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Question
Show that the relation R in the set A = {x ∈ Z : 0 ≤ x ≤ 12} given by R = {(a, b) : |a – b| is a multiple of 4} is an equivalence relation. Find the set of all elements related to 1.
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Solution
A = {x ∈ Z : 0 ≤ x ≤ 12} = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
R = {(a, b) : |a – b| is a multiple of 4}
(i) Reflexive:
For any element a ∈ A, we have (a, a) ∈ R as |a – a| = 0 is a multiple of 4.
∴ R is reflexive.
(ii) Symmetric:
Now, let (a, b) ∈ R
⇒ |a – b| is a multiple of 4.
⇒ |–(a – b)| = |b – a| is a multiple of 4.
⇒ (b, a) ∈ R
Thus (a, b) ∈ R
⇒ (b, a) ∈ R
∴ R is symmetric.
(iii) Transitive:
Now, let (a, b), (b, c) ∈ R.
⇒ |a – b| is a multiple of 4 and |b – c| is a multiple of 4.
⇒ |a – c| = |a – b + b – c| = |a – b| + |b – c|
⇒ (a – c) = (a – b) + (b – c) is a multiple of 4.
⇒ (a, c) ∈ R ...[∴ |a – b| is multiple of 4 and |b – c| is multiple of 4.]
∴ R is transitive.
Hence, R is an equivalence relation.
The set of elements related to 1 is {1, 5, 9} since
|1 – 1| = 0 is a multiple of 4.
|5 – 1| = 4 is a multiple of 4.
|9 – 1| = 8 is a multiple of 4.
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