# Multiplicative Proof

• October 14th 2010, 02:15 PM
Janu42
Multiplicative Proof
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.
• October 14th 2010, 02:21 PM
undefined
Quote:

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.

Doesn't this work?

Let a,b be coprime. So f(ab)=f(a)f(b) and g(ab)=g(a)g(b).

So (fg)(ab) = f(ab)g(ab) = f(a)f(b)g(a)f(b) = f(a)g(a)f(b)g(b) = (fg)(a)(fg)(b).

We just use multiplicative commutativity of integers.
• October 14th 2010, 03:29 PM
tonio
Quote:

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.

I think you're definitely overthinking it:

$(m,n)=1\Longrightarrow (fg)(mn):= f(mn)g(mn)=f(m)f(n)g(m)g(n) = f(m)g(m)f(n)g(n) = (fg)(m)(fg)(n)$

Tonio