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**Defunkt** If $\displaystyle a,b \in \mathbb{Z}$ and $\displaystyle a \sim b$ it means, by definition, that $\displaystyle \exists \ m,n \in \mathbb{Z}, \ n \neq 0,$ such that $\displaystyle a - b = \frac{m}{n} \in \mathbb{Q}$. What can you say about $\displaystyle b-a$ then?

Your proofs for symmetry and transitivity are both incorrect -- $\displaystyle a \sim b$ means that $\displaystyle a-b \in \mathbb{Q}$, but that doesn't necessarily mean that $\displaystyle a, b \in \mathbb{Q}$ -- take, for example, $\displaystyle a = \pi - 1, b = \pi$...