Let R be the set of real numbers
A = {(x, y) $\in$ R $\times$ R : y - x is an integer} is an equivalence relation on R.
B = {(x, y) $\in$ R $\times$ R : x = $\alpha$y for some rational number $\alpha$} is an equivalence relation on R.
Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1
Statement-1 is true, Statement-2 is false
Statement-1 is false, Statement-2 is true
Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation for statement -1
Show Solution
The Correct Option isB
Solution and Explanation
To determine whether the given statements about the sets A and B describe equivalence relations, we need to verify the defining properties of equivalence relations: reflexivity, symmetry, and transitivity for each set.
Checking Set A: \( A = \{(x, y) \mid (y - x) \text{ is an integer}\} \)
Reflexivity: For any real number \( x \), we have \( y - x = x - x = 0 \), which is an integer. Thus, each \( (x, x) \in A \).
Symmetry: If \( (x, y) \in A \), then \( y - x \) is an integer. Therefore, \( x - y = -(y - x) \) is also an integer, so \( (y, x) \in A \).
Transitivity: If \( (x, y) \in A \) and \( (y, z) \in A \), then \( y - x \) and \( z - y \) are integers. Hence, \( (z - x) = (z - y) + (y - x) \) is an integer, meaning \( (x, z) \in A \).
Since set A satisfies reflexivity, symmetry, and transitivity, it is an equivalence relation.
Checking Set B: \( B = \{(x, y) \mid x = \alpha y \text{ for some rational number } \alpha\} \)
Reflexivity: For any real number \( x \), we need \( x = \alpha x \) where \( \alpha \) is a rational number. This is only true if \( \alpha = 1 \), which is rational, so reflexivity holds.
Symmetry: If \( (x, y) \in B \), then \( x = \alpha y \). Let's express \( \alpha \) as a rational number \( \frac{p}{q} \), making \( x = \frac{p}{q}y \). For symmetry, we need \( y = \beta x \) and \(\beta\) should also be rational, implying \( \beta = \frac{q}{p} = \frac{1}{\alpha}\). However, \(\frac{1}{\alpha}\) being rational depends on \(\alpha\) not being zero, but this is not guaranteed symmetrically for all real \(x\) and \(y\). Therefore, symmetry does not hold generally.
Transitivity: If \( (x, y) \in B \) and \( (y, z) \in B \), then \( x = \alpha y \) and \( y = \beta z \). If both \(\alpha\) and \(\beta\) are rational, then \( x = \alpha(\beta z) = (\alpha \beta) z \); however, without ensuring total coverage for inverses, discarding zeroes, etc., general transitive handling gets complex and is not inherently closed.
Because B does not satisfy the symmetry property for all elements, it is not an equivalence relation.
Conclusion: Statement-1 is true as set A is an equivalence relation. Statement-2 is false because set B does not fulfill all conditions needed for an equivalence relation. Therefore, the correct answer is: Statement-1 is true, Statement-2 is false.
