Question

Prove that \( f : N \rightarrow N \) defined as \( f(x) = ax + b \) (\( a, b \in N \)) is one-one but not onto.

Hint:

To prove a function is one-one, show that \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \). To prove a function is not onto, find an element in the codomain for which there is no pre-image in the domain. For functions involving natural numbers, consider the properties of arithmetic progressions formed by \( ax + b \).

Answer 1

We are given a function \( f : N \rightarrow N \) defined by \( f(x) = ax + b \), where \( a \) and \( b \) are natural numbers (\( N = \{1, 2, 3, \dots\} \)). We need to prove that this function is one-one but not onto. Proof for One-One (Injective): A function \( f \) is one-one if for any \( x_1, x_2 \) in the domain, \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \). Let \( x_1, x_2 \in N \) such that \( f(x_1) = f(x_2) \). $$ax_1 + b = ax_2 + b$$ Subtract \( b \) from both sides: $$ax_1 = ax_2$$ Since \( a \in N \), \( a \neq 0 \). We can divide both sides by \( a \): $$x_1 = x_2$$ Thus, \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \), which means the function \( f \) is one-one. Proof for Not Onto (Not Surjective): A function \( f : A \rightarrow B \) is onto if for every \( y \in B \), there exists an \( x \in A \) such that \( f(x) = y \). In our case, \( A = N \) and \( B = N \). Consider an arbitrary \( y \in N \). We want to find if there exists an \( x \in N \) such that \( f(x) = y \). $$ax + b = y$$ $$ax = y - b$$ $$x = \frac{y - b}{a}$$ For \( f \) to be onto, for every \( y \in N \), the value of \( x = \frac{y - b}{a} \) must also be a natural number. Let's take a specific example. Let \( a = 2 \) and \( b = 1 \). Then \( f(x) = 2x + 1 \). The range of this function for \( x \in N \) is \( \{3, 5, 7, \dots\} \), which is the set of odd natural numbers greater than or equal to 3. Consider \( y = 1 \in N \). If \( f(x) = 1 \), then \( 2x + 1 = 1 \implies 2x = 0 \implies x = 0 \), which is not a natural number. Consider \( y = 2 \in N \). If \( f(x) = 2 \), then \( 2x + 1 = 2 \implies 2x = 1 \implies x = \frac{1}{2} \), which is not a natural number. In general, for \( f(x) = ax + b \), if we choose \( y \) such that \( y - b \) is not a positive multiple of \( a \), or if \( y - b \le 0 \), then \( x \) will not be a natural number. Since \( a \ge 1 \) and \( b \ge 1 \), let's consider \( y = 1 \in N \). If there exists \( x \in N \) such that \( ax + b = 1 \), then \( ax = 1 - b \). Since \( b \ge 1 \), \( 1 - b \le 0 \). If \( 1 - b = 0 \), then \( b = 1 \) and \( ax = 0 \), which implies \( x = 0 \) (not in \( N \)). If \( 1 - b<0 \), then \( ax \) is negative, and since \( a>0 \), \( x \) must be negative (not in \( N \)). Therefore, there exists \( y \in N \) (for example, \( y = 1 \)) for which there is no \( x \in N \) such that \( f(x) = y \). Hence, the function \( f \) is not onto.