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Consider the function \(f: \mathbb{Z} \rightarrow \mathbb{Z}\) (from integers to integers) defined as \(f(n) = n \mod 2\). In simple terms, the function outputs the remainder when \(n\) is divided by 2.

We need to verify if the function meets the three given conditions: 1. The domain of f should be the set of integers 2. The range of f should also be the set of integers 3. The function f should not be one-to-one Let's discuss each one of them.

By definition, the domain of \(f\) is the set of permissible input values for the function. In our case, the function has been defined for every integer \(n\), so the domain of the function \(f\) is the set of integers \(\mathbb{Z}\).

The range of a function is the set of possible output values. Let's consider some input values of our function and find their respective outputs using the given formula \(f(n) = n \mod 2\): - If \(n = 0\), then \(f(0) = 0 \mod 2 = 0\) - If \(n = 1\), then \(f(1) = 1 \mod 2 = 1\) - If \(n = 2\), then \(f(2) = 2 \mod 2 = 0\) - If \(n = 3\), then \(f(3) = 3 \mod 2 = 1\) - If \(n = -1\), then \(f(-1) = -1 \mod 2 = 1\) - If \(n = -2\), then \(f(-2) = -2 \mod 2 = 0\) Seeing this pattern, we can observe that the range of \(f\) is \(\{0,1\}\) and not the set of integers \(\mathbb{Z}\). So, we need to modify the function to fulfill this requirement. Let's redefine the function as follows: \(f(n) = \lfloor \frac{n}{2} \rfloor\) Now, let's check the conditions with this new function:

Since the new function is also defined for every integer \(n\), the domain of the function \(f\) is the set of integers \(\mathbb{Z}\).

Let's find the outputs for some inputs using the modified formula \(f(n) = \lfloor \frac{n}{2} \rfloor\): - If \(n = 0\), then \(f(0) = \lfloor \frac{0}{2} \rfloor = \lfloor 0 \rfloor = 0\) - If \(n = 1\), then \(f(1) = \lfloor \frac{1}{2} \rfloor = \lfloor 0.5 \rfloor = 0\) - If \(n = 2\), then \(f(2) = \lfloor \frac{2}{2} \rfloor = \lfloor 1 \rfloor = 1\) - If \(n = 3\), then \(f(3) = \lfloor \frac{3}{2} \rfloor = \lfloor 1.5 \rfloor = 1\) - If \(n = 4\), then \(f(4) = \lfloor \frac{4}{2} \rfloor = \lfloor 2 \rfloor = 2\) Similarly, - If \(n = -1\), then \(f(-1) = \lfloor \frac{-1}{2} \rfloor = \lfloor -0.5 \rfloor = -1\) - If \(n = -2\), then \(f(-2) = \lfloor \frac{-2}{2} \rfloor = \lfloor -1 \rfloor = -1\) - If \(n = -3\), then \(f(-3) = \lfloor \frac{-3}{2} \rfloor = \lfloor -1.5 \rfloor = -2\) Therefore, we can observe that the range of the function \(f\) is the set of integers \(\mathbb{Z}\).

For a function to be one-to-one, no two distinct elements in the domain have the same output in the range. Notice that in our function, many pairs of integers have the same output. For instance: - \(f(0) = f(1) = 0\) - \(f(2) = f(3) = 1\) - \(f(-1) = f(-2) = -1\) - \(f(-3) = f(-4) = -2\) We can see here that the function \(f\) is indeed not a one-to-one function. Thus, our function \(f(n) = \lfloor \frac{n}{2} \rfloor\) satisfies all the given conditions: 1. The domain of \(f\) is the set of integers \(\mathbb{Z}\). 2. The range of \(f\) is also the set of integers \(\mathbb{Z}\). 3. The function \(f\) is not one-to-one.