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

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(B) $1$. Reflexivity: For any $a \in R$,we know $a \ge a$ is always true. Thus,$(a, a) \in R_1$ for all $a \in R$. So,$R_1$ is reflexive.$2$. Symmetry

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Reflexive,transitive,but not symmetric

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(B) $1$. Reflexivity: For any $a \in R$,we know $a \ge a$ is always true. Thus,$(a, a) \in R_1$ for all $a \in R$. So,$R_1$ is reflexive.$2$. Symmetry: If $(a, b) \in R_1$,then $a \ge b$. This does not imply $b \ge a$ (e.g.,$2 \ge 1$ is true,but $1 \ge 2$ is false). Thus,$R_1$ is not symmetric.$3$. Transitivity: If $(a, b) \in R_1$ and $(b, c) \in R_1$,then $a \ge b$ and $b \ge c$. By the transitive property of inequality,$a \ge c$. Thus,$(a, c) \in R_1$. So,$R_1$ is transitive.Conclusion: $R_1$ is reflexive and transitive,but not symmetric.

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

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