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5 January, 03:30

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

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  1. 5 January, 06:30
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    (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.
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