A proof of the following result is given.Result Let n Z. If n4 is even, then 3n + 1 is

Chapter 4, Problem 4.84

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A proof of the following result is given.Result Let n Z. If n4 is even, then 3n + 1 is odd.Proof Assume that n4 = n22 is even. Since n4 is even, n2 is even. Furthermore, since n2 is even, n iseven. Because n is even, n = 2k for some integer k. Then3n + 1 = 3(2k) + 1 = 6k + 1 = 2(3k) + 1.Since 3k is an integer, 3n + 1 is odd.Answer the following questions.(1) Which proof technique is being used?(2) What is the starting assumption?(3) What must be shown to give a complete proof?(4) Give a reason for each of the following steps in the proof.(a) Since n4 is even, n2 is even.(b) Furthermore, since n2 is even, n is even.(c) Because n is even, n = 2k for some integer k.(d) Then 3n + 1 = 3(2k) + 1 = 6k + 1 = 2(3k) + 1.(e) Since 3k is an integer, 3n + 1 is odd.

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