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Let $A=\{a,b,c,d,e\}$. Suppose $R$ is an equivalence relation on $A$. Suppose also that $aRd$ and $bRc$, $eRa$ and $cRe$. How many equivalence classes does $R$ have?

My thoughts: (Not sure if I have the right idea...)

UPDATED/EDITED

Since $R$ is an equivalence relation on $A$ and $aRd$, $bRc$, $eRa$, and $cRe$, then

$$R=\{(a,d),(d,a),(a,a),(d,d),(b,c),(c,b),(c,c),\\ (b,b),(e,a),(a,e),(e,e),(c,e),(e,c)\}$$ (Did I miss any?)

So $R$ has $1$ equivalence class:

  • $[a]=[b]=[c]=[d]=[e]=\{a,b,c,d,e\}$
  • 1
    Well, R has to be transitive and $\,bRc\,\,,\,cRe\,$, so...2012-08-01
  • 0
    You only know that ${(a,d),(b,c),(e,a),(c,e)}\subset R$. The problmm does not state that this is the entire relationship. You need to apply the rules for equivalence relationships to extrapolate enough relationships.2012-08-01
  • 1
    You are TOLD that it is an equivalence relation, so you can assume the transitive, symmetric, and reflexive property all hold, and therefore any consequence of these properties and the given ordered pairs being in the relation.2012-08-01
  • 0
    You need to work on the transitivity to get the answer right.2012-08-01
  • 0
    Your conclusion about the number of equivalence classes is correct. Your list of the elements of $R$ is incomplete; $R$ is, in fact, *all* of $A\times A$.2012-08-01

6 Answers 6