By given relation, aRb if b=a+3k,k∈Z .
Reflexive: a=a+3(0) ⟹ aRa . So R is an reflexive relation.
Symmetric: Let aRb⟹b=a+3k where k is an integer.
"\\implies a = b - 3k = b + 3(-k) = b + 3k_1"
and "k_1 = -k \\in \\Z \\implies bRa" .
So. R is an symmetric relation.
Transitive: Let aRb,bRc "\\implies b = a+3k_1 , c =b +3k_2"
"\\implies c = a+3k_1 +3k_2 = a+3(k_1+k_2) = a+3k"
where "k = k_1+k_2 \\in \\Z."
Hence, aRc . So, R is an transitive relation.
Thus, R is an equivalence relation.
Equivalence class of 0 "= \\{3k: k\\in \\Z\\} = \\{ \\_ \\ \\_ \\ \\_, -6, -3, 0, 3 , 6, \\_ \\ \\_ \\ \\_ \\}"
Equivalence class of 1 "= \\{3k+1 : k\\in \\Z\\} = \\{ \\_ \\ \\_ \\ \\_, -5, -2, 1 , 4 , 7, \\_ \\ \\_ \\ \\_ \\}"
Equivalence class of 2 = "\\{3k+2 : k\\in \\Z\\} = \\{ \\_ \\ \\_ \\ \\_, -4, -1, 2 , 5 , 8, \\_ \\ \\_ \\ \\_ \\}"
These three distinct equivalence classes are possible
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