I ask this after trying to solve it myself unsuccessfully. It's part of a larger question I'm working on, but have separated this out so as to make it as clear and concise as possible.
Trying to solve this: $$\text{Let }r \in \, (0,1\, ) \text{ and } r \notin \mathbb{Q}$$ $$\text{Let }r’ \in \, (0,1\, ), r’>r, r’ \notin \mathbb{Q}, \text{ and } r’-r \notin \mathbb{Q}$$ $$\text{Let }q \in A = \{ r’-r+p : p \in \, (r,r+1\, ) \cap \mathbb{Q} \}$$ Finally, one of the two statements must be true: $$\text{1) } \exists q’ \in B = \, (r’,r’+1\, ) \cap \mathbb{Q} \text{ such that }\{q'-p : p \in B\} = \{q-p : p \in A\}$$ $$\text{2) } \nexists q’ \in B = \, (r’,r’+1\, ) \cap \mathbb{Q} \text{ such that }\{q'-p : p \in B\} = \{q-p : p \in A\}$$
Which statement is true?
The second statement is true.
Let $q'\in B$. Then $$\inf(\{p-q': p\in B\})-\inf(\{p-q:p\in A\})=(\inf B-q')-(\inf A-q)=(r'-q')-(r'-q)=q-q'\neq 0, $$ because $q'\in\mathbb{Q}$ and $q\notin\mathbb{Q}$. So, the sets are different as they have different infima.
Edit: The answer to the corrected question is the same, as $$\sup(\{q'-p:p\in B\})-\sup(\{q-p:p\in A\})=(q'-\inf B)-(q-\inf A)=q'-q. $$