To determine the transitivity of the relations \( R_1 \) and \( R_2 \), we will analyze each relation step by step.
### Step 1: Analyze Relation \( R_1 \)
**Definition of \( R_1 \):**
\[ R_1 = \{(a, b) : a, b \in \mathbb{R}, a^2 + b^2 \in \mathbb{Q}\} \]
**To check transitivity:**
We need to find if \( (a, b) \in R_1 \) and \( (b, c) \in R_1 \) implies \( (a, c) \in R_1 \).
**Choose elements:**
Let:
- \( a = 1 + \sqrt{2} \)
- \( b = 1 - \sqrt{2} \)
- \( c = 8^{1/4} = \sqrt{2} \)
**Check \( (a, b) \):**
Calculate \( a^2 + b^2 \):
\[
a^2 = (1 + \sqrt{2})^2 = 1 + 2 + 2\sqrt{2} = 3 + 2\sqrt{2}
\]
\[
b^2 = (1 - \sqrt{2})^2 = 1 + 2 - 2\sqrt{2} = 3 - 2\sqrt{2}
\]
\[
a^2 + b^2 = (3 + 2\sqrt{2}) + (3 - 2\sqrt{2}) = 6 \in \mathbb{Q}
\]
Thus, \( (a, b) \in R_1 \).
**Check \( (b, c) \):**
Calculate \( b^2 + c^2 \):
\[
b^2 = 3 - 2\sqrt{2}, \quad c^2 = (8^{1/4})^2 = \sqrt{8} = 2\sqrt{2}
\]
\[
b^2 + c^2 = (3 - 2\sqrt{2}) + (2\sqrt{2}) = 3 \in \mathbb{Q}
\]
Thus, \( (b, c) \in R_1 \).
**Check \( (a, c) \):**
Calculate \( a^2 + c^2 \):
\[
a^2 + c^2 = (3 + 2\sqrt{2}) + (2\sqrt{2}) = 3 + 4\sqrt{2}
\]
Since \( 4\sqrt{2} \) is irrational, \( a^2 + c^2 \notin \mathbb{Q} \).
Thus, \( (a, c) \notin R_1 \).
**Conclusion for \( R_1 \):**
Since \( (a, b) \in R_1 \) and \( (b, c) \in R_1 \) but \( (a, c) \notin R_1 \), the relation \( R_1 \) is **not transitive**.
### Step 2: Analyze Relation \( R_2 \)
**Definition of \( R_2 \):**
\[ R_2 = \{(a, b) : a, b \in \mathbb{R}, a^2 + b^2 \notin \mathbb{Q}\} \]
**To check transitivity:**
We need to find if \( (a, b) \in R_2 \) and \( (b, c) \in R_2 \) implies \( (a, c) \in R_2 \).
**Choose elements:**
Let:
- \( a = 1 + \sqrt{2} \)
- \( b = \sqrt{2} \)
- \( c = 1 - \sqrt{2} \)
**Check \( (a, b) \):**
Calculate \( a^2 + b^2 \):
\[
a^2 = (1 + \sqrt{2})^2 = 3 + 2\sqrt{2}, \quad b^2 = (\sqrt{2})^2 = 2
\]
\[
a^2 + b^2 = (3 + 2\sqrt{2}) + 2 = 5 + 2\sqrt{2}
\]
Since \( 5 + 2\sqrt{2} \) is irrational, \( (a, b) \in R_2 \).
**Check \( (b, c) \):**
Calculate \( b^2 + c^2 \):
\[
b^2 = 2, \quad c^2 = (1 - \sqrt{2})^2 = 3 - 2\sqrt{2}
\]
\[
b^2 + c^2 = 2 + (3 - 2\sqrt{2}) = 5 - 2\sqrt{2}
\]
Since \( 5 - 2\sqrt{2} \) is irrational, \( (b, c) \in R_2 \).
**Check \( (a, c) \):**
Calculate \( a^2 + c^2 \):
\[
a^2 + c^2 = (3 + 2\sqrt{2}) + (3 - 2\sqrt{2}) = 6 \in \mathbb{Q}
\]
Thus, \( (a, c) \notin R_2 \).
**Conclusion for \( R_2 \):**
Since \( (a, b) \in R_2 \) and \( (b, c) \in R_2 \) but \( (a, c) \notin R_2 \), the relation \( R_2 \) is **not transitive**.
### Final Conclusion:
Both relations \( R_1 \) and \( R_2 \) are not transitive.
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