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

Answer 1

Answer: (D)

(A), (C), and (D) only

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. 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:

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