Show that the relation R in the set Z of intergers given by
`R ={(a,b):2 ` divides a-b }
is an equivalence relation.

Explanation

`A= {1,2,3,4,5}`
`R={(a,b):|a-b|` is even }
It is clear that for any clement `a in A,` we have `|a - a| = 0`
(which is even).
Therefore, R is reflexive.
Let `(a,b) in R`.
`implies |a-b|` is even,
` implies |-(a-b)|=|b-a|` is also even
`implies (b,a) in R`
Therefore, R is symmetric.
Now, let `(a,b) in R ` and `(b,c) in R.`
` implies |a-b|` is even and `|b-c|` is even
`implies (a-b)` is even and `(b-c)` is even ` " " ` (assuming that `a gt b gt c`)
`implies (a-c)=(a-b)+(b-c)` is even ` " " ` [Sum of two even integers is even]
` implies |a-c|` is even
`implies (a,c) in R`
Therefore, R is transitive.
Hence, R is an equivalence relation.