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(A) $1$. Reflexivity: For any $x \in \mathbb{R}$, $x p x$ implies $x-x+\sqrt{2} = \sqrt{2}$. Since $\sqrt{2}$ is an irrational number, $x p x$ is true

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reflexive

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(A) $1$. Reflexivity: For any $x \in \mathbb{R}$, $x p x$ implies $x-x+\sqrt{2} = \sqrt{2}$. Since $\sqrt{2}$ is an irrational number, $x p x$ is true for all $x$. Thus, $p$ is reflexive.$2$. Symmetry: If $x p y$, then $x-y+\sqrt{2}$ is irrational. For symmetry, we check if $y p x$ holds, which means $y-x+\sqrt{2}$ is irrational. Let $x=0$ and $y=\sqrt{2}$. Then $x-y+\sqrt{2} = 0-\sqrt{2}+\sqrt{2} = 0$, which is rational. So $x p y$ is false. Let $x=1$ and $y=0$. Then $x-y+\sqrt{2} = 1-0+\sqrt{2} = 1+\sqrt{2}$, which is irrational. So $1 p 0$ is true. Now check $0 p 1$: $0-1+\sqrt{2} = \sqrt{2}-1$, which is irrational. So $0 p 1$ is true. However, consider $x=\sqrt{2}$ and $y=0$. $x p y = \sqrt{2}-0+\sqrt{2} = 2\sqrt{2}$ (irrational). $y p x = 0-\sqrt{2}+\sqrt{2} = 0$ (rational). Since $y p x$ is not necessarily true when $x p y$ is true, the relation is not symmetric.$3$. Transitivity: If $x p y$ and $y p z$, then $x-y+\sqrt{2} = i_1$ and $y-z+\sqrt{2} = i_2$, where $i_1, i_2$ are irrational. Adding these, $x-z+2\sqrt{2} = i_1+i_2$. This does not guarantee $x-z+\sqrt{2}$ is irrational. For example, let $x=1+\sqrt{2}$, $y=1$, and $z=1-\sqrt{2}$. $x p y = (1+\sqrt{2})-1+\sqrt{2} = 2\sqrt{2}$ (irrational). $y p z = 1-(1-\sqrt{2})+\sqrt{2} = 2\sqrt{2}$ (irrational). But $x p z = (1+\sqrt{2})-(1-\sqrt{2})+\sqrt{2} = 3\sqrt{2}$ (irrational). However, if we choose $x=\sqrt{2}, y=0, z=-\sqrt{2}$, then $x p y = 2\sqrt{2}$ (irrational), $y p z = 0-(-\sqrt{2})+\sqrt{2} = 2\sqrt{2}$ (irrational), but $x p z = \sqrt{2}-(-\sqrt{2})+\sqrt{2} = 3\sqrt{2}$ (irrational). Actually, consider $x=2\sqrt{2}, y=\sqrt{2}, z=0$. $x p y = 2\sqrt{2}-\sqrt{2}+\sqrt{2} = 2\sqrt{2}$ (irrational). $y p z = \sqrt{2}-0+\sqrt{2} = 2\sqrt{2}$ (irrational). $x p z = 2\sqrt{2}-0+\sqrt{2} = 3\sqrt{2}$ (irrational). The relation is not transitive in general. Thus, the relation is only reflexive.

For the real numbers $x$ and $y$, we define the relation $p$ as $x p y$ if $x-y+\sqrt{2}$ is an irrational number. Then the relation $p$ is

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