Let C 4 3 2 1 0 1 2 3 4 5 4 3 2 1 0 1 2 3

Let C : { 4 , 3 , 2 , 1 , 0 , 1 , 2 , 3 , 4 , 5} { 4 , 3 , 2 , 1 , 0 , 1 , 2 , 3 , 4 , 5 } be a relation defined as

xCy 3 | ( x y )

List the equivalence classes of C, or prove C is not an equivalence relation.

Solution

We are given

C : { 4 , 3 , 2 , 1 , 0 , 1 , 2 , 3 , 4 , 5} { 4 , 3 , 2 , 1 , 0 , 1 , 2 , 3 , 4 , 5 }

relation defined as

xCy 3 | ( x y )

3| (x-y) means x-y should be divisble by 3

so, firstly we will find relation

C={(-4, -4) (-4, -1) (-4,2) (-4,5) (-3, -3) (-3,0) (-3,3) (-2, -2) (-2,1)(-2,4) (-1,-4) (-1,-1) (-1,2) (-1,5) (0,-3) (0,3) (1,-2) (1,1) (1,4) (2,-1) (2,2) (2,5) (3,-3) (3,0) (3,3) (4,-2) (4,1) (4,4) (5,-4) (5,-1) (5,2) (5,5) }

now, we have to show whether its equivalnce relation

Reflexive:

We are given

C={(-4, -4) (-4, -1) (-4,2) (-4,5) (-3, -3) (-3,0) (-3,3) (-2, -2) (-2,1)(-2,4) (-1,-4) (-1,-1) (-1,2) (-1,5) (0,-3)(0,0) (0,3) (1,-2) (1,1) (1,4) (2,-1) (2,2) (2,5) (3,-3) (3,0) (3,3) (4,-2) (4,1) (4,4) (5,-4) (5,-1) (5,2) (5,5) }

We can see that our element as

(-4,-4) (-3,-3) (-2,-2) (-1,-1) (0,0) (1,1) (2,2) (3,3) (4,4) (5,5)

aRa

It means that C is reflexive relation

Symmetric :

if a R b

then , bRa

we can see that (-4,-1) and then (-1,-4)

so, it is symmetric relation

Transitive :

if a R b and b R c then a R c

we are given

C={(-4, -4) (-4, -1) (-4,2) (-4,5) (-3, -3) (-3,0) (-3,3) (-2, -2) (-2,1)(-2,4) (-1,-4) (-1,-1) (-1,2) (-1,5) (0,-3)(0,0) (0,3) (1,-2) (1,1) (1,4) (2,-1) (2,2) (2,5) (3,-3) (3,0) (3,3) (4,-2) (4,1) (4,4) (5,-4) (5,-1) (5,2) (5,5) }

We can see that

(-4, -1) and (-1,2) then (-4, 2)

so, it is transitive relation as well

Hence , this is equivalence relation ...........Answer