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Let $\mathrm{N}$ be the set of integers. A relation $\mathrm{R}$ on $\mathrm{N}$ is defined as $\mathrm{R}=\{(\mathrm{x}, \mathrm{y}): \mathrm{xy}>0, \mathrm{x}, \mathrm{y}, \in \mathrm{N}\}$. Then, which one of the following is correct?
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Verified Answer
The correct answer is:
$\mathrm{R}$ is an equivalence relation
A relation is equivalent if it is
(i) Reflexive
(ii)Symmetric and
(iii) Transitive
We check for the same, one by -one $\mathrm{x}, \mathrm{y} \in \mathrm{N} \Rightarrow \mathrm{x}>0, \mathrm{y}>0$
$\mathrm{R}=\{(\mathrm{x}, \mathrm{y}) \mid \mathrm{xy}>0, \mathrm{x}, \mathrm{y}, \in \mathrm{N}\}$
(i) Reflexive $\mathrm{x}, \mathrm{y} \in \mathrm{N}$
$\therefore \quad x, x, \in N \Rightarrow x^{2}>0$
$\therefore \quad \mathrm{R}$ is reflexive
(ii) Symmetric $\mathrm{x}, \mathrm{y} \in \mathrm{N}$
and $x y>0 \Rightarrow y x>0$
$\mathrm{R}$ is also symmetric
(iii) Transitive
$\therefore \quad x, y, z \in N$
$\Rightarrow x y>0, y z>0 \quad \Rightarrow x z>0$
$\therefore \quad \mathrm{R}$ is also transtive. Conclusion: $\mathrm{R}$ is an equivalence relation.
(i) Reflexive
(ii)Symmetric and
(iii) Transitive
We check for the same, one by -one $\mathrm{x}, \mathrm{y} \in \mathrm{N} \Rightarrow \mathrm{x}>0, \mathrm{y}>0$
$\mathrm{R}=\{(\mathrm{x}, \mathrm{y}) \mid \mathrm{xy}>0, \mathrm{x}, \mathrm{y}, \in \mathrm{N}\}$
(i) Reflexive $\mathrm{x}, \mathrm{y} \in \mathrm{N}$
$\therefore \quad x, x, \in N \Rightarrow x^{2}>0$
$\therefore \quad \mathrm{R}$ is reflexive
(ii) Symmetric $\mathrm{x}, \mathrm{y} \in \mathrm{N}$
and $x y>0 \Rightarrow y x>0$
$\mathrm{R}$ is also symmetric
(iii) Transitive
$\therefore \quad x, y, z \in N$
$\Rightarrow x y>0, y z>0 \quad \Rightarrow x z>0$
$\therefore \quad \mathrm{R}$ is also transtive. Conclusion: $\mathrm{R}$ is an equivalence relation.
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