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Let N denotes the set of all natural numbers and R be the relation on \( \Large N \times N \) defined by \( \Large \left(a,\ b\right) R \left(c,\ d\right)\ if\ ad \left(b+c\right)=bc \left(a+d\right) \), then R is:


A) symmetric only

B) reflexive only

C) transitive only

D) an equivalence relation

Correct Answer:
D) an equivalence relation

Description for Correct answer:
For \( \Large \left(a,\ b\right),\ \left(c,\ d\right)\ \epsilon\ N \times N \)

Let \( \Large \left(a,\ b\right)\ R\ \left(c,\ d\right) \)

=> \( \Large ad \left(b+c\right)=bc \left(a+d\right) \)

Reflexive: Since, \( \Large ab \left(b+a\right)=ba \left(a+b\right)\ A\ a,\ b\ \epsilon\ N \)

Therefore, \( \Large \left(a,\ b\right)\ R\ \left(a,\ b\right) \)is reflexive

Symmetric: For \( \Large \left(a,\ b\right),\ \left(c,\ d\right)\ \epsilon\ N \times N,\ Let\ \left(a,\ b\right)\ R\ \left(c,\ d\right) \)

Therefore, \( \Large ad \left(b+c\right)=bc \left(a+d\right) \)

=> \( \Large bc \left(a+d\right)=ad \left(b+c\right) \)

=> \( \Large cb \left(d+a\right)=da \left(c+b\right) \)

=> \( \Large \left(c,\ d\right)\ R\ \left(a,\ b\right) \)

R is symmetric

Transitive: For \( \Large \left(a,\ b\right),\ \left(c,\ d\right),\ \left(e,\ f\right)\ \epsilon\ N \times N \)

\( \Large \left(a,\ b\right)\ R\ \left(c,\ d\right),\ R\ \left(e,\ f\right) \)

\( \Large ad \left(b+c\right)=bc \left(a+d\right),\ cf \left(d+e\right)=de \left(f\right) \)

\( \Large adb+adc = bca+bcd \) ...(i)

\( \Large cfd+cfe = dec+def \) ...(ii)

Multiplying equation (i) by ef and (ii) by ab and adding equations (i) and (ii), we get

\( \Large adbef+adcef+cfdab+cfcab \) \( \Large =bcaef+bcdef+decab+defab \)

=> \( \Large adcf \left(b+e\right)=bcde \left(a+f\right) \)

=> \( \Large af \left(b+e\right)=be \left(a+f\right) \)

=> \( \Large \left(a,\ b\right)\ R\ \left(e,\ f\right) \)

Therefore, R is transitive

Hence R is an equivalence relation.

Part of solved Set theory questions and answers : >> Elementary Mathematics >> Set theory




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