If N2 = N x N, N is set of natural numbers and R is relation on N2, s.t. RC N2…
2023
If N2 = N x N, N is set of natural numbers and R is relation on N2, s.t. RC N2 x N2 i.e. <x,y>R<u,v> ↔ xv = yu, then which of the followings are TRUE ?
(A) Reflexive
(B) Symmetric
(C) Transitive
(D) Assymmetric
Choose the correct answer from the options given below :
- A.
(A) and (B) Only
- B.
(B) and (C) Only
- C.
(A), (C) and (D) Only
- D.
(A), (B) and (C) Only
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Correct answer: D
Solution:
Reflexive: For any pair (x,y) in N², x*y = y*x, so (x,y) is related to itself. Hence the relation is reflexive.
Symmetric: If (x,y) is related to (u,v), then x*v = y*u. By commutativity of multiplication this equality is identical to u*y = v*x, so (u,v) is related to (x,y). Thus the relation is symmetric.
Transitive: Suppose (x,y) is related to (u,v) so x*v = y*u, and (u,v) is related to (p,q) so u*q = v*p. Multiply the first equality by q: x*v*q = y*u*q. Substitute u*q = v*p to get x*v*q = y*v*p. Cancel the common factor v (valid for positive natural numbers) to obtain x*q = y*p, so (x,y) is related to (p,q). Hence the relation is transitive.
Not antisymmetric: Antisymmetry would require that whenever two pairs relate to each other, they must be equal. This is false: for example, (1,2) and (2,4) satisfy 1*4 = 2*2, so they are related both ways, but (1,2) ≠ (2,4). Therefore the relation is not antisymmetric.
Conclusion: The relation is reflexive, symmetric and transitive, and not antisymmetric. So the correct selection is the one listing reflexive, symmetric and transitive only.