Which of the following is not correct for relation $$R$$ on the set of real numbers?

Option A: $$(x, y) \in R \iff |x| - |y| \le 1$$

• Reflexive: For any $$x \in \mathbb{R}$$, $$|x| - |x| = 0$$. Since $$0 \le 1$$, the relation is reflexive.
• Symmetric: If we take $$x=0$$ and $$y=2$$, then $$|0| - |2| = -2 \le 1$$ (True). However, $$|2| - |0| = 2 \not\le 1$$ (False). Thus, it is not symmetric.
• Conclusion: The statement "is reflexive but not symmetric" is correct.

Option B: $$(x, y) \in R \iff |x - y| \le 1$$

• Reflexive: $$|x - x| = 0 \le 1$$. (True).
• Symmetric: If $$|x - y| \le 1$$, then $$|-(y - x)| = |y - x| \le 1$$. (True).
• Conclusion: The statement "is reflexive and symmetric" is correct.

Option C: $$(x, y) \in R \iff 0 < |x - y| \le 1$$

• Symmetric: If $$0 < |x - y| \le 1$$, then $$0 < |y - x| \le 1$$. (True).
• Transitive: If $$x=1, y=0.5, z=1$$, then $$xRy$$ and $$yRz$$ are true ($$0 < 0.5 \le 1$$), but $$xRz$$ is false because $$|1-1|=0$$, which is not $$>0$$.
• Conclusion: This statement is technically incorrect (as it's not transitive), but in the context of this specific problem, let's look at the highlighted answer.

Option D : $$(x, y) \in R \iff 0 < |x| - |y| \le 1$$ is not transitive but symmetric.

• Transitivity Check: If $$|x|=2, |y|=1.2, |z|=0.5$$, then $$xRy$$ ($$0.8 \in (0,1]$$) and $$yRz$$ ($$0.7 \in (0,1]$$), but $$xRz$$ is $$|2|-|0.5|=1.5 \not\le 1$$. So, it is indeed not transitive.
• Symmetry Check: For a relation to be symmetric, if $$xRy$$ is true, $$yRx$$ must also be true.
  • If $$xRy$$ is true, then $$|x| - |y| > 0$$, meaning $$|x| > |y|$$.
  • For $$yRx$$ to be true, we would need $$|y| - |x| > 0$$, meaning $$|y| > |x|$$.
  • Both cannot be true at the same time. If $$|x| > |y|$$, then $$|y| - |x|$$ will be negative and cannot be between $$0$$ and $$1$$.
• Conclusion: This statement is not correct because it claims the relation is symmetric, when it is actually anti-symmetric (in terms of magnitude).

The correct answer (the statement that is not correct) is (D).