While reading an introduction on elliptic curves, I stumbled upon something called reduction modulo $p$.
The definition states that we want to create a group homomorphism that maps an elliptic curve on the rational field to the integers modulo a prime $p$, i.e.: $$E(\mathbb{Q}) \rightarrow \tilde{E}(\mathbb{F}_p), \;\;\;\; P \mapsto \tilde{P}$$
Where $P = (x,y) \in E(\mathbb{Q}),$ and $\tilde{P} = (\tilde{x},\tilde{y}) \in \tilde{E}(\mathbb{F}_p)$.
Such a homomorphism is worked out trivially when $x$ and $y$ are in $\mathbb{Z}$, while in the rational case the reader has to write $x = \frac{a}{b}$. Then, if $p \nmid b$, then $\tilde{b}$ has an inverse in $\mathbb{F}_p$, hence $\tilde{x} = \tilde{a}\tilde{b}^{-1}$ in $\mathbb{F}_p$.
The last step is quite obscure to me. The paper provides some examples:
Let $E: y^2 = x^3 + 2x + 4.$
Then, $P = (2,4)$ and $Q = \left(\frac{1}{4},\frac{17}{8}\right)$ are points in $E(\mathbb{Q})$.
Then the reduction modulo 11 map $E(\mathbb{Q}) \rightarrow \tilde{E}(\mathbb{F}_{11})$ gives:
$\tilde{P} = (2,4)$ and $\tilde{Q} = (3,9)$.
I feel like I'm missing something trivial, but I really have no idea of how the author found the value of $\tilde{Q}$. Can someone help?