i have a question on $p$-adic numbers.
Does there exist a prime $p$ such that $\sqrt 2 \in \mathbb{Q}_p$?? And much more generally, is it true that for all prime $p$, $\mathbb{Q}_p \cap \mathbb{R} = \mathbb{Q}$?
My attempt: Assume that there is some prime $p$ such that $\alpha \in \mathbb{Q}_p$ satisfies $\alpha^2=2$. Then $|\alpha|_p^2 = |2|_p$, and $|2|_p = 1$ if $p \neq 2$ and $|2|_2 = 1/2$. So in both cases we conclude that $\alpha \in \mathbb{Z}_p$.
Write $\alpha = a_0 + a_1\cdot p + a_2\cdot p^2 +...$, with $a_i \in \{0,1,...,p-1\}$. Then $\alpha^2 = 2$ if and only if
$$a_0^2 + 2a_0a_1\cdot p + (2a_0a_2 + a_1^2)\cdot p^2 + ... = 2,$$
which gives a contradiction. Is this proof ok?
I don't know how to attack the other question...
Thanks in advance!
If $p\equiv\pm1\pmod8$ the congruence $x^2\equiv2\pmod p$ is soluble. By Hensel's Lemma it follows that the equation $x^2=2$ is soluble in $\Bbb Z_p$.
I'm not sure that $\Bbb R\cap\Bbb Q_p$ is very meaningful.