Question

Let `Z` be the set of integers. Show that the relation `R={(a , b): a , bZ` and `a+b` is even} is an equivalence relation on Z.

Answer 1

`R = {(a,b): a,b in Z and a+b` is even`}`
`a+a = 2a`, which is even. `:. (a,a) in R`
`:. R` is reflexive.

`(a,b) in R`.
It means `a+b` is even.
Thus, `(b,a) in R`
So, if `(a,b) in R`, then `(b,a) in R`
`:. R` is symmetric.

If `(a,b) in R`,
Then, `a+b` is even.
If `(b,c) in R`,
Then, `b+c` is even.
`:. a+b+b+c` is even.
`=>a+2b+c` is even.
As, `2b` is even, so, `a+c` is even.
`:. (a,c) in R`
`:. R` is transitive.
As `R` is reflexive, symmetric and transitive, `R` is an equivalence relation.