How can one find numbers in $\mathbb{Q}_p$ that aren't in $\mathbb{Q}$? I am reading Fernando Gouvea's An Introduction to p-adic Numbers; I have read through the construction of them as a completion of the $\mathbb{Q}$ via the p-adic absolute values. But I am struggling to figure out (come up with examples) numbers that are in the completions $\mathbb{Q}_p$ but are not in the $\mathbb{Q}$.
First off, it seems that $\mathbb{Q}_p$ is a set of sequences rather than actual numbers, and these sequences are made from rational numbers. But if I understand it correctly we should actually think about the limit of each sequence instead. So how can a sequence of rational numbers converge under the p-adic absolute values to a number that is not in $\mathbb{Q}$? Or would you be able to give an example of how to construct such a number?
 A: The expansions
$$
\sum_{k=0}^\infty a_k p^k
$$
with $a_k \in \{0,1,\dots,p-1\}$ all represent different $p$-adic numbers.  This defines an injection
$$
\{0,1,\dots,p-1\}^{\mathbb N} \to \mathbb Q_p
$$
It shows not only that $\mathbb Q_p$ is uncountable, but that it has (at least) cardinal $2^{\aleph_0}$.
A: 
A $p$-adic number
  $$\sum a_n p^n$$
  with $a_n$ taken from the set $\{0,1,2,3,\ldots,p-1\}$ is rational if and only if the sequence $a_n$ is eventually repeating (exactly as in the situation for $\mathbf{R}$ with decimal expansions).

The proof is easy. Take a rational $r$, and for convenience, assume that $r \in \mathbf{Z}_p$ (that is, $p$ does not occur in the denominator. Otherwise the $p$-adic expansion is just shifted to the left by this power of $p$)
To compute the $p$-adic expansion, one simply writes $r = r_0$ and then
$$r_0 = a_0 + p r_1,$$
for the unique $a \in \{0,1,\ldots,p-1\}$ with $a_0 \equiv r \pmod p$, and then repeat, with
$$r_{n} = a_n + p r_{n+1}.$$
By construction, $r_0 \equiv a_0 \pmod p$, so no factor of $p$ is introduced to the denominator. In particular, the denominator of $r_{n+1}$ is bounded by the denominator of $r_{n}$. On the other hand, let
$$M = \max\{|r_0|,1 \}.$$
Then $|r_n| \le M$ for all $n$ by induction, because
$$|r_{n+1}| = \left| \frac{r_n - a_n}{p} \right|
\le \frac{|r_n|}{p} + \frac{p-1}{p}
\le \frac{M}{p} + \frac{p-1}{p}$$
$$ = M + \frac{(M-1)(p-1)}{p} \le M.$$
But now this means that the $r_n$ are all rational numbers absolutely bounded by $M$ and whose denominators are not increasing. There are only finitely many such rational numbers, and hence they must repeat.
Conversely, if the $a_n$ do (eventually) repeat, then you have a geometric series which sums to a rational number. 
A: Any $p$-adic integer $x \in \mathbb{Z}_p$ can be uniquely written as a convergent sum
$$x = \sum_{n = 0}^{\infty} a_n \, p^n$$
with $a_n \in \{0, 1, \ldots,p-1\}$ (this means the sequence of partial sums converges to $x$ for the $p$-adic metric). The sequence $(a_n)_{n \ge 0}$ of coefficients is called the Hensel expansion of $x$. And conversely, any such sequence $(a_n)_{n \ge 0} \in \{0, 1, \ldots,p-1\}^{\mathbb{N}}$ defines a $p$-adic integer. We can show that $x \in \mathbb{Q}$, if and only if its Hensel expansion is eventually periodic (the proof is similar than in the real numbers). Hence, any sequence $(a_n)_{n \ge 0}$ that is not eventually periodic will yield an irrational number. You can for example take
$$x = \sum_{n = 0}^{+\infty} p^{n^2}$$
