Let l = `{0,pm1,pm2,pm3,pm4,...}` and `R={(a,b):(a-b)//4=k,kinl}` is an equivalence relation, find equivalence class[0].

Explanation

Given, `(a-b)/(4)=k`
implies a = 4k + b, where `0leblt4`
It is clear b has only value in 0, 1, 2, 3.
(i) Equivalence class of [0] = `{x:x inI and x~0}`
`={x:x-0=4k}={0,pm4,pm8,pm12,...}`
where, `k=0,pm1,pm2,pm3,...`
(ii) Equivalence class of [1] = `{x:x in I and x ~ 1}`
`={x:x-1=4k}={x:x=4k+1}`
`={...,-11,-7,-3,1,5,9,...}`
(iii) Equivalence class of [2] = `{x:x inI and x~2}`
`={x:x-2=4k}={x:x=4k+2}`
`={...,-10,-6,-2,2,6,10,...}`
(iv) Equivalence class of [3] = `{x:x in I and x ~ 3}`
`={x:x-3=4k}={x:x=4k+3}`
`={...,-9,-5,-1,5,9,13,...}`
Continue this process, we see that the equivalence class
`[4]=[0],[5]=[1],[6]=[2],[7]=[3],[8]=[0]`
Hence, total equivalence relations are [0], [1], [2], [3] and also clear
(i) `I=[0]uu[1]uu[2]uu[3]`
(ii) every equivalence is a non-empty
(iii) for any two equivalence classes `[a]nn[b]=phi`.