Final answer:
To show that the function f is not one-to-one, we can use mathematical induction on the variable n. We assume it is true for a base case n = 1 and then prove it for n = k + 1. By showing that f maps two elements from the domain to the same element in the codomain, we demonstrate that f is not one-to-one.
Step-by-step explanation:
To show that the function f is not one-to-one, we can use mathematical induction on the variable n. We will assume that the function is one-to-one for the base case (n = 1) and then prove that if it is true for n = k, it is also true for n = k + 1. Base case (n = 1): If n = 1, then the function f is a function from {1, 2,...,m} to {1}, meaning it can only map to a single element. In this case, it is trivially true that f is one-to-one. Inductive step: Suppose that f is one-to-one for some n = k, meaning that for any two distinct elements i and j in the domain {1, 2,...,m}, f(i) is not equal to f(j). We want to show that f is also one-to-one for n = k + 1. Now, let's consider the function f for n = k + 1. Since n = k + 1, we have an additional element in the codomain, which means we need to map it to an element in the domain. But since m > n, we have more elements in the domain than in the codomain, which means that at least two elements in the domain must be mapped to the same element in the codomain. This violates the definition of a one-to-one function, so f cannot be one-to-one for n = k + 1. Therefore, by mathematical induction, we have shown that f is not one-to-one for any positive integer value of n.