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

Answer

To determine whether the mapping \( f: \mathbb{N} \to A \) defined by \( f(2n-1) = 0 \) and \( f(2n) = 1 \) for \( n \in \mathbb{N} \) is many-one and onto, we will analyze the properties of the function step by step.

### Step 1: Understanding the Sets
- The set \( A = \{0, 1\} \) contains two elements.
- The set \( \mathbb{N} \) (natural numbers) contains elements like \( 1, 2, 3, 4, \ldots \).

### Step 2: Analyzing the Function
The function is defined as follows:
- For odd natural numbers (i.e., \( 1, 3, 5, \ldots \)), the function maps to \( 0 \).
- For even natural numbers (i.e., \( 2, 4, 6, \ldots \)), the function maps to \( 1 \).

### Step 3: Checking if the Function is Many-One
A function is many-one if two or more different elements in the domain map to the same element in the codomain.

- For example:
- \( f(1) = 0 \) (since \( 2n-1 \) for \( n=1 \) gives \( 1 \))
- \( f(3) = 0 \) (since \( 2n-1 \) for \( n=2 \) gives \( 3 \))

Since both \( 1 \) and \( 3 \) map to \( 0 \), the function is **not one-one**. Thus, it is many-one.

### Step 4: Checking if the Function is Onto
A function is onto if every element in the codomain has a pre-image in the domain.

- The codomain \( A \) has elements \( 0 \) and \( 1 \).
- We can find pre-images:
- For \( 0 \): \( f(1) = 0 \) (the pre-image is \( 1 \)).
- For \( 1 \): \( f(2) = 1 \) (the pre-image is \( 2 \)).

Since both elements of \( A \) have pre-images in \( \mathbb{N} \), the function is **onto**.

### Conclusion
The mapping \( f \) is many-one and onto. Therefore, the statement is **True**.