Question

Prove that the relation R in set of real number R defined as $R=\left\{\right(a,b):a\ge b\}$ is reflexive and transitive but not symmetric.

Answer 1

$aRb\iff a\ge b$

Reflexive:

$aRa\iff a\ge a$..... True

Therefore, the given relation $R$ is a reflexive relation.

Transitive:

$aRb\iff a\ge b\Rightarrow a-b=k$, where $k\ge 0$

$bRc\iff b\ge c\Rightarrow b-c=p$, where $p\ge 0$

Then,

$aRc\iff a\ge c$

Because $a-c=(a-b)+(b-c)=k+p\ge 0\Rightarrow a\ge c$.

Therefore, the given relation $R$ is a transitive relation.

Symmetric:

$aRb\iff a\ge b\Rightarrow a-b=k$, where $k\ge 0$

Then,

$bRa\iff b\ge a$

Because $b-a=-(a-b)=-k$ which implies that $b\le a$

Therefore, the given relation $R$ is not a symmetric relation.

Since, the given relation $R$ satisfies the reflexive and transitive relation properties but does not satisfies symmetric relation properties.