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ISRO | ISRO CS 2017 | Question 2
  • Last Updated : 15 Mar, 2018

Consider the set of integers I. Let D denote “divides with an integer quotient” (e.g. 4D8 but 4D7). Then D is

(A) Reflexive, not symmetric, transitive
(B) Not reflexive, not antisymmetric, transitive
(C) Reflexive, antisymmetric, transitive
(D) Not reflexive, not antisymmetric, not transitive

Answer: (B)

Explanation: Reflexibility: for all x ∈ I, R(x,x) is reflexive but here R(0,0) is a violation as 0 belongs to the set of integers but does not satisfy this relation.
Symmetric: for all x ∈ I, R(x,y)and R(y,x) is symmetric and clearly the above relation cannot be symmetric.Considering the example S={1,2}.
For this to be symmetric (1,2)(2,1) both ordered pair should be present but it is also a violation as 2 can be divided by 1 but 1 cannot be divided by 2 to give an integer quotient.
Antisymmetric: for all x ∈ I, R(x,y) and R(y,x) then x=y is antisymmetric. We can easily make a violation as R(-2,2) and R(2,-2) are not antisymmetric.
Transitivity:for all x ∈ I , R(x,y), R(y,z) then R(x,z). This is always true for the above divides relation.So The relation is not-reflexive, not-antisymmetric but transitive.

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