Question

\(\text{ If the function } f : \mathbb{N} \to \mathbb{N} \text{ is defined as } f(n) = \begin{cases} n - 1, & \text{if } n \text{ is even} \\ n + 1, & \text{if } n \text{ is odd} \end{cases} \text{, then:}\)
(A) f is injective
(B) f is into
(C) f is surjective
(D) f is invertible
Choose the correct answer from the options given below:

• A. (B) only
• B. (A), (B), and (D) only
• C. (A) and (C) only
• D. (A), (C), and (D) only

Hint:

When analyzing a function for injectivity, surjectivity, and invertibility, it is essential to check if the function satisfies the necessary conditions. For injectivity, ensure that different inputs map to different outputs. For surjectivity, verify that every element in the target set has a corresponding input in the domain. If both conditions hold, the function is invertible, and you can find the inverse function by solving for \( x \) in terms of \( y \). In this case, the function is defined piecewise for even and odd numbers, which makes it easy to handle.

Answer 1

The Correct Option is D

If n is even, \( f(n) = n - 1 \). If n is odd, \( f(n) = n + 1 \).

f is injective (one-to-one): Different inputs produce different outputs.

• For any two even numbers \( n_1 \) and \( n_2 \), or any two odd numbers \( n_1 \) and \( n_2 \), it holds that \( f(n_1) eq f(n_2) \).
• If \( n_1 \) is even and \( n_2 \) is odd, their outputs \( f(n_1) = n_1 - 1 \) and \( f(n_2) = n_2 + 1 \) are distinct.

Therefore, f is injective.

f is surjective (onto): Every natural number \( k \in \mathbb{N} \) is an output of f:

• Any odd number \( k \) is the output of \( f(k - 1) \), where \( k - 1 \) is even.
• Any even number \( k \) is the output of \( f(k + 1) \), where \( k + 1 \) is odd.

Therefore, f is surjective.

f is invertible: Since f is both injective and surjective, it is invertible. The inverse function \( f^{-1} \) is defined as:

\[f^{-1}(n) = \begin{cases} n + 1, & \text{if } n \text{ is odd} \\ n - 1, & \text{if } n \text{ is even} \end{cases}\]

Consequently, the function f satisfies properties (A), (C), and (D).

Answer:

\((A), (C), \text{ and } (D)\).