Q.Let A = R - {3}, B = R - {1}. Prove that the function f: A → B defined by f(x) = (x-2)/(x-3) is one-one and onto. (f is bijective)

Show injectivity by assuming $f(x_1)=f(x_2)$ and deriving $x_1=x_2$; show surjectivity by solving $y=f(x)$ for $x$ and checking the result stays in the domain.

$A = \mathbb R - \{3\}$, $B = \mathbb R - \{1\}$, $f(x) = \dfrac{x-2}{x-3}$.

One-one: Suppose $f(x_1) = f(x_2)$ for $x_1,x_2\in A$:

$\frac{x_1-2}{x_1-3} = \frac{x_2-2}{x_2-3}$

Cross-multiplying: $(x_1-2)(x_2-3) = (x_2-2)(x_1-3)$

$x_1x_2-3x_1-2x_2+6 = x_1x_2-3x_2-2x_1+6$

$-3x_1-2x_2 = -3x_2-2x_1 \ \Rightarrow\ -x_1 = -x_2 \ \Rightarrow\ x_1=x_2$

So $f$ is one-one.

Onto: Let $y\in B$ (so $y\ne 1$). Solve $y = \dfrac{x-2}{x-3}$ for $x$:

$y(x-3) = x-2 \ \Rightarrow\ yx-3y = x-2 \ \Rightarrow\ x(y-1) = 3y-2 \ \Rightarrow\ x = \frac{3y-2}{y-1}$

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