RD Chapter 1 Relations Ex 1.1 Solutions
Question - 11 : - 
Answer - 11 : -
Question - 12 : - 
Answer - 12 : -
Question - 13 : - 
Answer - 13 : -
Question - 14 : -
Answer - 14 : -
Question - 15 : -
Answer - 15 : -
Question - 16 : - Let A = {a, b, c} and the relation R be defined on A as follows:
R = {(a, a), (b, c), (a, b)}.
Then, write minimum number of ordered pairs to be added in R to make it reflexive and transitive.
Answer - 16 : -
A relation R in A is said to be reflexive if aRa for all a∈A
R is said to be transitive if aRb and bRc ⇒ aRc
for all a, b, c ∈ A.
Hence for R to be reflexive (b, b) and (c, c) must be there in the set R.
Also for R to be transitive (a, c) must be in R because (a, b) ∈ R and (b, c) ∈ R so (a, c) must be in R.
So at least 3 ordered pairs must be added for R to be reflexive and transitive.
Question - 17 : - Each of the following defines a relation on N:
• x > y, x, y ϵ N
• x + y = 10, x, y ϵ N
• xy is square of an integer, x, y ϵ N
• x + 4y = 10, x, y ϵ N
Determine which of the above relations are reflexive, symmetric and transitive.
Answer - 17 : -
A relation R in A is said to be reflexive if aRa for all a∈A, R is symmetric if aRb ⇒ bRa, for all a, b ∈ A and it is said to be transitive if aRb and bRc ⇒ aRc for all a, b, c ∈ A.
• x > y, x, y ϵ N
(x, y) ϵ {(2, 1), (3, 1).......(3, 2), (4, 2)....}
This is not reflexive as (1, 1), (2, 2)....are absent.
This is not symmetric as (2,1) is present but (1,2) is absent.
This is transitive as (3, 2) ϵ R and (2,1) ϵ R also (3,1) ϵ R ,similarly this property satisfies all cases.
• x + y = 10, x, y ϵ N
(x, y)ϵ {(1, 9), (9, 1), (2, 8), (8, 2), (3, 7), (7, 3), (4, 6), (6, 4), (5, 5)}
This is not reflexive as (1, 1),(2, 2)..... are absent.
This only follows the condition of symmetric set as (1, 9)ϵR also (9, 1)ϵR similarly other cases are also satisfy the condition.
This is not transitive because {(1, 9),(9, 1)}ϵR but (1, 1) is absent.
• xy is square of an integer, x, y ϵ N
(x, y) ϵ {(1, 1), (2, 2), (4, 1), (1, 4), (3, 3), (9, 1), (1, 9), (4, 4), (2, 8), (8, 2), (16, 1), (1, 16)...........}
This is reflexive as (1,1),(2,2)..... are present.
This is also symmetric because if aRb ⇒ bRa, for all a,bϵN.
This is transitive also because if aRb and bRc ⇒ aRc for all a, b, c ϵ N.
• x + 4y = 10, x, y ϵ N
(x, y) ϵ {(6, 1), (2, 2)}
This is not reflexive as (1, 1), (2, 2).....are absent.
This is not symmetric because (6,1) ϵ R but (1,6) is absent.
This is not transitive as there are only two elements in the set having no element common.