Key Concepts and Formulas

• Equivalence Relation: A relation $R$ on a set $A$ is an equivalence relation if it is reflexive, symmetric, and transitive.
  • Reflexivity: For all $a \in A$, $(a, a) \in R$.
  • Symmetry: For all $a, b \in A$, if $(a, b) \in R$, then $(b, a) \in R$.
  • Transitivity: For all $a, b, c \in A$, if $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$.
• Properties of Rational Numbers: The set of rational numbers $\mathbb{Q}$ is closed under addition, subtraction, multiplication, and division (by non-zero numbers).

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Step-by-Step Solution

We need to determine if the relations $R$ and $S$ are equivalence relations by checking for reflexivity, symmetry, and transitivity.

Analysis of Relation R

The relation $R$ is defined on the set of real numbers, $\mathbb{R}$, as: $R=\{(x, y) \mid x, y \in \mathbb{R} \text{ and } x=w y \text{ for some rational number } w \in \mathbb{Q}\}$

Step 1: Check Reflexivity for R

• Objective: To check if for every $x \in \mathbb{R}$, $(x, x) \in R$. This requires us to find a rational number $w$ such that $x = w x$.
• Working:
  • If $x \neq 0$, we can divide by $x$ to get $w=1$. Since $1 \in \mathbb{Q}$, the condition is satisfied.
  • If $x = 0$, the equation becomes $0 = w \cdot 0$, which is $0=0$. This is true for any rational number $w$ (e.g., $w=1$). Thus, $(0,0) \in R$.
• Conclusion: For all $x \in \mathbb{R}$, $(x, x) \in R$. So, $R$ is reflexive.

Step 2: Check Symmetry for R

• Objective: To check if for every $x, y \in \mathbb{R}$, if $(x, y) \in R$, then $(y, x) \in R$. This means if $x = w y$ for some $w \in \mathbb{Q}$, then $y = w' x$ for some $w' \in \mathbb{Q}$.
• Working:
  • Consider a counterexample. Let $x=0$ and $y=\sqrt{2}$.
  • Is $(0, \sqrt{2}) \in R$? Yes, because $0 = 0 \cdot \sqrt{2}$, and $0 \in \mathbb{Q}$.
  • Now, we check if $(\sqrt{2}, 0) \in R$. This would require $\sqrt{2} = w' \cdot 0$ for some $w' \in \mathbb{Q}$.
  • However, $w' \cdot 0 = 0$. So, the equation $\sqrt{2} = 0$ is false. There is no rational number $w'$ that satisfies this.
  • Therefore, $(\sqrt{2}, 0) \notin R$.
• Conclusion: Since $(0, \sqrt{2}) \in R$ but $(\sqrt{2}, 0) \notin R$, $R$ is not symmetric.

Step 3: Check Transitivity for R

• Objective: To check if for every $x, y, z \in \mathbb{R}$, if $(x, y) \in R$ and $(y, z) \in R$, then $(x, z) \in R$. This means if $x = w_1 y$ for some $w_1 \in \mathbb{Q}$ and $y = w_2 z$ for some $w_2 \in \mathbb{Q}$, then $x = w_3 z$ for some $w_3 \in \mathbb{Q}$.
• Working:
  • Given $x = w_1 y$ ($w_1 \in \mathbb{Q}$) and $y = w_2 z$ ($w_2 \in \mathbb{Q}$).
  • Substitute the second equation into the first: $x = w_1 (w_2 z) = (w_1 w_2) z$.
  • Since $w_1$ and $w_2$ are rational numbers, their product $w_1 w_2$ is also a rational number. Let $w_3 = w_1 w_2$.
  • Thus, $x = w_3 z$ where $w_3 \in \mathbb{Q}$.
• Conclusion: $R$ is transitive.

Overall for R: Since $R$ is reflexive and transitive but not symmetric, $R$ is not an equivalence relation.

Analysis of Relation S

The relation $S$ is defined on pairs of rational numbers: $S=\left\{\left(\frac{m}{n}, \frac{p}{q}\right) \mid m, n, p, q \in \mathbb{Z}, n, q \neq 0 \text{ and } q m=p n\right\}$ Let $x = m/n$ and $y = p/q$. The condition $qm = pn$ can be rewritten as $m/n = p/q$ (by dividing both sides by $nq$, which is non-zero). Thus, $S = \{(x, y) \mid x, y \in \mathbb{Q} \text{ and } x=y\}$. This is the equality relation on the set of rational numbers.

Step 4: Check Reflexivity for S

• Objective: To check if for every $x \in \mathbb{Q}$, $(x, x) \in S$. This requires $x=x$.
• Working: The equality $x=x$ is always true for any rational number $x$.
• Conclusion: $S$ is reflexive.

Step 5: Check Symmetry for S

• Objective: To check if for every $x, y \in \mathbb{Q}$, if $(x, y) \in S$, then $(y, x) \in S$. This means if $x=y$, then $y=x$.
• Working: If $x=y$, then it is always true that $y=x$.
• Conclusion: $S$ is symmetric.

Step 6: Check Transitivity for S

• Objective: To check if for every $x, y, z \in \mathbb{Q}$, if $(x, y) \in S$ and $(y, z) \in S$, then $(x, z) \in S$. This means if $x=y$ and $y=z$, then $x=z$.
• Working: If $x=y$ and $y=z$, then by the transitive property of equality, $x=z$.
• Conclusion: $S$ is transitive.

Overall for S: Since $S$ is reflexive, symmetric, and transitive, $S$ is an equivalence relation.

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Common Mistakes & Tips

• Division by Zero: Be extremely cautious when dividing by variables. If a variable can be zero, consider that case separately or use a method that avoids division (like multiplying).
• Counterexamples: For disproving symmetry or transitivity, a single, well-chosen counterexample is sufficient. Ensure your counterexample clearly violates the property.
• Simplifying Conditions: For relation $S$, simplifying the condition $qm=pn$ to $m/n = p/q$ greatly clarifies its nature as the equality relation.

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Summary

We analyzed relation $R$ and found it to be reflexive and transitive, but not symmetric, thus it is not an equivalence relation. We then analyzed relation $S$, which simplifies to the equality relation on rational numbers. This relation was found to be reflexive, symmetric, and transitive, making it an equivalence relation. Therefore, $S$ is an equivalence relation, and $R$ is not.

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