This one may be obvious, but I could not find a solution to verify my proof.
Question: Is the following an equivalence relation on R?
$aRb\space$ iff$\space (a-b)\in Q$
My Solution:
Reflexive: $\forall a\in
\mathbb{R},\space $$aRa \space[(a-a)\in Q]$
$a-a=0$ and $0\in Q$
Symmetric: $\forall a,b \in \mathbb{R},\space aRb \rightarrow bRa\space[(a-b)\in Q \rightarrow (b-a)\in Q]$
Let $a-b = c$, where $c\in Q$, then $b-a=-c$. Well if $c\in Q$, then $(-c)\in Q$ as well.
So $(b-a)\in Q$ as well.
Transitive: $\forall a,b,c \in \mathbb{R},\space aRb\wedge bRc\rightarrow aRc\space[((a-b)\in Q) \wedge ((b-c)\in Q)\rightarrow (a-c)\in Q]$
Let $a-b=d$, where $d\in Q$ and also let $b-c=e$, where $e\in Q$.
Consider $d+e$. Since rational numbers are closed under addition, then $(d+e)\in Q$
Which implies that $[(a-b)+(b-c)]\in Q \rightarrow (a-c)\in Q$
Since the relation is reflexive, symmetric, and transitive, then the relation is an equivalence relation.
Best Answer
Obviously $a-b$ is rational iff $b-a$ is rational. All we need to show that the equivalence class of $a$ is well defined (and thus we have an equivalence relation) is that if $b-c$ is rational then $a-c$ is rational (given $a-b$ is rational). But $a-c=a-b+(b-c)$ is rational by closure of the rationals with respect to addition.