To prove 3Z is closed under addition, we need to show 3*k_1 + 3*k_2 belongs to 3Z for arbitrary k_1 and k_2 in Z.

Since an interger ring Z distributes multiplication over addition, 3*k_1 +3*k_2 = 3*(k_1 +k_2).

Since k_1 and k_2 belongs to Z which is closed under addition, we can pick k_3 in Z satisfying k_3 = k_1 + k_2.

Now, 3*k_1 +3*k_2 = 3*(k_1+k_2) = 3*k_3, where k_1, k_2, k_3 belongs to Z.

Thus, 3Z is closed under addition.