Relations
Let R be a relation on Z (set of integers) defined by R = {(a, b) : a - b is divisible by 5}. Then R is
Select the correct option:
Solution
An equivalence relation
- Reflexive: a−a=0. Since 0 is divisible by any non-zero integer 5, (a,a)∈R. (Reflexive)
- Symmetric: If (a,b)∈R, then a−b=5k.
- Then b−a=−(a−b)=−5k=5(−k).
- Since −k is an integer, (b,a)∈R. (Symmetric)
- Transitive: If (a,b)∈R and (b,c)∈R, then a−b=5k1 and b−c=5k2.
- Summing these: (a−b)+(b−c)=5k1+5k2⟹a−c=5(k1+k2).
- Since k1+k2 is an integer, (a,c)∈R. (Transitive)
- Conclusion: R is an equivalence relation (the 'congruence modulo 5' relation).
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About This Question
- Subject
- mathematics
- Chapter
- sets, relations and functions
- Topic
- relations
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
An equivalence relation
- Reflexive: a−a=0. Since 0 is divisible by any non-zero integer 5, (a,a)∈R. (Reflexive)
- Symmetric: If (a,b)∈R, then a−b=5k.
- Then b−a=−(a−b)=−5k=5(−k).
- Since −k is an integer, (b,a)∈R. (Symmetric)
- Transitive: If (a,b)∈R and (b,c)∈R, then a−b=5k1 and b−c=5k2.
- Summing these: (a−b)+(b−c)=5k1+5k2⟹a−c=5(k1+k2).
- Since k1+k2 is an integer, (a,c)∈R. (Transitive)
- Conclusion: R is an equivalence relation (the 'congruence modulo 5' relation).
This medium 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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