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

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

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Grade:11

1 Answers

Abhinav kaushik
35 Points
4 years ago
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 R
now let 
a÷5 leaves a remainder say r
So again by definition of relation a÷5 will leave remainder r
So (a, a) element R for every a element N 
Thus it is a reflexive relation 
 
For symmetric i.e (a, b) element R implies
that (b, a) is also an element of R
Now let
a÷5 leave a remainder r
Then by defination of relation if (a, b) element R then
b÷5 must leave a remainder r
Thus 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 relation 
 
Now 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 r
So by definition b÷5 must leave a remainder r
And thus if (b, c) is an element of R then
c÷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 relation
AND SINCE THE RELATION IS reflexive, symmetric and transitive it is an EQUIVALENCE RELATION 
Hence proved. 
Hoping it would help
All the best!! 
 

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