## Answers

One-one: - Let f(n_1) = f(n_2) 3n_1 + 1 = 3n_2 + 1 3n_1 = 3n_2 n_1 = n_2 So, when f(n_1) = f(n_2), then n_1 = n_2. Hence, we can say that f is one-one. Onto: - The function f is not onto. as image of 0, -1, -4, does not belong to Z. as, put 3n + 1 = 0 = f(n) 3n = -1 or, n = -1/3 This implies, f(-1/3) = 0 and -1/3 notelement Z.

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Similarly, we can check for -1, -4. Hence, pre-image of some elements does not exists. So, f is not onto.

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