Suppose R and S are reflexive relations on set A and T is a transitive relation on set A. Prove or disprove each of these statements: a) R∪T must be transitive. b) R⊕S is irreflexive. c) R◦S is reflexive

(a) R∪T is not Transitive.

(b) R⊕S is irreflexive.

(c) R◦S is Reflexive

Step-by-step explanation:

Definition

Reflexive: A relation on a set A is reflexive if (a, a) ∈A for all a∈A.

Transitive: A relation on a set A is said to be transitive if (a, b) ∈A and (b, c) ∈A ⇒ (a, c) ∈A.

Union: For two sets A and B, A∪B means all elements are either in A or B.

Symmetric Difference A⊕B: These is the set of all elements either in A or in B but not in both.

Composite of A and B (A◦B) consists of all ordered pairs (a, c) such that (a, b) ∈B and (b, c) ∈A.

(a) R∪T is Transitive

Let a∈R, since R is Reflexive, (a, a) ∈R.

By the definition of T, (a, b) ∈T and (b, c) ∈T, (a, c) ∈T.

R∪T={ (a, a), (ac) }

R∪T is not transitive since there is no element b∈R∪T to satisfy the condition for transitivity.

(b) R⊕S is irreflexive

Let a∈R, since R is Reflexive, (a, a) ∈R.

Let a∈S, since S is Reflexive, (a, a) ∈S.

The symmetric difference contains all elements either in R or in S but not in both.

Therefore: a∉R⊕S.

Therefore R⊕S is irreflexive.

(c) R◦S is reflexive

Let Let a∈R, since R is Reflexive, (a, a) ∈R.

Let a∈S, since S is Reflexive, (a, a) ∈S.

By the definition of Composite, if

(a, a) ∈R and (a, a) ∈S, then (a, a) ∈R◦S.

Therefore, R◦S is Reflexive since (a, a) ∈R◦S for every a∈A.