Question

Let \(R_1\) and \(R_2\) be two relation defined as follows :
\(R_1=\{(a,b)\in\mathbb{R}^2 : a^2+b^2\in\mathbb{Q}\}\) and
\(R_2=\{(a,b)\in\mathbb{R}^2 : a^2+b^2\notin\mathbb{Q}\}\),
where \(\mathbb{Q}\) is the set of all rational numbers. Then :

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

Hint:

Look for counter-examples with irrational squares such as \(\sqrt2\) so that the sum of two squares is rational but another sum is not (and vice versa).

Answer 1

The Correct Option is A

\(R_1\) is not transitive. Take \(a=c=2^{1/4}\) and \(b=\sqrt{2-\sqrt2}\).
\(a^2+b^2=\sqrt2+2-\sqrt2=2\in\mathbb{Q}\), so \((a,b)\in R_1\); similarly \(b^2+c^2=2\in\mathbb{Q}\), so \((b,c)\in R_1\).
But \(a^2+c^2=2\sqrt2\notin\mathbb{Q}\), so \((a,c)\notin R_1\). Hence \(R_1\) is not transitive.
\(R_2\) is not transitive. Take \(a=1,\ b=2^{1/4},\ c=1\).
\(a^2+b^2=1+\sqrt2\notin\mathbb{Q}\Rightarrow(a,b)\in R_2\); \(b^2+c^2=\sqrt2+1\notin\mathbb{Q}\Rightarrow(b,c)\in R_2\).
But \(a^2+c^2=2\in\mathbb{Q}\Rightarrow(a,c)\notin R_2\). So \(R_2\) is not transitive either.
Answer: (1) Neither \(R_1\) nor \(R_2\) is transitive.