### Sets And Relations Question 7

#### Question 7 - 01 February - Shift 1

Let $R$ be a relation on $\mathbb{R}$, given by

$R=(a, b): 3 a-3 b+\sqrt{7}$ is an irrational number . Then $R$ is

(1) Reflexive but neither symmetric nor transitive

(2) Reflexive and transitive but not symmetric

(3) Reflexive and symmetric but not transitive

(4) An equivalence relation

## Show Answer

#### Answer: (1)

#### Solution:

#### Formula: Reflexive relation (iv), Symmetric relation (v), Transitive relation (vi), Equivalence relation (vii)

Check for reflexivity:

As 3(a-a) $+\sqrt{7}=\sqrt{7}$ which belongs to relation

so relation is reflexive

Check for symmetric:

Take $a=\frac{\sqrt{7}}{3}, b=0$

Now $(a, b) \in R$ but $(b, a) \notin R$

As $3(b-a)+\sqrt{7}=0$ which is rational so relation is not symmetric.

Check for Transitivity:

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

So now $(a, b) \in R ; and ; (b, c) \in R$ but $(a, c) \notin k$ which means relation is not transitive