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The p-adic field $\mathbb{Q}_p$ seems to be an infinite extension of $\mathbb{Q}$ but smaller than $\mathbb{R}$, so can you provide any example that are real number but not in the p-adic field $\mathbb{Q}_p$?

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    $\begingroup$ In what sense do you mean smaller then $\mathbb{R}$? They have the same cardinality. Neither one can be embedded in the other. $\endgroup$ – Gal Porat Apr 23 '17 at 17:42
  • $\begingroup$ $\sqrt{-1} \in \mathbb{Q}_5$. $\endgroup$ – user14972 Apr 23 '17 at 18:35
  • $\begingroup$ The answers here might be helpful, too. $\endgroup$ – Dietrich Burde Apr 23 '17 at 18:36
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$\mathbb{Q}_p$ and $\mathbb{R}$ aren't really comparable at all in any reasonable fashion; the question you ask doesn't really make sense.

To get a feel for the problems, observe that even some of the simplest questions you might ask are complicated.

For example, $\sqrt{1 + \sqrt{2}} \in \mathbb{R}$, assuming you take the positive square root of $2$. It's not if you take the negative square root of $2$.

What about in $\mathbb{Q}_7$? It also depends:

  • If you took $\sqrt{2} = \ldots 454_7$, then $\sqrt{1 + \sqrt{2}}$ is in $\mathbb{Q}_7$
  • If you took $\sqrt{2} = \ldots 213_7$, then $\sqrt{1 + \sqrt{2}}$ is not in $\mathbb{Q}_7$

Which of these two $7$-adic square roots of $2$ do you consider to be the "same" as the positive real square root of $2$?

So, you see, even for algebraic numbers there isn't a good answer to this.

And, while one could imagine making an infinite series of choices fixing which algebraic numbers are $7$-adic and which are not, and similarly for which algebraic numbers are real and which are not... you don't even have a basis for making comparisons when transcendental numbers are involved!


In fact, one can prove that there cannot exist any field homomorphisms $\mathbb{Q}_p \to \mathbb{R}$. For example, any such homomorphism would let us define an ordering on $\mathbb{Q}_p$ that makes it an ordered field... but no such ordering exists.

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There are plenty of algebraic numbers that are in $\Bbb R$ but not in a given $\Bbb Q_p$. For instance any $\sqrt a$ where $a$ is a quadratic nonresidue modulo $p$ (when $p$ is odd).

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