The function `f:N-{1}rarr N` , defined by `f(n)=` the highest prime factor of "n" ,is

Options

A. onto only
B. both one-one and onto
C. one-one only
D. neither one-one nor onto

Answer

To analyze the function \( f: \mathbb{N} \setminus \{1\} \to \mathbb{N} \) defined by \( f(n) \) as the highest prime factor of \( n \), we will determine whether the function is one-to-one (1-1) and onto (onto).

### Step 1: Understanding the Function
The function \( f(n) \) takes a natural number \( n \) (excluding 1) and returns the highest prime factor of \( n \). For example:
- \( f(4) = 2 \) (since the prime factorization of 4 is \( 2^2 \))
- \( f(16) = 2 \) (since the prime factorization of 16 is \( 2^4 \))
- \( f(15) = 5 \) (since the prime factorization of 15 is \( 3 \times 5 \))

### Step 2: Checking if the Function is One-to-One (1-1)
A function is one-to-one if different inputs map to different outputs. Let's check:
- \( f(4) = 2 \)
- \( f(16) = 2 \)

Here, both 4 and 16 give the same output (2). Therefore, the function is not one-to-one since two different inputs (4 and 16) yield the same output.

### Step 3: Checking if the Function is Onto (Onto)
A function is onto if every element in the codomain has a pre-image in the domain. The codomain here is \( \mathbb{N} \) (natural numbers). The highest prime factors can only be prime numbers, such as 2, 3, 5, 7, 11, etc. However, not every natural number is a prime number (for example, 1, 4, 6, etc. are not prime).

Thus, since there are natural numbers (like 1, 4, 6) that cannot be expressed as the highest prime factor of any natural number \( n \) (since they are not prime), the function is not onto.

### Conclusion
Since the function \( f(n) \) is neither one-to-one nor onto, we conclude that the correct option is:
- **Option 4: Neither one-to-one nor onto.**