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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 .

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    $\begingroup$ You have tested the function's output for 5 different input values. That not enough to prove the function is injective or surjective. To prove does not mean to test on some values, but rather demonstrate the required properties hold for any possible values. $\endgroup$ – CiaPan Apr 11 '18 at 15:26
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    $\begingroup$ Also those aren't the correct values at those five points. $\endgroup$ – Sam Cassidy Apr 11 '18 at 15:57
  • $\begingroup$ Should $A \to \Bbb{R}$ read $A \to A$? If you make that change, what you are trying to prove is true. Viewed as a function $A \to \Bbb{R}$, $f$ is injective but not surjective, since $f$ never takes the value $1$. $\endgroup$ – Rob Arthan Apr 11 '18 at 15:59
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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.

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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 $f(x)=1$.

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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 $a$.

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