Search any question & find its solution
Question:
Answered & Verified by Expert
If $\mathrm{R}$ be a relation defined as a $\mathrm{R} \mathrm{b}$ if $|\mathrm{a}-\mathrm{b}|>0$, then the relation is
Options:
Solution:
1205 Upvotes
Verified Answer
The correct answer is:
symmetric and transitive
We observe the following properties :
Reflexivity - Let a be an arbitrary element.
Then,
$|a-a|=0>0 \Rightarrow a \mathrm{R} \mathrm{a}$
This, $R$ is not reflexive on $R$.
Symmetry - Let a and b be two distinct elements, then $(a, b) \in R$
$\begin{array}{l}
\Rightarrow|\mathrm{a}-\mathrm{b}|>0 \Rightarrow \quad|\mathrm{b}-\mathrm{a}|>0 \\
\quad(\because|\mathrm{a}-\mathrm{b}|=|\mathrm{b}-\mathrm{a}|)
\end{array}$
Thus, $(a, b) \in R \Rightarrow(b, a) \in R$. So, $R$ is symmetric.
Transitivity $-$ Let $(a, b) \in R$ and $(b, c) \in R$.
Then $|a-b|>0$ and $|b-c|>0$
$\Rightarrow \quad|\mathrm{a}-\mathrm{c}|>0 \Rightarrow(\mathrm{a}, \mathrm{c}) \in \mathrm{R}$
So, $R$ is transitive.
Reflexivity - Let a be an arbitrary element.
Then,
$|a-a|=0>0 \Rightarrow a \mathrm{R} \mathrm{a}$
This, $R$ is not reflexive on $R$.
Symmetry - Let a and b be two distinct elements, then $(a, b) \in R$
$\begin{array}{l}
\Rightarrow|\mathrm{a}-\mathrm{b}|>0 \Rightarrow \quad|\mathrm{b}-\mathrm{a}|>0 \\
\quad(\because|\mathrm{a}-\mathrm{b}|=|\mathrm{b}-\mathrm{a}|)
\end{array}$
Thus, $(a, b) \in R \Rightarrow(b, a) \in R$. So, $R$ is symmetric.
Transitivity $-$ Let $(a, b) \in R$ and $(b, c) \in R$.
Then $|a-b|>0$ and $|b-c|>0$
$\Rightarrow \quad|\mathrm{a}-\mathrm{c}|>0 \Rightarrow(\mathrm{a}, \mathrm{c}) \in \mathrm{R}$
So, $R$ is transitive.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.