Key Concepts and Formulas

• Reflexive Relation: A relation $R$ on a set $A$ is reflexive if for every $a \in A$, $(a, a) \in R$.
• Symmetric Relation: A relation $R$ on a set $A$ is symmetric if for all $a, b \in A$, if $(a, b) \in R$, then $(b, a) \in R$.
• Transitive Relation: A relation $R$ on a set $A$ is transitive if for all $a, b, c \in A$, if $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$.
• Natural Numbers ($\mathbb{N}$): The set of positive integers $\{1, 2, 3, \ldots\}$.

Step-by-Step Solution

The relation $R$ is defined on the set $\mathbb{N} \times \mathbb{N}$. An element of this set is an ordered pair of natural numbers, e.g., $(x, y)$ where $x, y \in \mathbb{N}$. The relation is given by: $(x_1, y_1) \mathrm{R}\left(x_2, y_2\right) \quad \text{if and only if} \quad x_1 \leq x_2 \text{ or } y_1 \leq y_2$

Step 1: Check for Reflexivity To determine if $R$ is reflexive, we need to check if for any arbitrary element $(x, y) \in \mathbb{N} \times \mathbb{N}$, the condition $(x, y) \mathrm{R} (x, y)$ holds. According to the definition of $R$, $(x, y) \mathrm{R} (x, y)$ is true if $x \leq x$ or $y \leq y$. For any natural number $x$, the inequality $x \leq x$ is always true. Similarly, $y \leq y$ is always true for any natural number $y$. Since at least one of the conditions ($x \leq x$ or $y \leq y$) is true (in fact, both are true), the compound statement is true. Thus, for every $(x, y) \in \mathbb{N} \times \mathbb{N}$, $(x, y) \mathrm{R} (x, y)$ is true. Therefore, the relation $R$ is reflexive.

Step 2: Check for Symmetry To determine if $R$ is symmetric, we need to check if for any two elements $(x_1, y_1)$ and $(x_2, y_2)$ in $\mathbb{N} \times \mathbb{N}$, if $(x_1, y_1) \mathrm{R} (x_2, y_2)$ is true, then $(x_2, y_2) \mathrm{R} (x_1, y_1)$ must also be true. We can try to find a counterexample. Let's choose $(x_1, y_1) = (2, 1)$ and $(x_2, y_2) = (1, 2)$. First, let's check if $(2, 1) \mathrm{R} (1, 2)$ is true. The condition for this is $x_1 \leq x_2$ or $y_1 \leq y_2$. Substituting the values, we get $2 \leq 1$ or $1 \leq 2$. The statement $2 \leq 1$ is false. The statement $1 \leq 2$ is true. Since $1 \leq 2$ is true, the condition "$2 \leq 1$ or $1 \leq 2$" is true. So, $(2, 1) \mathrm{R} (1, 2)$ is true.

Now, let's check if $(1, 2) \mathrm{R} (2, 1)$ is true. The condition for this is $x_2 \leq x_1$ or $y_2 \leq y_1$. Substituting the values, we get $1 \leq 2$ or $2 \leq 1$. The statement $1 \leq 2$ is true. The statement $2 \leq 1$ is false. Since $1 \leq 2$ is true, the condition "$1 \leq 2$ or $2 \leq 1$" is true. So, $(1, 2) \mathrm{R} (2, 1)$ is true.

Let's try another counterexample to disprove symmetry. Let $(x_1, y_1) = (1, 2)$ and $(x_2, y_2) = (3, 1)$. Check $(1, 2) \mathrm{R} (3, 1)$: $1 \leq 3$ or $2 \leq 1$. Since $1 \leq 3$ is true, $(1, 2) \mathrm{R} (3, 1)$ is true. Check $(3, 1) \mathrm{R} (1, 2)$: $3 \leq 1$ or $1 \leq 2$. Since $1 \leq 2$ is true, $(3, 1) \mathrm{R} (1, 2)$ is true.

Let's try a counterexample where one condition holds for the first pair, and we want to see if the reverse holds. Consider $(x_1, y_1) = (1, 2)$ and $(x_2, y_2) = (3, 4)$. Check $(1, 2) \mathrm{R} (3, 4)$: $1 \leq 3$ or $2 \leq 4$. Both are true, so the statement is true. Now, check $(3, 4) \mathrm{R} (1, 2)$: $3 \leq 1$ or $4 \leq 2$. $3 \leq 1$ is false. $4 \leq 2$ is false. Since both parts of the 'or' are false, the statement "$3 \leq 1$ or $4 \leq 2$" is false. Thus, we have found a case where $(1, 2) \mathrm{R} (3, 4)$ is true, but $(3, 4) \mathrm{R} (1, 2)$ is false. Therefore, the relation $R$ is not symmetric.

Step 3: Check for Transitivity To determine if $R$ is transitive, we need to check if for any three elements $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ in $\mathbb{N} \times \mathbb{N}$, if $(x_1, y_1) \mathrm{R} (x_2, y_2)$ and $(x_2, y_2) \mathrm{R} (x_3, y_3)$ are both true, then $(x_1, y_1) \mathrm{R} (x_3, y_3)$ must also be true. We can try to find a counterexample. Let $(x_1, y_1) = (3, 9)$, $(x_2, y_2) = (4, 6)$, and $(x_3, y_3) = (2, 7)$.

First, check if $(x_1, y_1) \mathrm{R} (x_2, y_2)$, i.e., $(3, 9) \mathrm{R} (4, 6)$. The condition is $3 \leq 4$ or $9 \leq 6$. $3 \leq 4$ is true. $9 \leq 6$ is false. Since $3 \leq 4$ is true, the statement "$3 \leq 4$ or $9 \leq 6$" is true. So, $(3, 9) \mathrm{R} (4, 6)$ is true.

Next, check if $(x_2, y_2) \mathrm{R} (x_3, y_3)$, i.e., $(4, 6) \mathrm{R} (2, 7)$. The condition is $4 \leq 2$ or $6 \leq 7$. $4 \leq 2$ is false. $6 \leq 7$ is true. Since $6 \leq 7$ is true, the statement "$4 \leq 2$ or $6 \leq 7$" is true. So, $(4, 6) \mathrm{R} (2, 7)$ is true.

Now, we must check if $(x_1, y_1) \mathrm{R} (x_3, y_3)$ is true, i.e., $(3, 9) \mathrm{R} (2, 7)$. The condition is $3 \leq 2$ or $9 \leq 7$. $3 \leq 2$ is false. $9 \leq 7$ is false. Since both parts of the 'or' are false, the statement "$3 \leq 2$ or $9 \leq 7$" is false. So, $(3, 9) \mathrm{R} (2, 7)$ is false.

We have found a case where $(3, 9) \mathrm{R} (4, 6)$ and $(4, 6) \mathrm{R} (2, 7)$ are both true, but $(3, 9) \mathrm{R} (2, 7)$ is false. Therefore, the relation $R$ is not transitive.

Step 4: Evaluate the Statements Statement (I): $R$ is reflexive but not symmetric. From Step 1, $R$ is reflexive. From Step 2, $R$ is not symmetric. Thus, Statement (I) is true.

Statement (II): $R$ is transitive. From Step 3, $R$ is not transitive. Thus, Statement (II) is false.

Common Mistakes & Tips

• Confusing "and" and "or": The definition of the relation uses "or", which means the condition is satisfied if at least one part is true. This is crucial for checking transitivity.
• General vs. Specific: To prove reflexivity or transitivity, you must consider general cases. To disprove symmetry or transitivity, a single specific counterexample is sufficient.
• Natural Numbers: Remember that $\mathbb{N}$ usually refers to positive integers $\{1, 2, 3, \ldots\}$. The properties of inequalities hold for these numbers.

Summary We analyzed the given relation $R$ on $\mathbb{N} \times \mathbb{N}$ by checking its reflexivity, symmetry, and transitivity. We found that $R$ is reflexive because for any pair $(x, y)$, $x \leq x$ or $y \leq y$ is always true. We demonstrated that $R$ is not symmetric by providing a counterexample where $(1, 2) \mathrm{R} (3, 4)$ but $(3, 4) \mathrm{R} (1, 2)$ is false. Similarly, we showed $R$ is not transitive with a counterexample involving three pairs. Based on these findings, statement (I) "R is reflexive but not symmetric" is true, and statement (II) "R is transitive" is false. Therefore, only statement (I) is correct.

The final answer is $\boxed{D}$.