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​Let S be the set of all real numbers. Then, relation R = {(a, b ) : 1 + ab > 0} on S is
  • a)
    reflexive and symmetric but not transitive
  • b)
    reflexive and transitive but not symmetric
  • c)
    symmetric, transitive but not reflexive
  • d)
    None of the above is true
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
​Let S be the set of all real numbers. Then, relation R = {(a, b...
Since, 1 + a • a = 1 + a2 > 0,, Therefore, (a, a) ∈ R
So, R is reflexive.
Also, (a, b) ∈ R => 1 + ab > 0 => 1 + ba > 0
=> (b, a) ∈ R,
Hence, R is symmetric.
Since, (a, b) ∈ R and (b, c) ∈ R need
not imply (a, c) ∈ R.
Hence, R is not transitive
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Most Upvoted Answer
​Let S be the set of all real numbers. Then, relation R = {(a, b...
Explanation:

Reflexive:
- For a relation to be reflexive, (a, a) must be in the relation for all a in S.
- In this case, let's consider a = 0. When a = 0, we have 1 + 0*b = 1, which is greater than 0.
- Therefore, (0, 0) is in the relation R.
- Hence, the relation R is reflexive.

Symmetric:
- For a relation to be symmetric, if (a, b) is in the relation, then (b, a) must also be in the relation.
- Let's consider (1, 2) in R, since 1 + 1*2 = 3 > 0. Hence, (1, 2) is in R.
- However, 1 + 1*2 = 3 > 0, but 1 + 2*1 = 3 > 0, so (2, 1) is not in R.
- Therefore, the relation R is not symmetric.

Transitive:
- For a relation to be transitive, if (a, b) and (b, c) are in the relation, then (a, c) must also be in the relation.
- Let's consider (1, 2) and (2, 3) in R. We have 1 + 1*2 = 3 > 0 and 1 + 2*3 = 7 > 0.
- However, 1 + 1*3 = 4 > 0, but 4 is not greater than 0. Therefore, (1, 3) is not in R.
- Hence, the relation R is not transitive.
Therefore, the relation R = {(a, b) : 1 + ab > 0} on set S is reflexive and symmetric but not transitive, making option 'A' the correct answer.
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Community Answer
​Let S be the set of all real numbers. Then, relation R = {(a, b...
Since, 1 + a • a = 1 + a2 > 0,, Therefore, (a, a) ∈ R
So, R is reflexive.
Also, (a, b) ∈ R => 1 + ab > 0 => 1 + ba > 0
=> (b, a) ∈ R,
Hence, R is symmetric.
Since, (a, b) ∈ R and (b, c) ∈ R need
not imply (a, c) ∈ R.
Hence, R is not transitive
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​Let S be the set of all real numbers. Then, relation R = {(a, b ) : 1 + ab > 0} on S isa)reflexive and symmetric but not transitiveb)reflexive and transitive but not symmetricc)symmetric, transitive but not reflexived)None of the above is trueCorrect answer is option 'A'. Can you explain this answer?
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​Let S be the set of all real numbers. Then, relation R = {(a, b ) : 1 + ab > 0} on S isa)reflexive and symmetric but not transitiveb)reflexive and transitive but not symmetricc)symmetric, transitive but not reflexived)None of the above is trueCorrect answer is option 'A'. Can you explain this answer? for Mathematics 2024 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about ​Let S be the set of all real numbers. Then, relation R = {(a, b ) : 1 + ab > 0} on S isa)reflexive and symmetric but not transitiveb)reflexive and transitive but not symmetricc)symmetric, transitive but not reflexived)None of the above is trueCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for ​Let S be the set of all real numbers. Then, relation R = {(a, b ) : 1 + ab > 0} on S isa)reflexive and symmetric but not transitiveb)reflexive and transitive but not symmetricc)symmetric, transitive but not reflexived)None of the above is trueCorrect answer is option 'A'. Can you explain this answer?.
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