Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a "non trivial" solution of the Cauchy functional equation, i.e. $f$ is not of the form $f(x)=cx$ for any $c\in\mathbb{R}$ and satisfies the relation $$f(x+y)=f(x)+f(y)\quad\forall\,x,y\in\mathbb{R}$$ Define $$A=\{x\in\mathbb{R}:f(x)\geq 0\},\quad B=\{x\in\mathbb{R}:f(x)< 0\}$$ Now $\{A,B\}$ is a partition of $\mathbb{R}$ into two non empty, closed by sum subsets, and it's easy to see that it's different from the trivial ones $\left(\{\mathbb{R}_{\geq 0},\mathbb{R}^-\},\{\mathbb{R}_{\leq 0},\mathbb{R}^+\}\right)$. In particular $$\mathbb{R}^+=(A\cap\mathbb{R}^+)\cup(B\cap\mathbb{R}^+)$$ is a partition of $\mathbb{R}^+$ into two non empty, closed by sum subsets.

My question: **It's possible to partition $\mathbb{R}^+$ into two non empty set, both closed by sum and product?**

The analogous question for $\mathbb{R}$ has negative answer, and the argument is pretty easy: we can assume $0\in A$, and suppose that there exists $x\in B$. Now $-x\in A$ (otherwise $-x+x=0\in B$), and then $x^2=(-x)^2\in A$, which is a contradiction.

I guess the answer to my question is negative (it seems related to the absence of nontrivial field automorphism of $\mathbb{R}$), but i don't find any efficient argument.

Thank you for all suggestions.