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I want to determine wether the given relation is an equivalence relation on [$1$, $2$, $3$, $4$, $5$]. And If it is then list all the equivalence classes.

Relation: $\{(1, 1),(2, 2),(3, 3), (4, 4), (5, 5), (1, 5), (5, 1), (3, 5), (5, 3), (1, 3), (3, 1)\}$

My calculations:

It's reflexive since all numbers are related to themselves.

it's symetric.

I'm not sure if it's a transitive set. Since (4, 4), (2, 2) are alone and not connected to any others.

If it is transitive then how would you go about listing the equivalence classes?

Edit: I've gotten feedback now that says that it is transitive. So then I want to list the equivalence classes:

Some examples I have seen have definitions of the equivalence classes but this one doesn't so I assume all I gotta do is:

$[1] = \{ 1, 5, 3\}$

$[2] = \{ 2\}$

$[3] = \{ 3, 5, 1\}$

$[4] = \{ 4\}$

$[5] = \{ 5, 1, 3\}$

or something simillar?

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    Well, just by listing them. What, for example, is $1$ equivalent to? I see $1\sim \{1,3,5\}$, no? So that's an equivalence class. What's $2$ equivalent to? – 2017-02-19
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    ... but $(4,4),(2,2)$ is not of the form $(x,y),(y,z)$ so what does that have to do with transitivity? – 2017-02-19
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    It is indeed transitive. The only way that it would not be transitive is if there existed an $a,b,c$ such that $a\sim b$ and $b\sim c$ while $a\not\sim c$. Since that does not happen, it is transitive. In the statement of transitivity that $a\sim b$ and $b\sim c$ implies $a\sim c$, it is allowed that $a,b,c$ are all the same element. Here, $4\sim 4$ and $4\sim 4$ implies $4\sim 4$. – 2017-02-19
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    Reworded again, a transitive relation is one where any element that could be reached by another in two or more steps must also be reachable in only one step. – 2017-02-19
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    @JMoravitz, that's the answer. Why not post it as such? – 2017-02-19
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    Thank you for your answer! As a last request could you take a look at my equivalence classes and say if they look right or wrong? – 2017-02-19

1 Answers 1

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Yes the relation is transitive.

The only way that the relation would not be transitive, would be if there exists an $a,b,c$ such that $(a\sim b$ and $b\sim c),\,$ but $a \not \sim c$.

Equivalence Classes

Careful:

Each class must be pairwise disjoint. In your work, however, note that $[1] = [3]= [5] = \{1, 3, 5\}$

There should be only three equivalence classes: $[1] = \{1, 3, 5\}$, $\;[2] = \{2\},\;$ and $[4] = \{4\}$.

That's essentially what I think you are trying to say, But the relation defined in your question has only 3 unique (and disjoint) classes.

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    Could you explain more about the "pairwise disjoint" a bit further? English is not my first language so I don't think I know what you mean by that. I created a digraph of the relations and I can see that there are only 3 groups which explains why there shoud be only three equivalence classes right? This might be a dumb question but what determines which classes you choose, why did you pick classes [1], [2], [4] and not [5], [2] and [4]? – 2017-02-19
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    It simply means that the intersection of any two of the three equivalence classes is empty. $\{1,3, 5\} \cap \{2\} = \varnothing;\;\{1, 3, 5\} \cap \{4\} = \varnothing, \;\{2\} \cap \{4\} = \varnothing.$ – 2017-02-19
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    The equivalence class of $5$ is identical to the equivalence class of $1$, and both of those are also equivalent to the equivalence class of $3$. So we have one class $[1] = [3] = [5] = \{1, 3, 5\}$ (every element in $\{1, 3, 5\}$ is related to the other two. – 2017-02-19