Question:
Show that the relation '≥' on the set R of all real numbers is reflexive and transitive but not symmetric.
Solution:
Let $R$ be the set such that $R=\{(a, b): a, b \in R ; a \geq b\}$
Reflexivity:
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.
Symmetry:
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.
Transitivity:
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.