KCET · Maths · Functions
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
- A reflexive and transitive but not symmetric
- B an equivalence relation
- C symmetric and transitive but not reflexive
- D reflexive and symmetric but not transitive
Answer & Solution
Correct Answer
(D) reflexive and symmetric but not transitive
Step-by-step Solution
Detailed explanation
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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