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Let S be the set of all real numbers and let R be a relation on S, defined by a R b ⇔ |a – b| < 1. Then, R is
- a)reflexive and symmetric but not transitive
- b)reflexive and transitive but not symmetric
- c)symmetric and transitive but not reflexive
- d)an equivalence relation.
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Let S be the set of all real numbers and let R be a relation on S, def...
(i) |a – a| = 0 < 1 is always true
(ii) a R b ⇒ |a – b| < 1 ⇒ |-(a – b)| < ⇒ |b – a| < 1 ⇒ b R a.
(iii) 2R 1 and
But, 2 is not related to 1/2. So, R is not transitive.
(ii) a R b ⇒ |a – b| < 1 ⇒ |-(a – b)| < ⇒ |b – a| < 1 ⇒ b R a.
(iii) 2R 1 and
But, 2 is not related to 1/2. So, R is not transitive.
Most Upvoted Answer
Let S be the set of all real numbers and let R be a relation on S, def...
aRb <=> 1+ab >0
a. is reflexive because a*a for any real number except 0 will be positive hence >0, and if a=0 then a*a + 1 >0.
b. if a*b + 1 > 0 then b*a + 1 will also be > 0, hence symmetric.
c. a=-2 b=0, -2*0 + 1 >0, ab+1 > 0
b=0, and if c=4 then 0*4 + 1 > 0
but a is not related to c, because a=-2, c=4, and -2*4 + 1 < 0
Hence, the given relation is reflexive and symmetric but not transitive.
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Community Answer
Let S be the set of all real numbers and let R be a relation on S, def...
Explanation:
To determine whether the relation R defined on the set of real numbers S is reflexive, symmetric, and transitive, we need to analyze each property individually.
Reflexive:
A relation R is reflexive if every element a in S is related to itself, i.e., a R a. In this case, for any real number a, |a - a| = 0 which is not greater than 1. Therefore, the relation R is reflexive.
Symmetric:
A relation R is symmetric if for every pair of elements a and b in S, if a R b, then b R a. In this case, if |a - b| is less than or equal to 1, then it follows that |b - a| is also less than or equal to 1. Therefore, the relation R is symmetric.
Transitive:
A relation R is transitive if for every triple of elements a, b, and c in S, if a R b and b R c, then a R c. In this case, if |a - b| and |b - c| are both less than or equal to 1, it does not necessarily imply that |a - c| is also less than or equal to 1. Consider the example where a = 0, b = 0.5, and c = 1. Here, |a - b| = 0.5 and |b - c| = 0.5, both of which are less than or equal to 1. However, |a - c| = 1, which is greater than 1. Therefore, the relation R is not transitive.
Conclusion:
Based on the analysis of the properties, the relation R is reflexive and symmetric but not transitive. Therefore, the correct answer is option 'A' - reflexive and symmetric but not transitive.
Additional Information:
An equivalence relation is a relation that is reflexive, symmetric, and transitive. Since the relation R defined in this case is not transitive, it does not satisfy all the properties required for an equivalence relation. Hence, the relation R is not an equivalence relation.
To determine whether the relation R defined on the set of real numbers S is reflexive, symmetric, and transitive, we need to analyze each property individually.
Reflexive:
A relation R is reflexive if every element a in S is related to itself, i.e., a R a. In this case, for any real number a, |a - a| = 0 which is not greater than 1. Therefore, the relation R is reflexive.
Symmetric:
A relation R is symmetric if for every pair of elements a and b in S, if a R b, then b R a. In this case, if |a - b| is less than or equal to 1, then it follows that |b - a| is also less than or equal to 1. Therefore, the relation R is symmetric.
Transitive:
A relation R is transitive if for every triple of elements a, b, and c in S, if a R b and b R c, then a R c. In this case, if |a - b| and |b - c| are both less than or equal to 1, it does not necessarily imply that |a - c| is also less than or equal to 1. Consider the example where a = 0, b = 0.5, and c = 1. Here, |a - b| = 0.5 and |b - c| = 0.5, both of which are less than or equal to 1. However, |a - c| = 1, which is greater than 1. Therefore, the relation R is not transitive.
Conclusion:
Based on the analysis of the properties, the relation R is reflexive and symmetric but not transitive. Therefore, the correct answer is option 'A' - reflexive and symmetric but not transitive.
Additional Information:
An equivalence relation is a relation that is reflexive, symmetric, and transitive. Since the relation R defined in this case is not transitive, it does not satisfy all the properties required for an equivalence relation. Hence, the relation R is not an equivalence relation.
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Let S be the set of all real numbers and let R be a relation on S, defined by a R b ⇔ |a – b| <1. Then, R isa)reflexive and symmetric but not transitiveb)reflexive and transitive but not symmetricc)symmetric and transitive but not reflexived)an equivalence relation.Correct answer is option 'A'. Can you explain this answer?
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Let S be the set of all real numbers and let R be a relation on S, defined by a R b ⇔ |a – b| <1. Then, R isa)reflexive and symmetric but not transitiveb)reflexive and transitive but not symmetricc)symmetric and transitive but not reflexived)an equivalence relation.Correct answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Let S be the set of all real numbers and let R be a relation on S, defined by a R b ⇔ |a – b| <1. Then, R isa)reflexive and symmetric but not transitiveb)reflexive and transitive but not symmetricc)symmetric and transitive but not reflexived)an equivalence relation.Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let S be the set of all real numbers and let R be a relation on S, defined by a R b ⇔ |a – b| <1. Then, R isa)reflexive and symmetric but not transitiveb)reflexive and transitive but not symmetricc)symmetric and transitive but not reflexived)an equivalence relation.Correct answer is option 'A'. Can you explain this answer?.
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