Questions about p-adic numbers I got two questions about $p$-adic numbers:

  
*
  
*I often read that the field $\mathbb Q_p$ is much different than the field $\mathbb R$.
  

An element of $\mathbb Q_p$ is of the form $\sum_{i=-k}^{\infty}a_ip^i$ where $a_i\in \{0,...,p-1\}$.
But isn't this just a real number? So at least the elements of $\mathbb Q_p$ are a subset of $\mathbb R$? That would mean that these fields are especially different in terms of their operation?


  
*Let $x\in \mathbb Q_p^*$. Why $x$ can be written uniquely like this: $x=p^na$ where $a$ is an element of the $p$-adic integers?
  

Thanks in advance!
 A: As for the first question, the answer is no. For example, what real number should represent this
$$\sum_{n=0}^{\infty} 5^{n!}$$
$5$-adic number? Note that in real numbers this series is obviously divergent.
As for the second question: you should know that $\Bbb{Z}_p$ is a local ring whose unique maximal ideal is $p\Bbb{Z}_p$. This means that every $p$-adic integer which is not divisible by $p$ is invertible.
Moreover it is a UFD, and every element can be factorized as $up^k$ for some unit $u$, some $k \ge 0$.
So every element of $\Bbb{Q}_p$ has the form
$$\frac{up^k}{vp^h} = (uv^{-1})p^{k-h}$$
EDIT: The confusion comes to your mind, since you are thinking these numbers as they were real numbers: but they are not! Let's consider for example the sequence of integers (actual integers in $\Bbb{Z}$)
$$1, \ \ 1+5, \ \  1+5+5^2, \ \  1+5+5^2+5^3, \dots$$
in $\Bbb{R}$ these sequence diverges. However, if we think it inside $\Bbb{Q}_5$, this sequence converges to the $5$-adic number
$$A=\sum_{n=0}^{\infty} 5^n$$
actually, this is the inverse of $-4$ in $\Bbb{Q}_5$ since
$$-4A=A-5A = (1+5+5^2+5^3+5^4+ \dots)-(5+5^2+5^3+\dots) = 1$$
(all of this is not true in $\Bbb{Q}_p$ for $p \neq 5$, where the sequence diverges). So you have
$$\sum_{n=0}^{\infty} 5^n = -\frac{1}{4} \ \ \ \ \mbox{ in } \Bbb{Q}_5$$
This is possible because of the strange topological structure of $p$-adic integers.
A: *

*No, for example $x=\sqrt{-1}=i\in \mathbb{Q}_5$, but $x\not\in \mathbb{R}$.

*See here. 
