Show that each of the relation R in the set A = {х Z :0 ≤ x ≤ 12} ,given by R ={(a,b) : | a-b| is a multiple of 4 is an equlance relarion . Find the set of elements related to 1

solution in the text book could not be followed kindly explain in detail.

A = {1,2,3,4,5,6,7,8,9,10,11,12}

R = {(a,b) : Ia-bI = 4x, where x belongs to Whole numbers}

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For every 'a' belonging to A, (a,a) belongs to R, as Ia-aI = 0, which is 4(0), hence is a multiple of 4. Therefore R is reflexive.

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If (a,b) belongs to R, I(a-b)I = 4x, and also, I(b-a)I = 4x. Hence (b,a) also belongs to R, and therefore, R is symmetric.

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If (a,b) and (b,c) belongs to R, then I(a-b)I I(b-c)I is a multiple of 4.

Hence, (a-b) = 4x and (b-c) = 4y, where x and y belongs to Integers.[here its integers because when modulus is removed, the RHS can be +/- 4x. hence we can say so].

therefore, (a-b) + (b-c) = 4x + 4y = 4(x+y)

a-b + b- c = multiple of 4,

Hence, a-c = multiple of 4.

Therefore, (a,c) belongs to R and R is Transitive.

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Hence R is reflexive, symmetric and transitive and so, R is said to be an equivalence Relation.

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The set of elements related to 1 in R such that I(k-1)I is a multiple of 4, where k belongs to A is:-

I(1-1)I = 0, which is 4(0) and hence is a multiple of 4.

I(5-1)I = 4, which is 4(1) and hence is a multiple of 4.

I(9-1)I = 8,which is 4(2) and hence is a multiple of 4.

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beyond 9, we hav I(13-1)I = 12 = 4(3), but then 13 does not belong to A.

Hence the set of elements in A, related to 1 by R, is {1,5,9}.

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