# Equivalence relation that has 2 different classes of equivalence

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),(0,1),(0,2)} and ={(1,0),(1,2),(1,1)}, ={(3,3),(3,4)}, ={(4,3),(4,4)} and what now? is it ok?

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, $$ = \{0,1,2\}$$ $$ = \{0,1,2\}$$ $$ = \{0,1,2\}$$ $$ = \{3,4\}$$ $$ = \{3,4\}$$

And the distinct classes would be $ =  = $ and $ = $

• 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 ={all unique elements form (0,0),(0,1),(0,2)} – sswwqqaa Nov 30 '16 at 11:44
• 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. – Limzy Nov 30 '16 at 11:51
• For the 2nd question, yes as well. As long as $n\spaceS\spacea$, then $n\element[a]$ – Limzy Nov 30 '16 at 11:52
• Sorry, my TeX had some mistakes. As long as $n S a$, then $n\in[a]$ – Limzy Nov 30 '16 at 11:54
• 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. ={0,2,4} and ={1,3} – sswwqqaa Nov 30 '16 at 11:56