Let X = {1,2,3,4,5,6,7} and R = {x,y/x–y is divisible by 3} in x. Show that R is an equivalence relation.

Let $A = \{1, 2, 3, 4, 5, 6, 7\}$ and $R = \{(x, y) | x –y \text{ is divisible by }3\}$

$4-1=3$ is divisible by 3.

$5-2=3$ is divisible by 3.

$6-3=3$ is divisible by 3.

$7-4=3$ is divisible by 3.

And vice versa.

$1-4=-3$ is divisible by 3.

$2-5=-3$ is divisible by 3.

$3-6=-3$ is divisible by 3.

$4-7=-3$ is divisible by 3.

Also,

$1-1=0$ is divisible by 3.

$2-2=0$ is divisible by 3.

$3-3=0$ is divisible by 3.

$4-4=0$ is divisible by 3.

$5-5=0$ is divisible by 3.

$6-6=0$ is divisible by 3.

$7-7=0$ is divisible by 3.

$R=\{(4,1),(5,2),(6,3),(7,4),(1,4),(2,5),(3,6),(4,7), \\ (1,1),(2,2),(3,3),(4,4),(5,5),(6,6),(7,7)\}$

Reflexive:

Clearly, $\{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6),(7,7)\}$

So, $\{(a,a)\in R, \forall a\in A\}$

Hence, it is reflexive.

Symmetric:

Clearly, $\{(1,4),(4,1),(2,5),(5,2),(3,6),(6,3),(4,7),(7,4)\}$

So, $\{(a,b)\in R \Rightarrow (b,a)\in R, \forall a\in A\}$

Hence, it is symmetric.

Transitive:

Clearly,

$\{(1,4),(4,1),(1,1),(2,5),(5,2),(2,2),(3,6),(6,3),(3,3),(4,7),(7,4),(4,4)\}$

So, $\{(a,b)\in R, (b,c)\in R\Rightarrow (a,c)\in R,\forall a\in A\}$

Hence, it is transitive.

Thus, the given relation is an equivalence relation.