For any $r,s\in\mathbb{R}$, either $r+\mathbb{Q}=s+\mathbb{Q}$ or $\left( r+\mathbb{Q} \right) \cap \left( s+\mathbb{Q} \right) =\emptyset$. For $r\in\mathbb{R}$, let $r+\mathbb{Q}=\left\{ r+q:q\in \mathbb{Q}\right\}$. For any $r,s\in\mathbb{R}$, either $r+\mathbb{Q}=s+\mathbb{Q}$ or $\left( r+\mathbb{Q} \right) \cap \left( s+\mathbb{Q} \right) =\emptyset$.
Why? Can you explain? 
 A: Suppose $(r+\Bbb Q)\cap (s+\Bbb Q)\neq0$; so we can take $x\in (r+\Bbb Q)\cap (s+\Bbb Q)$ i.e. there exist $p,q\in\Bbb Q$ s.t. $x=r+p=s+q$; in particular from this we get $r=s+(q-p)\in s+\Bbb Q$, since clearly $q-p\in\Bbb Q$, hence $r+\Bbb Q\subseteq s+\Bbb Q$.
Similarly you can prove that $s+\Bbb Q\subseteq r+\Bbb Q$ and thus $r+\Bbb Q= s+\Bbb Q$.
A: Define an equivalence relation $\sim$ on $\mathbb{R}$ by letting $r \sim s$ if and only if $s=r+q$ for some $q \in \mathbb{Q}$. The equivalence classes of $\sim$ are precisely $r+\mathbb{Q}$ for $r \in \mathbb{R}$. Hence the sets $r+\mathbb{Q}$ partition $\mathbb{R}$, and the result follows.
A: Consider $r - s$. There are two possibilities:


*

*$r - s \in \mathbb Q$. In this case, $r + \mathbb Q = s + \mathbb Q$. (Indeed, if $x$ is in $r + \mathbb Q$, then $x$ can be written as $x = r + q$ for some $q \in \mathbb Q$. But this can be rewritten as $x = s + (r - s) + q$. Since $(r - s) \in \mathbb Q$, we have $(r - s) + q \in \mathbb Q$ too, so $x = s + (r - s) + q \in s + \mathbb Q$. Hence $r + \mathbb Q \subseteq s + \mathbb Q$. An almost identical argument gives $s + \mathbb Q \subseteq r + \mathbb Q$, hence $r + \mathbb Q = s + \mathbb Q$)

*$r - s \notin \mathbb Q$. In this case, $(r + \mathbb Q ) \cap (s + \mathbb Q) = \emptyset$. To prove this, we should prove that if $x \in r + \mathbb Q$, then $x \notin s + \mathbb Q$. The method is similar to the previous case. I'll leave this to you.
