This page contains the NCERT mathematics class 12 chapter Relations and Functions Exercise 1.1 Problem 3 Solution. Solutions for other problems are available at Exercise 1.1 Solutions
Exercise 1.1 Problem 3 Solution
3. Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b) : b = a + 1} is reflexive, symmetric or transitive.
To Check whether R is Reflexive: The relation R in the set A is reflexive if (a, a) ∈ R, for every a ∈ A.
In this case, for every a ∈ R, it should satisfy the relation, {a = a + 1}.
But as we know, a ≠ {a + 1}.
⇒ (a, a) ∉ R
∴ R is not Reflexive
To Check whether R is Symmetric: The relation R in the set A is symmetric if (a_1, a_2) ∈ R implies that (a_2, a_1) ∈ R, for all a_1, a_2 ∈ A
When {b = a + 1}, we know that a ≠ {b + 1} (∵ a = b - 1).
⇒ when (a, b) ∈ R, we’ve (b, a) ∉ R
∴ R is not symmetric
To Check whether R is Transitive: The relation R in the set A is transitive if (a_1, a_2) ∈ R and (a_2, a_3) ∈ R implies that (a_1, a_3) ∈ R, for all a_1, a_2, a_3 ∈ A
When {b = a + 1} and {c = b + 1},
we’ve c {= b + 1} {= (a + 1) + 1} {= a + 2} ≠ {a + 1}.
⇒ when (a, b) ∈ R and (b, c) ∈ R then (a, c) ∉ R.
∴ R is not transitive
∴ R is neither reflexive nor symmetric nor transitive.
