How many equivalence relations

Option B 5 is the correct choice. Number of equivalent relations will be 5. The transitive relation won't be satisfied if you choose to select the remaining three pairs using the following logic: Either include the pair or do not include the pair. Example: If you include the 2nd and 3rd pair i. Corresponding to each of which we will have an equivalence relation.. Login Register. Please log in or register to add a comment. Please log in or register to answer this question. Thank you LeenSharma for such a nice explanation:.

I'm correcting this in case someone like me gets stuck wrapping his head around this. Convinient method to find no. But there is a theorem which says : Corresponding to every equivalent relation there is an equivalent partitioning of the set.. We can do it by cases: 1 Everybody is in the same equivalence class. The pattern is more complicated, Bell numbers, please see the link in the answer of Dror , with no "closed form" formula.

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Since this part of the theorem is a disjunction, we will consider two cases: Either. Theorem 7. Consequences of these properties will be explored in the exercises. The following table restates the properties in Theorem 7. In Exercise 6 of Section 7. Consequently, each real number has an equivalence class. For this equivalence relation,. The results of Theorem 7. Because of the importance of this equivalence relation, these results for congruence modulo n are given in the following corollary.

Hence, Corollary 7. Technically, each pair of distinct subsets in the collection must be disjoint. We then say that the collection of subsets is pairwise disjoint.

We introduce the following formal definition. There is a close relation between partitions and equivalence classes since the equivalence classes of an equivalence relation form a partition of the underlying set, as will be proven in Theorem 7. The proof of this theorem relies on the results in Theorem 7. We will use Theorem 7. Part 1 of Theorem 7. That is, we need to show that any two equivalence classes are either equal or are disjoint.

However, this is exactly the result in Part 3 of Theorem 7. Note : Theorem 7. This process can be reversed. This will be explored in Exercise Is this set equal to any of the previous sets we have studied in this part?