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

For n even: \( f(n) = n - 1 \). For n odd: \( f(n) = n + 1 \).

f is injective: No two different inputs map to the same output. For example:

• If \( n_1 \) and \( n_2 \) are even or odd, the outputs \( f(n_1) \neq f(n_2) \).
• 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.

Hence, f is injective.

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

• For odd k, \( k = f(k - 1) \), where \( k - 1 \) is even.
• For even k, \( k = f(k + 1) \), where \( k + 1 \) is odd.

Hence, f is surjective.

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

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

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

Answer:

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

Answer 2

For \(n\) even: \( f(n) = n - 1 \). For \(n\) odd: \( f(n) = n + 1 \). 

f is injective: No two different inputs map to the same output. Let's verify:

• If \( n_1 \) and \( n_2 \) are both even or both odd, the outputs \( f(n_1) \neq f(n_2) \) because the function applies different operations depending on whether \(n\) is even or odd.
• If \( n_1 \) is even and \( n_2 \) is odd, their outputs are \( f(n_1) = n_1 - 1 \) and \( f(n_2) = n_2 + 1 \), which are distinct because \(n_1 - 1\) and \(n_2 + 1\) cannot be the same for any even and odd \(n_1, n_2\).

Hence, f is injective.

f is surjective: Every natural number \(k \in \mathbb{N}\) is an output of f: Let's verify:

• For odd \( k \), we have \( k = f(k - 1) \), where \( k - 1 \) is even.
• For even \( k \), we have \( k = f(k + 1) \), where \( k + 1 \) is odd.

Hence, f is surjective.

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

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

Conclusion: The function f satisfies the properties of being injective, surjective, and invertible. Therefore, the correct answer is:

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