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

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!!