Question

Let \( A = \mathbb{R} - \{5\} \) and \( B = \mathbb{R} - \{1\} \). Consider the function \( f : A \to B \), defined by \(f(x) = \frac{x - 3}{x - 5}\). Show that \( f \) is one-one and onto.
 

Hint:

To prove a function is one-one, show that \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \). To prove onto, solve \( f(x) = y \) and ensure \( x \) lies in the domain.

Answer 1

Step 1: Prove that \( f \) is one-one
Let \( f(x_1) = f(x_2) \), where \( x_1, x_2 \in A \): \[ \frac{x_1 - 3}{x_1 - 5} = \frac{x_2 - 3}{x_2 - 5} \] Cross-multiply: \[ (x_1 - 3)(x_2 - 5) = (x_2 - 3)(x_1 - 5) \] Simplify: \[ x_1x_2 - 5x_1 - 3x_2 + 15 = x_1x_2 - 5x_2 - 3x_1 + 15 \] \[ -5x_1 - 3x_2 = -5x_2 - 3x_1 \implies 5(x_2 - x_1) = 3(x_2 - x_1) \] If \( x_1 \neq x_2 \), this leads to a contradiction. Hence, \( x_1 = x_2 \), proving \( f \) is one-one. 

Step 2: Prove that \( f \) is onto
Let \( y \in B \). Solve \( f(x) = y \): \[ \frac{x - 3}{x - 5} = y \implies x - 3 = y(x - 5) \] \[ x - 3 = yx - 5y \implies x - yx = -5y + 3 \] \[ x(1 - y) = -5y + 3 \implies x = \frac{-5y + 3}{1 - y} \] For \( y \neq 1 \) (since \( y \in B \)), \( x \) exists in \( A \). Thus, \( f \) is onto. 

Conclude the function properties
Since \( f \) is both one-one and onto, \( f \) is a bijection.