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September 7, 2024Computer-Networks
September 7, 2024GATE 2001
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Question 2
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Consider the following relations:
- R1(a,b) iff (a+b) is even over the set of integers
R2(a,b) iff (a+b) is odd over the set of integers
R3(a,b) iff a.b > 0 over the set of non-zero rational numbers
R4(a,b) iff |a-b| <= 2 over the set of natural numbers
Which of the following statements is correct?
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R1 and R2 are equivalence relations, R3 and R4 are not
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R1 and R3 are equivalence relations, R2 and R4 are not
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R1 and R4 are equivalence relations, R2 and R3 are not
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R1, R2, R3 and R4 are all equivalence relations
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Question 2 Explanation:
R1:
a+a = 2a
The set (a,a) is reflexive.
The set representation (a,a), (b,b) is also Symmetric.
The set representation is Transitive.
So, this is Equivalence.
R2:
a+a ≠ 2a
The (a,a) is non-reflexive.
R3:
a⋅a>0 → Reflexive
a⋅b>0 and b⋅a>0 → Symmetric
a⋅b>0, b⋅c>0 then c⋅a>0 → Transitive
The relation R3 is equivalence relation.
R4:
|a – a| ≤ 2, which is not possible, not Reflexive.
R1 & R3 are equivalence,
R2 & R4 are not.
a+a = 2a
The set (a,a) is reflexive.
The set representation (a,a), (b,b) is also Symmetric.
The set representation is Transitive.
So, this is Equivalence.
R2:
a+a ≠ 2a
The (a,a) is non-reflexive.
R3:
a⋅a>0 → Reflexive
a⋅b>0 and b⋅a>0 → Symmetric
a⋅b>0, b⋅c>0 then c⋅a>0 → Transitive
The relation R3 is equivalence relation.
R4:
|a – a| ≤ 2, which is not possible, not Reflexive.
R1 & R3 are equivalence,
R2 & R4 are not.
Correct Answer: B
Question 2 Explanation:
R1:
a+a = 2a
The set (a,a) is reflexive.
The set representation (a,a), (b,b) is also Symmetric.
The set representation is Transitive.
So, this is Equivalence.
R2:
a+a ≠ 2a
The (a,a) is non-reflexive.
R3:
a⋅a>0 → Reflexive
a⋅b>0 and b⋅a>0 → Symmetric
a⋅b>0, b⋅c>0 then c⋅a>0 → Transitive
The relation R3 is equivalence relation.
R4:
|a – a| ≤ 2, which is not possible, not Reflexive.
R1 & R3 are equivalence,
R2 & R4 are not.
a+a = 2a
The set (a,a) is reflexive.
The set representation (a,a), (b,b) is also Symmetric.
The set representation is Transitive.
So, this is Equivalence.
R2:
a+a ≠ 2a
The (a,a) is non-reflexive.
R3:
a⋅a>0 → Reflexive
a⋅b>0 and b⋅a>0 → Symmetric
a⋅b>0, b⋅c>0 then c⋅a>0 → Transitive
The relation R3 is equivalence relation.
R4:
|a – a| ≤ 2, which is not possible, not Reflexive.
R1 & R3 are equivalence,
R2 & R4 are not.
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