How to prove one-to-one and onto? I am trying to prove $(-1, 1)$ is equinumerous to $\mathbb{R}$ by finding a bijection function from $(-1, 1)$ to $\mathbb{R}$.
I've found a function $f(x) = \frac{x}{x^2-1}$, and it's a bijection obviously from its graph, but I don't know how to prove it is bijection rigorously. Any help, thanks!
 A: Here is a simple proof that $f$ is injective without calculus:
$$
f(x) = f(y)
\iff \frac{x}{x^2-1} = \frac{y}{y^2-1} 
\iff (xy+1)(x-y)=0
\iff xy=-1 \text{ or } x=y
$$
But $|x|<1$, $|y|<1$ implies $|xy|=|x| |y| <1$ and so $xy$ is never $-1$.
A: The given function is continous and because of $$(\frac{x}{x^2-1})'=\frac{x^2-1-2x^2}{(x^2-1)^2}=\frac{-1-x^2}{(x^2-1)^2}<0$$ strictly decreasing in the interval $(-1,1)$ Additionally, we have $$\lim_{x\rightarrow -1+0} \frac{x}{x^2-1}=\infty$$ and $$\lim_{x\rightarrow 1-0} \frac{x}{x^2-1}=-\infty$$
So the range is $\mathbb R$ , hence $f^{-1}(x)$ exists and is defined everywhere.
A: Injective: compute the derivative. The function is derivable in the whole domain. 
It is $\frac{x^2-1-x\cdot 2x}{(x^2-1)^2}= \frac{-1-x^2}{(x^2-1)^2}<0$ for all $x$ in the domain. So the function is in fact strictly monotone decreasing.
Surjective: Compute the limits at the endpoints of the interval. 
$\lim\limits_{x\rightarrow -1} \frac{x}{x^2-1}= \infty$, $\lim\limits_{x\rightarrow 1} \frac{x}{x^2-1}= -\infty$. 
As the function is continuous, every real value is attained somewhere by the Bolzano theorem. 
A: To show that a function is a bijection you can simply show that there is an inverse function $f^{-1}:\mathbb{R}\rightarrow(-1,1)$ which can be found by setting 
\begin{align}
x & =\frac{y}{y^2-1}
\\ y&=\frac{1\pm\sqrt{1+4x^2}}{2x}  
\end{align}
We take the negative sign to stay within the domain, so we obtain $f^{-1}(x)=\frac{1-\sqrt{1+4x^2}}{2x}$. 
Lastly, we can confirm that $f^{-1}\circ f(x)=x$ for all $x\in(-1,1)$ and similarly that $f\circ f^{-1}(x)=x$ for all $x\in\mathbb{R}$.
