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Give an example of an equivalence relation defined on the set A={0,1,2,3,4} which has two different classes of equivalence.

I think I don't understand this topic, but i created something like this:

(0,0) (0,1) (1,0) (1,1) (0,2) (2,0) (2,2) (1,2) (2,1) and (3,3) (3,4) (4,3) (4,4) so equivalence class of 0 is: [0]={(0,0),(0,1),(0,2)} and [1]={(1,0),(1,2),(1,1)}, [3]={(3,3),(3,4)}, [4]={(4,3),(4,4)} and what now? is it ok?

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Your relation is fine but you might have a misconception about the definition of an equivalence class.

The definition of an equivalence class $[a]$ for a relation $S$ is given by, $$ [a] = \{x\space|\space x\space S \space a\}$$

Therefore, for the example you gave, the equivalence classes would be as follows, $$ [0] = \{0,1,2\}$$ $$ [1] = \{0,1,2\}$$ $$ [2] = \{0,1,2\}$$ $$ [3] = \{3,4\}$$ $$ [4] = \{3,4\}$$

And the distinct classes would be $[0] = [1] = [2]$ and $[3] = [4]$

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  • $\begingroup$ so this one has 5 classes of equivalence, but only 2 different classes? in the class we should list all the elements from the relation? like for [0]={all unique elements form (0,0),(0,1),(0,2)} $\endgroup$ – sswwqqaa Nov 30 '16 at 11:44
  • $\begingroup$ Yes, there are 5 equivalence classes but only 2 DISTINCT ones. In general, if the set that your relation acts on has $n$ elements, there will be $n$ equivalence classes. However, we cannot tell how many distinct ones there will be. $\endgroup$ – Limzy Nov 30 '16 at 11:51
  • $\begingroup$ For the 2nd question, yes as well. As long as $n\spaceS\spacea$, then $n\element[a]$ $\endgroup$ – Limzy Nov 30 '16 at 11:52
  • $\begingroup$ Sorry, my TeX had some mistakes. As long as $n S a$, then $n\in[a]$ $\endgroup$ – Limzy Nov 30 '16 at 11:54
  • $\begingroup$ thanks for your help :D One more question if we set our relation to mod(2), there will be only 2 equivalence classes and these 2 classes will be distinct. [0]={0,2,4} and [1]={1,3} $\endgroup$ – sswwqqaa Nov 30 '16 at 11:56

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