Let {R_1} be a relation defined by {R_1} = { | vedclass

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(B) For any $a \in R$,we have $a \ge a$,so $(a, a) \in R_1$. Thus,$R_1$ is reflexive.For symmetry,consider $(2, 1) \in R_1$ because $2 \ge 1$. However

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Reflexive,transitive but not symmetric

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(B) For any $a \in R$,we have $a \ge a$,so $(a, a) \in R_1$. Thus,$R_1$ is reflexive.For symmetry,consider $(2, 1) \in R_1$ because $2 \ge 1$. However,$(1, 2) 
otin R_1$ because $1 
ot\ge 2$. Thus,$R_1$ is not symmetric.For transitivity,let $(a, b) \in R_1$ and $(b, c) \in R_1$. This implies $a \ge b$ and $b \ge c$. By the transitive property of inequality,$a \ge c$,which means $(a, c) \in R_1$. Thus,$R_1$ is transitive.Therefore,$R_1$ is reflexive and transitive but not symmetric.

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

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