The image contains an exercise, specifically Exercise 3.2.7, which asks to prove that certain functions are one-to-one (injective) using the given definition and Proposition 3.2.4. The functions are:

(i) $f: [0, \infty) \to \mathbb{R}$, defined by $f(x) = x^2$.

(ii) $f: \mathbb{Z} \to \mathbb{Z}$, defined by $f(k) = 3k + 7$.

(iii) $f: \mathbb{N} \to \mathbb{N}$, defined by:

• $f(n) = n + 2$ if $n$ is odd,
• $f(n) = 2n$ if $n$ is even.

Solution Steps:

To show that a function is one-to-one (injective), we need to prove that: $f(a) = f(b) \implies a = b$ for all $a$ and $b$ in the domain of $f$.

(i) $f(x) = x^2$, with $f: [0, \infty) \to \mathbb{R}$

Here, $f(x) = x^2$ is defined on the domain $[0, \infty)$.

• Suppose $f(a) = f(b)$, i.e., $a^2 = b^2$.
• Since $a, b \geq 0$, we can take square roots on both sides: $a = b$. Thus, $f(x) = x^2$ is injective on $[0, \infty)$.

(ii) $f(k) = 3k + 7$, with $f: \mathbb{Z} \to \mathbb{Z}$

• Suppose $f(a) = f(b)$, i.e., $3a + 7 = 3b + 7$.
• Subtract 7 from both sides: $3a = 3b$.
• Divide by 3: $a = b$. Thus, $f(k) = 3k + 7$ is injective on $\mathbb{Z}$.

(iii) $f(n) = n + 2$ if $n$ is odd, and $f(n) = 2n$ if $n$ is even, with $f: \mathbb{N} \to \mathbb{N}$

We need to check injectivity separately for odd and even $n$.

• If $f(a) = f(b)$ where both $a$ and $b$ are odd, then $f(a) = a + 2$ and $f(b) = b + 2$.

  • Suppose $a + 2 = b + 2$, subtracting 2 from both sides gives $a = b$, so the function is injective for odd $n$.

• If both $a$ and $b$ are even, then $f(a) = 2a$ and $f(b) = 2b$.

  • Suppose $2a = 2b$, dividing by 2 gives $a = b$, so the function is injective for even $n$.

• Now, check injectivity across odd and even values:

  • For odd $a$, $f(a) = a + 2$ cannot equal $f(b) = 2b$ for even $b$, since one function is linear and the other is multiplicative. Thus, the function is injective for all $n \in \mathbb{N}$.

Let me know if you need further clarification or more details.

Related questions:

1. How does Proposition 3.2.4 apply to injectivity proofs in general?
2. Can a function be injective but not surjective?
3. How does the domain restriction to $[0, \infty)$ ensure injectivity for $f(x) = x^2$?
4. What happens if we remove the restriction $[0, \infty)$ in part (i)?
5. Can we generalize the function in part (ii) to a similar form for any linear transformation?

Tip:

When proving injectivity, always start by assuming $f(a) = f(b)$ and aim to show $a = b$.