Proving that an arithmetic function is multiplicative, but not completely multiplicative? 
Define an arithmetic function $\rho$ by $\rho(1)=1$ and $\rho(n)=2^m$ where m is the number of distinct prime numbers in the prime factorization of n. Prove that $\rho$ is multiplicative, but not completely multiplicative. 

This is my first introduction to arithmetic functions so I'm not quite sure how to prove these kinds of things. I can plug numbers into the $\rho$ function,  but how can I show this through a formal proof?

Let me start you off and see if you can give the proof yourself. Let $a$ and $b$ be positive integers, with $\gcd(a,b)=1$. Say $\rho(a)=2^n,\rho(b)=2^m$. You need to show that $\rho(ab)=2^{n+m}$. Consider what the prime factors of $ab$ are, with respect to $a$ and $b$.
To disprove complete multiplicativity, you need $a$ and $b$ with $\rho(ab)\neq\rho(a)\rho(b)$. Since you've proven that $\rho$ is multiplicative, your counterexample should be two positive integers $a$ and $b$ such that $\gcd(a,b)>1$. Try a few examples and see if you can find something.
