Question

Let \(\mathrm{R}_1\) and \(\mathrm{R}_2\) be two relations on \(\mathbb{R}^2\) defined as
\((a, b)\ \mathrm{R}_1\ (c, d)\) if \(ad - bc \geq 0\)
\((a, b)\ \mathrm{R}_2\ (c, d)\) if \(a + d \geq b + c\). Then :

• A. \(\mathrm{R}_1\) is transitive but \(\mathrm{R}_2\) is not transitive
• B. \(\mathrm{R}_2\) is transitive but \(\mathrm{R}_1\) is not transitive
• C. Both \(\mathrm{R}_1\) and \(\mathrm{R}_2\) are transitive
• D. Neither \(\mathrm{R}_1\) nor \(\mathrm{R}_2\) is transitive

Hint:

Rewrite \(\mathrm{R}_2\) as \(a-b \geq c-d\), which compares a single quantity, and test \(\mathrm{R}_1\) with a counter-example.

Answer 1

The Correct Option is B

R2: \(a + d \geq b + c\) is the same as \(a - b \geq c - d\). Define \(f(x,y)=x-y\). Then \((a,b)\ \mathrm{R}_2\ (c,d)\) means \(f(a,b)\geq f(c,d)\).
If \(f(a,b)\geq f(c,d)\) and \(f(c,d)\geq f(e,f)\), then \(f(a,b)\geq f(e,f)\). So \(\mathrm{R}_2\) is transitive.
R1: Take \((a,b)=(1,0)\), \((c,d)=(-1,1)\), \((e,f)=(-1,-1)\).
\((1,0)\ \mathrm{R}_1\ (-1,1)\): \(1\cdot 1-0\cdot(-1)=1\geq 0\), true.
\((-1,1)\ \mathrm{R}_1\ (-1,-1)\): \((-1)(-1)-(1)(-1)=2\geq 0\), true.
\((1,0)\ \mathrm{R}_1\ (-1,-1)\): \(1\cdot(-1)-0\cdot(-1)=-1<0\), false.
So \(\mathrm{R}_1\) is not transitive.
Hence \(\mathrm{R}_2\) is transitive but \(\mathrm{R}_1\) is not transitive.
Answer: (2) \(\mathrm{R}_2\) is transitive but \(\mathrm{R}_1\) is not transitive