JEE Mathematics 2023 Question with Solution

Given: $R = \{(a,b): 3a-3b+\sqrt{7} \text{ is irrational}\}$ on $\mathbb{R}$.

Find: Whether the relation is reflexive, symmetric, transitive, or an equivalence relation.

Reflexivity: For any $a \in \mathbb{R}$,

$$3(a-a)+\sqrt{7}=\sqrt{7}$$

which is irrational. Hence $(a,a) \in R$ for every $a$, so the relation is reflexive.

Symmetry: Take

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

Then

$$3a-3b+\sqrt{7} = 3\left(\frac{\sqrt{7}}{3}\right)-0+\sqrt{7} = 2\sqrt{7}$$

which is irrational, so $(a,b) \in R$. But for $(b,a)$,

$$3b-3a+\sqrt{7} = 0-\sqrt{7}+\sqrt{7} = 0$$

which is rational. Therefore $(b,a) \notin R$, so the relation is not symmetric.

Transitivity: Take

$$(a,b) = \left(\frac{\sqrt{7}}{3}, 1\right), \quad (b,c) = \left(1, \frac{2\sqrt{7}}{3}\right)$$

Then

$$3a-3b+\sqrt{7} = \sqrt{7}-3+\sqrt{7} = 2\sqrt{7}-3$$

which is irrational, so $(a,b) \in R$. Also,

$$3b-3c+\sqrt{7} = 3-2\sqrt{7}+\sqrt{7} = 3-\sqrt{7}$$

which is irrational, so $(b,c) \in R$. Now,

$$3a-3c+\sqrt{7} = \sqrt{7}-2\sqrt{7}+\sqrt{7} = 0$$

which is rational. Hence $(a,c) \notin R$. So the relation is not transitive.

Therefore, $R$ is reflexive but neither symmetric nor transitive. The correct option by the working is A.

The solution labels option C, but that conflicts with the extracted working and the listed options.

Given: $3a-3b+\sqrt{7}$ must be irrational.

Find: Which standard properties of relations are satisfied.

A number of the form

$$3(a-b)+\sqrt{7}$$

need not always remain irrational when $a$ and $b$ are interchanged or chained through a third element.

1. For reflexivity, substitute $b=a$.
2. For symmetry, compare $3(a-b)+\sqrt{7}$ with $3(b-a)+\sqrt{7}$.
3. For transitivity, construct examples where the first two expressions are irrational but the third becomes rational.

This gives:

• reflexive: yes
• symmetric: no
• transitive: no

Hence the correct option is A.