Fundamental Theorem of Arithmetic for Positive Rational Numbers and not just Integers? 
I'm using a picture because I want to reflect the book. I'm flustered. Why not introduce the Fundamental Theorem of Arithmetic  as Exercise 2.19,
for $\{$ positive rational numbers $\} \supseteq \mathbb{Z}>1$?  Why just $\mathbb{Z}>1$ first — and then make an exercise about it for $\{$ positive rational numbers $\}$ ? 
Wikipedia — Allowing negative exponents provides a canonical form for positive rational numbers. 
 A: $\newcommand{\ord}{\operatorname{ord}}$
Let $r$ be a non-zero rational number. Then, by definition, there exist integers $a,b \in \mathbb Z$, non-zero, such that $r = a/b$. By the fundamental theorem of arithmetic in $\mathbb Z$, we can write $$a = \pm\prod_{i=1}^\infty p_i^{\ord_{p_i}(a)},$$
and
$$b = \pm \prod_{i=1}^\infty p_i^{\ord_{p_i}(b)},$$
where $\{p_i\}$ is the set of all prime numbers in $\mathbb Z$ and $\ord_{p_i}(a),\ord_{p_i}(b) \in \mathbb N$ are zero for almost all indexes $i$ (except by a finite number). This decomposition is unique. So, if you define $\ord_{p_i}(r)=\ord_{p_i}(a)-\ord_{p_i}(b)$, then, since $r = a/b$, we have
$$ r = {\left(\pm\prod_{i=1}^\infty p_i^{\ord_{p_i}(a)}\right) \over \left(\pm \prod_{i=1}^\infty p_i^{\ord_{p_i}(b)}\right)} = \pm\prod_{i=1}^\infty p_i^{\ord_{p_i}(r)},$$
where, now, $\ord_{p_i}(r) \in \mathbb Z$ is zero for almost all indexes. This shows the existence of prime factorization in $\mathbb Q$. If you are interested, verify this factorization is unique. 
