Let m be a positive integer with m>1. Find out whether or not the subsequent relation is an equivalent relation.
R = {(a,b)|a ≡ b (mod m)}
Ans: Relation R is illustrated as ≡m (congruence modulo m) on the set of positive integers. Let us check if it is an equivalence relation.
Reflexivity: Let x ∈ Z+ be any integer, after that x ≡m x since both yields similar remainder when divided by m. So (x, x) ∈ R ∀ x ∈ Z. ∴R is a reflexive relation.
Symmetry: Let x and y be any two integers and (x, y) ∈ R. This depicts that x ≡m y and therefore y ≡m x. So, (y, x) ∈ R. ∴ R is a symmetric relation.
Transitivity: Let x, y and z be any three elements of Z like that (x, y) and (y, z) ∈ R. So, we have x ≡m y and y ≡m z. It entails that (x-y) and (y-z) are divisible by m. Hence, (x - y) + (y - z) = (x - z) is as well divisible by m that is x ≡m z.
∴ (x, y) and (y, z) ∈ R ⇒ (x, z) ∈ R. That is R is a transitive relation.
Ans: Relation R is illustrated as ≡m (congruence modulo m) on the set of positive integers. Let us check if it is an equivalence relation.
Reflexivity: Let x ∈ Z+ be any integer, after that x ≡m x since both yields similar remainder when divided by m. So (x, x) ∈ R ∀ x ∈ Z. ∴R is a reflexive relation.
Symmetry: Let x and y be any two integers and (x, y) ∈ R. This depicts that x ≡m y and therefore y ≡m x. So, (y, x) ∈ R. ∴ R is a symmetric relation.
Transitivity: Let x, y and z be any three elements of Z like that (x, y) and (y, z) ∈ R. So, we have x ≡m y and y ≡m z. It entails that (x-y) and (y-z) are divisible by m. Hence, (x - y) + (y - z) = (x - z) is as well divisible by m that is x ≡m z.
∴ (x, y) and (y, z) ∈ R ⇒ (x, z) ∈ R that is R is a transitive relation.
Hence R is an equivalence relation.