Let `f=NtoN` be defined by `f(x)=x^(2)+x+1,x inN`, then prove that f is one-one but not onto.

Answer

To prove that the function \( f: \mathbb{N} \to \mathbb{N} \) defined by \( f(x) = x^2 + x + 1 \) is one-one but not onto, we will follow these steps:

### Step 1: Prove that \( f \) is one-one

To show that \( f \) is one-one, we need to prove that if \( f(x_1) = f(x_2) \), then \( x_1 = x_2 \).

1. Start with the equation:
\[ f(x_1) = f(x_2) \]
This means:
\[ x_1^2 + x_1 + 1 = x_2^2 + x_2 + 1 \]

2. Simplify the equation:
\[ x_1^2 + x_1 = x_2^2 + x_2 \]
Rearranging gives:
\[ x_1^2 - x_2^2 + x_1 - x_2 = 0 \]

3. Factor the left-hand side:
\[ (x_1 - x_2)(x_1 + x_2) + (x_1 - x_2) = 0 \]
This can be factored further:
\[ (x_1 - x_2)(x_1 + x_2 + 1) = 0 \]

4. From this product, we have two cases:
- \( x_1 - x_2 = 0 \) which implies \( x_1 = x_2 \)
- \( x_1 + x_2 + 1 = 0 \) which cannot happen since \( x_1 \) and \( x_2 \) are natural numbers (they are positive integers).

Thus, the only solution is \( x_1 = x_2 \), proving that \( f \) is one-one.

### Step 2: Prove that \( f \) is not onto

To show that \( f \) is not onto, we need to demonstrate that there exists at least one natural number in the codomain \( \mathbb{N} \) that is not in the range of \( f \).

1. Calculate the minimum value of \( f(x) \):
- For \( x = 1 \):
\[ f(1) = 1^2 + 1 + 1 = 3 \]
- For \( x = 2 \):
\[ f(2) = 2^2 + 2 + 1 = 7 \]

2. As \( x \) increases, \( f(x) \) will continue to increase since it is a quadratic function that opens upwards.

3. The range of \( f \) starts from 3 and includes all values greater than or equal to 3. Thus, the range of \( f \) is:
\[ \{3, 4, 5, 6, 7, \ldots\} \]

4. However, the codomain \( \mathbb{N} \) includes all natural numbers:
\[ \{1, 2, 3, 4, 5, 6, 7, \ldots\} \]

5. Since \( 1 \) and \( 2 \) are natural numbers that are not in the range of \( f \), we conclude that \( f \) is not onto.

### Conclusion

We have shown that \( f \) is one-one because \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \). We have also shown that \( f \) is not onto because there are natural numbers (specifically 1 and 2) that are not in the range of \( f \).

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