Ex. 4. Suppose that R1 and R2 are equivalence relations on the set S. Determine whether each of these combinations of R1 and R2 must be an equivalence relation. a) R1∪R2 b) R1∩R2 c) R1⊕R2

Answer:a) R1∪R2 may not be an equivalence relation.b) R1∩R2 is an equivalence relation.c) R1⊕R2 is not an equivalence relation.Step-by-step explanation:a) R1∪R2 may not be an equivalence relation. Counter-example S={1,2,3} R1={(1,1),(2,2),(3,3),(1,2),(2,1)} R2={(1,1),(2,2),(3,3),(1,3),(3,1)} R1 and R2 are  equivalence relations in S. Let us see R1∪R2 is not. (2,1)∈ R1∪R2 and (1,3)∈ R1∪R2. If  R1∪R2 were an equivalence relation in S it should be transitive, so (2,3) should belong to  R1∪R2. But (2,3)∉ R1∪R2 and  R1∪R2 is not an equivalence relation b) R1∩R2 is an equivalence relation.  Let us prove is reflexive: (a,a)∈R1 for every a in S, (a,a)∈R2 for every a in S, so (a,a)∈R1∩R2 is symmetric: If (a,b)∈R1∩R2 then (a,b)∈R1 and (a,b)∈R2. Since R1 and R2 are  equivalence relations, (b,a)∈R1 and (b,a)∈R2, hence (b,a)∈R1∩R2 is transitive: If (a,b)∈R1∩R2 and (b,c)∈R1∩R2 then (a,b)∈R1 and (b,c)∈R1 and (a,b)∈R2 and (b,c)∈R2, so (a,c)∈R1 and (a,c)∈R2, hence (a,c)∈R1∩R2. c) R1⊕R2 is not an equivalence relation. Since R1⊕R2 = (R1∪R2)-(R1∩R2) the intersection of R1 and R2 is not in R1⊕R2, so the diagonal (the elements of the form (a,a) with a∈S) are not in R1⊕R2 hence it is not reflexive.