Let N be the set of natural numbers and f: N `rarr` N be a function given by fox)=x+1 . Which one of the following is correct?

Options

A. A. f is one-one and onto
B. B. f is one-one but not onto
C. C. f is only onto
D. D. f is neither one-one nor onto

Answer

To solve the problem, we need to analyze the function \( f: \mathbb{N} \to \mathbb{N} \) defined by \( f(x) = x + 1 \) and determine whether it is one-to-one (injective) and/or onto (surjective).

### Step 1: Check if the function is one-to-one (injective)
A function \( f \) is one-to-one if for any two different values \( x_1 \) and \( x_2 \) in the domain, \( f(x_1) \neq f(x_2) \).

1. Assume \( f(x_1) = f(x_2) \).
2. Then, \( x_1 + 1 = x_2 + 1 \).
3. By subtracting 1 from both sides, we get \( x_1 = x_2 \).

Since we assumed \( f(x_1) = f(x_2) \) and derived that \( x_1 = x_2 \), this confirms that the function is one-to-one.

### Step 2: Check if the function is onto (surjective)
A function \( f \) is onto if for every element \( y \) in the co-domain \( \mathbb{N} \), there exists an element \( x \) in the domain \( \mathbb{N} \) such that \( f(x) = y \).

1. Let \( y \) be any natural number in the co-domain.
2. We need to find \( x \) such that \( f(x) = y \).
3. From the function definition, we have \( f(x) = x + 1 \).
4. Setting \( x + 1 = y \) gives us \( x = y - 1 \).

Now, we need to check if \( x = y - 1 \) is a natural number:
- If \( y = 1 \), then \( x = 1 - 1 = 0 \), and 0 is not a natural number.
- Therefore, there is no corresponding \( x \) in the domain for \( y = 1 \).

Since we found at least one \( y \) (specifically \( y = 1 \)) in the co-domain for which there is no corresponding \( x \) in the domain, the function is not onto.

### Conclusion
The function \( f(x) = x + 1 \) is one-to-one but not onto.

### Final Answer
The correct option is: **1-1 but not onto**.

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