Show that the relation $R$ in the set $R$ of real numbers, defined as $R = {(a, b) : a \leq b^{2}}$ is neither reflexive nor symmetric nor transitive

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Solution

$a R b \Leftrightarrow a \leq b^{2}$

Reflexive:
$a R a \Leftrightarrow a \leq a^{2}$ ..... False
Because relation $R$ varies on real numbers
Therefore, when $a \in (0, 1)$ it fails $(∵ a > a^{2}, a \in (0, 1))$
Therefore, the given relation $R$ is not a reflexive relation.

Transitive:
$a R b \Leftrightarrow a \leq b^{2}$
$b R c \Leftrightarrow b \leq c^{2}$
Then,
$a R c \Leftrightarrow a \leq c^{2}$
Because relation $R$ varies on real numbers
Lets assume, $a = 2, b = - 4, c = 1$
$a R b$ and $b R c$ are satisfied as
$a R b : 2 < 16,$ and $b R c : - 4 < 1$
But, $a R c : 2 > 1$
Therefore, the given relation $R$ is not a transitive relation.

Symmetric:
$a R b \Leftrightarrow a \leq b$
Then,
$b R a \Leftrightarrow b \leq a$
Because relation $R$ varies on real numbers
Lets assume, $a = - 2, b = 5$
$a R b : - 2 < 25$
But, $b R a : 5 > 4$
Therefore, the given relation $R$ is not a symmetric relation.

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

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