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Let $\mathbf{R}$ be the relation on $Z \times (Z \setminus \{0\})$ given by $m \mathrel{\mathbf{R}} n$ iff $m - n =2k$ for some $k \in\mathbb Z$. I have proven that this is indeed an equivalence relation by meeting the three required properties, yet I am having trouble understanding the meaning of and determining the equivalence classes of $\mathbf{R}$. Any pointers would be a great help.

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  • $\begingroup$ Perhaps you should try writing out the equivalence classes and attempting to understand what you see, and how it relates to the definition of the relation. $\endgroup$ – rogerl Feb 2 '17 at 23:38
  • $\begingroup$ The way the question is worded, it seems you want the difference of two ordered pairs to equal an integer? I think you must have made a mistake by posting this. $\endgroup$ – AMPerrine Feb 4 '17 at 23:46
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I'm emitting a doubt about the symetrical property of $\mathbf R$ because $0$ is forbidden as a second argument. So $0$ is in relation to the left with some numbers, but there is no description of what happen to the right. I would have defined it on $\mathbb Z^2$ without restriction.

Anyway, two numbers $m$ and $n$ are in the same equivalence class if they differ by and even number.

So $\mathbf R$ is splitting $\mathbb Z$ into odd and even numbers, and there are two classes $\tilde 1$ (odd numbers) and $\tilde 2$ (even numbers). [and I still don't know what to think about $\tilde 0$, which without the restriction should be equal to $\tilde 2$].

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  • $\begingroup$ Okay that definitely clears up for me how to think of an equivalence class. Thank you sir. $\endgroup$ – urbjhawk Feb 3 '17 at 0:48
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I don't think your relation is symmetric since $0R0$ is not true. Remark that $R$ is defined over $\mathbb{Z}\times (\mathbb{Z}\setminus\{0\})$.

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