A) A partial order relation
B) An equivalence relation
C) An identity relation
D) None of these
Correct Answer: B
Solution :
| [b] We observe the following properties: |
| Reflexivity: Let (a, b) be an arbitrary element of\[N\times N\]. |
| Then, \[(a,b)\in N\times N\Rightarrow a,b\in N\] |
| \[\Rightarrow ab(b+a)=ba(a+b)\Rightarrow (a,b)R(a,b)\] |
| Thus, \[(a,b)R(a,b)\]for all \[(a,b)\in N\times N.\] |
| So, R is reflexive on\[N\times N\]. |
| Symmetry: Let \[(a,b),(c,d)\in N\times N\]be such that \[(a,b)R(c,d).\] Then, \[(a,b)R(c,d)\] |
| \[\Rightarrow ad(b+c)=bc(a+d)\] |
| \[\Rightarrow \,\,\,cb(d+a)=da(c+b)\Rightarrow (c,d)R(a,b)\] |
| Thus, \[(a,b)R(c,d)\Rightarrow (c,d)R(a,b)\] for all \[(a,b),(c,d)\in N\times N.\] |
| So, R is symmetric on \[N\times N.\] |
| Transitivity: Let \[(a,b),(c,d),(e,f)\in N\times N\]such that \[(a,b)R(c,d),and(c,d)R(e,f).\] |
| Then, \[(a,b)R(c,d)\Rightarrow ad(b+c)=bc(a+d)\] |
| \[\Rightarrow \frac{b+c}{bc}=\frac{a+d}{ad}\Rightarrow \frac{1}{b}+\frac{1}{c}=\frac{1}{a}+\frac{1}{d}\] ?. (i) |
| And \[(c,d)R(e,f)\Rightarrow cf(d+e)=de(c+f)\] |
| \[\Rightarrow \frac{d+e}{de}=\frac{c+f}{cf}\Rightarrow \frac{1}{d}+\frac{1}{e}=\frac{1}{c}+\frac{1}{f}\] ?. (ii) |
| Adding (i) and (ii), we get |
| \[\left( \frac{1}{b}+\frac{1}{c} \right)+\left( \frac{1}{d}+\frac{1}{e} \right)=\left( \frac{1}{a}+\frac{1}{d} \right)+\left( \frac{1}{c}+\frac{1}{f} \right)\] |
| \[\Rightarrow \frac{1}{b}+\frac{1}{e}=\frac{1}{a}+\frac{1}{f}\Rightarrow \frac{b+e}{be}=\frac{a+f}{af}\] |
| \[\Rightarrow af(b+e)=be(a,f)\Rightarrow (a,b)R(e,f)\] |
| Thus, \[(a,b)R(c,d)\] and \[(c,d)R(e,f)\] |
| \[\Rightarrow (a,b)R(e,f)\]For all \[(a,b),(c,d),(e,f)\in N\times N.\] |
| So, R is transitive on\[N\times N\]. |
| Hence. R being reflexive, symmetric and transitive, is an equivalence relation on \[N\times N\]. |
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