Let X={1,2,3,4,5,6,7} and R={(x,y)∣x−y is divisible by 3} in X. Let us show that R is an equivalence relation. Since x−x=0 is divisible by 3 for any x∈X, we conclude that (x,x)∈R for any x∈X, and hence R is a reflexive relation. If x,y∈X and (x,y)∈R, then x−y is divisible by 3. It follows that y−x=−(x−y) is also divisible by 3, and hence (y,x)∈R.
We conclude that the relation R is symmetric. If x,y,z∈X and (x,y)∈R, (y,z)∈R, then x−y is divisible by 3 and y−z is divisible by 3. It follows that x−z=(x−y)+(y−z) is also divisible by 3, and hence (x,z)∈R. We conclude that the relation R is transitive. Consequently, R is an equivalence relation.
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