non-isomorphic completions of $\mathbb{Q}$

No field ${\mathbb{Q}}_p$ of the $p$-adic numbers ($p$-adic rationals (http://planetmath.org/PAdicIntegers)) is isomorphic with the field $ℝ$ of the real numbers.

Proof.  Let’s assume the existence of a field isomorphism  $f:ℝ\to {\mathbb{Q}}_p$  for some positive prime number $p$.  If we denote  $f(\sqrt{p})=a$,  then we obtain

$$a^2={(f(\sqrt{p}))}^2=f({(\sqrt{p})}^2)=f(p)=p,$$

because the isomorphism maps the elements of the prime subfield on themselves.  Thus, if  $|\cdot {|}_p$  is the normed $p$-adic valuation (http://planetmath.org/PAdicValuation) of $\mathbb{Q}$ and of ${\mathbb{Q}}_p$, we get

$${|a|}_p=\sqrt{{|a^2|}_p}=\sqrt{{|p|}_p}=\sqrt{\frac{1}{p}},$$

which value is an irrational number as a square root of a non-square (http://planetmath.org/SquareRootOf2IsIrrationalProof) rational.  But this is impossible, since the value group of the completion ${\mathbb{Q}}_p$ must be the same as the value group ${|\mathbb{Q}\setminus \{0\}|}_p$ which consists of all integer powers of $p$.  So we conclude that there can not exist such an isomorphism.