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GATE | GATE-CS-2016 (Set 2) | Question 36

  • Difficulty Level : Easy
  • Last Updated : 28 Jun, 2021

A binary relation R on N x N is defined as follows:

(a, b) R (c, d) if a <= c or b <= d.

Consider the following propositions:

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P: R is reflexive
Q: R is transitive

Which one of the following statements is TRUE?
(A) Both P and Q are true.
(B) P is true and Q is false.
(C) P is false and Q is true.
(D) Both P and Q are false.

Answer: (B)

Explanation: THEORY:


A relation ‘R’ on a set ‘A’ is said to be reflexive if, (xRx) for every x£A.
Ex. If A= {1,2}
And R, and P be a relation on AxA, defined as,
R= {(2,2),(1,1)} => R is reflexive as it contains all pair of type (xRx).
P= {(1,1)} => P is not reflexive relation on A, as it doesn’t contain (2,2).



A relation ‘R’ on set ‘A’ is said to be transitive if (xRy) and (yRz), then (xRz) for every x,y,z £A.
Ex: if A= {1,2}
Let R be a relation on AxA, defined as,
R= {(1,1),(1,2),(2,1)} => R is transitive.



Given, (a, b) R (c, d) if a <= c or b <= d

i.Check for reflexivity:

if an element of set be (a,b) then, (a,b)R(a,b) should hold true.
Here, a<=a or b<=b.
So, (a,b)R(a,b) holds true.

Hence, ‘R’ is reflexive.

ii. Check for transitivity:

 if elements of set be (2,3),(3,1) and(1,1)
Then, (2,3)R(3,1) as 2<=3
And (3,1)R(1,1) as 1<=1
But (2,3)R(1,1) doesn’t hold true as 2>=1 and 3>=1.

Hence, R is reflexive but not transitive.


This solution is contributed by Sandeep pandey.

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