Relations
The relation R on the set {1, 2, 3, ..., 10} defined by R = {(x, y) : x and y have exactly one common factor other than 1} is
Select the correct option:
Solution
Symmetric
- Test Reflexive: For (x,x) to be in R, x must have exactly one factor other than 1. This would only be true for prime numbers. For x=4, factors are {1,2,4} (two factors other than 1). Thus, not reflexive for all x.
- Test Symmetric: If (x,y) is a pair where they share exactly one prime factor, then surely (y,x) shares that same property. Symmetric (Yes).
- Test Transitive:
- Let x=4,y=6. Common factor is {2}. So (4,6)∈R.
- Let y=6,z=9. Common factor is {3}. So (6,9)∈R.
- For transitivity, (4,9) must be in R. But 4,9 have zero common factors other than 1. Not transitive.
- Conclusion: Symmetric only.
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About This Question
- Subject
- mathematics
- Chapter
- sets, relations and functions
- Topic
- relations
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
Symmetric
- Test Reflexive: For (x,x) to be in R, x must have exactly one factor other than 1. This would only be true for prime numbers. For x=4, factors are {1,2,4} (two factors other than 1). Thus, not reflexive for all x.
- Test Symmetric: If (x,y) is a pair where they share exactly one prime factor, then surely (y,x) shares that same property. Symmetric (Yes).
- Test Transitive:
- Let x=4,y=6. Common factor is {2}. So (4,6)∈R.
- Let y=6,z=9. Common factor is {3}. So (6,9)∈R.
- For transitivity, (4,9) must be in R. But 4,9 have zero common factors other than 1. Not transitive.
- Conclusion: Symmetric only.
This hard difficulty mathematics question is from the chapter sets, relations and functions, covering the topic of relations. It appeared in the 2025 exam.
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