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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.

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    $\begingroup$ What are we supposed to do, check your answers? $\endgroup$ – Kenny Lau May 12 '17 at 17:10
  • $\begingroup$ Your "reflexive" part is wrong. Check the definition of "reflexive". $\endgroup$ – Kenny Lau May 12 '17 at 17:11
  • $\begingroup$ Reflexive means that $aRa$ is true. $\endgroup$ – Michael Burr May 12 '17 at 17:11
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    $\begingroup$ Even after the edit, your reflexive part has typos / misconceptions. Reflexive: $a\in \Bbb R\implies a\mathcal{R}a$. There shouldn't be any mention of $b$ at all. Further, your transitive section has a typo in it still $[(a-b)\in\Bbb Q\wedge (b-\color{red}{c})\in\Bbb Q)\implies \dots)$ $\endgroup$ – JMoravitz May 12 '17 at 17:15
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    $\begingroup$ Please, James, see @JMoravitz 's comment immediately above. $\endgroup$ – amWhy May 12 '17 at 17:17
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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.

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    $\begingroup$ More generally, if $G$ is any abelian group and $H$ is any subgroup, then $aRb \iff a-b\in H$ is an equivalence relation on $G$, using the same proof. (In this question, $G=\Bbb R$ and $H=\Bbb Q$.) This fact is part of the proof that $G/H$ is itself a group. $\endgroup$ – Greg Martin May 12 '17 at 18:50

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