Since x – x = 0 is multiple of 2, (x, x) ∈ R
Therefore reflexive.
If y – x is a multiple of 2 then x – y is also a multiple of 2. Therefore (x, y) ∈ R ⇒ (y, x) ∈ R. Hence symmetric.
If y – x is a multiple of 2 and z-y is a multiple of 2, then their sum y – x + z – y = z – x is a multiple of 2.
Therefore (x, y), (y, z) ∈ R ⇒ (x, z) ∈ R
Hence transitive.
Therefore R is an equivalance relation.