# Relations, Equivalence Relations, Partitions [closed]

Consider A=$\mathbb{Z}$. For each integer $n$, define

$$B_n = \{{m\in \mathbb{Z}\ | \ (\exists q)(m=n+5q)}\}.$$

Prove that $\{{B_n}\}_{n\in\mathbb{Z}}$ is a partition of $\mathbb{Z}$. Identify the equivalence classes.

## closed as off-topic by Andrés E. Caicedo, José Carlos Santos, Lee Mosher, zipirovich, JackDec 9 '17 at 21:40

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• Are $B_n$ and $B_{n-5}$ disjoint . Because $m = n + 5q = n-5 + 5(q+1)$. So they are not disjoint. Unless in the definition something is missing. – Chirantan Chowdhury Dec 9 '17 at 17:57
• Welcome to stackexchange. You are more likely to get answers rather than votes to close if you edit the question to show us what you have tried and where you are stuck. – Ethan Bolker Dec 9 '17 at 18:00
• @ChirantanChowdhury Those sets are identical, which is not a problem. (If they weren't identical they would have to be disjoint and your objection would be relevant.) – Ethan Bolker Dec 9 '17 at 18:02
• So we need to consider only $B_1,B_2,B_3,B_4,B_5$ which are disjoint by the division algorithm applied to a number divided by 5. – Chirantan Chowdhury Dec 9 '17 at 18:06
• This is NOT a do-my-homework-for-me site. Within an hour you've posted two questions that are just copy-and-pasted texts of probably homework questions, without adding anything from yourself at all. If you're unwilling to make any effort on your homework, why should other people do that for you? – zipirovich Dec 9 '17 at 18:13

$B_n$ is the set of all the numbers that is congruent to $n$ modulo $5$. Now obviously
1. Reflexive: $a\equiv a~(mod~5)$,
2. Transitive: $a\equiv b~(mod~5)$ and $b\equiv c~(mod~5)$ implies $a\equiv c~(mod~5)$,
3. Symmetric: $a\equiv b~(mod~5)$ implies $b\equiv a~(mod~5)$