JEE Main 2023 — Mathematics Algebra

Given:

$$R={(a,b):3a-3b+\sqrt{7}\mathrm{is}\mathrm{irrational}}$$R=(a,b):3a−3b+7​isirrational

Reflexive:

$3a-3a+\sqrt{7}=\sqrt{7}$3a−3a+7​=7​ is irrational.

So, $(a,a)\in R$(a,a)∈R, hence $R$R is reflexive.

Given:

$$R={(a,b):3a-3b+\sqrt{7}\mathrm{is}\mathrm{irrational}}$$R=(a,b):3a−3b+7​isirrational

Symmetric:

Let $(a,b)\in R$(a,b)∈R, then

$3a-3b+\sqrt{7}$3a−3b+7​ is irrational.

So, $3b-3a+\sqrt{7}$3b−3a+7​ is not necessarily irrational, since if $3a=\sqrt{7}$3a=7​ and $3b\in Z$3b∈Z, then $3a-3b+\sqrt{7}$3a−3b+7​ is irrational., but $3b-3a+\sqrt{7}$3b−3a+7​ is not irrational.

So, $(a,b)\in R\Rightarrow (b,a)\in R$(a,b)∈R⇒(b,a)∈R, hence $R$R is not symmetric

Transitive:

Take $(a,b)$(a,b) as $(\frac{\sqrt{7}}{3},1)$(37​​,1) and $(b,c)$(b,c) as $(1,\frac{2\sqrt{7}}{3})$(1,327​​)

So, $(a,b)\in R$(a,b)∈R and $(b,c)\in R$(b,c)∈R but $(a,c)\notin R$(a,c)∈/R which means relation is not transitive.