Step 1 (Reflexive, general argument): For any real $x$, $x-x+\sqrt{2}=\sqrt{2}$ is always irrational, so $xRx$ holds for every $x$ with no need for an example; R is reflexive by direct substitution.
Step 2 (Symmetry, algebraic argument): Let $d=x-y$. $xRy$ fails exactly when $d+\sqrt{2}$ is rational, which happens only when $d$ itself supplies a $-\sqrt{2}$ term, i.e. $d = q-\sqrt{2}$ for rational $q$. Now pick $d=\sqrt{2}$ (so $d+\sqrt{2}=2\sqrt{2}$ is irrational and $xRy$ holds). Then $yRx$ depends on $-d+\sqrt{2} = -\sqrt{2}+\sqrt{2}=0$, which is rational, so $yRx$ fails. This shows a case where $xRy$ is true while $yRx$ is false, so symmetry cannot hold in general.
Step 3 (Transitivity, algebraic argument): Suppose $x-y+\sqrt{2}$ and $y-z+\sqrt{2}$ are both irrational. Note that $(x-z+\sqrt{2}) = (x-y+\sqrt{2})+(y-z+\sqrt{2})-\sqrt{2}$. Even though the two bracketed terms are irrational, subtracting $\sqrt{2}$ can cancel the irrational part and leave a rational number. Taking $x-y=\sqrt{3}$ and $y-z=-\sqrt{3}-\sqrt{2}$ makes $x-z+\sqrt{2}=0$, a rational number, so $xRz$ fails even though $xRy$ and $yRz$ both hold. Hence R is not transitive.
Step 4: Combining both findings, R is reflexive for every real number, but it is neither symmetric nor transitive in general.
\[\boxed{\text{Reflexive but not symmetric and transitive}}\]