For natural numbers $a$ and $b$, show that $a \Bbb Z + b \Bbb Z = \gcd(a, b)\Bbb Z $

I just basically said that the gcd of a and b { written as C } obviously divides aZ and bZ, therefore it can be rewritten as

CdZ + CeZ -> C[dZ + eZ] -> CZ[d+e] where Cd=a and Ce=b. 

if a and b and C are natural numbers, then so are d and e. the sum of d and e is an integer, that when multiplied by an integer produces an integer. therefore

CZ[d+e] = CZ, so aZ + bZ = GCD(a,b)Z

{ Where C is the gcd(a,b) }