Check my proof. Is this a group? On the set $I=(0,1)=\{x\in\Bbb R\mid 0<x<1\}$, the operation $*$ is defined by: $$\forall s,t \in I, s*t= \frac{1}{1+\frac{(1-s)}{s} \frac{(1-t)}{t}}$$ Operations are usual ones with $\Bbb R$. We have been told to assume that $*$ is associative.
I believe the identity is $e=\frac{1}{2}$. But I cannot not find the inverse. I'm assuming it doesn't have have one. Thus it would not be a group.
Also, I need to show that $I$ is closed under $*$. I believe that I can claim closure. If $s,t \in\Bbb R$ s.t. $0<s,t<1$ then $a*b$ is closed because the fraction will always have values between $(0,1)$.
Yes? No? Prettier ways to say it?
 A: First note that the group operation is commutative, by symmetry of its definition.
You can rewrite:
$$s*t = \frac{1}{1+\frac{1-s-t+st}{st}} = \frac{1}{2+\frac{1-s-t}{st}}$$
From which: $$s*\frac{1}{2} = \frac{1}{2+\frac{\frac{1}{2}-s}{\frac{1}{2}s}} = \frac{1}{2+\frac{1-2s}{s}}=\frac{1}{\frac{1}{s}} = s$$
So $e=\frac{1}{2}$ is indeed the identity element.
Now, given any element $s \in S$, you want to find $t$ such that $s*t = \frac{1}{2}$. In other words, you have to solve the equation in $t$:
$$\frac{1}{2+\frac{1-s-t}{st}} = \frac{1}{2}$$
$$\frac{st}{2st+1-s-t} = \frac{1}{2}$$
$$2st = 2st + 1 - s -t$$
$$t = 1-s$$
Just to verify, let's compute:
$$s*(1-s) = \frac{1}{2+\frac{1-s-(1-s)}{s(1-s)}} = \frac{1}{2+\frac{0}{s(1-s)}} = \frac{1}{2}$$
For closure, you need to show that $s*t < 1$ (it is obviously positive).
$$\frac{1}{2+\frac{1-s-t}{st}} <^? 1$$
$$1 <^? 2+\frac{1-s-t}{st}$$
$$-st <^? 1-s-t$$
$$-(1-s)(1-t)<^? 0$$
which holds, since $1-s$ and $1-t$ are elements of $H$.
We are done!
A: You can note that
$$
\frac{1-s}{s}=\frac{1}{s}-1
$$
and that the map $f\colon I\to(0,\infty)$, $s\mapsto s^{-1}-1$ is bijective, with
$$
f^{-1}(x)=\frac{1}{1+x}
$$
Now consider that $s*t=1/(1+f(s)f(t))$ and so
$$
s*t=f^{-1}(f(s)f(t))
$$
which means that
$$
f(s*t)=f(s)f(t)
$$
Thus the operation is just the transfer to $I$ of the multiplication on $(0,\infty)$ through $f^{-1}$. Since $(0,\infty)$ is a group, so is $I$.
What's the neutral element? $f^{-1}(1)=1/2$. What's the inverse of $s$? It's
$$
f^{-1}((f(s))^{-1})=\frac{1}{1+(f(s))^{-1}}=1-s
$$
