Let $\mathbb Q_p$ be $p$-adic numbers field. I know that the cardinal of $\mathbb Z_p$ (interger $p$-adic numbers) is continuum, and every $p$-adic number $x$ can be in form $x=p^nx^\prime$, where $x^\prime\in\mathbb Z_p$, $n\in\mathbb Z$.
So the cardinal of $\mathbb Q_p$ is continuum or more than that?
Best Answer
The field $\mathbb Q_p$ is the fraction field of $\mathbb Z_p$.
Since you already know that $|\mathbb Z_p|=2^{\aleph_0}$, let us show that this is also the cardinality of $\mathbb Q_p$:
Note that every element of $\mathbb Q_p$ is an equivalence class of pairs in $\mathbb Z_p$, much like the rationals are with respect to the integers.
Since $\mathbb Z_p\times\mathbb Z_p$ is also of cardinality continuum, we have that $\mathbb Q_p$ can be injected into this set either by the axiom of choice, or directly by choosing representatives which are co-prime.
This shows that $\mathbb Q_p$ has at most continuum many elements, since $\mathbb Z_p$ is a subset of its fraction field, then the $p$-adic field has exactly $2^{\aleph_0}$ many elements.