Can fractions be relatively prime?

Two numbers are relatively prime if they do not share any factors, other than 1. Is it possible for fractions to be relatively prime? To reword this, do fractions even have factors?

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One can extend gcd and lcm to rationals. – Bill Dubuque May 18 '14 at 0:44

Every rational number is of the form $\prod_{n=1}^\infty p_n^{e_n}$ where $p_n$ is the $n$th prime number and each $e_n$ is some integer. The difference between that statement and a similar one about integers is that with integers one has $e_n\ge0$ for every value of $n$. Two integers are relatively prime if the set of primes with non-zero exponents for one of them is disjoint from the set of primes with nonzero exponents for the other. One could use that same definition with rational numbers and say, for example, that $8/35$ and $11/9$ are relatively prime because $$\frac{8}{35} = 2^3 5^{-1} 7^{-1} \text{ and }\frac{11}{9} = 3^{-2} 11^1$$ and $$\{2,5,7\} \cap \{3,11\}=\varnothing.$$
$\frac{1}{3}, \frac{2}{5}$ become $\frac{5}{15}, \frac{6}{15}$ which have relatively prime numerators. In this sense fractions are just integers split up into a common number of parts.