The exercise shown in the image (Exercise 3.2.7) asks to demonstrate that the following functions are one-to-one (injective), using the definition of one-to-one and Proposition 3.2.4.

The functions are as follows:

1. $f: [0, \infty) \to \mathbb{R}, \, f(x) = x^2$
2. $f: \mathbb{Z} \to \mathbb{Z}, \, f(k) = 3k + 7$
3. $f: \mathbb{N} \to \mathbb{N}, \, f(n) = \begin{cases} n+2 & \text{if } n \text{ is odd} \\ 2n & \text{if } n \text{ is even} \end{cases}$

To prove injectivity (one-to-one), we need to show that for any two inputs $x_1$ and $x_2$, if $f(x_1) = f(x_2)$, then it must follow that $x_1 = x_2$.

Approach:

1. For $f(x) = x^2$, $x \in [0, \infty)$: $f(x_1) = f(x_2) \implies x_1^2 = x_2^2$ Since $x_1$ and $x_2$ are non-negative (due to the domain being $[0, \infty)$), this implies: $x_1 = x_2$ Therefore, $f(x) = x^2$ is injective on $[0, \infty)$.

2. For $f(k) = 3k + 7$, $k \in \mathbb{Z}$: $f(k_1) = f(k_2) \implies 3k_1 + 7 = 3k_2 + 7$ Simplifying: $3k_1 = 3k_2 \implies k_1 = k_2$ Therefore, $f(k) = 3k + 7$ is injective on $\mathbb{Z}$.

3. For the piecewise function:

   • For odd $n$, $f(n) = n + 2$: $f(n_1) = f(n_2) \implies n_1 + 2 = n_2 + 2 \implies n_1 = n_2$ Hence, the function is injective for odd $n$.

   • For even $n$, $f(n) = 2n$: $f(n_1) = f(n_2) \implies 2n_1 = 2n_2 \implies n_1 = n_2$ Therefore, the function is injective for even $n$.

Since both parts of the piecewise function are injective, the whole function is injective.

Would you like more detailed explanations on any part? Here are some related questions to expand your understanding:

1. What is the definition of a one-to-one (injective) function?
2. How does the injectivity of $f(x) = x^2$ change if the domain is the entire real line?
3. Why is the function $f(k) = 3k + 7$ injective for all integers?
4. Can a function be injective but not surjective? Provide examples.
5. What are the implications of injectivity in the context of inverse functions?

Tip: To check if a function is injective, always set $f(x_1) = f(x_2)$ and try to show that this implies $x_1 = x_2$.