Let S be the set of all sets and let R = {(A, B): (A ⊂ B)}, i.e. A i

1.	Let S be the set of all sets and let R = {(A, B): (A ⊂ B)}, i.e. A is a proper subset of B. Show that R is (i) transitive (ii) not reflexive (iii) not symmetric.
Answer» (i) Transitive Consider A, B and C ∈ S, where (A, B) and (B, C) ∈ R We get (A, B) ∈ R => A ⊂ B ……. (1) (B, C) ∈ R => B ⊂ C …….. (2) Using both the equations we get A ⊂ C => (A, C) ∈ R Hence, R is a transitive relation S. (ii) Non reflexive We know that A ⊄ A where (A, A) ∈ R Hence, R is non reflexive. (iii) Non symmetric Consider A ⊂ B where (A, B) ∈ R We know that B ⊄ A So (B, A) ∉ R We get (A, B) ∈ R and (B, A) ∉ R Hence, R is non symmetric.

Let S be the set of all sets and let R = {(A, B): (A ⊂ B)}, i.e. A is a proper subset of B. Show that R is (i) transitive (ii) not reflexive (iii) not symmetric.

Answer»

(i) Transitive

Consider A, B and C ∈ S, where (A, B) and (B, C) ∈ R

We get

(A, B) ∈ R => A ⊂ B ……. (1)

(B, C) ∈ R => B ⊂ C …….. (2)

Using both the equations we get

A ⊂ C => (A, C) ∈ R

Hence, R is a transitive relation S.

(ii) Non reflexive

We know that

A ⊄ A where (A, A) ∈ R

Hence, R is non reflexive.

(iii) Non symmetric

Consider A ⊂ B where (A, B) ∈ R

We know that B ⊄ A

So (B, A) ∉ R

We get (A, B) ∈ R and (B, A) ∉ R

Hence, R is non symmetric.

Discussion

No Comment Found

Related InterviewSolutions

In fig. shows a relationship between the sets P and Q. Write this relation in :a. Set builder form b. Roster form c. What is its domain and range?
Let A = {1, 2, 3, 4, 5, 6}. Let R be a relation on A defined by R = {(a, b): a, b A, b is exactly divisible by a} i. Write R in roster form ii. Find the domain of R iii. Find the range of R.
Let `A = {1, 2, 3, 4, 6}`. Let R be the relation on A defined by `{(adot b): a , b in A , b`is exactly divisible by a}.(i) Write R in roster form(ii) Find the domain of R(iii) Find the range of R.
Let `A={1,2,3,5}" and "B={4,6,9}.` Define a relation from A to B, given by `R={(a,b):a inA,b inB" and "(a-b)" is odd"}.` (i) Write R in roster form. (ii) Find dom (R) and range (R).
Show that the relation R on the set Z of integers, given by R = {(a, b) : 2 divides a – b}, is an equivalence relation.
Let W denote the words in the English dictionary. Let the relation R be defined by R = {(x, y) ∈ W × W : the words x and y have at least one letter in common}. Then R is (a) Reflexive and transitive, not symmetric (b) Reflexive and symmetric and not transitive (c) Symmetric and transitive, not reflexive (d) Reflexive, symmetric and transitive
Given the relation R = {(1, 2), (2, 3)} on the set A = {1, 2, 3}, the minimum number of ordered pairs which can be added to R to make it an equivalence relation is(a) 2 (b) 3 (c) 5 (d) 7
If R be a relation defined as a R b ⇔ | a | < b, then R is(a) Reflexive only (b) Symmetric only (c) Transitive only (d) Reflexive and transitive but not symmetric
Let I be the set of integers and R be a relation on I defined by R = {(x, y) : (x – y) is divisible by 11, x, y ∈ I}. Then R is(a) An equivalence relation (b) Symmetric only (c) Reflexive only (d) Transitive only
Let Z be the set of integers. Then the relation R = {(a, b) : a, b ∈ Z and (a + b) is even} defined on Z is(a) Only symmetric (b) Symmetric and transitive only (c) An equivalence relation (d) None of the above

Let S be the set of all sets and let R = {(A, B): (A ⊂ B)}, i.e. A is a proper subset of B. Show that R is (i) transitive (ii) not reflexive (iii) not symmetric.

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment