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 \( f \) is injective
Assume \( f(x_1) = f(x_2) \) for \( x_1, x_2 \in A \): \[ \frac{x_1 - 3}{x_1 - 5} = \frac{x_2 - 3}{x_2 - 5} \] Applying 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 yields: \[ -5x_1 - 3x_2 = -5x_2 - 3x_1 \] Rearranging terms: \( 5(x_2 - x_1) = 3(x_2 - x_1) \)
If \( x_1 eq x_2 \), this equation results in a contradiction. Therefore, it must be that \( x_1 = x_2 \), which proves that \( f \) is injective.

Step 2: Demonstrate \( f \) is surjective
For any \( y \in B \), we aim to find an \( x \in A \) such that \( f(x) = y \). Set up the equation: \[ \frac{x - 3}{x - 5} = y \] Isolate \( x \): \( x - 3 = y(x - 5) \)
\( x - 3 = yx - 5y \)
\( x - yx = -5y + 3 \)
\( x(1 - y) = -5y + 3 \)
\( x = \frac{-5y + 3}{1 - y} \)
Provided \( y eq 1 \) (which is true for \( y \in B \)), a corresponding \( x \) exists in \( A \). Consequently, \( f \) is surjective. 

Conclusion on function properties
As \( f \) is proven to be both injective and surjective, \( f \) is classified as a bijection.