Key Concepts and Formulas

• Equivalence Relation Properties: A relation $R$ on a set $S$ is an equivalence relation if it is:
  • Reflexive: For all $a \in S$, $(a, a) \in R$.
  • Symmetric: For all $a, b \in S$, if $(a, b) \in R$, then $(b, a) \in R$.
  • Transitive: For all $a, b, c \in S$, if $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$.
• Properties of Integers and Rational Numbers:
  • The sum, difference, and product of two integers are integers.
  • The sum, difference, and product of two rational numbers are rational numbers.
  • The reciprocal of a non-zero rational number is a rational number.

Step-by-Step Solution

Analysis of Statement I:

Statement I defines a relation $A$ on the set of real numbers $R$ as $A=\{(x, y) \in R \times R: y-x \text{ is an integer}\}$.

• Step 1: Check for Reflexivity.

  • What we are doing: We need to verify if for every real number $x$, the pair $(x, x)$ belongs to relation $A$.
  • Math: According to the definition of $A$, $(x, x) \in A$ if $x-x$ is an integer. $x-x = 0$.
  • Reasoning: Since $0$ is an integer, the condition $x-x$ is an integer is satisfied for all $x \in R$.
  • Conclusion: Relation $A$ is reflexive.

• Step 2: Check for Symmetry.

  • What we are doing: We need to verify if for any two real numbers $x$ and $y$, if $(x, y) \in A$, then $(y, x) \in A$.
  • Math: Assume $(x, y) \in A$. This implies that $y-x = k$ for some integer $k$. We want to check if $(y, x) \in A$, which means checking if $x-y$ is an integer. From $y-x = k$, we can multiply by $-1$ to get $-(y-x) = -k$, which simplifies to $x-y = -k$.
  • Reasoning: Since $k$ is an integer, its additive inverse, $-k$, is also an integer. Thus, if $y-x$ is an integer, then $x-y$ is also an integer.
  • Conclusion: Relation $A$ is symmetric.

• Step 3: Check for Transitivity.

  • What we are doing: We need to verify if for any three real numbers $x, y,$ and $z$, if $(x, y) \in A$ and $(y, z) \in A$, then $(x, z) \in A$.
  • Math: Assume $(x, y) \in A$ and $(y, z) \in A$. This means $y-x = k_1$ for some integer $k_1$ (Equation 1). And $z-y = k_2$ for some integer $k_2$ (Equation 2). We want to check if $(x, z) \in A$, which means checking if $z-x$ is an integer. Adding Equation 1 and Equation 2: $(y-x) + (z-y) = k_1 + k_2$ $z-x = k_1 + k_2$.
  • Reasoning: Since $k_1$ and $k_2$ are integers, their sum $k_1+k_2$ is also an integer. Therefore, if $y-x$ and $z-y$ are integers, then $z-x$ is also an integer.
  • Conclusion: Relation $A$ is transitive.

• Overall Conclusion for Statement I: Since relation $A$ satisfies reflexivity, symmetry, and transitivity, Statement I is True.

Analysis of Statement II:

Statement II defines a relation $B$ on the set of real numbers $R$ as $B=\{(x, y) \in R \times R: x=\alpha y \text{ for some rational number } \alpha\}$.

• Step 4: Check for Reflexivity.

  • What we are doing: We need to verify if for every real number $x$, the pair $(x, x)$ belongs to relation $B$.
  • Math: According to the definition of $B$, $(x, x) \in B$ if $x=\alpha x$ for some rational number $\alpha$. If $x=0$, then $0 = \alpha \cdot 0$. This holds for any rational number $\alpha$ (e.g., $\alpha = 1$). If $x \neq 0$, we can choose $\alpha = 1$. Since $1$ is a rational number, $x = 1 \cdot x$ is true.
  • Reasoning: For any real number $x$, we can find a rational number $\alpha$ (specifically, $\alpha=1$) such that $x=\alpha x$.
  • Conclusion: Relation $B$ is reflexive.

• Step 5: Check for Symmetry.

  • What we are doing: We need to verify if for any two real numbers $x$ and $y$, if $(x, y) \in B$, then $(y, x) \in B$.

  • Math: Assume $(x, y) \in B$. This means $x=\alpha y$ for some rational number $\alpha$. We need to check if $(y, x) \in B$, which means checking if $y=\beta x$ for some rational number $\beta$.

    Consider the case where $y \neq 0$. If $x=\alpha y$, then $y = \frac{1}{\alpha} x$. If $\alpha$ is a non-zero rational number, then $\frac{1}{\alpha}$ is also a non-zero rational number. So, we can set $\beta = \frac{1}{\alpha}$.

    However, we must consider the case where $\alpha=0$. If $\alpha=0$, then $x = 0 \cdot y = 0$. In this situation, $(x, y) = (0, y)$ is in $B$ if $0 = \alpha y$ for some rational $\alpha$. This is true if $\alpha=0$. Now, we check for symmetry: if $(0, y) \in B$, is $(y, 0) \in B$? For $(y, 0)$ to be in $B$, we need $y = \beta \cdot 0$ for some rational $\beta$. This implies $y=0$. So, if $(0, y) \in B$ where $y \neq 0$, then $(y, 0)$ is not necessarily in $B$. For example, let $x=0$ and $y=1$. Then $(0, 1) \in B$ because $0 = 0 \cdot 1$ and $0$ is rational. However, $(1, 0) \notin B$ because $1 = \beta \cdot 0$ has no solution for $\beta$.

  • Reasoning: The symmetry property fails when one of the numbers is zero and the other is non-zero.

  • Conclusion: Relation $B$ is not symmetric.

• Step 6: Conclusion for Statement II.

  • What we are doing: Based on the checks for reflexivity, symmetry, and transitivity, we determine if statement II is true or false.
  • Reasoning: Since relation $B$ is not symmetric, it is not an equivalence relation.
  • Conclusion: Statement II is False.

Common Mistakes & Tips

• Careful with Zero: When dealing with ratios or division in relations, always pay special attention to cases involving zero. Division by zero is undefined, and relations like $x = \alpha y$ can behave differently when $x$ or $y$ is zero.
• Distinguish Rational and Real: The definition of relation $B$ specifically requires $\alpha$ to be a rational number. This is crucial for checking the properties.
• Equivalence Relation Definition: Remember that all three properties (reflexive, symmetric, transitive) must hold for a relation to be an equivalence relation. If any one property fails, the relation is not an equivalence relation.

Summary

Statement I defines a relation where the difference between the two real numbers is an integer. This relation is proven to be reflexive, symmetric, and transitive, making it an equivalence relation. Statement II defines a relation where one real number is a rational multiple of the other. While this relation is reflexive, it fails the symmetry property when one number is zero and the other is non-zero, thus it is not an equivalence relation. Therefore, Statement I is true and Statement II is false.

The final answer is $\boxed{B}$.