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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 \( n \in \mathbb{N} \).

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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 \( a = b \).

Show that the following functions are one-to-one using the definition and Proposition 3.2.4. The functions are: (i) f(x) = x^2 for f: [0, ∞) → ℝ, (ii) f(k) = 3k + 7 for f: ℤ → ℤ, (iii) f(n) = n + 2 if n is odd, and f(n) = 2n if n is even, for f: ℕ → ℕ.

Explore the proof of injectivity (one-to-one property) for three functions using Proposition 3.2.4. Functions include f(x) = x^2 for x ≥ 0, f(k) = 3k + 7, and a mixed function based on odd and even numbers. Learn how to apply injectivity conditions in real and integer domains.

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