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Let R be a relation on the set N of natural numbers defined by nRm
⇔ n is a factor of m (i.e. n(m). Then R is 
  • a)
    Reflexive and symmetric
  • b)
    Transitive and symmetric
  • c)
    Equivalence
  • d)
    Reflexive, transitive but not symmetric
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
Let R be a relation on the set N of natural numbers defined by nRm&hAr...
Since n | n for all n in N,
therefore R is reflexive.
Since 2 | 6 but 6 | 2, therefore R is not symmetric.
Let n R m and m R p ⇒ n|m and m|p ⇒ n|p ⇒ nRp
So. R is transitive. 
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Let R be a relation on the set N of natural numbers defined by nRm⇔ n is a factor of m (i.e. n(m). Then R isa)Reflexive and symmetricb)Transitive and symmetricc)Equivalenced)Reflexive, transitive but not symmetricCorrect answer is option 'D'. Can you explain this answer?
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Let R be a relation on the set N of natural numbers defined by nRm⇔ n is a factor of m (i.e. n(m). Then R isa)Reflexive and symmetricb)Transitive and symmetricc)Equivalenced)Reflexive, transitive but not symmetricCorrect answer is option 'D'. Can you explain this answer? for CA Foundation 2024 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about Let R be a relation on the set N of natural numbers defined by nRm⇔ n is a factor of m (i.e. n(m). Then R isa)Reflexive and symmetricb)Transitive and symmetricc)Equivalenced)Reflexive, transitive but not symmetricCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for CA Foundation 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let R be a relation on the set N of natural numbers defined by nRm⇔ n is a factor of m (i.e. n(m). Then R isa)Reflexive and symmetricb)Transitive and symmetricc)Equivalenced)Reflexive, transitive but not symmetricCorrect answer is option 'D'. Can you explain this answer?.
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