Final answer:
v ∈ im(f) if and only if v + 5 is even.
Step-by-step explanation:
To prove this statement, let's consider the function f: Z → Z defined by f(x) = x + 5. We are given that v ∈ Z and v > 5.
First, assume v ∈ im(f). This means there exists an integer x in Z such that f(x) = v. Substituting the definition of f, we have x + 5 = v. Rearranging, we find that x = v - 5. Since v is greater than 5, x is also an integer. This implies that v + 5 = (v - 5) + 10 = f(x) + 10. As f(x) is even, adding 10 to an even number results in an even number. Therefore, v + 5 is even.
Conversely, assume v + 5 is even. Then, v + 5 = 2k for some integer k. Rearranging, we get v = 2k - 5. Define x = k - 3. Now, f(x) = x + 5 = (k - 3) + 5 = k + 2, which is exactly v. Thus, v ∈ im(f).
In summary, we have shown that v ∈ im(f) if and only if v + 5 is even, establishing the equivalence between the two conditions.