Partitioning the irrational numbers in a special way Can the set of  irrational numbers be partitioned into two nonempty subsets , each of which is closed under addition ?    
I can show that if $A \subseteq \mathbb R$ has the property that $A$ is closed under addition and there is at least two elements in $A$ that can be extended to a Hamel basis of $\mathbb R$ over $\mathbb Q$ then $A$ can be partitioned into two nonempty subsets , each of which is closed under addition . Unfortunately for me , the set of  irrational numbers is not closed under addition .
Please help . Thanks in advance 
 A: Let $V\subset\mathbb{R}$ be a vector subspace over $\mathbb{Q}$ such that $\mathbb{R}=V\oplus\mathbb{Q}$ (e.g., extend $\{1\}$ to a basis of $\mathbb{R}$ over $\mathbb{Q}$ and let $V$ be the span of all the basis elements except $1$).  We identify $\mathbb{R}$ with the set of ordered pairs $(v,q)$ where $v\in V$ and $q\in \mathbb{Q}$.  Let $A$ be the set of $(v,q)$ such that $v>0$ and $B$ be the set of $(v,q)$ such that $v<0$.  Then $A$ and $B$ are a partition of the irrational numbers, since $(v,q)$ is rational iff $v=0$.  Furthermore, $A$ and $B$ are clearly closed under addition and nonempty.
More generally, if $\preceq$ is any total order on $V$ compatible with the group structure, we could let $A$ be the set of $(v,q)$ such that $v\succ 0$ and $B$ be the set of $(v,q)$ such that $v\prec 0$.  Conversely, every partition $A\cup B$ of the irrationals into two sets closed under addition arises from a total order on $V$ in this way.  Indeed, given such a partition, we get an ordering $\preceq$ on $V$ by saying $V\cap A$ is the set of positive elements of $V$.  For any $v\succ 0$ and $q\in \mathbb{Q}$, we then have $(-v,0)\in B$ and therefore $(v,q)\in A$ since $(-v,0)+(v,q)=(0,q)\not\in B$.  Thus $A$ contains all $(v,q)$ such that $v\succ 0$, and similarly $B$ contains all $(v,q)$ such that $v\prec 0$.
