The relation R on the set of real numbers defined by $R = \{(a, b): a \leq b^2\}$ is

The relation R on the set of real numbers is defined as $R = \{(a, b): a \leq b^2\}$.

Two numbers $a$ and $b$ are related if $a$ is less than or equal to $b$ squared.

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A relation is reflexive if $(a, a) \in R$ for all real numbers $a$.

For the relation R, this requires $a \leq a^2$ to be true for all real numbers.

Testing with $a = 0.5$:

$0.5 \leq (0.5)^2$

$0.5 \leq 0.25$

This is false.

The relation is not reflexive.

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A relation is symmetric if whenever $(a, b) \in R$, then $(b, a) \in R$.

For the relation R, if $a \leq b^2$, then $b \leq a^2$ must also be true.

Testing with $a = 1$ and $b = 2$:

Check $(1, 2)$: $1 \leq 2^2 = 4$ is true.

Check $(2, 1)$: $2 \leq 1^2 = 1$ is false.

Since $(1, 2) \in R$ but $(2, 1) \notin R$, the relation is not symmetric.

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A relation is transitive if whenever $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$.

For the relation R, if $a \leq b^2$ and $b \leq c^2$, then $a \leq c^2$ must be true.

Testing with $a = 4$, $b = -2$, $c = 1$:

Check $(4, -2)$: $4 \leq (-2)^2 = 4$ is true.

Check $(-2, 1)$: $-2 \leq 1^2 = 1$ is true.

Check $(4, 1)$: $4 \leq 1^2 = 1$ is false.

Since $(4, -2) \in R$ and $(-2, 1) \in R$, but $(4, 1) \notin R$, the relation is not transitive.

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The relation R is neither reflexive nor symmetric nor transitive.