In Cassels book "Lectures on Elliptic Curves", he defines the $p$-adic integers as:
$\quad \mathbb{Z}_p = \{\alpha \in \mathbb{Q}_p \mid |\alpha|_p \leq 1\}$
He latter states that the $p$-adic integers are exactly the elements of the form:
$\quad b = \frac{u}{v} \quad \quad u, v \in \mathbb{Z}, \; p \nmid v$
However he also shows that, for some values of $p$ prime, $\mathbb{Q}_p$ contains irrational and complex numbers. What is their $p$-adic valuation?
Presumably, it must be greater than 1? Otherwise, they would be in $\mathbb{Z}_p$.
edit: punctuation
"Originally", we have the $p$-adic valuation on $\mathbb Q$. Using it as a metric, we can redo the construction with which we obtained $\mathbb R$ from $\mathbb Q$, namely by taking Cauchy sequences (with respect to the standard absolute value) and "making them converge" by creating $\mathbb R$ out of these. Doing the same construction with $|\;|_p$ creates $\mathbb Q_p$. By the pecularities of $p$-adic valuations, if $q_n\to q$ with $q_n\in\mathbb Q$ and $q\in\mathbb Q_p$, then $|q_n-q|_p\to 0$ implies that for $n$ big enough $|q_n-q|_p<|q|_p$ (provided $|q|_p>0$). And this implies $|q|_p=|q_n|_p$. That is, any element of $\mathbb Q_p$ is the quotient of an element of $\mathbb Z_p$ and a power of $p$. Since we use a completely different interpretation of distance than what we are used to from $\mathbb R$ and $\mathbb C$, it is not suitable to talk about complex numbers being elements of $\mathbb Q_p$. Of course all elements of $\mathbb Q_p$ that are not in $\mathbb Q$ are literally irrational. But there is no canonical identification of elements of $\mathbb Q_p\setminus \mathbb Q$ with elements of $\mathbb R\setminus \mathbb Q$. Also, we may be able to solve $x^2+1=0$ in $\mathbb Q_p$, but I would not be too eagerly try to identify them with $\pm i\in\mathbb C$ (which would raise the question which of the two solutions in $\mathbb Q_p$ is $i$ and which is $-i$). In fact, the solution(s) of $x^2+1=0$ as well as things like $x^2-2=0$ (if they exist, depending on $p$) are already in $\mathbb Z_p$, so with the same right you could say that $\mathbb Z_p$ contains irrational and complex numbers!