Having come across Monsky's theorem today (that a square cannot be cut into an odd number of triangles of equal area), I was reading his proof. He introduces an "ultranorm" (or non-Archimidean norm), which is a function with the properties

  1. $\|xy\| = \|x\|\cdot\|y\|$
  2. $\|x + y\| \le \max\{\|x\|, \|y\|\}$
  3. $\|x\| = 0$ if and only if $x = 0$

He notes that the $2$-adic norm on the rationals can be extended to an ultranorm on the entire reals. Then he continues his proof with an arbitrary ultranorm having the property $\|2\| < 1$.

He then divides points $(x,y)\in \Bbb R^2$ into three sets: $$\begin{align}\mathscr A &= \{(x,y)\mid \|x\| < 1 \text{ and } \|y\| < 1\}\\ \mathscr B &= \{(x,y)\mid \|x\| \ge 1 \text{ and } \|x\| \ge \|y\|\}\\ \mathscr C &= \{(x,y)\mid \|y\| \ge 1 \text{ and } \|x\| < \|y\|\}\end{align}$$

He then proves that translation by a point in $\mathscr A$ preserves all three sets.

The key element of the proof is that there has to be at least one triangle $T$ with vertices from all three sets. From this he deduces that $\|\operatorname{area}T\| > 1$:

For we may assume that the vertex of $T$ of type $\mathscr A$ is $(0, 0)$. Let $(x, y)$ and $(x',y')$ be the vertices of type $\mathscr B$ and $\mathscr C$. Then $\operatorname{area}T$, up to sign, is equal to $\frac12(xy'-x'y)$. But $\|xy'\| > \|x'y\|$. So $\|\operatorname{area}T\| = \left\|\frac12\right\|\cdot\|x\|\cdot\|y'\| > 1$.

Now by property 2 of the ultranorm, I see that $$\|\operatorname{area}T\| = \left\|\frac12\right\|\cdot\|xy' - x'y\| \le \left\|\frac12\right\|\max\{\|xy'\|, \|x'y\|\} = \left\|\frac12\right\|\cdot\|x\|\cdot\|y'\| > 1$$

but I don't see what justifies replacing the $\le$ with $=$.

What am I missing?

  • 3
    $\begingroup$ In fact $\|x+y\|=\max\{\|x\|,\|y\|\}$ whenever $x\ne y$; see this question and answer. $\endgroup$ – Brian M. Scott Mar 24 '16 at 22:36
  • $\begingroup$ That's it. I had forgotten that. Thanks. $\endgroup$ – Paul Sinclair Mar 24 '16 at 22:58
  • $\begingroup$ You’re welcome. $\endgroup$ – Brian M. Scott Mar 24 '16 at 22:59
  • $\begingroup$ And now that you point it out to me, I notice that Monsky mentions that after giving the properties as well. I apparently had skipped that line. $\endgroup$ – Paul Sinclair Mar 24 '16 at 23:01

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