Can the set of positive rational numbers be decomposed into two non-empty, disjoint parts such that closed by the addition for?

Question

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Can the set of positive rational numbers be decomposed into two non-empty, disjoint parts such that closed by the addition for?

Answer 1

I assume you are asking whether $\mathbb{Q}^+$ can be partitioned into two nonempty disjoint sets, i.e., $\mathbb{Q}^+=A\cup B$ and $A\cap B=\varnothing$ and $A,B$ are closed under addition. If so, then the answer is no, you cannot do that. Suppose you can, then $1$ is either in $A$ or $B$. WLOG, assume $1\in A$, then $1+1=2\in A$ and keep doing this you can see all positive integers should be in $A$. For any rational number $q$, it can be written as $q=m/n$ for positive coprime integers $m,n$, so if $q\in B$, then $q+q=2q\in B$, and if you keep doing this, $nq\in B$. However, $nq=m$, so $m\in A\cap B$, which is a contradiction to $A\cap B=\varnothing$. This shows any rationals should be in $A$, which implies $B=\varnothing$, again contradiction...