I'm quite confused about the existence of prime elements in the ring $R=\mathbb{Q}.$ We know that $r \in R$ is a prime iff $r$ is a nonzero, nonunit of $R$ and $r|ab \implies r|a \ \text{or} \ \ r|b \ (a,b \in R).$ But given any nonzero $r\in R,$ $\frac{1}{r}$ will always exist, i.e. $r(\frac{1}{r})=(\frac{1}{r})r =1).$ Hence $r$ is a unit and $R$ has no prime elements?
Appreciate if someone can correct my misconceptions. Thank you.
Your reasoning is correct. Note that when you go from the integral domain $\mathbb{Z}$ to the field of fractions $\mathbb{Q}$, you lose all your prime elements.
In general, if $F$ is a field, then $F$ has no prime elements.