Let R_{1} and R_{2} be two relations defined | vedclass

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(D) relation $R$ is transitive if $(a, b) \in R$ and $(b, c) \in R$ implies $(a, c) \in R$.For $R_{1}$: Let $a = 2 + \sqrt{3}$,$b = 2 - \sqrt{3}$,and

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Neither $R_{1}$ nor $R_{2}$ is transitive

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(D) relation $R$ is transitive if $(a, b) \in R$ and $(b, c) \in R$ implies $(a, c) \in R$.For $R_{1}$: Let $a = 2 + \sqrt{3}$,$b = 2 - \sqrt{3}$,and $c = 1 + 2\sqrt{3}$.Then $a^{2} + b^{2} = (7 + 4\sqrt{3}) + (7 - 4\sqrt{3}) = 14 \in \mathbb{Q}$. So $(a, b) \in R_{1}$.Also $b^{2} + c^{2} = (7 - 4\sqrt{3}) + (13 + 4\sqrt{3}) = 20 \in \mathbb{Q}$. So $(b, c) \in R_{1}$.However,$a^{2} + c^{2} = (7 + 4\sqrt{3}) + (13 + 4\sqrt{3}) = 20 + 8\sqrt{3} 
otin \mathbb{Q}$.Thus,$(a, c) 
otin R_{1}$,so $R_{1}$ is not transitive.For $R_{2}$: Let $a^{2} = 1$,$b^{2} = \sqrt{3}$,and $c^{2} = 2 - \sqrt{3}$.Then $a^{2} + b^{2} = 1 + \sqrt{3} 
otin \mathbb{Q}$. So $(a, b) \in R_{2}$.Also $b^{2} + c^{2} = \sqrt{3} + (2 - \sqrt{3}) = 2 \in \mathbb{Q}$.Wait,for $(b, c) \in R_{2}$,we need $b^{2} + c^{2} 
otin \mathbb{Q}$. Let $b^{2} = \sqrt{3}$ and $c^{2} = 1 + \sqrt{3}$.Then $b^{2} + c^{2} = 1 + 2\sqrt{3} 
otin \mathbb{Q}$. So $(b, c) \in R_{2}$.Now check $a^{2} + c^{2} = 1 + (1 + \sqrt{3}) = 2 + \sqrt{3} 
otin \mathbb{Q}$.However,if we choose $a^{2} = 1$,$b^{2} = \sqrt{3}$,$c^{2} = 2$,then $a^{2} + b^{2} 
otin \mathbb{Q}$ and $b^{2} + c^{2} 
otin \mathbb{Q}$,but $a^{2} + c^{2} = 3 \in \mathbb{Q}$.Thus,$(a, c) 
otin R_{2}$,so $R_{2}$ is not transitive.Therefore,neither $R_{1}$ nor $R_{2}$ is transitive.

Let $R_{1}$ and $R_{2}$ be two relations defined as follows:
$R_{1} = \{(a, b) \in \mathbb{R}^{2} : a^{2} + b^{2} \in \mathbb{Q}\}$ and $R_{2} = \{(a, b) \in \mathbb{R}^{2} : a^{2} + b^{2} \notin \mathbb{Q}\}$
where $\mathbb{Q}$ is the set of all rational numbers. Then:

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Let $R_{1}$ and $R_{2}$ be two relations defined as follows:$R_{1} = \{(a, b) \in \mathbb{R}^{2} : a^{2} + b^{2} \in \mathbb{Q}\}$ and $R_{2} = \{(a, b) \in \mathbb{R}^{2} : a^{2} + b^{2} \notin \mathbb{Q}\}$where $\mathbb{Q}$ is the set of all rational numbers. Then: