Search any question & find its solution
Question:
Answered & Verified by Expert
If N denote the set of all natural numbers and $\mathrm{R}$ be the relation on $N \times N$ defined by $(a, b) R(c, d)$, if $a d(b+c)=b c(a+d)$, then $R$ is
Options:
Solution:
1491 Upvotes
Verified Answer
The correct answer is:
an equivalence relation
$\begin{array}{l}\text { (a,b) } \mathrm{R}(\mathrm{c}, \mathrm{d}) \\ \Rightarrow \mathrm{ad}(\mathrm{b}+\mathrm{c})=\mathrm{bc}(\mathrm{a}+\mathrm{d}) \\ \text { Reflexive }: \mathrm{ab}(\mathrm{b}+\mathrm{a})=\mathrm{ba}(\mathrm{a}+\mathrm{b}), \forall \mathrm{ab} \subset \mathrm{N} \\ \therefore(\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{a}, \mathrm{b}) \\ \text { So, Ris reflexive, } \\ \text { Symmetric }: \mathrm{ad}(\mathrm{b}+\mathrm{c})=\mathrm{bc}(\mathrm{a}+\mathrm{d}) \\ \Rightarrow \mathrm{bc}(\mathrm{a}+\mathrm{d})=\mathrm{ad}(\mathrm{b}+\mathrm{c}) \\ \Rightarrow \mathrm{cd}(\mathrm{d}+\mathrm{a})=\mathrm{da}(\mathrm{c}+\mathrm{b}) \\ \Rightarrow(\mathrm{c}, \mathrm{d}) \mathrm{R}(\mathrm{a}, \mathrm{b})\end{array}$
So, R is symmetric.
Transitive : For $(\mathrm{a}, \mathrm{b}),(\mathrm{c}, \mathrm{d}),(\mathrm{e}, \mathrm{f}) \in \mathrm{N} \times \mathrm{N}$
Let $(\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{c}, \mathrm{d}),(\mathrm{c}, \mathrm{d}) \mathrm{R}(\mathrm{e}, \mathrm{f})$
$\therefore \mathrm{ad}(\mathrm{b}+\mathrm{c})=\mathrm{bc}(\mathrm{a}+\mathrm{d}), \mathrm{cf}(\mathrm{d}+\mathrm{e})=\mathrm{de}(\mathrm{c}+\mathrm{f})$
$\Rightarrow \mathrm{adb}+\mathrm{adc}-\mathrm{bca}+\mathrm{bcd} \ldots(\mathrm{i})$
and cfd $+\mathrm{cfe}=\mathrm{dec}+\mathrm{def} \ldots(\mathrm{ii})$
On multiplying eq. $(\mathrm{i})$ by ef and eq. (ii) by
ab and then adding, we have
adbef $+\mathrm{adcef}+\mathrm{cfdab}+\mathrm{cfeab}$
$=\mathrm{bcaef}+\mathrm{bcdef}+\mathrm{decab}+\mathrm{defab}$
$\Rightarrow \mathrm{adcf}(\mathrm{b}+\mathrm{e})=\mathrm{bcde}(\mathrm{a}+\mathrm{f})$
$\Rightarrow \mathrm{af}(\mathrm{b}+\mathrm{e})=$ be $(\mathrm{a}+\mathrm{f})$
$\Rightarrow(\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{e}+\mathrm{f}$
So, R is symmetric.
Transitive : For $(\mathrm{a}, \mathrm{b}),(\mathrm{c}, \mathrm{d}),(\mathrm{e}, \mathrm{f}) \in \mathrm{N} \times \mathrm{N}$
Let $(\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{c}, \mathrm{d}),(\mathrm{c}, \mathrm{d}) \mathrm{R}(\mathrm{e}, \mathrm{f})$
$\therefore \mathrm{ad}(\mathrm{b}+\mathrm{c})=\mathrm{bc}(\mathrm{a}+\mathrm{d}), \mathrm{cf}(\mathrm{d}+\mathrm{e})=\mathrm{de}(\mathrm{c}+\mathrm{f})$
$\Rightarrow \mathrm{adb}+\mathrm{adc}-\mathrm{bca}+\mathrm{bcd} \ldots(\mathrm{i})$
and cfd $+\mathrm{cfe}=\mathrm{dec}+\mathrm{def} \ldots(\mathrm{ii})$
On multiplying eq. $(\mathrm{i})$ by ef and eq. (ii) by
ab and then adding, we have
adbef $+\mathrm{adcef}+\mathrm{cfdab}+\mathrm{cfeab}$
$=\mathrm{bcaef}+\mathrm{bcdef}+\mathrm{decab}+\mathrm{defab}$
$\Rightarrow \mathrm{adcf}(\mathrm{b}+\mathrm{e})=\mathrm{bcde}(\mathrm{a}+\mathrm{f})$
$\Rightarrow \mathrm{af}(\mathrm{b}+\mathrm{e})=$ be $(\mathrm{a}+\mathrm{f})$
$\Rightarrow(\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{e}+\mathrm{f}$
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.