Let p be a prime number and let a be an integer that is not divisible by p. Then { [1]p,[2]p.........[p-1]p} = {[a.1]p,[a.2]p,.......[a.(p-1)]p} (Hint: Note that the map is a one-to-one and onto. , A=B means A subset of B and B subset of A) Solution since a is not divisibe by p for 1<=r [x-y] = [0] so [x] = [y] since the cardinality of domain and codomain is same proving one-one is sufficient to show range = codomain =completeset we can prove it onto we have [a^-1] = [n] for some 1<=n.

1. Let p be a prime number and let a be an integer that is not divisible by p. Then {
[1]p,[2]p.........[p-1]p} = {[a.1]p,[a.2]p,.......[a.(p-1)]p} (Hint: Note that the map is a one-to-one
and onto. , A=B means A subset of B and B subset of A)
Solution
since a is not divisibe by p
for 1<=r
[x-y] = [0]
so [x] = [y]
since the cardinality of domain and codomain is same proving one-one is sufficient to show
range = codomain =completeset
we can prove it onto
we have [a^-1] = [n] for some 1<=n