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Question
Show that each of 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 in each case.
Solution
A ={x in Z : 0 <= <= 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 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 ) in 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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