show that each of the relation in the set A = { x E Z : 0x12 }, given by ,(a) R = {(a,b) : Ia-bI is a multiple of 4} . (b) R ={(a,b): a = b} is an equivalrnce relation - i understood this part, but the next one i didnt get it -Find the set of all elements related to 1 in each case...pls answer sir srry for my previous mistake i'll not repeat that again..

A = { x ∊ Z ,0 ≾ x ≾ 12}
1).R = {(a,b) : |a-b| is a multiple of 4
A = { x ∊ Z ,0 ≾ x ≾ 12}=  { 0,1,2,3,4,5,6,7,8,9,10,11,12}
For any element a∊A, we have (a, a) = |a-a| = 0 is a multiple of 4
Hence R is reflexive

Now let (a,b)∊ R, hence |a-b| is a multiple of 4
So |-(a-b)| = |b-a | is multiple of 4.
So (b,a)∊R
Hence R  is symmetric.

Now let (a,b) and (b,c)∊R
Hence |a-b | is a multiple of 4, |b-c | is also a multiple of 4
a-c = a-b + b-c is multiple of 4.
Hence |a-c | is also a multiple of 4
Hence (a,c)∊R
So R is transitive.

Thus R is an equivalence relation.
The set of elements related to 1 is {1,5,9}
So  |1-1 | = 0 is a multiple of 4
|5-1 | = 4 is a multiple of 4
|9-1 | = 8 is a multiple of 4

Hope you have got it.