On the set of all real numbers,a relation R i | vedclass

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(A) $1$. Reflexivity: For any $a \in \mathbb{R}$,$|a - a| = 0 \le 1$. Thus,$a \, R \, a$ holds for all $a \in \mathbb{R}$. So,$R$ is reflexive. $2$. S

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Reflexive and symmetric

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(A) $1$. Reflexivity: For any $a \in \mathbb{R}$,$|a - a| = 0 \le 1$. Thus,$a \, R \, a$ holds for all $a \in \mathbb{R}$. So,$R$ is reflexive. $2$. Symmetry: If $a \, R \, b$,then $|a - b| \le 1$. Since $|a - b| = |b - a|$,it follows that $|b - a| \le 1$,which means $b \, R \, a$. So,$R$ is symmetric. $3$. Transitivity: Consider $a = 1, b = 1.5, c = 2$. Here,$|1 - 1.5| = 0.5 \le 1$ $(1 \, R \, 1.5)$ and $|1.5 - 2| = 0.5 \le 1$ $(1.5 \, R \, 2)$. However,$|1 - 2| = 1 \le 1$ holds,but if we take $a = 1, b = 1.8, c = 2.6$,then $|1 - 1.8| = 0.8 \le 1$ and $|1.8 - 2.6| = 0.8 \le 1$,but $|1 - 2.6| = 1.6 > 1$. Thus,$R$ is not transitive. Conclusion: $R$ is reflexive and symmetric.

On the set of all real numbers,a relation $R$ is defined as $a \, R \, b$ if and only if $|a - b| \le 1$. Then $R$ is:

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