Prove that the relation given by $a\sim b\Leftrightarrow a-b\in\mathbb{Z}$ is a congruence relation on the additive group $\mathbb{Q}$. Q: Prove that the relation given by $a\sim b\Leftrightarrow a-b\in\mathbb{Z}$ is a congruence relation on the additive group $\mathbb{Q}$. A: Maybe...

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*$a\sim a\Leftrightarrow a-a=0\in \mathbb{Z}$ ✓

*$a\sim b\Leftrightarrow a-b\in \mathbb{Z}$. $a\in \mathbb{Z}\Rightarrow -a\in \mathbb{Z}$ and $-b\in\mathbb{Z}\Rightarrow b\in \mathbb{Z}$, yielding $a-b+(-2a+2b)\in\mathbb{Z}\Leftrightarrow -a+b\in \mathbb{Z}\Leftrightarrow b\sim a.$ ✓

*$a\sim b$ and $b\sim c$ so $a-b\in \mathbb{Z}$ and $b-c\in \mathbb{Z}$ so $a-b+b-c\in\mathbb{Z}\Leftrightarrow a-c\in\mathbb{Z}\Leftrightarrow a\sim c.$ ✓

*$a_1\sim a_2$ and $b_1\sim b_2$. So $(a_1-a_2)(b_1-b_2)\in \mathbb{Z}\Leftrightarrow a_1b_1-a_1b_2-a_2b_1+a_2b_2\in\mathbb{Z}\Leftrightarrow a_1b_1+a_2b_2\in\mathbb{Z}$$\Leftrightarrow a_1b_1\sim a_2b_2$ ✓


The last bullet shows that the equivalence relation is a congruence relation on $\mathbb{Q}$.
A: Your proof that $a \sim b$ implies $b \sim a$ is a bit shaky. It is not necessarily true that $a \in \Bbb{Z}$ and $b \in \Bbb{Z}$ (let $a = 1/2$, $b = -1/2$, for example). You can get this more easily by simply noting that $b-a = -(a-b)$. Your proofs of reflexivity and transitivity are fine.
For the last property, you actually need to show that given $a_1 \sim a_2$, $b_1 \sim b_2$, we have $a_1 + b_1 \sim a_2 + b_2$. In other words, we must show that $a_2+b_2 - (a_1+b_1)$ is an integer. Hopefully this is clear. If not, try rewriting it as $a_2 - a_1 + b_2 - b_1$.
A: The second bullet is wrong. It should be $a \sim b \iff a-b \in \mathbb{Z} \iff b-a \in \mathbb{Z} \iff b \sim a$.
