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...
- $a\sim a\Leftrightarrow a-a=0\in \mathbb{Z}$ ✓
- $a\sim b\Leftrightarrow a-b\in \mathbb{Z}$. $a\in \mathbb{Z}\Rightarrow -a\in \mathbb{Z}$ 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 $\mathbb{Q}$.