Question

Prove that the function \( f: \mathbb{N} \to \mathbb{N} \) defined by \( f(x) = x - 1 \), when \( x>2 \), and \( f(1) = f(2) = 1 \), is onto but not one-to-one.

Answer 1

Function \( f: \mathbb{N} \to \mathbb{N} \) is onto but not one-to-one

The function \( f: \mathbb{N} \to \mathbb{N} \) is defined as:

\[ f(x) = \begin{cases} x - 1 & \text{if } x > 2, \\ 1 & \text{if } x = 1 \text{ or } x = 2. \end{cases} \]

Step 1: Prove that \( f \) is onto

To show that \( f \) is onto, we must prove that for every element \( y \in \mathbb{N} \), there exists some \( x \in \mathbb{N} \) such that \( f(x) = y \).

Consider any \( y \in \mathbb{N} \). We have two cases to consider:

• If \( y = 1 \), then \( f(1) = 1 \) and \( f(2) = 1 \), so \( y = 1 \) is mapped by \( x = 1 \) or \( x = 2 \).
• If \( y > 1 \), then for \( x = y + 1 \), we have \( f(x) = x - 1 = y \). Therefore, every natural number \( y > 1 \) is the image of some \( x \in \mathbb{N} \), where \( x > 2 \).

Thus, the function \( f \) is onto.

Step 2: Prove that \( f \) is not one-to-one

To show that \( f \) is not one-to-one, we need to find at least two different values \( x_1 \) and \( x_2 \) such that \( f(x_1) = f(x_2) \) but \( x_1 \neq x_2 \).

Consider \( f(1) = 1 \) and \( f(2) = 1 \). Here, \( f(1) = f(2) \), but clearly, \( 1 \neq 2 \).

Therefore, \( f \) is not one-to-one because \( f(1) = f(2) \) even though \( 1 \neq 2 \).

Conclusion: The function \( f \) is onto but not one-to-one.