Kenneth Rosen Edition 6th Exercise 7.1 Question 48 (Page No. 473)

Question

Suppose that $R$ and $S$ are reflexive relations on a set A.Are the below statements true or false?

(a) $R\, \cup \, S$ is reflexive

(b)$R\, \cap \, S$ is reflexive

(c)$R\, \oplus \, S$ is irreflexive

(d)$R\, - \, S$ is irreflexive

(e)$SoR$ is reflexive.

My Answers are

(a)-(e)-All true.
Are my answers correct?

Comments

R={(1,1),(2,2),(3,3)} and S={(1,1)}
 

S is still not reflexive.

For a Relation $R$ on set $A$ to be Reflexive, $(a,a) \in R$ , $\forall a \in A$ 

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I  understood, as both relation are on A,so both relation contain all elements of A too.

It is not possible one relation contain some element of A and other contain some elements of A

I mistaken because I thought , as both R and S different relation, their dependency on set A could be different.

But that is not true.

Dependency on set A will be same for both the relation

right?

@Deepak

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yes @srestha mam,  those two relations R and S formed on A only

Answer 1

Lets take A ={1,2,3,4}

R will be = {(1,1),(2,2),(3,3),(4,4)} as minimum elements

S will be = {(1,1),(2,2),(3,3),(4,4)} as minimum elements

 

(a) R∪S is reflexive:True since, it will defiantly contain {(1,1),(2,2),(3,3),(4,4)}. 

(b) R∩S is reflexive: True since, it will defiantly contain {(1,1),(2,2),(3,3),(4,4)}. 
 

(c) R⊕S is irreflexive:  Since common elements in both R and S {(1,1),(2,2),(3,3),(4,4)} will be removed , Which is basic requirement of Reflexive relation , Hence, irreflexive .  
 

(d) R−S is irreflexive:  Since common elements in both R and S {(1,1),(2,2),(3,3),(4,4)} will be removed , Which is basic requirement of Reflexive relation , Hence, irreflexive . 
 

(e) SoR is reflexive:  will contain {(1,1),(2,2),(3,3),(4,4)}  Which is basic requirement of Reflexive relation , Hence, Reflexive .