Show that the relation


Show that the relation '≥' on the set R of all real numbers is reflexive and transitive but not symmetric.


Let $R$ be the set such that $R=\{(a, b): a, b \in R ; a \geq b\}$


Let $a$ be an arbitrary element of $R$.

$\Rightarrow a \in R$

$\Rightarrow a=a$

$\Rightarrow a \geq a$ is true for $a=a$

$\Rightarrow(a, a) \in R$

Hence, $R$ is reflexive.


Let $(a, b) \in R$

$\Rightarrow a \geq b$ is same as $b \leq a$, but not $b \geq a$

Thus, $(b, a) \notin R$

Hence, $R$ is not symmetric.


Let $(a, b)$ and $(b, c) \in R$

$\Rightarrow a \geq b$ and $b \geq c$

$\Rightarrow a \geq b \geq c$

$\Rightarrow a \geq c$

$\Rightarrow(a, c) \in R$

Hence, $R$ is transitive.

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