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  • Among the relations <img alt="\mathrm{S=\left \{ \left ( a,b \right ):a,b\in\mathbb{R}-\left \{ 0 \right \},2+ \frac{a}{b}> 0\right \}}" src="https://entrancecorner.oncodecogs.com/gif.

Among the relations

\mathrm{S=\left \{ \left ( a,b \right ):a,b\in\mathbb{R}-\left \{ 0 \right \},2+ \frac{a}{b}> 0\right \}} and \mathrm{T=\left \{ \left ( a,b \right ):a,b\in\mathbb{R},a^{2}-b^{2}\in\mathbb{Z} \right \},}

Option: 1

neither \mathrm{S} nor \mathrm{T} is transitive


Option: 2

\mathrm{S} is transitive but \mathrm{T} is not


Option: 3

\mathrm{T} is symmetric but \mathrm{S} is not


Option: 4

both \mathrm{S} and \mathrm{T} are symmetric


Answers (1)

best_answer

\begin{aligned} & \mathrm{S}=\left\{(\mathrm{a}, \mathrm{b}) \mid \mathrm{a}, \mathrm{b} \in \mathrm{R}-\{0\}, 2+\frac{\mathrm{a}}{\mathrm{b}}>0\right\} \& \\ \end{aligned}

\begin{aligned} & \mathrm{~T}=\left\{(\mathrm{a}, \mathrm{b}) \mid \mathrm{a}, \mathrm{b} \in \mathrm{R}, \mathrm{a}^2-\mathrm{b}^2 \in \mathrm{Z}\right\} \end{aligned}
For \mathrm{S}, 2+\frac{\mathrm{a}}{\mathrm{b}}>0 \Rightarrow \frac{\mathrm{a}}{\mathrm{b}}>-2
Let (-1,2) \in \mathrm{S}\left(\therefore-\frac{1}{2}>-2\right)
\&(2,-1) \notin s\left(\therefore \frac{2}{-1} \text { not greater than }-2\right)
So, \mathrm{S} is not symmetric

For T,

\begin{aligned} & \text { If }(a, b) \in T \Rightarrow a^2-b^2 \in Z \\ \end{aligned}

\begin{aligned} & \Rightarrow-\left(a^2-b^2\right) \in Z \\ \end{aligned}

\begin{aligned} & \Rightarrow b^2-a^2 \in Z \\ & \Rightarrow(b, a) \in T \end{aligned}
So, \mathrm{T} is symmetric

Posted by

HARSH KANKARIA

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