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(B) Reflexive: $A$ relation $r$ is reflexive if $(a, a) \in r$ for all $a \in R$. This means $a \cdot a = a^2$ must be an irrational number for all $a

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symmetric only

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(B) Reflexive: $A$ relation $r$ is reflexive if $(a, a) \in r$ for all $a \in R$. This means $a \cdot a = a^2$ must be an irrational number for all $a \in R$. If $a = 1$,$a^2 = 1$,which is rational. Thus,$r$ is not reflexive.Symmetric: $A$ relation $r$ is symmetric if $(a, b) \in r \implies (b, a) \in r$. If $ab$ is an irrational number,then $ba$ is also an irrational number because multiplication is commutative in $R$. Thus,$r$ is symmetric.Transitive: $A$ relation $r$ is transitive if $(a, b) \in r$ and $(b, c) \in r \implies (a, c) \in r$. Consider $a = 1, b = \sqrt{2}, c = 2$. Here,$ab = 1 \cdot \sqrt{2} = \sqrt{2}$ (irrational) and $bc = \sqrt{2} \cdot 2 = 2\sqrt{2}$ (irrational). However,$ac = 1 \cdot 2 = 2$,which is a rational number. Thus,$(1, \sqrt{2}) \in r$ and $(\sqrt{2}, 2) \in r$,but $(1, 2) 
otin r$. Therefore,$r$ is not transitive.Conclusion: The relation $r$ is symmetric only.

Let $r$ be a relation from $R$ (set of real numbers) to $R$ defined by $r = \{(x, y) \mid x, y \in R \text{ and } xy \text{ is an irrational number}\}$. Then,the relation $r$ is:

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