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

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.