Key Concepts and Formulas

• Reflexive Relation: A relation $R$ on a set $S$ is reflexive if for every $x \in S$, $(x, x) \in R$.
• Symmetric Relation: A relation $R$ on a set $S$ is symmetric if for every $x, y \in S$, if $(x, y) \in R$, then $(y, x) \in R$.
• Transitive Relation: A relation $R$ on a set $S$ is transitive if for every $x, y, z \in S$, if $(x, y) \in R$ and $(y, z) \in R$, then $(x, z) \in R$.
• Properties of Absolute Value: $|a| \ge 0$ for all real $a$. $|a-b| = |b-a|$. $|a+b| \le |a| + |b|$ (Triangle Inequality).

Step-by-Step Solution

The question asks us to identify the statement that is not correct regarding the given relations on the set of real numbers ($\mathbb{R}$). We will analyze each option by checking the properties of the defined relation.

Option (A): $(x, y) \in R \Leftrightarrow 0 < |x| - |y| \le 1$. The statement claims this relation "is neither transitive nor symmetric."

• Checking for Symmetry: Assume $(x, y) \in R$. This means $0 < |x| - |y| \le 1$. This implies $|x| - |y| > 0$, so $|x| > |y|$. For symmetry, we would need $(y, x) \in R$, which means $0 < |y| - |x| \le 1$. This implies $|y| - |x| > 0$, so $|y| > |x|$. The conditions $|x| > |y|$ and $|y| > |x|$ cannot hold simultaneously. Let $x=2$ and $y=1.5$. Then $|x|-|y| = 2-1.5 = 0.5$, which satisfies $0 < 0.5 \le 1$. So, $(2, 1.5) \in R$. However, for $(y, x) = (1.5, 2)$, $|y|-|x| = 1.5-2 = -0.5$, which does not satisfy $0 < -0.5 \le 1$. Thus, $(1.5, 2) \notin R$. Therefore, the relation is not symmetric.

• Checking for Transitivity: Assume $(x, y) \in R$ and $(y, z) \in R$. This means $0 < |x| - |y| \le 1$ and $0 < |y| - |z| \le 1$. From these inequalities, we have $|x| > |y|$ and $|y| > |z|$, which implies $|x| > |z|$, so $|x| - |z| > 0$. Let's check if $|x| - |z| \le 1$. Consider $x=3, y=2.2, z=1.5$. For $(3, 2.2)$: $|3| - |2.2| = 3 - 2.2 = 0.8$. This satisfies $0 < 0.8 \le 1$, so $(3, 2.2) \in R$. For $(2.2, 1.5)$: $|2.2| - |1.5| = 2.2 - 1.5 = 0.7$. This satisfies $0 < 0.7 \le 1$, so $(2.2, 1.5) \in R$. Now consider $(x, z) = (3, 1.5)$: $|3| - |1.5| = 3 - 1.5 = 1.5$. This value $1.5$ does not satisfy $0 < 1.5 \le 1$. Thus, $(3, 1.5) \notin R$. Therefore, the relation is not transitive.

  Since the relation is neither symmetric nor transitive, the statement "is neither transitive nor symmetric" is correct.

Option (B): $(x, y) \in R \Leftrightarrow 0 < |x - y| \le 1$. The statement claims this relation "is symmetric and transitive."

• Checking for Symmetry: Assume $(x, y) \in R$. This means $0 < |x - y| \le 1$. Since $|x - y| = |-(y - x)| = |y - x|$, we have $0 < |y - x| \le 1$. Thus, $(y, x) \in R$. Therefore, the relation is symmetric.

• Checking for Transitivity: Assume $(x, y) \in R$ and $(y, z) \in R$. This means $0 < |x - y| \le 1$ and $0 < |y - z| \le 1$. We need to check if $(x, z) \in R$, i.e., if $0 < |x - z| \le 1$. Consider $x=2, y=1, z=0$. For $(2, 1)$: $|2 - 1| = 1$. This satisfies $0 < 1 \le 1$, so $(2, 1) \in R$. For $(1, 0)$: $|1 - 0| = 1$. This satisfies $0 < 1 \le 1$, so $(1, 0) \in R$. Now consider $(x, z) = (2, 0)$: $|2 - 0| = 2$. This value $2$ does not satisfy $0 < 2 \le 1$. Thus, $(2, 0) \notin R$. Therefore, the relation is not transitive.

  The statement claims the relation "is symmetric and transitive". Since the relation is not transitive, this statement is incorrect.

Option (C): $(x, y) \in R \Leftrightarrow |x| - |y| \le 1$. The statement claims this relation "is reflexive but not symmetric."

• Checking for Reflexivity: For $(x, x) \in R$, we need $|x| - |x| \le 1$. This simplifies to $0 \le 1$, which is true for all $x \in \mathbb{R}$. Therefore, the relation is reflexive.

• Checking for Symmetry: Assume $(x, y) \in R$. This means $|x| - |y| \le 1$. For symmetry, we would need $(y, x) \in R$, which means $|y| - |x| \le 1$. Consider $x=1$ and $y=3$. Then $|x|-|y| = 1-3 = -2$, which satisfies $-2 \le 1$. So, $(1, 3) \in R$. However, for $(y, x) = (3, 1)$, $|y|-|x| = 3-1 = 2$. This value $2$ does not satisfy $2 \le 1$. Thus, $(3, 1) \notin R$. Therefore, the relation is not symmetric.

  The statement claims the relation "is reflexive but not symmetric". Our analysis confirms this. Thus, the statement is correct.

Option (D): $(x, y) \in R \Leftrightarrow |x - y| \le 1$. The statement claims this relation "is reflexive and symmetric."

• Checking for Reflexivity: For $(x, x) \in R$, we need $|x - x| \le 1$. This simplifies to $0 \le 1$, which is true for all $x \in \mathbb{R}$. Therefore, the relation is reflexive.

• Checking for Symmetry: Assume $(x, y) \in R$. This means $|x - y| \le 1$. Since $|x - y| = |y - x|$, we have $|y - x| \le 1$. Thus, $(y, x) \in R$. Therefore, the relation is symmetric.

  The statement claims the relation "is reflexive and symmetric". Our analysis confirms this. Thus, the statement is correct.

Common Mistakes & Tips

• When checking for transitivity, remember that if the premise " $(x, y) \in R$ and $(y, z) \in R$ " is never met for any $x, y, z$, the relation is vacuously transitive. However, in these options, counterexamples can be found.
• Be careful with the strict inequality ($>$) versus non-strict inequality ($\ge$) in the definitions of the relations.
• Always test with specific numerical examples to confirm or deny properties like symmetry and transitivity.

Summary

We analyzed each of the four given relations and their associated statements for reflexivity, symmetry, and transitivity. Option (A) correctly states that the relation is neither transitive nor symmetric. Option (C) correctly states that the relation is reflexive but not symmetric. Option (D) correctly states that the relation is reflexive and symmetric. Option (B) claims the relation is symmetric and transitive. However, we found a counterexample showing it is not transitive, making this statement incorrect. The question asks for the not correct statement.

The final answer is $\boxed{B}$ which corresponds to option (B).