Can somebody give a neat stepwise solution for th

Question

Can somebody give a neat stepwise solution for the above attached image for ncert exams

Abhinav kaushik · Accepted Answer

A relation which is reflexive, symmetric and transitive is called an equivalence relation Thus for reflexive i.e (a, a) element R for all a element Rnow let a÷5 leaves a remainder say rSo again by definition of relation a÷5 will leave remainder rSo (a, a) element R for every a element N Thus it is a reflexive relationFor symmetric i.e (a, b) element R impliesthat (b, a) is also an element of RNow leta÷5 leave a remainder rThen by defination of relation if (a, b) element R thenb÷5 must leave a remainder rThus if b and a leaves same remainder on division by 5 then for every (a, b) element R there is (b, a) element R Thus it is a symmetric relationNow for transitive i.e for all (a, b) element R and (b, c) element R there is (a, c) element R Now let a÷5 leaves a remainder rSo by definition b÷5 must leave a remainder rAnd thus if (b, c) is an element of R thenc÷5 must also leave a reminder r So it is evident that for every (a, b) element R and (b, c) element R there is every (a, c) element R Thus it is a transitive relationAND SINCE THE RELATION IS reflexive, symmetric and transitive it is an EQUIVALENCE RELATION Hence proved. Hoping it would helpAll the best!!

Differential Calculus> Can somebody give a neat stepwise solutio...

Can somebody give a neat stepwise solution for the above attached image for ncert exams

supraja venkatraman , 5 Years ago

Abhinav kaushik

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Differential Calculus> Can somebody give a neat stepwise solutio...

Can somebody give a neat stepwise solution for the above attached image for ncert exams

supraja venkatraman , 5 Years ago

Abhinav kaushik

Provide a better Answer & Earn Cool Goodies

LIVE ONLINE CLASSES

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