Let us show that a relation '∼' on R by 'a∼b if a−b is an integer' is an equivalence relation, that is reflexive, symmetric and transitive.
For each a∈R, a−a=0∈Z , and thus a∼a each a∈R, and the relation is reflexive.
If a∼b, then a−b∈Z. It follows that b−a=−(a−b)∈Z, and thus b∼a. Consequently, the relation is symmetric.
Let a∼b and b∼c. Then a−b∈Z, b−c∈Z. It follows that a−c=(a−b)+(b−c)∈Z, and thus a∼c. Therefore, the relation is transitive.
By defenition, the equivalence class of an element a is [a]={b∈R ∣ b−a∈Z}.
Let us give the equivalence classes of 0 and 2 :
[0]={b∈R ∣ b−0∈Z}={b∈R ∣ b∈Z}=Z,
[2]={b∈R ∣ b−2∈Z}={b∈R ∣ b=2+n, n∈Z}=2+Z.
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