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Please Solve RD Sharma Class 12 Chapter Relation Exercise 1.2 Question 15 Subquestion (ii) Maths Textbook Solution.

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Given:Hence \: proved, R\cup S\: is \: re\! flexive.
Hint:(a, a) \, \epsilon \, R \, \forall\, a \, \epsilon\, X\, or\, as\, I \subseteq R \ , \text{where, I, is the identity relation on A.}
Given:   R and S are two relations on A such that R is reflexive.
To Prove:R\cup S\: is\: re\! f\! lexive.
Suppose \: R\cup S\: is\: not\: re\! f\! lexive.
\\\text{This means that there is and }R\cup S\: \text{ such that }(a, a) \: \notin\: R\cup S       
Since, a \, \epsilon \: R\cup S,
a \: \epsilon \: R \: or \: a\: \epsilon \: S
I\! f \: a \: \epsilon \: R, then (a, a) \: \epsilon \: R [R\: is\: re\! flexive]
(a,a) \: \epsilon \: R\cup S
Hence, R\cup S\: is \: re\! flexive.

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