DM - Functions & Relations -Q17

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Let "Closure of a Relation R with respect to a Property P" be defined as follows:

let R be a relation on a set A. Then a relation R' is the closure of the relation R with respect to property P if and only if

1. R' satisfy the property P, and

2. R\subseteq R', and

3. for any(every) relation R'' if R\subseteq R'' and R'' satisfies the property R, then R'\subseteq R''

i.e if there is a relation R' with property P containing R such that R' is a subset of every relation R'' with property P containing R, then R' is called the closure of R with repect to P

 

Which of the following is False ?

 

(A). For any non-empty relation R on a non-empty set A, reflexive closure i.e. closure of R with respect to property “reflexivity”, can always exist.

(B). For any non-empty relation R on a non-empty set A, symmetric closure i.e. closure of R with respect to property “symmetricity”, can always exist.

(C). For any non-empty relation R on a non-empty set A, irreflexive closure i.e. closure of R with respect to property “irreflexivity”, can always exist.

(D). For any non-empty relation R on a non-empty set A, transitive closure i.e. closure of R with respect to property “transitivity”, can always exist.

asked Jun 24 in Discrete Maths by gbeditor (44,500 points)
edited 4 days ago by deepak-gatebook

1 Answer

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Best answer

Statement 1 :

For any non-empty relation R on a non-empty set A, reflexive closure i.e. closure of R with respect to property “reflexivity”, can always exist.

It is true. Reflexive closure of R can be found like this 

R' = R \cup \{ (a,a) | a \in A \}


Statement 2 :

For any non-empty relation R on a non-empty set A, symmetric closure i.e. closure of R with respect to property “symmetricity”, can always exist.

It is true. Symmetric closure of R can be found like this 

R' = R \cup R^{-1}


Statement 3 :

For any non-empty relation R on a non-empty set A, irreflexive closure i.e. closure of R with respect to property “irreflexivity”, can always exist.

It is false. Consider this relation : R = \{ (1,1) \}

Now, we can not make it irreflexive by ADDING something to it. 


Statement 4 :

 For any non-empty relation R on a non-empty set A, transitive closure i.e. closure of R with respect to property “transitivity”, can always exist.

This is true. 

https://en.wikipedia.org/wiki/Transitive_closure#:~:text=Informally%2C%20the%20transitive%20closure%20gives,337).

answered 4 days ago by deepak-gatebook (112,390 points)
Answer:
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