Competency-Based
Assessment Made Simple.

Reflexive:

For $R$ to be reflexive, $(x, x) \in R$ for all $x \in \mathbb{R}$. This means $x \cdot x = x^2$ must be irrational for all real numbers $x$. However, if $x = 1$, then $x^2 = 1$, which is rational. Therefore, $R$ is not reflexive.

Symmetric:

For $R$ to be symmetric, if $(x, y) \in R$, then $(y, x) \in R$. This means if $x \cdot y$ is irrational, then $y \cdot x$ must also be irrational. Since $x \cdot y = y \cdot x$, if $x \cdot y$ is irrational, then $y \cdot x$ is also irrational. Therefore, $R$ is symmetric.

Transitive:

For $R$ to be transitive, if $(x, y) \in R$ and $(y, z) \in R$, then $(x, z) \in R$. This means if $x \cdot y$ is irrational and $y \cdot z$ is irrational, then $x \cdot z$ must be irrational. Let's consider a counterexample: Let $x = \sqrt{2}$, $y = \sqrt{2}$, and $z = \sqrt{2}$. Then $x \cdot y = \sqrt{2} \cdot \sqrt{2} = 2$, which is rational. So, $(x, y) \notin R$. Let $x = \sqrt{2}$, $y = \sqrt{3}$, and $z = \sqrt{2}$. Then $x \cdot y = \sqrt{6}$ (irrational) and $y \cdot z = \sqrt{6}$ (irrational). But $x \cdot z = \sqrt{2} \cdot \sqrt{2} = 2$ (rational). Thus, $(x, z) \notin R$. Another counterexample: Let $x = \sqrt{2}$, $y = 1/\sqrt{2}$, and $z = \sqrt{3}$. $x \cdot y = \sqrt{2} \cdot (1/\sqrt{2}) = 1$ (rational), so $(x, y) \notin R$. Let $x = \sqrt{2}$, $y = \sqrt{2}$, $z = \sqrt{3}$. Consider $x = \sqrt{2}$, $y = \sqrt{3}$, $z = \sqrt{2}$. $xy = \sqrt{6}$ (irrational), $yz = \sqrt{6}$ (irrational), but $xz = 2$ (rational). Thus, $R$ is not transitive.