Extending 2-adic valuation to real numbers

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?

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. A proof of the fact that $\mathbb{C}$ and $\mathbb{C}_p$ are isomorphic can be found on page 83 of the book: Non-Archimedean Functional Analysis - [A.C.M. van Rooij] - 1978.

More generally, it has been proven (see here) that any two algebraically closed fields of the same characteristic and cardinality are isomorphic.

$\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.