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Question:

Let $ \ R$ be an equivalence relation on $ \mathbb{Z}$ defined as follows:

$ \ (a,b) \in R \iff 2a + 3b$ is divisible $ 5$.

What is the equivalence class of $ 0$.

My attempt:

$[0] = { \{ y \in \mathbb{Z}: 5 | \ 3y}\} = { \{ 0,5,-5,10,-10,.......} \} $

I am new to equivalence relations. Is this equivalence class correct? In general do we write it as a set that satires certain conditions and then find all the elements?

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  • $\begingroup$ it is okay, but $5\mid 3y$ and also $5\mid 2y$, so $5\mid y$... $\endgroup$ – MAN-MADE Aug 17 '17 at 4:35
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Yes, your answer is correct. But it can be written in a shorter form. Notice that this set consists precisely of all integer multiples of $5$, so $$[0]=\{0,5,-5,10,-10,\ldots\}=\{5k\mid k\in\mathbb{Z}\}=5\mathbb{Z}.$$

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