For real numbers a and b, define aRb if $b-a +\sqrt{5}$ is an irrational number. Then the relation R is :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Reflexive
$aRb=b-a +\sqrt{5}$
R is reflexive → as for all $a∈R$, (a, a) ∈ Relation
as $a-a+\sqrt{5}=\sqrt{5}$ (always irrational)
R not symmetric → let $b=\sqrt{5},a=2$
so (a, b) ∈ Relation as $\sqrt{5}-2+\sqrt{5}$ is irrational
but (b, a) ∉ Relation as $2-\sqrt{5}+\sqrt{5}=2$ is not irrational
R is not transitive →
$(1+\sqrt{5},4)∈R$ as $1+\sqrt{5}-1+\sqrt{5}=-3+2\sqrt{5}$ (irrational)
$(4,2\sqrt{5})∈R$ as $4-2\sqrt{5}+\sqrt{5}=4-\sqrt{5}$ (irrational)
$(1+\sqrt{5},2\sqrt{5})∉R$ as $1+\sqrt{5}-2\sqrt{5}=0$ (not irrational)