Solve this question...and explain the equivalence class part.
Dear Student,
A = { x ∊ Z ,0 x 12}
1).S = {(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 3
Hence S is reflexive.
2) Now let (a,b)∊ S, hence |a-b| is a multiple of 3
So |-(a-b)| = |b-a | is multiple of 3.
So (b,a)∊S
Hence S is symmetric.
3) Now let (a,b) and (b,c)∊S
Hence |a-b | is a multiple of 3
So, a-b=±3k------(1)
|b-c | is also a multiple of 3
b-c=±3m------(1)
a-c = a-b + b-c =±3(k+m)-----{from 1 and 2}
|a-c|=3(k+m).
Hence |a-c | is also a multiple of 3
Hence (a,c)∊S
So S is transitive.
Thus S is an equivalence relation.
The set of elements related to 0 is {0,3,6,9,12}
So |0-0 | = 0 is a multiple of 3
|3-0 | = 3 is a multiple of 3
|6-0 | = 6 is a multiple of 3
|9-0 | = 9 is a multiple of 3
|12-0 | = 12 is a multiple of 3.
Regards!
A = { x ∊ Z ,0 x 12}
1).S = {(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 3
Hence S is reflexive.
2) Now let (a,b)∊ S, hence |a-b| is a multiple of 3
So |-(a-b)| = |b-a | is multiple of 3.
So (b,a)∊S
Hence S is symmetric.
3) Now let (a,b) and (b,c)∊S
Hence |a-b | is a multiple of 3
So, a-b=±3k------(1)
|b-c | is also a multiple of 3
b-c=±3m------(1)
a-c = a-b + b-c =±3(k+m)-----{from 1 and 2}
|a-c|=3(k+m).
Hence |a-c | is also a multiple of 3
Hence (a,c)∊S
So S is transitive.
Thus S is an equivalence relation.
The set of elements related to 0 is {0,3,6,9,12}
So |0-0 | = 0 is a multiple of 3
|3-0 | = 3 is a multiple of 3
|6-0 | = 6 is a multiple of 3
|9-0 | = 9 is a multiple of 3
|12-0 | = 12 is a multiple of 3.
Regards!