Consider the following statements :
1. A function `f:Z to Z`, defined by `f(x) = x+1`, is one-one as well as onto.
2. A function `f:N to N`, defined by `f(x) = x +1`, is one-one but not onto.
Which of the above statements is/are correct?

Options

A. 1 only
B. 2 only
C. Both 1 and 2
D. Neither 1 nor 2

Answer

To analyze the given statements, we will evaluate each function step by step.

### Step 1: Evaluate Statement 1
**Statement**: A function \( f: \mathbb{Z} \to \mathbb{Z} \), defined by \( f(x) = x + 1 \), is one-one as well as onto.

1. **Check if the function is one-one**:
- A function is one-one (injective) if \( f(a) = f(b) \) implies \( a = b \).
- Let's assume \( f(a) = f(b) \).
- Then, \( a + 1 = b + 1 \).
- Subtracting 1 from both sides gives \( a = b \).
- Hence, the function is one-one.

2. **Check if the function is onto**:
- A function is onto (surjective) if for every element \( y \in \mathbb{Z} \), there exists an \( x \in \mathbb{Z} \) such that \( f(x) = y \).
- Let \( y \) be any integer. We need to find \( x \) such that \( f(x) = y \).
- Setting \( f(x) = y \) gives \( x + 1 = y \).
- Thus, \( x = y - 1 \), which is also an integer.
- Therefore, for every integer \( y \), there exists an integer \( x \) (specifically \( y - 1 \)) such that \( f(x) = y \).
- Hence, the function is onto.

**Conclusion for Statement 1**: The function \( f(x) = x + 1 \) is both one-one and onto.

### Step 2: Evaluate Statement 2
**Statement**: A function \( f: \mathbb{N} \to \mathbb{N} \), defined by \( f(x) = x + 1 \), is one-one but not onto.

1. **Check if the function is one-one**:
- Similar to the first statement, we check if \( f(a) = f(b) \) implies \( a = b \).
- If \( f(a) = f(b) \), then \( a + 1 = b + 1 \).
- Subtracting 1 from both sides gives \( a = b \).
- Hence, the function is one-one.

2. **Check if the function is onto**:
- We need to see if for every \( y \in \mathbb{N} \), there exists an \( x \in \mathbb{N} \) such that \( f(x) = y \).
- Let \( y \) be any natural number (starting from 1).
- We need to find \( x \) such that \( f(x) = y \).
- Setting \( f(x) = y \) gives \( x + 1 = y \).
- Thus, \( x = y - 1 \).
- However, if \( y = 1 \), then \( x = 1 - 1 = 0 \), and 0 is not a natural number.
- Therefore, there is no \( x \in \mathbb{N} \) such that \( f(x) = 1 \).
- Hence, the function is not onto.

**Conclusion for Statement 2**: The function \( f(x) = x + 1 \) is one-one but not onto.

### Final Conclusion
- **Statement 1** is correct.
- **Statement 2** is also correct.

Thus, both statements are correct.