5

⇒ b = 5 + 1

⇒ b = 6

⇒ (a, b) ≡ (5, 6)

Put a = 6

⇒ b = 6 + 1

⇒ b = 7

⇒ (a, b) ≠ (6, 7) [∵ 7 ∉ A]

Hence, R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}

Check for Reflexivity:

For 1, 2, …, 6 ∈ A [∵ A = {1, 2, 3, 4, 5, 6}]

(1, 1) ∉ R

(2, 2) ∉ R

…

(6, 6) ∉ R

So, ∀ a ∈ A, then (a, a) ∉ R.

⇒ R is not reflexive.

∴ R is not reflexive.

Check for Symmetry:

∀ 1, 2 ∈ A [∵ A = {1, 2, 3, 4, 5, 6}]

If (1, 2) ∈ R

Then, (2, 1) ∉ R

[∵ R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}]

So, if (a, b) ∈ R, then (b, a) ∉ R

∀ a, b ∈ A

⇒ R is not symmetric.

∴ R is not symmetric.

Check for Transitivity:

∀ 1, 2, 3 ∈ A

If (1, 2) ∈ R and (2, 3) ∈ R

Then, (1, 3) ∉ R

[∵ R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}]

So, if (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∉ R.

∀ a, b, c ∈ A

⇒ R is not transitive.

∴ R is not transitive.

Given : n(A) = 3 and n(B) = 2 

To find : distinct elements of set A and B 

Also, 

It is given that, 

{(x, 1), (y, 2), (z, 1)} A × B 

Set A has 3 elements whereas.

Set B has 2 elements. 

Also, 

A × B = {(a, b): a ∈ A and b ∈ B} 

Therefore, 

A ∈ {x, y, z} and B ∈ {1, 2}

Given : A = {−1, 1}

To find : A × A × A 

So, 

A × A = {(−1, −1), (−1, 1), (1, −1), (1, 1)} 

And, 

A × A × A = {(−1, −1, −1), (−1, −1, 1), (−1, 1, −1), (−1, 1, 1), (1, −1, −1), (1, −1, 1), (1, 1, −1), (1, 1, 1)}

Given : A = {1, 2, 3, 4} and R = {(a, b): a∈ A, b ∈ A, a divides b} 

To find : set R 

Both elements of R, a and b, belongs to set A and relation between and elements a and b is that a divides b.

So,

1 divides 1, 2, 3 and 4. 

2 divides 2 and 4. 

3 divides 3. 

4 divides 4. 

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

Given,

S = {a, b) : b = |a-1|, a Z and |a| ≤ 3} 

Z denotes integer which can be positive as well as negative 

Now, 

|a| ≤ 3 and b = |a-1| 

∴ a {-3, -2, -1, 0, 1, 2, 3} 

S = {a, b): b = |a-1|, a Z and |a| ≤ 3} 

⇒ S = {a, |a-1|): b = |a-1|, a Z and |a| ≤ 3} 

⇒ S = {(-3, |-3 – 1|), (-2, |-2 – 1|), (-1, |-1 – 1|), (0, |0 – 1|), (1, |1 – 1|), (2, |2 – 1|), (3, |3 – 1|)} 

⇒ S = {(-3, |-4|), (-2, |-3|), (-1, |-2|), (0, |-1|), (1, |0|), (2, |1|), (3, |2|)} 

⇒ S = {(-3, 4), (-2, 3), (-1, 2), (0, 1), (1, 0), (2, 1), (3, 2)} 

So, 

Domain of relation S = {-3, -2, -1, 0, 1, 2, 3} 

Range of relation S = {0, 1, 2, 3, 4}

Given,

R = {(x, x³) : x is a prime number less than 10} 

Prime numbers less than 10 are 2, 3, 5 and 7 

∴ R = {(2, 2³), (3, 3³), (5, 5³), (7, 7³)} 

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

So, 

Domain of relation R = {2, 3, 5, 7} 

Range of relation R = {8, 27, 125, 343}

A = {5} 

B = {5, 6} 

A × B = {(5, 5), (5, 6)} 

All the possible subsets of A × B are, 

{} 

{(5, 5)} 

{(5, 6)} 

{(5, 6), (5, 6)}

Given, 

Relation R from A to B by R = {(x, y): the difference between x and y is odd, xA, y B} 

A = {1, 2, 3, 5} and B = {4, 6, 9} 

For x = 1 :

y – x = 4 – 1 = 3 which is odd 

⇒ 4 y 

y – x = 6 – 1 = 5 which is odd 

⇒ 6 y 

y – x = 9 – 1 = 8 which is even 

⇒ 8 ∉ y 

For x = 2 :

 y – x = 4 – 2 = 2 which is even 

⇒ 4 ∉ y 

y – x = 6 – 2 = 4 which is even 

⇒ 6 ∉ y y – x = 9 – 2 = 7 which is odd 

⇒ 8 y 

For x = 3 : 

y – x = 4 – 3 = 1 which is odd 

⇒ 4 y 

y – x = 6 – 3 = 3 which is odd 

⇒ 6 y 

y – x = 9 – 3 = 6 which is even 

⇒ 8 ∉ y 

For x = 5 : 

x – y = 5 – 4 = 1 which is odd 

⇒ 4 y 

y – x = 6 – 5 = 4 which is even 

⇒ 6 ∉ y 

y – x = 9 – 4 = 4 which is even 

⇒ 8 ∉ y 

∴ R = {(1, 4), (1, 6), (2, 8), (3, 4), (3, 6), (5, 4)} 

NOTE : 

Domain of relation R= {1, 2, 3, 5} 

Range of relation R= {4, 6, 8}

i) {(x, x²) : x is a prime number less than 10}.

Roster form: R = {(1, 1), (2, 4), (3, 9), (5, 25), (7, 49)} 

(ii) The domain of R is the set of first co-ordinates of R 

Dom(R) = {1, 2, 3, 5, 7} 

The range of R is the set of second co-ordinates of R 

Range(R) = {1, 4, 9, 25, 49}

Any subset of (A × A) is called a binary relation to A. Here, (A × A) is the cartesian product of A with A. 

Let A = {4, 5, 6) and R = {(4, 5), (6, 4), (5, 6)} 

Here, R is a binary relation to A. 

The domain of R is the set of first co-ordinates of R 

Dom(R) = {4, 6, 5} 

The range of R is the set of second co-ordinates of R 

Range(R) = {5, 4, 6}

Given: A = {2, 3, 4, 5} and B = {3, 6, 7, 10} 

(i) R = {(x, y), : x ϵ A, y ϵ B and x is relatively prime to y} 

So, R in Roster Form, 

R = {(2, 3), (2, 7), (3, 7), (3, 10), (4, 3), (4, 7), (5, 3), (5, 6), (5, 7)} 

(ii) Dom(R) = {2, 3, 4, 5} 

Range(R) = {3, 6, 7, 10}

Given: A = {3, 4} and B = {7, 9} 

R = {(a, b): a ϵ A, b ϵ B and (a – b) is odd} 

So, R = {(4, 7), (4, 9)} 

An empty relation means there is no elements in the relation set. 

Here we get two relations which satisfy the given conditions. 

Therefore, the given relation is not an Empty Relation. 

The given relation would be an Empty Relation if, 

(1) A = {3} or, 

(2) A = {3, any odd number} or,

The number of relations from set A to set B =

2^n(A)xn(B)

n(A) = Number of elements in set A 

n(B) = Number of elements in set B 

Here,

n(A) = 3

n(B) = 1 

Total number of relations =

2^3x1

= 8

C. transitive only

According to the question:

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

Thus R = {(b, c)} can only be transitive.

i.e (a,b)∈ R and (b,c)∈ R → (a,c)∈ R

Given: A = {2, 4, 5, 7} and b = {1, 2, 3, 4, 5, 6, 7, 8} 

(i) R = {(x, y) x ϵ A, y ϵ B and x divides y} 

So, R in Roster Form, 

R = {(2, 2), (2, 4), (2, 6), (2, 8), (4, 4), (4, 8), (5, 5), (7, 7)} 

(ii) Dom(R) = {2, 4, 5, 7} 

Range(R) = {2, 4, 5, 7, 6, 7, 8}

Given: R = {(a, b): a, b ϵ Z and (a – b) is an integer 

The condition satisfies for all the values of a and b to be any integer. 

So, R = {(a, b): for all a, b ϵ (-∞, ∞)} 

Dom(R) = {-∞, ∞} 

Range(R) = {-∞, ∞}

Given: A = {1, 2, 3, 4, 6} 

(i) R = {(a, b) : a, b ϵ A, and a divides b} 

R is Foster Form is, 

R = {(1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)} 

(ii) Dom(R) = {1, 2, 3, 4, 6} 

Range(R) = {2, 3, 4, 6}

Given: A = {2, 3} and B= {3, 5}

(i) (A × B) = {(2, 3), (2, 5), (3, 3), (3, 5)} 

Therefore, n(A × B) = 4 

(ii) No. of relation from A to B is a subset of Cartesian product of (A × B). 

Here no. of elements in A = 2 and no. of elements in B = 2. 

So, (A × B) = 2 × 2 = 4 

So, the total number of relations can be defined from A to B is 

= 2⁴ = 16

Given: R = {(x, y): x, y ϵ Z and x² + y² ≤ 4} 

(i) R is Foster Form is, 

R = {(-2, 0), (-1, -1), (-1, 0), (-1, 1), (0, -2), (0, -1), (0, 0), (0, 1), (0, 2), (1, -1), (1, 0), (1, 1), (2, 0)} 

(ii) Dom(R) = {-2, -1, 0, 1, 2} 

Range(R) = {-2, -1, 0, 1, 2}