Question

Let $R$ be a relation on $R$, given by $R=\{(a, b): 3 a-3 b+\sqrt{7}$ is an irrational number $\}$Then $R$ is

• A. reflexive and symmetric but not transitive
• B. reflexive and transitive but not symmetric
• C. reflexive but neither symmetric nor transitive
• D. an equivalence relation

Answer 1

The Correct Option is C

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 $\left(\frac{\sqrt{7}}{3},1\right)$
$\&(b,c)$ as $\left(1,\frac{2\sqrt{7}}{3}\right)$
So now (a, b) $\in R\&(b,c)\in R$ but $(a,c)\notin R$ which means relation is not transitive

Answer 2

Check for reflexivity: 
Since \[ 3(a - a) + \sqrt{7} = \sqrt{7} \] which is irrational, the relation is reflexive. 

Check for symmetry: 
Let \( a = \frac{\sqrt{7}}{3}, b = 0 \). Then \( (a, b) \in R \) but \( (b, a) \notin R \). 
Since \[ 3(b - a) + \sqrt{7} = 0 \] which is rational, the relation is not symmetric. 

Check for transitivity: 
Let \( (a, b) = \left( 1, \frac{2\sqrt{7}}{3} \right) \) and \( (b, c) = \left( 1, \frac{2\sqrt{7}}{3} \right) \). Then, \( (a, b) \in R \) and \( (b, c) \in R \), but \( (a, c) \notin R \), so the relation is not transitive.