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Let \( S \) be the set of all real numbers. A relation \( R \) has been defined on \( S \) by \( a R b \Leftrightarrow |a-b| \leq 1 \),
then \( R \) is
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then \( R \) is
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Verified Answer
The correct answer is:
reflexive and symmetric but not transitive
Given that $a R b \Leftrightarrow|a-b| \leq 1$
Now, $a R a \Leftrightarrow|a-a|=0 \leq 1$
Therefore, $\mathrm{R}$ is reflexive
Again, $a R b \Leftrightarrow|a-b| \leq 1$
Then, $b R a \Leftrightarrow|b-a| \leq 1$
$\Rightarrow|a-b| \leq 1$, which is true.
Therefore, $\mathrm{R}$ is symmetric
Take $\mathrm{a} \mathrm{b}==12$, Then, $|a-b|=|1-2|=1=1$
Take $\mathrm{b} \mathrm{c}==23$ and . Then, $|b-c|=|2-3|=1$
But aRc $|a-c|=|1-3|=2>1$
Therefore, $\mathrm{R}$ not transitive
Hence, $\mathrm{R}$ is reflexive and symmetric but not transitive.
Now, $a R a \Leftrightarrow|a-a|=0 \leq 1$
Therefore, $\mathrm{R}$ is reflexive
Again, $a R b \Leftrightarrow|a-b| \leq 1$
Then, $b R a \Leftrightarrow|b-a| \leq 1$
$\Rightarrow|a-b| \leq 1$, which is true.
Therefore, $\mathrm{R}$ is symmetric
Take $\mathrm{a} \mathrm{b}==12$, Then, $|a-b|=|1-2|=1=1$
Take $\mathrm{b} \mathrm{c}==23$ and . Then, $|b-c|=|2-3|=1$
But aRc $|a-c|=|1-3|=2>1$
Therefore, $\mathrm{R}$ not transitive
Hence, $\mathrm{R}$ is reflexive and symmetric but not transitive.
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