No, symmetric property means that a R b implies b R a. But if there is no such b element, we can not use the transitive property to get reflexive property.
March 12, 2007
Exercises B.2-4
R is equivalence relation, so R is reflexive, symmetric and transitive. But R is antisymmetric, so there is no distinct element a, b in S such that a R b. Otherwise b R a should be in R, since symmetric property. Hence there are only a R a in relation R.
Exercises B.2-2
a ≡ a (mod n) -> it is reflexive.
a ≡ b (mod n) then b ≡ a(mod n) -> it is symmetric
a ≡ b (mod n) and b ≡ c(mod n) then a ≡ c(mod n) -> it is transitive
The ≡ relation partition integer into n parts.
Exercises B.2-1
For any subset x, y of Z, ⊆ is reflexive because x⊆x. If x is not equal to y, then it is impossible x ⊆ y and y ⊆ x, so ⊆ is antisymmetric. If x ⊆ y and y ⊆ z, then x ⊆ z, hence ⊆ is transitive. Because of these three property, ⊆ in Z is partial order.
