Show that open interval $(-1,1)$ is isomorphic to $(\mathbb{R},+)$ Define group structure on $G=(-1,1)$ by
$$a*b=\frac{a+b}{1+ab}$$
for any $a,b\in G$. Show that $G$ is isomorphic to $\mathbb{R}$ under addition.
I've tried the obvious maps $f:G\rightarrow \mathbb{R}$ such as $f(x)=\tan(\pi x/2)$ or $f(x)=\frac{x}{1-x^2}$ but none of these seem to work.
Any help would be appreciated, thanks.
 A: Just for good measure, let's prove it:


*

*$\tanh(x+y)=\tanh(x) * \tanh(y)$


It is well-known that by the $\tanh$ addition identity, $\tanh(x+y)=\frac{\tanh(x)+\tanh(y)}{1+\tanh(x)\tanh(y)}$, which is equivalent to the statement above.


*

*$\tanh$ is bijective from $\Bbb{R}$ to $(-1, 1)$


We have $\tanh x=\frac{e^{2x}-1}{e^{2x}+1}$. We have the following:
$$e^{2x}-1 < e^{2x}+1 \implies \frac{e^{2x}-1}{e^{2x}+1}=\tanh x < 1$$
Also:
$$0 < e^{2x} \implies 1 < e^{2x}+1 \implies 1-e^{2x} < 1 < e^{2x}+1$$
$$\implies \frac{1-e^{2x}}{e^{2x}+1}=-\tanh x < 1 \implies \tanh x > -1$$
Thus, for all $x \in \Bbb{R}$, we have $\tanh x \in (-1, 1)$. Now, let's say $\tanh x=\tanh y$:
$$\frac{e^{2x}-1}{e^{2x}+1}=\frac{e^{2y}-1}{e^{2y}+1} \implies e^{2(x+y)}+e^{2x}-e^{2y}+1=e^{2(x+y)}-e^{2x}+e^{2y}+1$$
$$\implies e^{2x}-e^{2y}=e^{2y}-e^{2x} \implies 2e^{2x}=2e^{2y} \implies e^{2x}=e^{2y}$$
Thus, $\tanh x=\tanh y \implies x=y$ and its range is $(-1, 1)$, so we have injection.
Now, for any $y \in (-1, 1)$, we have the following equation:
$$y=\frac{e^{2x}-1}{e^{2x}+1} \implies e^{2x}y+y=e^{2x}-1 \implies e^{2x}(y-1)=-(y+1)$$
$$e^{2x}=-\frac{y-1}{y+1}=\frac{1-y}{y+1}$$
Now, since $y < 1$, we have $1-y > 0$ and since $y > -1$, we have $y+1 > 0$, so $\frac{1-y}{y+1} > 0$, meaning we can take the $\ln$ of both sides:
$$2x=\ln\left(\frac{1-y}{y+1}\right) \implies x=\frac{\ln\left(\frac{1-y}{y+1}\right)}{2}$$
Thus, for any $y \in (-1, 1)$, there exists an $x \in \Bbb{R}$ such that $\tanh x=y$, so we have surjection.
With injection and surjection, we can now deduce bijection.
A: You need a function $f\colon\mathbb{R}\to(-1,1)$ such that, for every $x,y\in\mathbb{R}$, 
$$
f(x+y)=f(x)*f(y)=\frac{f(x)+f(y)}{1+f(x)f(y)}
$$
Since the neutral element in $\bigl((-1,1),{*}\bigr)$ is $0$, you also know that $f(0)=0$.
Let's look for a differentiable $f$. Thus, differentiating with respect to $x$, we have
$$
f'(x+y)=
\frac{f'(x)(1+f(x)f(y))-(f(x)+f(y))f'(x)f(y)}{(1+f(x)f(y))^2}=
\frac{f'(x)(1-f(y)^2)}{{(1+f(x)f(y))^2}}
$$
If we set $y=-x$, we get $f(-x)=-f(x)$, so
$$
f'(0)=\frac{f'(x)(1-f(x)^2)}{(1-f(x)^2)^2}=\frac{f'(x)}{1-f(x)^2}
$$
Thus
$$
xf'(0)=\int_0^x \frac{f'(t)}{1-f(t)^2}\,dt=
\int_0^{f(x)}\frac{1}{1-u^2}\,du=
\frac{1}{2}\left[\log\frac{1+u}{1-u}\right]_0^{f(x)}
$$
Setting $k=f'(0)$ we have
$$
e^{2kx}=\frac{1+f(x)}{1-f(x)}
$$
and so
$$
f(x)=\tanh(kx)
$$
This is bijective whenever $k\ne0$. Indeed, for $k>0$ the function is increasing and
$$
\lim_{x\to-\infty}\tanh(kx)=-1,
\quad
\lim_{x\to\infty}\tanh(kx)=1
$$
It is reversed if $k<0$.
