show that f is bijective function

Question

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 .

Answer 1

Computing $f(x)$ for finitely many values $x$ can, at most, be used to prove that $f$ is not injective. But it can not be used to prove that it is injective. In order to do that, you will have to prove that, whenever $x,y\in A$ and $x\neq y$, then $f(x)\neq f(y)$.

And your function is not surjective, since there is no $x\in A$ such that $f(x)=1$.

Answer 2

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.

Answer 3

Hints (assuming you actually want to show is that $f$ bijective when viewed as a function $A \to A$$.

First of all you need to prove that $f$ actually is a function from $A$ to $A$. I.e., you have to prove that $f(x) = 1$ has no solution. I.e., you have to derive a contradiction from $\frac{x}{x-1} = 1$ and to do that you just multiply through by the denominator of the l.h.s., to get $x = x - 1$.

Now you need to show that $f$ is injective, i.e., that $f(x) = f(y)$ implies $x = y$. So assume $\frac{x}{x-1} = \frac{y}{y-1}$, and multiply by the denominators to get $x(y-1) = (x - 1)y$, from which you can derive $x = y$ (do you see how?).

Finally you need to show that $f$ is surjective, i.e., that for every $a \in A$, $f(x) = a$ has a solution in $A$, i.e., that there is an $x \in A$ such that $\frac{x}{x-1} = a$, i.e., such that $x = a(x-1) = ax - a$, i.e., such that $(a-1) x = a$, from which you can work out what $x$ has to be in terms of $a$.