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Let X = {1, 2, 4, 7, 8, 10, 12, 14, 15, 18, 22, 23}. Consider the relation R on X given by (a, b) ∈ R if a − b is divisible by 7.

(a) Show that R is an equivalence relation on X.

(b) Determine the equivalence class [1].

(c) Determine the partition of X into equivalence classes.

edit 1) I need help on C, I get a) is congruent modulo and therefore b) is "the quotient of X modulo R" by my understanding but I don't quite understand how to write out the notation of the partition in this regard. Thank you for the quick answers, sorry I was not immediately specific.

edit 2) Thank for the explanations and help in the comments, especially JMoravitz and Lee Mosher. Much appreciated!

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  • $\begingroup$ Welcome to Maths SX! It is the restriction of the congruence modulo $7$ relation to a subset of $\mathbf Z$. $\endgroup$
    – Bernard
    Commented Nov 10, 2020 at 13:20
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    $\begingroup$ This should be straightforward so long as you understand the definitions involved. Surely you have some thoughts of your own about this, yes? Share your attempt. If you don't share your attempt we have to assume you didn't try anything at all and aren't willing to try anything, in which case we aren't likely to help. On the other hand, if you show your attempt, we can see what difficulty you are having and actually help address that specific lack of understanding. If where you are stuck is definitions, that can be answered by a book rather than by a human. $\endgroup$
    – JMoravitz
    Commented Nov 10, 2020 at 13:28
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    $\begingroup$ If you say what you have tried and where you are stuck, you are likely to get better help. $\endgroup$
    – paw88789
    Commented Nov 10, 2020 at 13:29
  • $\begingroup$ "I need help on C" The partitions are simply those numbers who are all multiples of seven, those numbers who are all one more than a multiple of seven, those numbers who are all two more than a multiple of seven, etc... ignoring those classes who happen to be empty due to the specific choice of $X$. So, you have $\{1,8,15,22\},\{2,23\},\color{grey}{\{\},}\{4,18\},\dots$ $\endgroup$
    – JMoravitz
    Commented Nov 10, 2020 at 13:44

1 Answer 1

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To answer C, you can do things one step at a time.

Starting with $1 \in X$, to determine the equivalence class $[1]$, ask yourself this question:

  • What elements $a \in X$ are equivalent to $1$?

Then translate:

  • For what elements $a \in X$ is it true that $(a,1) \in R$?

Then translate again:

  • For what elements $a \in X$ is it true that $a-1$ is divisible by $7$?

Now go through the elements $a \in X$ one-by-one, answering this question for each in turn:

  • For $a=1$, is $a-1=1-1=0$ divisible by $7$? Yes it is, so $1 \in [a]$.
  • For $a=2$, is $a-1=2-1=1$ divisible by $7$? No it is not, so $2 \not\in [a]$.
  • For $a=4$, is $a-1=4-1=3$ divisible by $7$? No it is not, so $4 \not\in [a]$.

And so on.

Along the way, you may have a leap of understanding which let's you cut off all of this one-by-one stuff. But until you do, keep doing the one-by-one stuff.

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