Let $n$ be an integer bigger than $0$. Define an arithmetic function $\rho$, by $\rho(1)=1$ and $\rho(2)=2^m$ and $m$ is the number of prime numbers in the prime factorization of $n$.
Prove that $\rho$ is multiplicative but not completely multiplicative.
I think we need to show that $m$ and $n$ are relatively prime? I do not know how to prove it?
Any input is appreciated.
If you need $2^{f(n)}$ to be multiplicative, then you need $f(n)$ to take multiplication (of relatively prime numbers) to addition. That's because $2^{f(pq)} = 2^{f(p) + f(q)} = 2^{f(p)} \cdot 2^{f(q)}$ when $p$ and $q$ are relatively prime numbers.
We will think of your $m$ as $f(n)$ here. What is $f(6)$? Well we want to show that $f(2\cdot 3) = f(2) + f(3)$. $6$ has two prime factors, $2$ and $3$. $f(2)=1$ and $f(3)=1$ since $2$ and $3$ are prime. So indeed $f(6)=2=f(2)+f(3)$.
Now let's consider $f(12)$. Can we write $f(2\cdot 6) = f(2) + f(6) = 1 + 2 = 3$? The answer is no. In fact, $f(12) = f(2^2 \cdot 3) = 2$. What happened here is that $6$ and $12$ share prime factors.
What about $f(30)$? $f(30)=f(5 \cdot 6) = f(5 \cdot 3 \cdot 2) = 3$. We arrive to that answer since we can count the number of prime factors. Moreover, $f(30) = f(5) + f(6) = 1 + 2$. This works since $5$ is relatively prime to $6$. That is, $5$ is a prime that doesn't already divide $6$.
If $m$ and $n$ are relatively prime, then they have two distinct sets of prime divisors. Say $p_1, ..., p_s$ are the prime divisors of $m$, then $m=p_1^{a_1} \cdots p_s^{a_s}$ for integers $a_i \in \mathbb{N}$. Similarly, let $q_1,..., q_r$ be the prime divisors of $n$. Thus $n = q_1^{b_1} \cdots q_r^{b_1}$ for $b_i \in \mathbb{N}$.
Two questions: What is $f(n)$ and $f(m)$? What is $f(n\cdot m)$?