Question

Let $f: A \to B$ be defined by $f(x) = \frac{x - 2}{x - 3}$, where $A = \mathbb{R} - \{3\}$ and $B = \mathbb{R} - \{1\}$. Discuss the bijectivity of the function.

Hint:

To prove a function is bijective, check that it is both injective (one-to-one) and surjective (onto). Use algebraic manipulation to verify injectivity and show that every element in the codomain has a pre-image for surjectivity.

Answer 1

To discuss the bijectivity of the function, we need to check if it is both injective (one-to-one) and surjective (onto). 1. Injectivity (One-to-one): A function is injective if distinct elements in the domain map to distinct elements in the codomain. Assume $f(x_1) = f(x_2)$, i.e., \[ \frac{x_1 - 2}{x_1 - 3} = \frac{x_2 - 2}{x_2 - 3} \] Cross multiplying gives: \[ (x_1 - 2)(x_2 - 3) = (x_2 - 2)(x_1 - 3) \] Simplifying the equation: \[ x_1 x_2 - 3x_1 - 2x_2 + 6 = x_2 x_1 - 3x_2 - 2x_1 + 6 \] After canceling terms and simplifying, we get: \[ x_1 = x_2 \] Thus, the function is injective. 2. Surjectivity (Onto): A function is surjective if every element in the codomain has a corresponding element in the domain. We need to show that for every $y \in B = \mathbb{R} - \{1\}$, there exists an $x \in A = \mathbb{R} - \{3\}$ such that $f(x) = y$. Starting from: \[ y = \frac{x - 2}{x - 3} \] Multiplying both sides by $(x - 3)$: \[ y(x - 3) = x - 2 \] Expanding and simplifying: \[ yx - 3y = x - 2 \] \[ x(y - 1) = 3y - 2 \] \[ x = \frac{3y - 2}{y - 1} \] This shows that for any $y \in \mathbb{R} - \{1\}$, we can find an $x \in \mathbb{R} - \{3\}$ such that $f(x) = y$. Thus, the function is surjective. Since the function is both injective and surjective, it is bijective.