No field $\mathbb{Q}_p$ of the $p$ adic numbers ($p$ adic rationals) is isomorphic with the field $\mathbb{R}$ of the real numbers.

Proof. Let's assume the existence of a field isomorphism $f:\,\mathbb{R}\to \mathbb{Q}_p$ , 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 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 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.