Let a relation R on N times N be defined as: | vedclass

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(B) $1$. Reflexivity: For any $(x, y) \in N 	imes N$,we have $x \leq x$ or $y \leq y$. Thus,$((x, y), (x, y)) \in R$. So,$R$ is reflexive.$2$. Symmet

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Only $(I)$ is correct.

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(B) $1$. Reflexivity: For any $(x, y) \in N 	imes N$,we have $x \leq x$ or $y \leq y$. Thus,$((x, y), (x, y)) \in R$. So,$R$ is reflexive.$2$. Symmetry: Consider $((1, 2), (2, 1))$. Here $1 \leq 2$ is true,so $((1, 2), (2, 1)) \in R$. However,for $((2, 1), (1, 2))$,$2 \leq 1$ is false and $1 \leq 2$ is true. Wait,let us check another example: $((1, 3), (2, 2))$. $1 \leq 2$ is true,so it is in $R$. But for $((2, 2), (1, 3))$,$2 \leq 1$ is false and $2 \leq 3$ is true. Let us take $((2, 3), (1, 1))$. $2 \leq 1$ is false and $3 \leq 1$ is false. So $((2, 3), (1, 1)) 
otin R$. Since $((1, 1), (2, 3)) \in R$ but $((2, 3), (1, 1)) 
otin R$,$R$ is not symmetric.$3$. Transitivity: Consider $A = (2, 4)$,$B = (3, 3)$,and $C = (1, 3)$.$(A, B) \in R$ because $2 \leq 3$ is true.$(B, C) \in R$ because $3 \leq 3$ is true.However,for $(A, C) = ((2, 4), (1, 3))$,$2 \leq 1$ is false and $4 \leq 3$ is false. Thus,$(A, C) 
otin R$. So,$R$ is not transitive.Therefore,only statement $(I)$ is correct.

Let a relation $R$ on $N \times N$ be defined as: $(x_1, y_1) R (x_2, y_2)$ if and only if $x_1 \leq x_2$ or $y_1 \leq y_2$. Consider the two statements:
$(I)$ $R$ is reflexive but not symmetric.
$(II)$ $R$ is transitive.
Then which one of the following is true?

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Let a relation $R$ on $N \times N$ be defined as: $(x_1, y_1) R (x_2, y_2)$ if and only if $x_1 \leq x_2$ or $y_1 \leq y_2$. Consider the two statements:$(I)$ $R$ is reflexive but not symmetric.$(II)$ $R$ is transitive.Then which one of the following is true?