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: Demonstrate injectivity of \( f \)
Assume \( f(x_1) = f(x_2) \), for \( x_1, x_2 \in A \): \[ \frac{x_1 - 3}{x_1 - 5} = \frac{x_2 - 3}{x_2 - 5} \] Performing cross-multiplication: \[ (x_1 - 3)(x_2 - 5) = (x_2 - 3)(x_1 - 5) \] Expanding both sides: \[ x_1x_2 - 5x_1 - 3x_2 + 15 = x_1x_2 - 5x_2 - 3x_1 + 15 \] Simplifying the equation leads to: \[ -5x_1 - 3x_2 = -5x_2 - 3x_1 \] Rearranging terms: \( 5x_2 - 3x_2 = 5x_1 - 3x_1 \implies 2x_2 = 2x_1 \implies x_2 = x_1 \). This confirms that if \( f(x_1) = f(x_2) \), then \( x_1 = x_2 \), thus \( f \) is one-to-one.
Step 2: Demonstrate surjectivity of \( f \)
For any \( y \in B \), we seek an \( x \in A \) such that \( f(x) = y \). Setting up the equation: \[ \frac{x - 3}{x - 5} = y \] Solving for \( x \): \[ x - 3 = y(x - 5) \] \[ x - 3 = yx - 5y \] Grouping \( x \) terms: \( x - yx = 3 - 5y \) \[ x(1 - y) = 3 - 5y \implies x = \frac{3 - 5y}{1 - y} \] Provided \( y eq 1 \) (which is true for \( y \in B \)), a corresponding \( x \) exists in \( A \). Therefore, \( f \) is onto. 
Step 3: Classify the function
As \( f \) has been shown to be both one-one and onto, it is classified as a bijection.