Key Concepts and Formulas

• Transitivity of a Relation: A relation $R$ on a set $A$ is transitive if for all $a, b, c \in A$, whenever $(a, b) \in R$ and $(b, c) \in R$, it implies that $(a, c) \in R$.
• Disproving Transitivity: To show a relation is not transitive, a single counterexample is sufficient. This means finding elements $a, b, c$ such that $(a, b) \in R$ and $(b, c) \in R$, but $(a, c) \notin R$.
• Rational and Irrational Numbers: A number is rational if it can be expressed as a fraction $p/q$ where $p$ and $q$ are integers and $q \neq 0$. Otherwise, it is irrational. The sum or product of rational and irrational numbers can result in either rational or irrational numbers, depending on the specific values.

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Step-by-Step Solution

The problem defines two relations, $R_1$ and $R_2$, on the set of real numbers $\mathbb{R}$, where $\mathbb{Q}$ denotes the set of rational numbers. We need to determine if these relations are transitive.

1. Analysis of Relation $R_1$

The relation $R_1$ is defined as $R_1 = \{(a, b) \in \mathbb{R}^2 : a^2 + b^2 \in \mathbb{Q}\}$. To check for transitivity, we need to see if there exist $a, b, c \in \mathbb{R}$ such that $(a, b) \in R_1$, $(b, c) \in R_1$, and $(a, c) \notin R_1$. This translates to:

1. $a^2 + b^2 \in \mathbb{Q}$
2. $b^2 + c^2 \in \mathbb{Q}$
3. $a^2 + c^2 \notin \mathbb{Q}$

• Step 1.1: Select specific real numbers $a, b, c$ to test the transitivity of $R_1$. We aim to construct a scenario where the sum of squares is rational for the first two pairs, but irrational for the third. Let's choose numbers involving square roots that can combine to form rational numbers. Let $a = 1 + \sqrt{2}$, $b = 1 - \sqrt{2}$, and $c = \sqrt[4]{8}$.

• Step 1.2: Calculate the squares of these numbers. $a^2 = (1 + \sqrt{2})^2 = 1^2 + (\sqrt{2})^2 + 2(1)(\sqrt{2}) = 1 + 2 + 2\sqrt{2} = 3 + 2\sqrt{2}$ $b^2 = (1 - \sqrt{2})^2 = 1^2 + (\sqrt{2})^2 - 2(1)(\sqrt{2}) = 1 + 2 - 2\sqrt{2} = 3 - 2\sqrt{2}$ $c^2 = (\sqrt[4]{8})^2 = (8^{1/4})^2 = 8^{2/4} = 8^{1/2} = \sqrt{8} = 2\sqrt{2}$

• Step 1.3: Check if $(a, b) \in R_1$. This requires checking if $a^2 + b^2$ is rational. $a^2 + b^2 = (3 + 2\sqrt{2}) + (3 - 2\sqrt{2}) = 6$ Since $6$ is a rational number, $(a, b) \in R_1$.

• Step 1.4: Check if $(b, c) \in R_1$. This requires checking if $b^2 + c^2$ is rational. $b^2 + c^2 = (3 - 2\sqrt{2}) + 2\sqrt{2} = 3$ Since $3$ is a rational number, $(b, c) \in R_1$.

• Step 1.5: Check if $(a, c) \in R_1$. This requires checking if $a^2 + c^2$ is rational. $a^2 + c^2 = (3 + 2\sqrt{2}) + 2\sqrt{2} = 3 + 4\sqrt{2}$ Since $3 + 4\sqrt{2}$ is an irrational number, $(a, c) \notin R_1$.

• Step 1.6: Conclude on the transitivity of $R_1$. We have found $a, b, c \in \mathbb{R}$ such that $(a, b) \in R_1$ and $(b, c) \in R_1$, but $(a, c) \notin R_1$. Therefore, $R_1$ is not transitive.

2. Analysis of Relation $R_2$

The relation $R_2$ is defined as $R_2 = \{(a, b) \in \mathbb{R}^2 : a^2 + b^2 \notin \mathbb{Q}\}$. To check for transitivity, we need to see if there exist $a, b, c \in \mathbb{R}$ such that $(a, b) \in R_2$, $(b, c) \in R_2$, and $(a, c) \notin R_2$. This translates to:

1. $a^2 + b^2 \notin \mathbb{Q}$
2. $b^2 + c^2 \notin \mathbb{Q}$
3. $a^2 + c^2 \in \mathbb{Q}$

• Step 2.1: Select specific real numbers $a, b, c$ to test the transitivity of $R_2$. We aim to construct a scenario where the sum of squares is irrational for the first two pairs, but rational for the third. This can be achieved if the irrational parts in the intermediate sums do not cancel out, but do cancel out in the final sum. Let $a = 1 + \sqrt{2}$, $b = \sqrt{2}$, and $c = 1 - \sqrt{2}$.

• Step 2.2: Calculate the squares of these numbers. $a^2 = (1 + \sqrt{2})^2 = 1^2 + (\sqrt{2})^2 + 2(1)(\sqrt{2}) = 1 + 2 + 2\sqrt{2} = 3 + 2\sqrt{2}$ $b^2 = (\sqrt{2})^2 = 2$ $c^2 = (1 - \sqrt{2})^2 = 1^2 + (\sqrt{2})^2 - 2(1)(\sqrt{2}) = 1 + 2 - 2\sqrt{2} = 3 - 2\sqrt{2}$

• Step 2.3: Check if $(a, b) \in R_2$. This requires checking if $a^2 + b^2$ is irrational. $a^2 + b^2 = (3 + 2\sqrt{2}) + 2 = 5 + 2\sqrt{2}$ Since $5 + 2\sqrt{2}$ is an irrational number, $(a, b) \in R_2$.

• Step 2.4: Check if $(b, c) \in R_2$. This requires checking if $b^2 + c^2$ is irrational. $b^2 + c^2 = 2 + (3 - 2\sqrt{2}) = 5 - 2\sqrt{2}$ Since $5 - 2\sqrt{2}$ is an irrational number, $(b, c) \in R_2$.

• Step 2.5: Check if $(a, c) \in R_2$. This requires checking if $a^2 + c^2$ is irrational. $a^2 + c^2 = (3 + 2\sqrt{2}) + (3 - 2\sqrt{2}) = 6$ Since $6$ is a rational number, $a^2 + c^2 \in \mathbb{Q}$. This means $(a, c) \notin R_2$.

• Step 2.6: Conclude on the transitivity of $R_2$. We have found $a, b, c \in \mathbb{R}$ such that $(a, b) \in R_2$ and $(b, c) \in R_2$, but $(a, c) \notin R_2$. Therefore, $R_2$ is not transitive.

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Common Mistakes & Tips

• Proof by Counterexample: When disproving a property like transitivity, a single, well-chosen counterexample is sufficient and definitive. Avoid making general statements about all possible cases.
• Strategic Selection of Numbers: For problems involving rationality and irrationality of sums or products, numbers containing square roots (e.g., $x + y\sqrt{k}$) are often effective for constructing counterexamples, as their combinations can lead to either rational or irrational results.
• Careful Calculation: Always double-check your arithmetic and algebraic manipulations, especially when dealing with square roots, to ensure the correct classification of numbers as rational or irrational.

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Summary

We analyzed the transitivity of relations $R_1$ and $R_2$. For $R_1$, we found a counterexample where $(a, b) \in R_1$ and $(b, c) \in R_1$, but $(a, c) \notin R_1$, proving $R_1$ is not transitive. Similarly, for $R_2$, we found a counterexample where $(a, b) \in R_2$ and $(b, c) \in R_2$, but $(a, c) \notin R_2$, proving $R_2$ is also not transitive. Therefore, neither relation is transitive.

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