Question:

Step 1: Understand the function definition. The function $f$ is defined as $f: \mathbb{N} \setminus \{1\} \to \mathbb{N}$, where $f(n)$ is the highest prime factor of $n$. The domain of the function is $\mathbb{N} \setminus \{1\} = \{2, 3, 4, \ldots\}$. The codomain of the function is $\mathbb{N} = \{1, 2, 3, 4, \ldots\}$. Step 2: Check if the function is one-one (injective). A function is one-one if distinct elements in the domain always map to distinct elements in the codomain. To show that $f$ is not one-one, we need to find two different numbers $n_1, n_2$ in the domain such that $f(n_1) = f(n_2)$. Consider the values $n_1 = 2$ and $n_2 = 4$. The highest prime factor of $2$ is $2$, so $f(2) = 2$. The prime factorization of $4$ is $2^2$, and its highest prime factor is $2$, so $f(4) = 2$. Since $2 \neq 4$ but $f(2) = f(4) = 2$, the function $f$ is not one-one. Step 3: Check if the function is onto (surjective). A function is onto if its range is equal to its codomain. The codomain of $f$ is $\mathbb{N}$. The range of $f$ consists of all possible output values of $f(n)$. For any $n \in \mathbb{N} \setminus \{1\}$, $f(n)$ is defined as the highest prime factor of $n$. This means that the output of the function must always be a prime number. For example, the range includes values like: $f(2) = 2$ $f(3) = 3$ $f(4) = 2$ $f(5) = 5$ $f(6) = 3$ (since $6 = 2 \times 3$) The set of all prime numbers $\{2, 3, 5, 7, \ldots\}$ is the range of $f$. The codomain is $\mathbb{N} = \{1, 2, 3, 4, \ldots\}$. Since the range of $f$ (the set of prime numbers) does not include all natural numbers (e.g., $1, 4, 6, 8, 9, \ldots$ are in the codomain but not in the range), the function $f$ is not onto. Specifically, there is no $n \in \mathbb{N} \setminus \{1\}$ such that $f(n) = 1$, because the highest prime factor of any integer $n > 1$ must be a prime number $\ge 2$. Also, there is no $n \in \mathbb{N} \setminus \{1\}$ such that $f(n) = 4$, as $4$ is not a prime number. Step 4: Conclude the nature of the function. Based on the analysis in Step 2, the function is not one-one. Based on the analysis in Step 3, the function is not onto. Therefore, the function $f$ is neither one-one nor onto. The final answer is $\boxed{\text{Neither one-one nor onto}}$.