Yesterday I was working through this simple proof of what is sometimes called Cauchy's theorem, that if a prime 𝑝 divides a group 𝐺's size, then 𝐺 has an element of order 𝑝.

The idea is to build a set 𝑆 of all 𝑝-tuples from elements of 𝐺 such that each element of 𝑆 multiplies out to 1, i.e.

𝑆:={(𝑔₁,𝑔₂,…,𝑔ₚ)|∏𝑔ᵢ=1}

and then count the number of elements of 𝑆 in the following way: gather together all elements of 𝑆 that are cyclically permuted, and notice that each such class contains exactly 1 or 𝑝 elements (and all cyclic permutations of an element of 𝑆 are also in 𝑆 because if 𝑎𝑏=1 then 𝑏𝑎=1).

The classes that contain a single element correspond to an element of order 𝑝 as desired, so to show that such a class exist, observe that there is at least one such class, namely the one of (1,1,…,1), and since #𝑆=(#𝐺)ᵖ⁻¹ (every component of an element of 𝑆 is free except the last one due to the defining equation that the product equals 1), then 𝑝|#𝑆.

Thus only way that S can be decomposed into many sets of either size 1 or 𝑝 is if there is more than one set of size 1, meaning, an element of order 𝑝 exists in 𝐺.

A weak ordering (https://en.wikipedia.org/wiki/Weak_ordering) is just a linear order with ties. On $n$ items, there are obviously at least $n!$ of them (don't use ties); here's a simple combinatorial proof that there are at most $(n+1{)}^{n-1}$, also proving the inequality $n!\le (n+1{)}^{n-1}$. Tighter but messier upper bounds on weak orders are known.

In the complete graph $K_{n+1}$, choose one vertex as root. By Cayley's formula (https://en.wikipedia.org/wiki/Cayley%27s_formula), it has exactly $(n+1{)}^{n-1}$ spanning trees. Each spanning tree can be used to weak order the non-root vertices by their distance from the root. Each weak order on non-root vertices arises from a rooted spanning tree in this way by setting the parent of each non-root vertex to be any of its immediate predecessors, or the root if it has none.

#ProofInAToot

Anyone happen to know whether this proof has been published anywhere? I know of a reference for the weaker bound $(n+1{)}^n$ but its proof is non-combinatorial and ugly.

I'm curious, has anyone seen this one before:

Lemma: Suppose that three given points in ${\mathbb{R}}^d$ are disjoint from an open convex set $U$. Then two of the points can be connected by a curve disjoint from $U$ of length at most twice their Euclidean distance.

For instance, in the plane, two nearby points can be separated by a thickened line or line segment, so they have no short connecting curve, but three points cannot all be made far from each other by a single convex obstacle.

Proof: Consider the plane containing the three points; if $U$ does not intersect this plane then the points can be connected directly. Within this plane draw a line through each point, disjoint from $U$. If the three lines form a triangle or unbounded three-sided region containing $U$, with the three points on its sides, then one of its internal angles is at least ${60}^{\circ }$, and the curve connecting two points through this angle has length at most twice the distance between the points (worst case: $U$ is an equilateral triangle and the three points are its edge midpoints). If one of the three points is not on a side of the region containing $U$, it can see the entire line through another of the points, and be connected directly to that point.

@Scmbradley #ProofInAToot!

Let ${\epsilon }_{AJ},{\epsilon }_{JK},{\epsilon }_{AK}$ be pairwise timelike events in a flat region, and event ${\epsilon }_{AP}$ timelike and straight wrt. ${\epsilon }_{AJ},{\epsilon }_{AK}$ and lightlike wrt. ${\epsilon }_{JK}$. Consequently, these four events are plane wrt. each other:
$$0=\left|\begin{array}{ccccc}0& 1& 1& 1& 1\\ 1& 0& s^2[{\epsilon }_{AJ},{\epsilon }_{AP}]& s^2[{\epsilon }_{AJ},{\epsilon }_{AK}]& s^2[{\epsilon }_{AJ},{\epsilon }_{JK}]\\ 1& s^2[{\epsilon }_{AJ},{\epsilon }_{AP}]& 0& (\sqrt{s^2[{\epsilon }_{AJ},{\epsilon }_{AK}]}-\sqrt{s^2[{\epsilon }_{AJ},{\epsilon }_{AP}]}{)}^2& 0\\ 1& s^2[{\epsilon }_{AJ},{\epsilon }_{AK}]& (\sqrt{s^2[{\epsilon }_{AJ},{\epsilon }_{AK}]}-\sqrt{s^2[{\epsilon }_{AJ},{\epsilon }_{AP}]}{)}^2& 0& s^2[{\epsilon }_{JK},{\epsilon }_{AK}]\\ 1& s^2[{\epsilon }_{AJ},{\epsilon }_{JK}]& 0& s^2[{\epsilon }_{JK},{\epsilon }_{AK}]& 0\end{array}\right|.$$
Thus:$$\frac{s^2[{\epsilon }_{AJ},{\epsilon }_{JK}]}{s^2[{\epsilon }_{AJ},{\epsilon }_{AK}]}=1-\sqrt{\frac{s^2[{\epsilon }_{AJ},{\epsilon }_{AP}]}{s^2[{\epsilon }_{AJ},{\epsilon }_{AK}]}}+\frac{s^2[{\epsilon }_{JK},{\epsilon }_{AK}]}{s^2[{\epsilon }_{AJ},{\epsilon }_{AK}]}(1-\sqrt{\frac{s^2[{\epsilon }_{AJ},{\epsilon }_{AK}]}{s^2[{\epsilon }_{AJ},{\epsilon }_{AP}]}}).$$

Now adding the following inequality:
$$2\sqrt{\frac{s^2[{\epsilon }_{JK},{\epsilon }_{AK}]}{s^2[{\epsilon }_{AJ},{\epsilon }_{AK}]}}\le \frac{s^2[{\epsilon }_{JK},{\epsilon }_{AK}]}{s^2[{\epsilon }_{AJ},{\epsilon }_{AK}]}\sqrt{\frac{s^2[{\epsilon }_{AJ},{\epsilon }_{AK}]}{s^2[{\epsilon }_{AJ},{\epsilon }_{AJ}]}}+\sqrt{\frac{s^2[{\epsilon }_{AJ},{\epsilon }_{AP}]}{s^2[{\epsilon }_{AJ},{\epsilon }_{AK}]}}$$

and collecting yields:
$$\frac{s^2[{\epsilon }_{AJ},{\epsilon }_{JK}]}{s^2[{\epsilon }_{AJ},{\epsilon }_{AK}]}+2\sqrt{\frac{s^2[{\epsilon }_{JK},{\epsilon }_{AK}]}{s^2[{\epsilon }_{AJ},{\epsilon }_{AK}]}}\le 1+\frac{s^2[{\epsilon }_{JK},{\epsilon }_{AK}]}{s^2[{\epsilon }_{AJ},{\epsilon }_{AK}]},$$

Therefore
$$\sqrt{\frac{s^2[{\epsilon }_{AJ},{\epsilon }_{JK}]}{s^2[{\epsilon }_{AJ},{\epsilon }_{AK}]}}+\sqrt{\frac{s^2[{\epsilon }_{JK},{\epsilon }_{AK}]}{s^2[{\epsilon }_{AJ},{\epsilon }_{AK}]}}\le 1,$$

a.k.a. reverse triangle inequality of (timelike) Lorentzian distances, https://en.wikipedia.org/wiki/Triangle_inequality#Reversal_in_Minkowski_space

@elboadi #ProofInAToot !

$t^{\ast }:=(L/c)-(c/a)(\sqrt{1+((a/c)t^{\ast }{)}^2}-1),$

$(a/c)T_{ABA}:=2\text{ArcSinh}[(a/c)t^{\ast }].$

$t_m-t_i:=$
$(c/b)(\sqrt{1+((b/c)t_m{)}^2}-1)+(L/c)-(c/a)(\sqrt{1+((a/c)t_i{)}^2}-1),$

$t_f-t_m:=$
$(c/b)(\sqrt{1+((b/c)t_m{)}^2}-1)+(L/c)-(c/a)(\sqrt{1+((a/c)t_f{)}^2}-1).$

$(a/c)t_f:=\text{Sinh}[\text{ArcSinh}[(a/c)t_i]+(a/c)T_{ABA}]$

$⟹(b/a)=\frac{1}{1+(a/c^2)L}=\text{Exp}[-a/(2c)T_{ABA}].$

#ProofInAToot ? #ClaimInAToot #Rigid #Relativity

Simplemente maravilloso: #proofinatoot

Assume $\mathbb{R}$ is enumerable.

But $\mathrm{\forall }$ enumerations of $\mathbb{R}$, $\mathrm{\exists }r\in \mathbb{R}$ s.t. $\mathrm{\forall }i,digi{t}_i(r)==(digi{t}_i(x_i)+1\mathrm{mod}10)$.

Ergo $\mathbb{R}$ is not enumerable.

#ProofInAToot

This proof has now driven 2 people insane.