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Let $X = \mathbb{R} \setminus\{1\}$. Let $f : X → X$ be defined by:

$f(x)=\frac {x+1}{x-1}$

How would I prove that $f$ is a bijection? I've started by showing that it is injective since:

$(\frac {x_1+1}{x_1-1} = \frac {x_2+1}{x_2-1}) ⇒ (x_1=x_2)$

Where do I go from here?

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    $\begingroup$ Show that it is surjective. Most easily by computing the inverse. $\endgroup$ – Daniel Fischer Nov 3 '13 at 15:35
  • $\begingroup$ Show that for every element $x_0\in X$ there is an element $x_1\in X$ such that $f(x_1)=x_0$. $\endgroup$ – abiessu Nov 3 '13 at 15:38
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Note that $$f(x) =\frac{x+1}{x-1} = \frac{x-1+2}{x-1} = 1 + \frac{2}{x-1}.$$ From this you should have no pain proving it is injective and surjective on $\mathbb R\setminus\{1\}$.

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Suppose that we are given $y \in \mathbb{R} \setminus \{1\}$ and we want to find $x \in \mathbb{R} \setminus \{1\}$ such that $f(x) = y$. To do this, just write down what it means to have $f(x) = y$ and then solve for $x$ in terms of $y$.

$$ \frac{x+1}{x-1} =y \iff x+1 = xy-y \iff \cdots \iff x = \cdots$$

Remark: Because $f$ is in fact a bijection, you will get a unique solution for $x$ in terms of $y$. Considering this $x$ as a function of $y$, say $x = g(y)$, this will mean that you have found the inverse function $g$ of $f$.

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To prove that $f$ is a bijection you first show its one to one which is also called an injection. Suppose that $f(x_1)=f(x_2)$ where $x_1,x_2\in X$ and $x_1,x_2\neq 1$. It follows that $\frac{x_1+1}{x_1-1}=\frac{x_2+1}{x_2-1}$ Show now that $x_1=x_2$. Now show that $f$ is onto. Let $y\in X$ and $y\neq 1$ why?. Now show that there exists a $x\in X$ such that $f(x)=y$. Now figure out what you are going to let that $x$ be in order to get back $y$.

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