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Official PYQ · 2006

GATE CS 2006 Q4 — Sets, Relations and Functions

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A relation RR is defined on ordered pairs of integers as follows: (x,y)R(u,v) if x<u and y>v(x,y)R(u,v) \text{ if } x<u \text{ and } y>v Then RR is:

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Answer: A

To determine the nature of the relation RR defined on Z×Z\mathbb{Z} \times \mathbb{Z}, let us analyze its properties.

The relation is defined as:
(x,y)R(u,v)  ⟺  x<u and y>v(x,y)R(u,v) \iff x < u \text{ and } y > v

1. Reflexivity

For RR to be reflexive, every element (x,y)(x, y) must be related to itself, i.e., (x,y)R(x,y)(x, y) R (x, y) must hold for all x,y∈Zx, y \in \mathbb{Z}.

Applying the definition:
(x,y)R(x,y)  ⟺  x<x and y>y(x, y) R (x, y) \iff x < x \text{ and } y > y

Since x<xx < x is always false for any integer xx, the relation RR is not reflexive (in fact, it is irreflexive).

Conclusion on Reflexivity:

  • A Partial Order must be reflexive, antisymmetric, and transitive.
  • An Equivalence Relation must be reflexive, symmetric, and transitive.

Since RR is not reflexive, it can be neither a partial order nor an equivalence relation.

2. Transitivity (Optional verification)

If (x,y)R(u,v)(x, y) R (u, v) and (u,v)R(w,z)(u, v) R (w, z), then:
x<u and y>vx < u \text{ and } y > v
u<w and v>zu < w \text{ and } v > z

By the transitive property of inequality on integers:
x<u<w  ⟹  x<wx < u < w \implies x < w
y>v>z  ⟹  y>zy > v > z \implies y > z

Thus, (x,y)R(w,z)(x, y) R (w, z), which means RR is transitive. However, the lack of reflexivity prevents it from being a partial order or an equivalence relation.

Therefore, the correct option is A.

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