Q32: Is $R = \{(x, y) : x\cdot y$ is irrational$\}$ reflexive, symmetric, transitive?

R is reflexive if $(x, x) \in R$ for every real number $x$⁠. To show it is not, one real number $x$ with $(x, x) \notin R$ is enough.

$x = 1$ works, because $1\cdot 1 = 1$ is rational.

Watch out: some pairs $(x, x)$ are in R. For example $x = \sqrt[4]{2}$ gives $x\cdot x = \sqrt2$⁠, which is irrational. That does not help, because reflexive needs every $x$⁠.

R is symmetric if $(x, y) \in R$ always gives $(y, x) \in R$⁠. So start with any pair in R and test the reversed pair.

Multiplication of real numbers is commutative, so $y\cdot x$ and $x\cdot y$ are the same number. If one is irrational, so is the other.

R is transitive if $(x, y) \in R$ and $(y, z) \in R$ always give $(x, z) \in R$⁠. To show it is not, one counterexample is enough.

We need $xy$ and $yz$ irrational but $xz$ rational. An irrational number in the middle and rational numbers at the two ends does this.

$\sqrt2$ is irrational (a standard result). If $2\sqrt2$ were a rational number $r$⁠, then $\sqrt2 = \frac{r}{2}$ would be rational too, which is false. So $2\sqrt2$ is irrational.

$(1, \sqrt2)$ and $(\sqrt2, 2)$ are in R, but $(1, 2)$ is not. The rule must hold for all pairs, so this one example is enough to show that R is not transitive.