Respuesta :
Answer:
A, B, D, H are the correct statements
Step-by-step explanation:
A) If R1 were symmetric, since we have the relation (2,4), we should have for symmetry the relation (4,2), which is not the case. Therefore R1 is not symmetric and A is True.
B) There are only 2 relations in R3, (2,4) and (4,2). One relation is the symmetric relation of the other, therefore R3 is symmetric and B is True.
C) If R3 were transitive it should satysfy the transitive rule. Since 2 is related with 4 and 4 is related with 2, then for transitivity 2 should be related with itself, which is not true. As a consecuence, R3 is not transitive and C is False.
D) The elements that appear in R2 are 1, 2, 3 and 4. R2 is reflexive because we have all the relations (1,1), (2,2), (3,3) and (4,4) that relate an element with itself. We conclude that D is True.
E) R4 is not symmetric because we have the relation (1,2) but we dont have the symmetric relation (2,1) in R4. Therefore E is False.
F) The element 4 appears in R1 but we dont have the relation (4,4) in R1. This means that R1 in not reflexive, so F is False.
G) Similar to what happened with R2, the elements 1,2,3 and 4 appear in R5 and all the relations (1,1), (2,2), (3,3) and (4,4) are present in R5. As a consecuence, every element is related to itself, which means that R5 IS reflexive, and because of that, G is False.
H) All the realtions in R5 are of the form '(a,a)' for a in {1,2,3,4}. Suppose we have a chain of relations (a,b), and (b,c) in R5, and we want to know if (a,c) is in R5. Necessarily, a = b, and b = c. Then, a and c are equal, which means that the relation (a,c) is the relation (a,a), which is in R5 for any a in {1,2,3,4}. This proves the transitivity of R5 and therefore, H is True.
I) R3 is not reflexive because 2 is an element of the set and the relation (2,2) is not in R3. I is False
