The number of solutions for the equation $ x*x = 2f(x) $

Question

Let $ (G,*) $ be a group so that the function $ f:\mathbb{R} \rightarrow \mathbb{R} $ , $ f(x)=\frac{x}{1+|x|} $ is an izomorphism between $ (R,+) $ and $ (G,*) $

$ 1) $ The number of solutions for the equation $ x*x = 2f(x) $ is ?

$ 2) $ Calculate $ \frac{1}{2}*\frac{1}{2}*\cdots*\frac{1}{2} $, where $ \frac{1}{2} $ appears $ 10 $ times .

The answer for the first one should be $\infty $ and for the second one $ \frac{10}{11} $

I haven't encountered problems like these because we aren't actually given the $ * $ law . Also for the second part the inverse of $ f $ which is $ f(x)=\frac{x}{1-|x|} $ could help us finding the expression.

I also want to ask some mods to delete my last question because it appeared to be off-topic and I'm sorry for that.

Answer 1

For me, the question seems poorly posed, but maybe that is deliberate. Perhaps you are meant to stretch your thinking.

From the definition, $f:\mathbb{R} \rightarrow \mathbb{R}$ a group isomorphism, $(\mathbb{R},+) \rightarrow (G,*)$ it is clear that $G \subset \mathbb{R}$, $G = f(\mathbb{R})$ with $*$ defined on $G$ by $x_1*x_2 = f\left(f^{-1}(x_1) + f^{-1}(x_2)\right)$.

1) The left-hand side of $x*x = 2f(x)$ indicates the $x\in G$ since that is where $*$ is defined, while the right-hand side indicates that $x\in \mathbb{R}$ since that is the domain of $f$. The only way out of the conundrum is to identify the elements of $G$ with elements of $\mathbb{R}$ using $G \subset \mathbb{R}$. $$ x*x = f\left(f^{-1}(x) + f^{-1}(x)\right) = f\left(\frac{2x}{1 - |x|}\right) = \frac{\frac{2x}{1 - |x|}}{1+\left|\frac{2x}{1 - |x|}\right|} = \frac{2x}{1+|x|} $$ using that $|x| < 1$ for $x \in G$. Since $2f(x)$ also equals $\frac{2x}{1+|x|}$ considering $x\in\mathbb{R}$. $x*x = 2f(x)$ for any of the infinite number of $x\in G$.

Answer 2

HINT:

Although you aren't given the $*$ law explicitly, it is forced by the given information! Likewise, the underlying set of $G$ is forced. Because you know $f$ is an isomorphism (and in particular is surjective onto $G$!), it must be that the underlying set of $G$ is the image of $f$. This is the open interval $(-1,1)$. Meanwhile, the fact that $f$ is an isomorphism with $(\mathbb{R},+)$ means you can deduce a formula for $*$! You know that

$$f(a+b) = f(a) * f(b)$$

because $f$ is an isomorphism (and in particular a homomorphism). Therefore, if you have some $x,y\in G = (-1,1)$, and you want to calculate $x*y$, if you were able to identify $a,b\in\mathbb{R}$ such that $x = f(a)$ and $y=f(b)$, you would be able to use the above formula to find $x*y = f(a)*f(b)$. Can you take it from here?