Answer: 1) R = {(a, a), (а,b), (b, a), (b, b), (с, с), (b, с), (с, b)}.
It is clearly not transitive since (a, b) ∈ R and (b, c) ∈ R whilst (a, c) ¢ R. On the other hand, it is reflexive since (x, x) ∈ R for all cases of x: x = a, x = b, and x = c. Likewise, it is symmetric since (а, b) ∈ R and (b, а) ∈ R and (b, с) ∈ R and (c, b) ∈ R.
2) Let S=Z and define R = x and y have the same parity
i.e., x and y are either both even or both odd.
The parity relation is an equivalence relation.
a. For any x ∈ Z, x has the same parity as itself, so (x,x) ∈ R.
b. If (x,y) ∈ R, x and y have the same parity, so (y,x) ∈ R.
c. If (x.y) ∈ R, and (y,z) ∈ R, then x and z have the same parity as y, so they have the same parity as each other (if y is odd, both x and z are odd; if y is even, both x and z are even), thus (x,z)∈ R.
3) A reflexive relation is a serial relation but the converse is not true. So, for number 3, a relation that is reflexive but not transitive would also be serial but not transitive, so the relation provided in (1) satisfies this condition.
Explanation:
1) By definition,
a) R, a relation in a set X, is reflexive if and only if ∀x∈X, xRx ---> xRx.
That is, x works at the same place of x.
b) R is symmetric if and only if ∀x,y ∈ X, xRy ---> yRx
That is if x works at the same place y, then y works at the same place for x.
c) R is transitive if and only if ∀x,y,z ∈ X, xRy∧yRz ---> xRz
That is, if x works at the same place for y and y works at the same place for z, then x works at the same place for z.
2) An equivalence relation on a set S, is a relation on S which is reflexive, symmetric and transitive.
3) A reflexive relation is a serial relation but the converse is not true. So, for number 3, a relation that is reflexive but not transitive would also be serial and not transitive.
QED!