Answer and Step-by-step explanation:
To determine whether the function f: X → X defined as f(n) = 2n mod 5 is surjective, we need to check if every element in the codomain X has a corresponding pre-image in the domain X.
In this case, the domain X is {0, 1, 2, 3, 4}, and the codomain X is also {0, 1, 2, 3, 4}. To show that the function f is surjective, we need to demonstrate that for every element y in the codomain X, there exists an element x in the domain X such that f(x) = y.
Let's analyze each element in the codomain X and find its corresponding pre-image in the domain X:
1. For y = 0: f(x) = 2x mod 5. The only x in the domain X that satisfies f(x) = 0 is x = 0. Thus, there exists a pre-image for y = 0.
2. For y = 1: f(x) = 2x mod 5. The only x in the domain X that satisfies f(x) = 1 is x = 3. Thus, there exists a pre-image for y = 1.
3. For y = 2: f(x) = 2x mod 5. The only x in the domain X that satisfies f(x) = 2 is x = 1. Thus, there exists a pre-image for y = 2.
4. For y = 3: f(x) = 2x mod 5. The only x in the domain X that satisfies f(x) = 3 is x = 4. Thus, there exists a pre-image for y = 3.
5. For y = 4: f(x) = 2x mod 5. The only x in the domain X that satisfies f(x) = 4 is x = 2. Thus, there exists a pre-image for y = 4.
Since we have found a pre-image in the domain X for each element in the codomain X, we can conclude that the function f is surjective.