NCERT Solutions
Class 12 Maths
Relations and Functions
Ex. 1.1 Q9
Show that each of the relation R in the set A = {x ∈ Z: 0 ≤ x ≤ 12}, given by
(i) R = {(a, b) : |a – b| is a multiple of 4} (ii) R = {(a, b) : a = b}
is an equivalence relation. Find the set of all elements related to 1 in each case.
Given, A = {x ∈ Z: 0 ≤ x ≤ 12} = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
(i) R = {(a, b) : |a – b| is a multiple of 4}
For any element a ∈ A, we have (a, a) ∈ R as |a – a| = 0 is a multiple of 4.
So, R is reflexive.
Now, let (a, b) ∈ R
=> |a – b| is a multiple of 4.
=>| – (a – b)| = |b – a| is a multiple of 4.
=> (b, a) ∈ R
So, R is symmetric.
Now, let (a, b), (b, c) ∈ R.
=> |a – b| is a multiple of 4 and |b – c| is a multiple of 4.
=> (a – b) is a multiple of 4 and (b – c) is a multiple of 4.
=> (a – c) = (a – b) + (b – c) is a multiple of 4.
=> |a – c| is a multiple of 4.
=> (a, c) ∈ R
So, R is transitive.
Hence, R is an equivalence relation.
The set of elements related to 1 is {1, 5, 9} as
|1 – 1| = 0 is a multiple of 4.
|5 – 1| = 4 is a multiple of 4.
|9 – 1| = 8 is a multiple of 4.
(ii) R = {(a, b): a = b}
For any element a ∈ A, we have (a, a) ∈ R, since a = a.
So, R is reflexive.
Now, let (a, b) ∈ R.
=> a = b
=> b = a
=> (b, a) ∈ R
So, R is symmetric.
Now, let (a, b) ∈ R and (b, c) ∈ R.
=> a = b and b = c
=> a = c
=> (a, c) ∈ R
So, R is transitive.
Hence, R is an equivalence relation.
The elements in R that are related to 1 will be those elements from set A which are equal to 1.
Hence, the set of elements related to 1 is {1}.