Question

Let $R_1$ and $R_2$ be two relations defined as follows:

$R_1=\left\{\left(a,b\right)\in R^2:a^2+b^2\in \mathrm{\mathbb{Q}}\right\}$and

$R_2=\left\{\left(a,b\right)\in R^2:a^2+b^2\notin \mathrm{\mathbb{Q}}\right\}$, where $\mathrm{\mathbb{Q}}$ is the set of all rational numbers. Then :

• A. $R_1$ is transitive but $R_2$ is not transitive
• B. $R_1$ and $R_2$ are both transitive
• C. $R_2$ is transitive but $R_1$ is not transitive
• D. Neither $R_1$ nor $R_2$ is transitive

Answer 1

The correct option is D

Neither $R_1$ nor $R_2$ is transitive

Explanation for the correct option:

Checking transitivity:

For $R_1$:

Consider $a=1+\sqrt{3},b=1-\sqrt{3},c={12}^{\frac{1}{4}}$

$(a,b)\in R_1$ because:

$$\begin{gathered} a^2+b^2={\left(1+\sqrt{3}\right)}^2+{\left(1-\sqrt{3}\right)}^2 \\ =1+3+2\sqrt{3}+1+3-2\sqrt{3} \\ =8\in \mathrm{\mathbb{Q}} \end{gathered}$$

And $(b,c)\in R_1$because

$$\begin{gathered} b^2+c^2={\left(1-\sqrt{3}\right)}^2+{\left({12}^{\frac{1}{4}}\right)}^2 \\ =1+3-\sqrt{12}+\sqrt{12} \\ =4\in \mathrm{\mathbb{Q}} \end{gathered}$$

But $(a,c)\notin R_1$because

$$\begin{gathered} a^2+c^2={\left(1+\sqrt{3}\right)}^2+{\left({12}^{\frac{1}{4}}\right)}^2 \\ =1+3+2\sqrt{12}+\sqrt{12} \\ =4+3\sqrt{12}\notin \mathrm{\mathbb{Q}} \end{gathered}$$

Hence $R_1$is not transitive

For $R_2$:

Consider $a=1+\sqrt{3},b=\sqrt{3},c=1-\sqrt{3}$

$(a,b)\in R_2$ because:

$$\begin{gathered} a^2+b^2={\left(1+\sqrt{3}\right)}^2+{\left(\sqrt{3}\right)}^2 \\ =7+2\sqrt{3}\notin \mathrm{\mathbb{Q}} \end{gathered}$$

$\left(b,c\right)\in R_2$ since

$$\begin{gathered} b^2+c^2={\left(\sqrt{3}\right)}^2+{\left(1-\sqrt{3}\right)}^2 \\ =7-2\sqrt{3}\notin \mathrm{\mathbb{Q}} \end{gathered}$$

But$(a,c)\notin R_2$ because:

$$\begin{gathered} a^2+c^2={\left(1+\sqrt{3}\right)}^2+{\left(1-\sqrt{3}\right)}^2 \\ =8\in \mathrm{\mathbb{Q}} \end{gathered}$$

Hence $R_2$is not transitive

Hence,the correct option is (D)