Let $\mathfrak{R}$ be a fixed o-minimal expansion of an ordered field $(R,<,+,\cdot,-,0,1)$ , A power function is a definable endomorphism of the multiplicative group $(Pos(R) , \cdot ,1)$ .
Claim(1) : if $f:Pos(R)\rightarrow R$ is definable, not identically $0$, and $f(xy)=f(x)f(y)$ for all $x,y\in Pos(R)$ , then $f$ is a power function. that is $f(x)>0$ for all $x\in Pos(R)$
Claim(2) : $(Pos(R) , \cdot ,1)$ is divisible, that is for any $x\in Pos(R)$ and for any $n\in\mathbb{N}$ , there is some $y\in Pos(R)$ such that $y^n=x$
Claim (3) : $x\rightarrow x^q$ where $q\in\mathbb{Q}$ is a power function. I have a problem with raising to rational powers
some useful facts about o-minimal expansion of an ordered field
fact(1) - the monotonicity theorem : Every unary definable function is ultimately differentiable (with ultimately definable derivative), ultimately either constant or strictly monotone, and has a limit at $+\infty$ in $R \cup\{-\infty, +\infty\}$
fact(2) : The mean value theorem and its usual consequences hold for definable functions.
fact(3) : The groups $(R,+)$ and $(\operatorname{Pos}(R), \cdot)$ have no proper, nontrivial definable subgroups.
my attempt in showing that claim(3) follows from claim(2) - assume that $(Pos(R) , \cdot ,1)$ is divisible consider the map $x\mapsto x^n$ . its kernel is a definable subgroup of $(Pos(R) , \cdot ,1)$ . by fact(3) it is injective and by divisibility it is surjective. thus we can define the nth root function, hence the rational exponents
any hints to why the three claims hold would be appreciated
Thanks to @Atticus Stonestrom
By theorem 2.3 in the article DEFINABLE SETS IN ORDERED STRUCTURES.I by ANAND PILLAY AND CHARLES STEINHORN, $R$ is a real closed field, in particular every positive element is a square. then $f(x)=f(y\cdot y)=(f(y))^2\geq 0$.
Assume for a contradiction that $f(x)=0$ for some $x>0$. Then for any $y\in Pos(R)$ $f(y)=f(y/x\cdot x)=f(y/x)f(x)=0$ which is absurd.
consider the map $x\mapsto x^n$. restricted to $Pos(R)$, this is a definable group endomorphism of $(Pos(R),\cdot)$. certainly it has non-trivial image, and its image is a definable subgroup of $(Pos(R),\cdot)$, since $(Pos(R),\cdot)$ have no proper, non trivial definable subgroup, its image is everything.
We can use the same argument on the kernel of $x\mapsto x^n$ as in Claim 2 to establish the infectivity of that map. Or we can use the fact that every ordered group is torsion free.
Next, for every $n\in\mathbb{N}$ define $x^{\frac{1}{n}}$ to be the unique element $y$ in $Pos(R)$ such that $y^n=x$. For $q=\frac{n}{m}\in\mathbb{Q}^+$ define $x^{\frac{n}{m}}$ to be $(x^{\frac{1}{m}})^n$ and for $q \in\mathbb{Q}^-$ define $x^q$ to be $(x^{-1})^{-q}$. It remains to verify that it is well defined and have the same propreties as in the case of $\mathbb{R}$. in particular it verifies : $(x\cdot y)^q=x^q\cdot y^q$.
Finally, the map $x\mapsto x^q$ is definable:
For $q\in\mathbb{Q}^+$, the map $x\mapsto x^{\frac{n}{m}}$ where $n,m\in\mathbb{N}$ is defined by the formula $\phi_{n,m}(x,y)$ : $x^n=y^m \land 0<x \land 0<y$. And for $q<0$ $x^q=(x^{-1})^{-q}$, and since the map $x\mapsto x^{-1}$ is definable by the formula $\xi(x,y)$ ; $x\cdot y=1$. we can see that the map $x\mapsto x^q$ is definable as the composition of two definable maps.