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Step-by-Step Solution
1. Understanding the Concept of Transitivity
A relation $R$ on a set $S$ is said to be transitive if for any three elements $x, y, z$ in $S,$ whenever $(x, y) \in R$ and $(y, z) \in R,$ it must follow that $(x, z) \in R.$
2. Defining the Relations
We have two relations on the set of real numbers $R:$
$R_{1} = \{(a, b) \in \mathbb{R}^{2} : a^{2} + b^{2} \in \mathbb{Q}\},$
$R_{2} = \{(a, b) \in \mathbb{R}^{2} : a^{2} + b^{2} \notin \mathbb{Q}\}.$
Here, $\mathbb{Q}$ is the set of rational numbers. We want to check whether $R_{1}$ or $R_{2}$ (or both) is transitive.
3. Checking Transitivity of $R_{1}$
Choose specific real numbers
Select three real numbers $a, b, c$ such that we can test the condition for transitivity. Let:
$a = 1 + \sqrt{2}$
$b = 1 - \sqrt{2}$
$c = 8^{\tfrac{1}{4}}$ (the fourth root of 8)
Compute $a^2 + b^2$
$ a = 1 + \sqrt{2}, \quad b = 1 - \sqrt{2} $
So,
$
a^2 = (1 + \sqrt{2})^2 = 1 + 2\sqrt{2} + 2, \quad
b^2 = (1 - \sqrt{2})^2 = 1 - 2\sqrt{2} + 2.
$
Then
$
a^2 + b^2 = (1 + 2\sqrt{2} + 2) + (1 - 2\sqrt{2} + 2) = 6,
$
which is a rational number.
Hence, $(a, b) \in R_{1}$ since $a^2 + b^2 \in \mathbb{Q}.$
Compute $b^2 + c^2$
$ b = 1 - \sqrt{2}, \quad c = 8^{\tfrac{1}{4}} = \sqrt{\sqrt{8}} = 2^{3/4}. $
Then
$
b^2 = 3 - 2\sqrt{2}, \quad c^2 = \bigl(2^{3/4}\bigr)^2 = 2^{3/2} = 2\sqrt{2}.
$
Hence,
$
b^2 + c^2 = (3 - 2\sqrt{2}) + 2\sqrt{2} = 3,
$
which is rational.
Thus, $(b, c) \in R_{1}$ as well.
Compute $a^2 + c^2$
Recall
$
a^2 = 3 + 2\sqrt{2}, \quad c^2 = 2\sqrt{2}.
$
Hence,
$
a^2 + c^2 = (3 + 2\sqrt{2}) + 2\sqrt{2} = 3 + 4\sqrt{2}.
$
This is not a rational number because $\sqrt{2}$ is irrational and the sum $3 + 4\sqrt{2}$ remains irrational.
Therefore, $(a, c) \notin R_{1}.$
Conclusion for $R_{1}$
We found that $(a, b) \in R_{1}$ and $(b, c) \in R_{1}$, but $(a, c) \notin R_{1}.$ That violates the transitivity condition. Hence, $R_{1}$ is not transitive.
4. Checking Transitivity of $R_{2}$
Choose specific real numbers
Select another set of real numbers $a, b, c$:
$a = 1 + \sqrt{2}$
$b = \sqrt{2}$
$c = 1 - \sqrt{2}$
Compute $a^2 + b^2$
$ a = 1 + \sqrt{2}, \quad b = \sqrt{2}. $
Then
$
a^2 = (1 + \sqrt{2})^2 = 1 + 2\sqrt{2} + 2, \quad
b^2 = (\sqrt{2})^2 = 2.
$
Hence,
$
a^2 + b^2 = (3 + 2\sqrt{2}) + 2 = 5 + 2\sqrt{2},
$
which is not rational.
Thus, $(a, b) \in R_{2}$.
Compute $b^2 + c^2$
$ b = \sqrt{2}, \quad c = 1 - \sqrt{2}. $
Then
$
b^2 = 2, \quad c^2 = (1 - \sqrt{2})^2 = 1 - 2\sqrt{2} + 2 = 3 - 2\sqrt{2}.
$
Hence,
$
b^2 + c^2 = 2 + (3 - 2\sqrt{2}) = 5 - 2\sqrt{2},
$
which is not rational.
Thus, $(b, c) \in R_{2}$.
Compute $a^2 + c^2$
Recall
$
a^2 = 3 + 2\sqrt{2}, \quad c^2 = 3 - 2\sqrt{2}.
$
Hence,
$
a^2 + c^2 = (3 + 2\sqrt{2}) + (3 - 2\sqrt{2}) = 6,
$
which is rational.
Therefore, $(a, c) \notin R_{2}$ because their sum of squares is in $\mathbb{Q}.$
Conclusion for $R_{2}$
We found that $(a, b) \in R_{2}$ and $(b, c) \in R_{2}$, but $(a, c) \notin R_{2}$. This violates the transitivity condition. Hence, $R_{2}$ is not transitive.
5. Final Answer
Neither $R_{1}$ nor $R_{2}$ is transitive.
