Originally Posted by
Janu42 I'm definitely overthinking this...
Show that if f and g are multiplicative functions, then fg is also multiplicative, where (fg)(n) = f(n)g(n) for every positive integer n.
Here's what I'm thinking:
If n is prime, then n = 1*n, and f(1*n) = f(1)f(n), and same for g.... 1 and n are clearly relatively prime, so the definition of multiplicative applies and I can finish it out.
If n is composite, n =p1p2....pk for k distinct primes.
f(n) = f(p1)f(p2)f(p3)....etc... and same for g since all the pi's are relatively prime to each other, thus letting the definition for multiplicative work.
So no matter if n is composite or prime, each time it works.