Prove that the following set is equinumerous by showing that

Prove that the following set is equinumerous by showing that there is a bijection:

z and the set S of all integers congruent to 1 modulo 3

In order to prove that the set is equinumerous, we need to find a bijective function which implies there exists two injective f: A->B and f:B->A

If both the function exists, then the bijection exists and the function will be equinumerous

f(x) = 3x + 1

since (3x+1) modulo (3) = (3x) modulo 3 + 1 modulo 3 = 0 mod 3 + 1 mod 3 = 1 mod 3

Now checking for reverse mapping i.e. f:B->A

3x + 1 = y

x = (y-1)/3

Hence there is a function y = (x-1)/3 which make sure that every element in S is mapped to the set Z.

Now we have proved that there exists two injective functions both from f:A-B and f:B-A, hence now using Schr