Let `A = {0, 1}` and the set of all natural numbers.Then the mapping `f: N rarr A` defined by `f(2n-1) = 0, f(2n) = 1, AA n in N,` is

Explanation

To solve the problem, we need to analyze the mapping \( f: \mathbb{N} \rightarrow A \) defined by the following rules:

1. \( f(2n - 1) = 0 \) for odd natural numbers (where \( n \in \mathbb{N} \)).
2. \( f(2n) = 1 \) for even natural numbers (where \( n \in \mathbb{N} \)).

### Step-by-Step Solution:

**Step 1: Identify the Domain and Codomain**
- The domain of the function \( f \) is the set of all natural numbers \( \mathbb{N} = \{ 1, 2, 3, \ldots \} \).
- The codomain of the function \( f \) is the set \( A = \{ 0, 1 \} \).

**Step 2: Determine the Mapping for Odd and Even Natural Numbers**
- For odd natural numbers (e.g., 1, 3, 5, ...):
- \( f(1) = f(2 \cdot 1 - 1) = 0 \)
- \( f(3) = f(2 \cdot 2 - 1) = 0 \)
- \( f(5) = f(2 \cdot 3 - 1) = 0 \)
- In general, \( f(2n - 1) = 0 \) for all \( n \in \mathbb{N} \).

- For even natural numbers (e.g., 2, 4, 6, ...):
- \( f(2) = f(2 \cdot 1) = 1 \)
- \( f(4) = f(2 \cdot 2) = 1 \)
- \( f(6) = f(2 \cdot 3) = 1 \)
- In general, \( f(2n) = 1 \) for all \( n \in \mathbb{N} \).

**Step 3: Analyze the Function**
- The function maps all odd natural numbers to 0 and all even natural numbers to 1.
- This means:
- The image of all odd natural numbers is \( \{0\} \).
- The image of all even natural numbers is \( \{1\} \).

**Step 4: Determine the Type of Function**
- A function is called **many-one** if distinct elements in the domain map to the same element in the codomain. Here, all odd numbers map to 0 and all even numbers map to 1, which means:
- Multiple inputs (odd numbers) yield the same output (0).
- Multiple inputs (even numbers) yield the same output (1).
- Therefore, \( f \) is a **many-one function**.

**Step 5: Check if the Function is Onto**
- A function is onto if every element in the codomain has a pre-image in the domain.
- Here, both elements of the codomain \( A = \{0, 1\} \) have pre-images:
- 0 is the image of all odd natural numbers.
- 1 is the image of all even natural numbers.
- Thus, \( f \) is also an **onto function**.

### Final Conclusion:
The mapping \( f: \mathbb{N} \rightarrow A \) defined by the given rules is a **many-one onto function**.