Can someone please help me understand how to do this exercise?
Show that multiplicative group of positive rationals is a free abelian group of countably infinite rank.
So, I know i can begin with: Let Q be multiplicative group of positive rationals. Let r $\in$ Q that $r = \frac{a}{b} \in Q$.
I have no idea what next? :/
On
For any $\frac{m}{n}$ with $m,n\in\Bbb N$ write $m=p_1^{r_1}....p_k^{r_k}$ and $n=p_1^{s_1}...p_k^{s_k}$ for $s_i,r_i\in\{0,1,2,...\}$ and $p_i$ primes. Then, $$\frac{m}{n}=p_1^{r_1-s_1}...p_k^{r_k-s_k}.$$ So, $\{p^t|t\in\Bbb N\text{ and }p \text{ prime }\}$ is a basis. Note that if $p$ is in basis then, $1/p$ cannot be in basis. Also, $p^0=1$.
Hint: Fundamental theorem of arithmetic.
Full solution:
Every positive rational can be written as a unique product $$n=p_1^{a_{n1}}p_2^{a_{n2}}p_3^{a_{n3}}\ldots$$ where $p_i$ is the $i$th prime, and $a_{nj}\in\mathbb Z$ is eventually constant and $0$. Thus we can write a group homomorphism $$\phi:(\mathbb Q,\times)\to(\mathbb Z^\omega,+)$$ sending $n\to(a_{nj})_{j\in\mathbb{N}}$. But because each product also uniquely determines a rational, this map is actually a bijection and thus $$(\mathbb Q,\times)\sim(\mathbb Z^\omega,+)$$ which gives the result.