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I was recently introduced to the idea of 2-adic norms for numbers in $\mathbb{Q}$. I was told that the rules can be expanded into $\mathbb{R}$, but I am unclear of how to evaluate them. For example, what would $|12|_{2}$ evaluate to? or $|\sqrt{2}|_{2}$

This is to help me understand (in some small way) Monsky's Theorem for my combinatorics class.

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    12 is a rational number. You can take the metric completion of the rationals under the p-adic norm to get the p-adic numbers, which are similar to but different from the real numbers. – 2017-02-09
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    Who told you it could be expanded to the reals? Pretty sure that's wrong. There is a way to complete the rationals via this norm which is similar to the way we complete the rationals with the standard $|x|$ norm, but there is no way to expand to the reals. – 2017-02-09
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    @mathematician I'm afraid I don't know what "metric completion" means. This is my first upper level math course after undergraduate differential equations, if that helps you understand the limits of my math vocabulary. – 2017-02-09
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    @ThomasAndrews my combinatorics professor. – 2017-02-09
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    You can just think of it as the same thing you do to construct the reals from the rationals using the Euclidean distance, but using the p-adic norm instead. The p-adic numbers are things that the rationals converge to under the p-adic norm. – 2017-02-09
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    There is a deep result that says that the a certain field is isomorphic to the complex numbers, but the isomorphism is not constructive, so it depends on what you mean by "can be extended." It can't be extended in a way that we can compute (as far as anybody knows.) However, we can find the $p$-adic valuation of any algebraic number by considering its minimum polynomial. – 2017-02-09
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    If $\alpha$ is a algebraic number with minimum rational polynomial $x^n+a_{n-1}x^{n-1}+...+a_0$, then $|\alpha|_p=|a_0|_p^{1/n}$, I believe. So, for $\sqrt{2}$, that means $|\sqrt{2}|_p = |2|_p^{1/2}$. – 2017-02-09
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    Consider my mind blown. I have lots of hours on wikipedia ahead. – 2017-02-09

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The result is technical and non-constructive - you can't actually define an extension of the $p$-adic norm to all real numbers, but you can prove (at least if you assume the Axiom of Choice) that there exists such an extension. And if there is an extension, then there are a huge number of different extensions.

So there isn't a natural value for $|\pi|_p$.

However, given an algebraic number, $\alpha$, there is a well-defined value for $|\alpha|_p$. Specifically, if $\alpha$ is the root of the minimal rational polynomial:

$$x^n+a_{n-1}x^{n-1}+\cdots a_1x+a_0$$

Then there is a formula for $|\alpha|_p$ in terms of tha $a_i$.

I think that formula is $|a_0|^{1/n}$, but I could be wrong about that.

For algebraic numbers of the form $\sqrt[k]{q}$ it is definitely the case that $|\sqrt[k]q|_p=|q|_p^{1/k}$.

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    For $\alpha$ algebraic, each of the finitely many primes of $K=\mathbb{Q}(\alpha)$ lying over $p$ determines a different extension of $|\cdot|_p$ to $K$. An example to show the norm cannot be read off of the minimal polynomial: consider $\alpha_1 = 2+\sqrt{7}$, $\alpha_2=2-\sqrt{7}$, which have the same minimal polynomial. Then $\alpha_1\cdot\alpha_2 = -3$, so if $|\alpha_1|_3=|\alpha_2|_3$ then $|\alpha_1|_3=|\alpha_2|_3=|3|_3^{1/2}$. This would violate the ultrametric inequality, as $|\alpha_1+\alpha_2|_3=|4|_3=1$. – 2017-02-18
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    Ah, interesting. I thought I might have been wrong about that. :) @JulianRosen – 2017-02-18