Partition of $\mathbb{R}^+$ into two semigroups

The Solution: Notice that $\mathbb{R}$ is a vector space over $\mathbb{Q}$. Choose some linear function $f:\mathbb{R}\rightarrow\mathbb{Q}$ over this vector space for which there exist an $x,y\in\mathbb{R}^+$ such that $f(x)\geq 0$ and $f(y)<0$. Then, the sets $A=\{x:f(x)\geq 0\}$ and $B=\{x:f(x)<0\}$ for $x\in\mathbb{R}^+$ partition the space and are semigroups.

Some notes on it: The existence of such an $f$ is guaranteed by the axiom of choice, which is equivalent to the statement that all vector spaces have a basis. Once we have a basis $S$, we can change to the basis $S'=\{|s|:s\in S\}$ and define $f$ on $S'$ such that it takes a negative value for at least one $s\in S'$ and takes a non-negative value for some other $s\in S'$.

A more general statement we could make is that we could partition $\mathbb{R}^+$ into as many subsemigroups as we want by choosing more linear functions - for instance, if we had a $f$ and $g$ so that none of the following were empty, we could partition into three sets as: $$\{x:f(x)\geq 0\}$$ $$\{x:f(x)< 0\wedge g(x)\geq 0\}$$ $$\{x:f(x)< 0\wedge g(x) < 0\}$$ (We could also partition into the sets for which a single $f$ is positive, negative, or zero, but this doesn't generalize well)

Another note is that $f$ could be $\mathbb{R}\rightarrow\mathbb{R}$ not $\mathbb{R}\rightarrow\mathbb{Q}$ - I used the latter to emphasize that I want $f$ to be linear in the rationals - that $f(kx)=kf(x)$ is only necessary when $k$ is rational.

The answer is YES.

We shall follow the idea which appeared here, and I will explain the final step in detail.

Let $$\mathscr F=\{(S,T)\,:\,S,T\subset \mathbb R^+,\,S\cap T=\varnothing\,\,\& \,\,S,T\,\,\text{semigroups}\}.$$ Then $\mathscr F\ne\varnothing$, as $\,\big(\mathbb Q^+,\sqrt{2}\,\mathbb Q^+\big)\in\mathscr F$, and if we define "$\preceq$" in $\mathscr F$ as $$ (S_1,T_1)\,\preceq\,(S_2,T_2), $$ if and only if $S_1\subset S_2$ and $T_1\subset T_2$, then clearly "$\preceq$" is a partial ordering.

We shall use Zorn's Lemma to obtain a maximal element $(S_{\mathrm{max}},T_{\mathrm{max}})\in\mathscr F$, i.e., an element not possessing a strictly larger, in the $\preceq$-sense, and show that $S_{\mathrm{max}} \cup T_{\mathrm{max}}=\mathbb R^+$.

In order to apply Zorn's Lemma and extract this maximal element, we need to establish that every chain has an upper bound. A chain $\mathscr C$ is a non-empty subset of $\mathscr F$, which is totally ordered by $\preceq$, i.e., if $(S_1,T_1),\,(S_2,T_2)\in\mathscr C$, then either $(S_1,T_1)\,\preceq\,(S_2,T_2)$ or $(S_2,T_2)\,\preceq\,(S_1,T_1)$. Upper bound of $\mathscr C$ is an element $(S,T)\in\mathscr F$, larger than all the elements of $\mathscr C$, i.e., $(S',T')\preceq (S,T)$, for all $\,(S',T')\in \mathscr C$. If we set $$ \hat S=\bigcup\{S: (S,T)\in \mathscr C,\,\,\text{for some semigroup}\,\,T\subset\mathbb R^+ \}, \\ \hat T=\bigcup\{T: (S,T)\in \mathscr C,\,\,\text{for some semigroup}\,\,S\subset\mathbb R^+ \}, $$ we shall show that $(\hat S,\hat T)\in\mathscr F$. Indeed, if $x,y\in \hat S$, then $x\in S_1$ and $y\in S_2$, for some $(S_1,T_1),(S_2,T_2)\in\mathscr C$. But as $\mathscr C$ is a chain, then either $S_1\subset S_2$ or $S_2\subset S_1$. Say the first happens, then $x,y\in S_2$, and as $S_2$ is a semigroup, then $x+y\in S_2\subset \hat S$. Similarly, if $S_2\subset S_1$, and similarly $\hat T$ is a semigroup. Also, $\hat S\cap\hat T=\varnothing$, for if $x\in \hat S\cap\hat T$, then $x\in S_1$, for some $(S_1,T_1)\in\mathscr C$ and $x\in T_2$, for some $(S_2,T_2)\in\mathscr C$, and as $\mathscr C$ is a chain, assuming for example that $(S_1,T_1)\preceq(S_2,T_2)$, we obtain that $x\in S_2\cap T_2$, which is a contradiction, as $S_2,T_2$ are assumed disjoint. Also, $(\hat S,\hat T)$ is an upper bound of $\mathscr C$, as by its definition $S\subset \hat S$ and $T\subset\hat T$, for all $(S,T)\in\mathscr C$.

Hence, every chain in $\mathscr C$ has an upper bound and thus Zorn's Lemma is applicable, which means that $\mathscr F$ possesses a maximal element $(S_{\mathrm{max}},T_{\mathrm{max}})$. As $S_{\mathrm{max}},T_{\mathrm{max}}\subset\mathbb R^+$ are two disjoint semigroups, it only remains to show that $S_{\mathrm{max}} \cup T_{\mathrm{max}}=\mathbb R^+$.

Suppose not, and that $z\in \mathbb R^+\smallsetminus(S_{\mathrm{max}} \cup T_{\mathrm{max}})$ and let $S'$ be the semigroup generated by $\hat S\cup\{z\}$ and $T'$ the semigroup generated by $\hat T\cup\{z\}$. We shall obtain a contradiction by showing that either $(S',\hat T)\in\mathscr F$ or $(\hat S,T')\in\mathscr F$, each of which is strictly larger (with respect to $\preceq$) than the maximal element $(\hat S,\hat T)$.

The elements of $S'$ are of the form $x+nz$, where $x\in \hat S$ and $n\in\mathbb N$, and similarly the elements of $T'$ are of the form $y+mz$, where $y\in \hat T$and $m\in\mathbb N$. If $S'\cap \hat T\ne\varnothing$, then $$ x+nz=y, \quad\text{for some $x\in\hat S,\,y\in\hat T,\,n\in\mathbb N$}, \tag{1} $$ and if $\hat S\cap T'\ne\varnothing$, then $$ x'=y'+n'z, \quad\text{for some $x'\in\hat S,\,y'\in\hat T,\,n'\in\mathbb N$}. \tag{2} $$ Now $(1)$ and $(2)$ imply that $$ nn'z=n'y-n'x=nx'-ny', $$ and hence $nx'+n'x=ny'+n'y\in \hat S\cap \hat T$, which is not possible. Thus, either $S'\cap \hat T=\varnothing$ or $\hat S\cap T'=\varnothing$, which implies that $(\hat S,T')\in\mathscr F$ or $(S',\hat T)\in \mathscr F$, which contradicts the fact that $(\hat S,\hat T)$ is maximal.

Thus $\hat S\cup\hat T=\mathbb R^+$ and we are done.

Note. Although I find Meelo's answer extremely elegant, I believe that this answer has certain advantages:

a. Using the same argument we can partition $\mathbb R^+$ in $k$ semigroups, for every $k\in\mathbb N$.

b. This method works for other commutative groups, $G$, possessing a subset $T$, which is a semigroup and $G=T\cup\{0\}\cup (-T)$.

Let $H=\{h_i:i\in I\}$ be a Hamel basis for $\mathbb R$ over $\mathbb Q$, where the index set $I$ is well-ordered. Thus each real number $x$ has a unique expression as a finite linear combination of elements of $H$ with rational coefficients, i.e., $x=\sum_{i\in I}x_ih_i$ where $x_i\in\mathbb Q$ and $\{i\in I:x_i\ne0\}$ is finite. If $x\ne0$, define $f(x)$ to be the first nonzero coefficient in this expression, i.e., $f(x)=x_i\ne0$, and $x_j=0$ for all $j\lt i$.

Let$$A=\{x\in\mathbb R^+:f(x)\gt0\}$$and let$$B=\{x\in\mathbb R^+:f(x)\lt0\}.$$

It is easy to see that $A$ and $B$ are disjoint nonempty sets whose union is $\mathbb R^+$, and that each is closed under addition.