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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.

Alexander Burstein
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mdcs_mdcs
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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 09 '17 at 17:57
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    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 09 '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 09 '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 09 '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 09 '17 at 18:13
  • I saw those questions from the book. I'm trying to answer them but I'm stuck with these questions. I tried answering them but with all my effort, I really can't. it's frustrating :( – mdcs_mdcs Dec 09 '17 at 18:22
  • What is your have you done to solve the problem? You need to show your effort. – Alexander Burstein Dec 09 '17 at 20:17

1 Answers1

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$B_n$ 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 $b\equiv a~(mod~5)$
QED
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