Finding an example of a non-rational p-adic number We know that every rational number can be written as a $p$-adic integer with expansion $\sum\limits_{n=-m}^\infty a_n p^n$, where $a_n\in\{0,\dots,p-1\}$ and $m\in\mathbb{N}$; therefore there exists an injection $\mathbb{Q} \hookrightarrow \mathbb{Q}_p$.
But how do I show that $\mathbb{Q}_p$ is bigger? How do I find an example of a $p$-adic number which is not rational?
I heard $p$-adic numbers and even $p$-adic integers being uncountable, is the proof easy?
 A: HINT: 
A rational number has a periodic $p$-adic expansion. So, an example would be 
 $$\sum_{n\ge 0} p^{n^2}$$
A: Take a prime $p>2$, and $m$ an integer with $m\equiv1\pmod p$. Then $\sqrt m\in\Bbb Z_p$.
For $p=2$, if $m\equiv1\pmod8$, then $\sqrt m\in\Bbb Z_2$.
Finally, $\Bbb Z_p$ contains all $p-1$ of the $(p-1)$-th roots of unity. Uninteresting for $p=2$ and $p=3$, but very interesting for bigger primes.
A: First of all, it is every integer (not every rational) that can be written as you have presented it as a sequence $a_n\in\{0,\ldots, p-1\}$. The number $1/p$ cannot be so represented.
To answer your cardinality question, there are at least two arguments that $\mathbb Z_p$ is uncountable, and thus that $\mathbb Q_p$ is uncountable. 
From your presentation of the $p$-adics as the series  $\sum a_n p^n$, we see that the $\mathbb Z_p$ has the cardinality of the set of maps $\mathbb N^+\rightarrow \{0,\ldots, p-1\}$, (i.e., $n\mapsto a_n$) and hence uncountable.
Another argument is to start with the fact that $\mathbb Z_p$ is infinite and (Hausdorff) compact, e.g., because $\mathbb Z_p=  \lim \mathbb Z/p^{n}$, so a closed set of a compact set, hence compact (and infinite). By Baire's theorem, in a compact Hausdorff space, the countable union of nowhere dense closed sets cannot contain a non-empty open set. In particular, an infinite countable set cannot be (Hausdorff and) compact. 
