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
