Let R be the relation on the set mathbb{R} of | 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 $a \in \mathbb{R}$. Therefore,$R$ is reflexive.$2

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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 $a \in \mathbb{R}$. Therefore,$R$ 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 $b \ R \ a$. Therefore,$R$ is symmetric.$3$. Transitivity: Consider $a = 1, b = 1.5, c = 2$. Here,$|1 - 1.5| = 0.5 \le 1$ (so $1 \ R \ 1.5$) and $|1.5 - 2| = 0.5 \le 1$ (so $1.5 \ R \ 2$). However,$|1 - 2| = 1 \le 1$ is true,but consider $a = 1, b = 2, c = 3$. $|1 - 2| = 1 \le 1$ and $|2 - 3| = 1 \le 1$,but $|1 - 3| = 2 > 1$. Thus,$1 \ R \ 2$ and $2 \ R \ 3$ does not imply $1 \ R \ 3$. Therefore,$R$ is not transitive.Conclusion: $R$ is reflexive and symmetric.

Let $R$ be the relation on the set $\mathbb{R}$ of all real numbers defined by $a \ R \ b$ if $|a - b| \le 1$. Then $R$ is

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