-1
$\begingroup$

I'm struggling to find an equilateral triangle in $\Bbb Z_2$ so I wondered if there aren't any.

Suppose the triangle $\{x,y,z\}$

If this is to be equilateral then $\lvert x-y\rvert_p=\lvert z-y\rvert_p=\lvert x-z\rvert_p=r$

The strong triangle inequality quickly gives us that every triangle is isosceles.

When I try to choose a 3rd point such that the triangle is equilateral I always seem to be frustrated, but I can't put my finger on the algebraic proof that to do so is impossible. It seems I should think about the balls radius $r$ around $x,y$ then show no $z$ exists in both balls which is not closer than $r$ to either $x$ or $y$.

$\endgroup$
3
$\begingroup$

By translation, we may assume wlog that $x=0$, and by scaling we may asssume that one of $y,z$ is an odd integer. Then for an equilateral triangle, we Need that both $y,z$ are odd - but that make $y.z$ even.

Put differently, we have $|a+b|_2\le \max\{|a|_2,|b|_2\}$ with equality iff $|a|_2\ne |b|_2$.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Are you Hagen, or his "Doppelgänger"? $\endgroup$ – Dietrich Burde Dec 19 '18 at 9:44
  • 2
    $\begingroup$ @i am his Doppelgänger who forgot to take his correct logon cedentials for vacatopn :) $\endgroup$ – Hagen von Eitzen Dec 19 '18 at 9:48
3
$\begingroup$

As Hagen von Eitzen pointed out, this cannot be done in $\Bbb{Z}_2$. If you are given three integers some two of them will be congruent modulo two (and the same also holds in the 2-adic completion). Basically the pigeonhole principle in action.

Of course, if $p>2$ then the choices $x=0,y=1,z=2$ will yield an equilateral $p$-adic triangle.

With $p=2$ you need to go to an unramified extension to do the same. Say, instead of $\Bbb{Z}_2$ let's look at $\Bbb{Z}_2[\omega]$ with $\omega$ a primitive third root of unity (and a solution of $\omega^2+\omega+1=0$). Then the (extensions of) 2-adic distances between all four of $0,1,\omega,\omega^2$ are all equal to one.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Ah good, so I wasn't doing anything wrong by failing to derive it from the metric space rules alone. $\endgroup$ – samerivertwice Dec 19 '18 at 9:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.