Let R_1 be a relation defined by R_1 = {(a, b | vedclass

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(B) $1$. Reflexivity: For any $a \in \mathbb{R}$,we have $a \geq a$. Thus,$(a, a) \in R_1$,so $R_1$ is reflexive.$2$. Symmetry: Consider $(2, 1) \in R

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Reflexive,transitive but not symmetric

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(B) $1$. Reflexivity: For any $a \in \mathbb{R}$,we have $a \geq a$. Thus,$(a, a) \in R_1$,so $R_1$ is reflexive.$2$. Symmetry: Consider $(2, 1) \in R_1$ because $2 \geq 1$. However,$(1, 2) 
otin R_1$ because $1 
geq 2$. Thus,$R_1$ is not symmetric.$3$. Transitivity: Let $(a, b) \in R_1$ and $(b, c) \in R_1$. This implies $a \geq b$ and $b \geq c$. By the transitive property of inequality,$a \geq c$. Therefore,$(a, c) \in R_1$,so $R_1$ is transitive.Conclusion: $R_1$ is reflexive and transitive,but not symmetric.

Let $R_1$ be a relation defined by $R_1 = \{(a, b) | a \geq b, a, b \in \mathbb{R}\}$. Then $R_1$ is

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