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.
