Let us determine if the following is an equivalence relation on X={1,2,3,4,5}. If the following are equivalence relation, then let us enumerate its equivalence classes.
1. R={(1,1),(2,2),(3,3),(4.4),(5.5),(1,3),(3,1)}
Taking into account that {(1,1),(2,2),(3,3),(4.4),(5.5)}⊂R, we conclude that the relation is reflexive. Since R−1={(1,1),(2,2),(3,3),(4.4),(5.5),(3,1),(1,3)}=R, the relation R is symmetric. Taking into account that (a,b),(b,c)∈R implies (a,c)∈R for all (a,b),(b,c)∈R, we conclude that this relation is transitive.
Let us enumerate its equivalence classes. Recall that [a]={x∈X:(a,x)∈R}. It follows that there are 4 different equivalence classes: [1]={1,3}, [2]={2}, [4]={4}, [5]={5}.
2. R={(1,1),(2,2),(3,3),(4,4),(5,5),(1,5),(5,1),(3,5),(5,3),(1,3),(3,1)}
Taking into account that {(1,1),(2,2),(3,3),(4.4),(5.5)}⊂R, we conclude that the relation is reflexive. Since R−1={(1,1),(2,2),(3,3),(4,4),(5,5),(5,1),(1,5),(5,3),(3,5),(3,1),(1,3)}=R,
the relation R is symmetric. Taking into account that (a,b),(b,c)∈R implies (a,c)∈R for all (a,b),(b,c)∈R, we conclude that this relation is transitive.
Let us enumerate its equivalence classes. It follows that there are 3 different equivalence classes: [1]={1,3,5}, [2]={2}, [4]={4}.
3. R={(x,y):13 divides x+y}
Taking into account that 13 does not divide 2=1+1, we conclude that (1,1)∈/R, and hence R is not reflexive. Therefore, R is not an equivalence relation.
Comments