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.