If.R1.and. R2 are two equivalence relation then R1 U R2 is not transitive ? Why ?

Dear Student,

$$\begin{gathered} Given: R_1, R_2 are equivalence relations \\ For transitivity of R_1 \cup R_2, there are two cases possible: \\ \left(1\right) When \left(x, y\right) and \left(y, z\right) belong to the same relation eg. R_1. Since R_1 is transitive, so, \left(x, z\right) \in R_1 and R_1 \cup R_2. \\ Thus, the union is transitive. \\ \left(2\right) When \left(x, y\right) \in R_1 - \left(R_1 \cap R_2\right) and \left(y, z\right) \in R_2 - \left(R_1 \cap R_2\right), then the union might not be transitive. \\ Eg. \\ Let A = \left\{5, 6, 7, 8\right\} \\ {R}_1 = \left\{\left(5, 5\right), \left(5, 6\right), \left(5, 7\right), \left(6, 5\right), \left(6, 6\right), \left(6, 7\right), \left(7, 5\right), \left(7, 6\right), \left(7, 7\right), \left(8, 8\right)\right\} \\ {R}_2 = \left\{\left(5, 5\right), \left(5, 6\right), \left(5, 8\right), \left(6, 5\right), \left(6, 6\right), \left(6, 8\right), \left(7, 7\right), \left(8, 5\right), \left(8, 6\right), \left(8, 8\right)\right\} \\ Then, R_1 and R_2 are both equivalence relations. However, \left(7, 5\right) and \left(5, 8\right) \in R_1 \cup R_2, but \left(7, 8\right) \notin R_1 \cup R_2. \end{gathered}$$

​Hope this information will clear your doubts about topic.
​
​If you have any more doubts, please ask here on the forum and our experts will try and help you out as soon as possible.

Regards