For real numbers x and y, we write $x\,R\,y⇔|x-y+\sqrt{2}|$ is an irrational number. Then the relation R is:
Answer & explanation
Correct answer: option 1
For $x ∈ R$, $|x-x+\sqrt{2}|=\sqrt{2}$ → irrational ⇒ reflexive
but $(1,\sqrt{2})∉R$
as $|1-\sqrt{2}+\sqrt{2}|=1$ ⇒ not irrational or nor symmetric
$(\sqrt{2},1)∈ R,\,(1,2\sqrt{2})∈ R$
but $(\sqrt{2},2\sqrt{2})∉R$
as $|\sqrt{2}-2\sqrt{2}+\sqrt{2}|=0$ Not irrational or not transitive