Q:

show that f is bijective function

Let $A=\{x \in \mathbb{R}:x\neq 1\}$. Define $f:A\to \mathbb{R}$ by $f(x) = \frac{x}{x-1}$. Show that $f$ is bijective.

As I learn, a function is bijective if it is injective and surjective at the same time. The way I prove these statement is by substituting $x=2,3,4,5,6\ldots$ into the function $f(x)$. Then I got a map function $f:[2,3,4,5,6]\to[2 , \frac 32 , 2 , \frac 52 , 3]$. Then I draw a one to one relationship between them to show these functions are injective and the element is the codomain also always occupied, so we can say this function is surjective too . So, my conclusion is the function $f$ is bijective .

A:

To prove bijection, you need to show that your function is 1-1 and onto.
For injectivity:
Suppose $f(a)=f(b)$
$\implies \frac{a}{a-1}=\frac{b}{b-1}$
$\implies a(b-1)=b(a-1) \implies ab-a =ba-b \implies a=b$ 
Hence, the function is injective.
For Surjectivity, it has already been pointed out that it is not surjective.

