in Set Theory & Algebra
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3 votes
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Suppose that A is a nonempty set, and f is a function that has A as its domain. Let R be the relation on A consisting of all ordered pairs (x, y) such that f (x) = f (y).  What are the equivalence classes of R?
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4 Comments

@srestha. The equivalence classes would consist of individual elements of the non-empty set A since the relation consists of pairs like (x, x) only.
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ok then it will be not transitive and symmetric. rt?
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@srestha. if the relation consists of pairs like (1,1), (2,2).......and so on then its transitive and symmetric as well.
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1 Answer

2 votes
2 votes
Equivalence classes will be the sets of elements of A which have the same value for function f.

eg: if A={1,2,3,4,5,6,7,8} and f={(1,a),(2,b),(3,a),(4,d),(5,a),(6,b),(7,c),(8,d)}

then R={(1,1),(2,2),(3,3),(4,4),(5,5),(6,6),(7,7),(8,8)(1,5),(5,1),(1,3),(3,1),(3,5),(5,3),(2,6),(6,2),(4,8),(8,4)}

then the equivalence classes are:{1,3,5},{2,6},{7},{4,8}
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2 Comments

A correction maybe in R: it must also contain pair (4,8) and (8,4)

So equivalence class {8} will become {4,8}

is it correct?
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Hi,

Yes, you are correct.I missed that. Now, I have made the changes. Thanks a lot. :)
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