Let R be a relation on N x N defined by (a, b) R(c, d) if and only if a d ( b - c ) = b c ( a - d ) . Then R is [2023]

Question

Let R be a relation on N x N defined by (a, b) R(c, d) if and only if $a d (b - c) = b c (a - d)$ . Then R is [2023]

Smart Achievers · Accepted Answer

(2)

We have, $(a, b) R (c, d) \Leftrightarrow a d (b - c) = b c (a - d)$

$\Rightarrow \frac{b - c}{b c} = \frac{a - d}{a d} and \frac{1}{c} - \frac{1}{b} = \frac{1}{d} - \frac{1}{a} \Rightarrow \frac{1}{a} - \frac{1}{b} = \frac{1}{d} - \frac{1}{c}$

For reflexive: $(a, b) R (a, b) \Rightarrow \frac{1}{a} - \frac{1}{b} = \frac{1}{b} - \frac{1}{a}$ which is false.

Hence, it is not reflexive.

For symmetric: $(a, b) R (c, d) \Rightarrow \frac{1}{a} - \frac{1}{b} = \frac{1}{d} - \frac{1}{c}$

$\Rightarrow \frac{1}{c} - \frac{1}{d} = \frac{1}{b} - \frac{1}{a} \Rightarrow (c, d) R (a, b)$

Hence, it is symmetric.

For transitive: $(a, b) R (c, d) \Rightarrow \frac{1}{a} - \frac{1}{b} = \frac{1}{d} - \frac{1}{c}$

$and (c, d) R (e, f) \Rightarrow \frac{1}{c} - \frac{1}{d} = \frac{1}{f} - \frac{1}{e}$

$\Rightarrow \frac{1}{a} - \frac{1}{b} = \frac{1}{e} - \frac{1}{f} \neq (a, b) R (e, f)$

Hence, it is not transitive.

Solution

Q.
Let R be a relation on N x N defined by (a, b) R(c, d) if and only if $a d (b - c) = b c (a - d)$ . Then R is [2023]

1 transitive but neither reflexive nor symmetric

2 symmetric but neither reflexive nor transitive

3 symmetric and transitive but not reflexive

4 reflexive and symmetric but not transitive

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