check whether the relation R in …

Let's check the properties of the relation $R$ defined in $\mathbb{R}$ (the set of real numbers) by $R = \{(a, b) : a \times b \text{ is an irrational number}\}$: 1. **Reflexive:** A relation $R$ is reflexive if for every element $a$ in the set $A$, the pair $(a, a)$ belongs to $R$. In this case, for any real number $a$, $a \times a = a^2$ is always a real number, not necessarily irrational. So, $R$ is not reflexive. 2. **Symmetric:** A relation $R$ is symmetric if for every pair $(a, b)$ in $R$, the pair $(b, a)$ also belongs to $R$. If $a \times b$ is irrational, it doesn't imply that $b \times a$ is irrational. For example, $\sqrt{2} \times \sqrt{3}$ is irrational, but $\sqrt{3} \times \sqrt{2}$ is also irrational. So, $R$ is symmetric. 3. **Transitive:** A relation $R$ is transitive if for every pair $(a, b)$ and $(b, c)$ in $R$, the pair $(a, c)$ also belongs to $R$. If $a \times b$ and $b \times c$ are irrational, it doesn't necessarily imply that $a \times c$ is irrational. For example, $\sqrt{2} \times \sqrt{3}$ and $\sqrt{3} \times \sqrt{2}$ are both irrational, but $\sqrt{2} \times \sqrt{2} = 2$ is rational. So, $R$ is not transitive. In summary: - $R$ is not reflexive. - $R$ is symmetric. - $R$ is not transitive.