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$E={1,2,3,4,5,6,7,8}$ Defines the product set E × E the relation R: $(p, q) R (p_0, q_0) $if $ p-p_0$ even and $q-q_0 $divisible by 3

Question :How many equivalence classes are there

My attempt :

$p-p_0$is even $\Rightarrow$$p-p_0=2k$$\Rightarrow$$p=p_0[2]$ So for that exist two equivalent classes 0 and 1

$q-q_0$is divisible by 3 $\Rightarrow$$q-q_0=3k$$\Rightarrow$$q=q_0[3]$ So for that exist three equivalent classes 0 and 1 and 2

So finally :

exist 6 equivalent classes :(1, 0),(1,1),(1,2),(0,0)(0,1)(0,2)$

Is that correct?

Bill Dubuque
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    So long as you are understanding that your "equivalence class $(1,0)$" is in fact in reference to the set ${(1,3),(3,3),(5,3),(7,3),(1,6),(3,6),(5,6),(7,6)}$ – JMoravitz Jan 19 '21 at 03:20
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    Do you understand what an equivalence class is? It is a maximal collection of elements in your domain who are all related to one another. $(1,3)$ is related to each of the pairs in the set I wrote above. It is related to itself, it is related to $(1,6)$, it is related to $(3,3)$, etc... Note that $0$ is not an element of $E$ and so $(1,0)$ is not actually an element of $E\times E$. – JMoravitz Jan 19 '21 at 03:33
  • OK, thank you sir. –  Jan 19 '21 at 03:37
  • yes, you´re right –  Jan 19 '21 at 03:40
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    @BrienNavarro he's right in that there are six. He is incorrect in what they look like in the post, but hopefully from my comments before that confusion should be cleared. – JMoravitz Jan 19 '21 at 03:40
  • Yes, sorry for don´t write clear, it´s correct your afirmation@JMoravitz –  Jan 19 '21 at 03:43
  • @JMoravitz Can you show me the steps for your solution? –  Jan 19 '21 at 04:37
  • @JMoravitz, Supposon the question is $(p,q)R(p_0,q_0) $if $p−p_0$event and$ q-q_0 $ divisible by n.. class (0)=class (n)? If 0 isn't exist in E we can change class (0) by class(n)? –  Jan 19 '21 at 14:43

1 Answers1

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As JMoravitz commented, $0$ is not an element of your set.
So you seem to change notation between the question and your tentative of answering it.
To keep it tidy, I'll reformulate the question.

We have a set $E = \{ 1, 2, 3, 4, 5, 6, 7, 8\}$ and a binary relation $R$ defined on $E\times E$ by $$(p,q) R (r,s) \quad\text{ iff }\quad 2|p-r \;\text{ and }\; 3|q-s.$$ Apparently, you already concluded that $R$ is an equivalence relation on $E\times E$, and just want to know what are the equivalence classes.
These are the following $6$: $$(1,1)/R, (1,2)/R, (1,3)/R, (2,1)/R, (2,2)/R, (2,3)/R.$$ (Here, I'm using the notation in which, for an equivalence relation $\sim$ on a set $X$, we denote by $x/{\sim}$ the equivalence class of $x \in X$; another common notation would be $[x]_{\sim}$, but I'll use the previous one.)

To show that those are exactly the equivalence classes, let us start by describe the class of the element $(1,1)$, that is, to list the elements which are $R$-related with $(1,1)$.
Given the definition, these are the elements whose first coordinate is odd (so that the difference with $1$ is even) and the second coordinate is $1$, $4$ or $7$ (so that subtracting $1$ we get a multiple of $3$).
Hence $$(1,1)/R = \{ (1,1), (1,4), (1,7), (3,1), (3,4), (3,7), (5,1), (5,4), (5,7), (7,1), (7,4), (7,7) \}.$$ Analogously you can conclude that \begin{align} (1,2)/R &= \{ (1,2), (1,5), (1,8), (3,2), (3,5), (3,8), (5,2), (5,5), (5,8), (7,2), (7,5), (7,8) \},\\ (1,3)/R &= \{ (1,3), (1,6), (3,3), (3,6), (5,3), (5,6), (7,3), (7,6) \},\\ (2,1)/R &= \{ (2,1), (2,4), (2,7), (4,1), (4,4), (4,7), (6,1), (6,4), (6,7), (8,1), (8,4), (8,7) \},\\ (2,2)/R &= \{ (2,2), (2,5), (2,8), (4,2), (4,5), (4,8), (6,2), (6,5), (6,8), (8,2), (8,5), (8,8) \},\\ (2,3)/R &= \{ (2,3), (2,6), (4,3), (4,6), (6,3), (6,6), (8,3), (8,6) \}. \end{align} Now, just to confirm, you can count that there are $64$ elements listed in those classes (without repetitions), and since that is the same number of elements in $E \times E$, we know that we didn't forget any one.

(These means that you are right, if we increment by one each coordinate of each pair in your list of equivalence classes. Notice, however, that even so, there would be a problem in your answer: for example $(1,1)$ is not an equivalence class; what is correct is that $(1,1)/R$ is an equivalence class.)

amrsa
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  • Well, if there is no $0$ element in the set, there is no class of $0$ either. If $(p,q)R(r,s)$ iff $p-r$ is even and $q-s$ divisible by $n$, then we would have $2n$ equivalence classes. – amrsa Jan 19 '21 at 14:45
  • In general, if $\sim_1$ is an equivalence relation on $X$, with $n_1$ equivalence classes, and $\sim_2$ an equivalence relation on $Y$, with $n_2$ equivalence classes, then the relation on $\sim$ on $X\times Y$ defined by $$(x_1,y_1)\sim(x_2,y_2);\text{ iff };x_1\sim_1 x_2\text{ and }y_1\sim_2 y_2$$ is an equivalence relation with $n_1n_2$ classes. How does this apply to your case? You have $X=Y=E$, $x\sim_1 x'$ iff $x-x'$ is even and $y\sim_2 y'$ iff $y-y'$ is a multiple of $3$. – amrsa Jan 19 '21 at 14:49
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    for example $(4,1)/R=(2,1)/R$. This is because $(2,1)R(4,1)$. All the elements are there... – amrsa Jan 19 '21 at 14:51
  • I can't. I'll have to go now. I'll come back and delete most of these comments... – amrsa Jan 19 '21 at 14:54
  • Why you will delete these comments ? –  Jan 19 '21 at 14:57