# multiplicative functions

• August 4th 2012, 06:17 PM
alyssams
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• August 4th 2012, 06:22 PM
richard1234
Re: multiplicative functions
Quote:

Originally Posted by alyssams
I know that given m, n with gcd(m,n)=1 you can split k into two relatively prime factors in a suitable way depending on m and n.
Anyway, help is appreciated!

How do you know that the domain is the set of integers?

If f is multiplicative, then $f(k)f(n) = f(kn)$ for all k,n. Then $f(n) = \frac{f(kn)}{f(k)}$, which is also multiplicative. Also we know that $f(k) \neq 0$ so we don't have to worry about dividing by zero or anything like that...