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When proving Monsky's theorem, one of the steps, which, from what I have so far seen, no proof can avoid, is extending the 2-adic valuation to all real numbers, so that it still satisfies $|xy|_2=|x|_2|y|_2$ and $|x+y|_2\leq\max\{|x|_2,|y|_2\}$. However, no proof of Monsky's theorem which I have seen explained why this extension exists. One of the proofs has mentioned that the proof of existence requires axiom of choice. This made me think that standard argument using Zorn's lemma would do the trick, but I have realized that giving arbitrary value to an irrational will not always lead to consistent valuation (e.g. we can't set $|\sqrt{2}|_2=1$, as this would lead to $|2|_2=|\sqrt{2}\cdot\sqrt{2}|_2=|\sqrt{2}|_2^2=1\neq\frac{1}{2}=|2|_2$).

How does one prove existence of an extension of $|\cdot|_2$ as specified above?

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  • $\begingroup$ Looks to me, with my untrained eyes (and they are not well-trained in judging the complexity of these sort of statements), that for a fixed $n$, the statement is $\Pi^1_2$, therefore absolute between $L$ and $V$. So for each odd $n$ the statement is provable without choice, therefore it is provable without choice. Why is this tagged as [axiom-of-choice] and not [p-adic-number-theory] for example? $\endgroup$ – Asaf Karagila Jul 5 '15 at 21:29
  • $\begingroup$ @AsafKaragila I have added [p-adic-number-theory] tag. As for axiom of choice - as I have mentioned in the body of my question, I heard that the (standard) proof of Monsky actually does use choice. If that's avoidable - great enough, if not - let it be. $\endgroup$ – Wojowu Jul 5 '15 at 21:37
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    $\begingroup$ Do you know the standard proof? It's easy to ask "Hey, I heard about this theorem, and it seems to me that it is using the axiom of choice, can it be done without it?", but if you don't know the usual proof, then perhaps it is worth learning that part first. The [axiom-of-choice] tag is for questions specifically about the role of the axiom in proofs, and what can happen in its absence. Since you don't know the usual proof, it seems strange that you'd ask for the role of choice in that proof. $\endgroup$ – Asaf Karagila Jul 5 '15 at 21:45
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    $\begingroup$ @AsafKaragila No, I don't know a standard proof using choice (indeed, my question asks for details the part of proof which uses choice). I guess that I indeed should put away the part of the question which asks about necessity of choice, and maybe make it into another question later on. $\endgroup$ – Wojowu Jul 5 '15 at 22:03
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    $\begingroup$ can't $\Bbb R$ (even $\Bbb C$) be embedded as a subfield of $\overline{\Bbb Q_2}$ ? $\endgroup$ – mercio Jul 6 '15 at 13:24
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$\Large A $fter digging up Lang and related sources, which says the 2-adic valuation on $\mathbb{Q}$ can be extended to $\mathbb{R}$.

Let $(K, | \cdot|)$ be an ordered field and $L/K$ be an extension. If either of the following holds, then there is a norm on $L$ extending $|\cdot|$.

  • $L/K$ is Algebraic
  • $(K, |\cdot|)$ is non-Archimedian

Thus $(\mathbb{R}, |\cdot|_2)$ is very different from the familiar Euclidean $(\mathbb{R}, ||\cdot||)$. E.g. $\sqrt{2}$ can no longer be approximated as the limit of rational numbers; it's only definition is as the positive solution of $x^2 = 2$. My guess is that you simply adjoin elements inductively. Whenever you find an element in $\mathbb{R}$ not in your extension.

This may be an important insight into Monsky's problem that the natural topology on $[0,1]^2 \subset \mathbb{R}^2$ for this problem. Using the 2-adic norm, Monsky paritions the unit square into 3 subsets:

  • $A = \{ (x,y): ||x|| < 1 \text{ and } ||y|| < 1\;\;\;\;\,\}$
  • $B = \{ (x,y): ||x|| > 1 \text{ and } ||x|| > ||y||\}$
  • $C = \{ (x,y): ||y|| > 1 \text{ and } ||y|| > ||x||\}$

In Euclidean norm, these subsets (representing vertices of triangles) are dense in the unit square

Think about it -- it's very easy to draw a partition of a square into triangles. How can it be that none of them have equal area? You could try to draw an arrangement of triangles with some parameters; then they have some relationship which you can solve for. Monsky shows even with those equations you can find no solution.


From Comments:

I don't see what's the problem with defining R as a completion of Q under archimedean metric, and afterwards asking for an extension of non-archimedean metric to this completion. Note that we are not asking R to be complete w.r.t. to this extended metric.

You can take 2-adic completion of any extension $K/\mathbb{Q}$ but this extension may not be unique. To ensure the existence of a completion over $\mathbb{R}$, the axiom of choice may be required.

As your example suggests, clearly the roots of 2 need special attention possibly $|\sqrt[n]{2}|_2 = \frac{1}{2^{1/n}}$ but there are weirder ones as well:

$$ 2 = (3 + \sqrt{7})(3 - \sqrt{7}) \text{ so that } |3 + \sqrt{7}|_2 |3 - \sqrt{7}|_2 = \tfrac{1}{2}$$

Probably they are both $\frac{1}{\sqrt{2}}$. So this valuation has to be extended real algebraic integers $\overline{\mathbb{Q}} \cap \mathbb{R}$.

Monsky's paper says you need only extend $\mathbb{Q}$ by the coordinates of the $m$ vertices. This is a finite extension, although you won't know which ones they are a priori.


We should also note that even dissections are possible, but that's not the issue you are worried about.

Wikipedia has discussions on some equidissection problems and the Dehn Invariant related to Hilbert's 3rd problem.

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  • $\begingroup$ Actually, with 2-adic, we don't have $2^n\rightarrow\infty$ - indeed, we have $2^n\rightarrow 0$, because $|2^n-0|_2=\frac{1}{2^n}$ which can be smaller than any $\varepsilon$ if $n$ are large enough. $\endgroup$ – Wojowu Jul 6 '15 at 6:55
  • $\begingroup$ Also, I highly doubt that $\sum\frac{1}{4^n}$ is convergent. In 2-adic metric, sequence of partial sums is not even Cauchy, as the difference of consecutive terms is $\frac{1}{4^n}$, and this sequence doesn't tend to 0, because taking $\varepsilon=1$ we will never have $|\frac{1}{4^n}-0|_2<1$. $\endgroup$ – Wojowu Jul 6 '15 at 6:57
  • $\begingroup$ The extension only has to satisfy 3 properties, and I can't see any reason, intuitive or formal, which would make existence of this extension implausible or impossible. We can in exactly the same way ask question about extension of 2-adic metric to any field containing Q. Also remember that there is more than one definition of R, so we can easily ask the same question without ever mentioning Cauchy sequences and alike. $\endgroup$ – Wojowu Jul 6 '15 at 11:19
  • $\begingroup$ To comment on one more thing from your answer - I don't see what's the problem with defining R as a completion of Q under archimedean metric, and afterwards asking for extension of non-archimedean metric to this completion. Note that we are not asking R to be complete w.r.t. to this extended metric. $\endgroup$ – Wojowu Jul 9 '15 at 9:41
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    $\begingroup$ About the norms of $3+\sqrt7$ and $3-\sqrt7$: they must both be equal to $1/\sqrt2$. First, $|3\pm\sqrt7|>1/2$, since otherwise $1/2=|3+\sqrt7||3-\sqrt7|\leq1/4$. Then $|3-\sqrt7|>1/2=1/2\cdot 1=|2\sqrt7|$ implies $|3+\sqrt7|=|(3-\sqrt7)+2\sqrt7|=|3-\sqrt7|$. $\endgroup$ – Jose Brox Jan 15 '18 at 10:01
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How does one prove the existence of an extension of $|\cdot|_2$ as specified above?

Any p-adic valuation on $\mathbb{Q}$ can be extended to $\mathbb{R}$ or $\mathbb{C}$. The construction is as follows: since there exists a field isomorphism $T:\mathbb{C}\to\mathbb{C}_p$ that extends the identity map in $\mathbb{Q}$, you can define the valuation $|\cdot|_*:\mathbb{C}\to\mathbb{R}$ by $|x|_*=|T(x)|_p$. It is not hard to see that $|x+y|_*\leq\max\{|x|_*,|y|_*\}$ for all $x,y\in\mathbb{C}$. Then, the restriction of $|\cdot|_*$ to $\mathbb{R}$ is a valuation on $\mathbb{R}$ that extends the $p$-adic valuation of $\mathbb{Q}$. Of course, this construction cannot be done explicitly since there is not an explicit formula for $T$. We only know about its existence. For a proof of the fact that $\mathbb{C}$ and $\mathbb{C}_p$ are isomorphic you can see page 83 of the book: Non-Archimedean Functional Analysis - [A.C.M. van Rooij] - 1978.

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