DM - Functions & Relations -Q18

+1 vote

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

Consider the following statements :

1. The reflexive closure R' of a relation R can be obtained as :

R' = R \cup I, where I = \{ (a,a) | a \in A \}

2. The symmetric closure R’ of a relation R can be obtained as :

R' = R \cup R^{-1}, where R^{-1} = \{ (b,a) | (a,b) \in R \}

3. The transitive closure R' of a relation R can be obtained as :

R' = R \cup T , where T = \{ (a,c) | (a,b) \in R \text{ and (b,c) }\in R \}

4. The symmetric closure R' of a relation R can be obtained as :

R' = R \cup T , where T = \{ (b,a) | (a,b) \in R \text{ And (b,a) }\notin R \}

Which of the above statements are correct ?

(A). Only 1

(B). Only 1,4

(C). Only 1,2,4

(D). All

asked Jun 24 in Discrete Maths by gbeditor (44,490 points)
reshown 6 days ago by gbeditor

1 Answer

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Best answer

Statement 2 and 4 are equivalent. 

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

Symmetric closure of R can be found like this 

R' = R \cup R^{-1}


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

Reflexive closure of R can be found like this 

R' = R \cup \{ (a,a) | a \in A \}


Statemetn 3 is false. 

For example, consider : R = \{ (1,1),(2,3),(3,1),(1,2) \}

Now, statement 3 says that we add pairs (2,1),(3,2) and (1,3) to make it transitive. But after adding these pairs to R, we don't get Transitive relation. i.e. R' = \{ (1,1),(2,3),(3,1),(1,2),(2,1),(3,2),(1,3) \}

R' here is Not transitive because (3,3) is not present in R'. 

 

answered 3 days ago by deepak-gatebook (112,390 points)
can you explain the significance of (b,a) does not belong to R, why it is given?
Say, you have (2,3) and (3,2) in R then none of these pairs will be in T. Also, assume you have (2,2) in R then also (2,2) will not be in T. But assume you have (2,3) in R but (3,2) is not in R then (3,2) will be in T.
So, T has all those pairs (b,a) such that (a,b) is in R but (b,a) is Not in R.
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